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Team PrabhaT
15 Mock testcoMBINeD
pHYsIcs, cHeMIstrY AND MAtHeMAtIcs
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IIT-JEE JEE MAIN AND ADVANCED15 MoCk TEsT
CoMBINEDpHYsICs, CHEMIsTrY AND MATHEMATICs
by Team Prabhat
Published by prABHAT prAkAsHAN4/19 Asaf Ali Road, New Delhi–110 002
ISBN ISBN 978-93-5322-575-9
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l Mock Test – 1 Physics ...................................................................................................................................1-18
l Mock Test – 2 Physics .................................................................................................................................19-32
l Mock Test – 3 Physics .................................................................................................................................33-44
l Mock Test – 4 Physics .................................................................................................................................45-58
l Mock Test – 5 Physics .................................................................................................................................59-72
l Mock Test – 1 Chemistry ............................................................................................................................73-91
l Mock Test – 2 Chemistry ..........................................................................................................................92-102
l Mock Test – 3 Chemistry ........................................................................................................................103-113
l Mock Test – 4 Chemistry ........................................................................................................................114-124
l Mock Test – 5 Chemistry ........................................................................................................................125-134
l Mock Test – 1 Mathematics ....................................................................................................................135-154
l Mock Test – 2 Mathematics ....................................................................................................................155-168
l Mock Test – 3 Mathematics ....................................................................................................................169-182
l Mock Test – 4 Mathematics ....................................................................................................................183-196
l Mock Test – 5 Mathematics ....................................................................................................................197-212
CONTENTS
1Mock Test-1
Mock Test “JEE-Main”
Do not open this Test Booklet until you are asked to do so.
Read carefully the Instructions on the Back Cover of this Test Booklet.
Important Instructions:
1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly
prohibited.
2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer
Sheet and fill in the particulars carefully.
3. The test is of 3 hours duration.
4. The Test Booklet consists of 75 questions. The maximum marks are 300.
5. There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 25
questions in each part of equal weightage.
6. Each question is allotted 4 (four) marks for each correct response. For MCQs - 4 Marks will be awarded for every correct
answer and 1 Mark will be deducted for every incorrect answer.
7. For answer with numeric value - 4 Marks will be awarded for every correct answer and 0 Mark will be deducted for every
incorrect answer.
8. Use Blue/Black Ball Point Pen only for writing particulars/ marking responses on Side-1 and Side-2 of the Answer Sheet.
Use of pencil is strictly prohibited.
9. No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any
electronic device, etc., except the Admit Card inside the examination hall/room.
10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the
bottom of each page and at the end of the booklet.
11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall.
However, the candidates are allowed to take away this Test Booklet with them.
12. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial
number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate
should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet.
13. Do not fold or make any stray marks on the Answer Sheet.
Name of the Candidate (in Capital letters):
Roll Number : in figures
in words
Examination Centre Number:
Name of Examination Centre (in Capital letters):
Candidate’s Signature: Invigilator’s Signature:
A Test Booklet Code
Physics 2
Read the following instructions carefully:
1. The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with
Blue/Black Ball Point Pen.
2. For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.
3. The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test
Booklet/Answer Sheet.
4. Out of the four options given for each question, only one option is the correct answer.
5. For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted for
MCQs and No deduction for numeric questions from the total score. No deduction from the total score,
however, will be made if no response is indicated for an item in the Answer Sheet.
6. Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in
Test Booklet Code and Answer Sheet Code), another set will be provided.
7. The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All
calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself,
marked ‘Space for Rough Work’. This space is given at the bottom of each page and at the end of the booklet.
8. On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the
Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
9. Each candidate must show on demand his/her Admit Card to the Invigilator.
10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.
11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the
Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance
Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair
means case. The candidates are also required to put their left hand THUMB impression in the space
provided in the Attendance Sheet.
12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited
13. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the
Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board.
14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager,
mobile phone, electronic device or any other material except the Admit Card inside the examination
hall/room.
3 Mock Test-1
JEE-MAIN: PHYSICS MOCK TEST-1
SECTION 1 (Multiple Choice Question)
1. If a particle moves from point P (2,3,5) to point Q (3,4,5).
Its displacement vector be:
a. ˆˆ ˆ 10i j k+ + b. ˆˆ ˆ 5i j k+ +
c. ji ˆˆ + d. ˆˆ ˆ2 4 6i j k+ +
2. Long horizontal rod has a bead which can slide along its
length and is initially placed at a distance L from one end
A of the rod. The rod is set in angular motion about A with
a constant angular acceleration α. If the coefficient of
friction between the rod and bead is µ, and gravity is
neglected, then the time after which the bead starts
slipping is:
a.
µ
α b.
µ
α
c.
1
µα d. infinitesimal
3. A wind-powered generator converts wind energy into
electric energy. Assume that the generator converts a
fixed fraction of the wind energy intercepted by its blades
into electrical energy. For wind speed v, the electrical
power output will be proportional to
a. v b. v2
c. v3 d. v
4
4. A spherical hollow is made in a lead sphere of radius R,
such that its surface touches the outside surface of lead
sphere and passes through the center. What is the shift in
the center of mass of lead sphere due to the hollowing?
a. 7
R b.
14
R
c. 2
R d. R
5. The bob of a simple pendulum executes simple harmonic
motion in water with a period t, while the period of
oscillation of the bob is 0t in air. Neglect frictional force
of water and given that the density of the bob is
(4 /3) 1000× kg/m3. What relationship is true between t
and 0t ?
a. 0t t= b. 0 / 2t t=
c. 02t t= d. 04t t=
6. A capillary tube A is dipped in water. Another identical
tube B is dipped in a soap-water solution. Which of the
following shows the relative nature of the liquid columns
in the two tubes?
a. b.
c. d.
7. Two rods of different materials having coefficients of
linear expansion 1 2,α α and Young’s
moduli 1 2,Y Y respectively are fixed between two rigid
massive walls. The rods are heated such that they undergo
the same increase in temperature. There is no bending of
the rods. If 1 2: 2 : 3,α α = the thermal stresses developed
in the two rods are equal provided 1 2:Y Y is equal to:
a. 2 : 3 b. 1 : 1
c. 3 : 2 d. 4 : 9
8. Three liquids of equal volumes are thoroughly mixed. If
their specific heats are 1 2 3, ,s s s and their temperature
1 2 3, ,θ θ θ and their densities 1 2 3, ,d d d respectively then
the final temperature of the mixture is:
a. 1 1 2 2 3 3
1 1 2 2 3 3
s s s
d s d s d s
θ θ θ+ ++ +
b. 1 1 1 2 2 2 3 3 3
1 1 2 2 3 3
d s d s d s
d s d s d s
θ θ θ+ ++ +
c. 1 1 1 2 2 2 3 3 3
1 1 2 2 3 3
d s d s d s
d d d
θ θ θθ θ θ
+ ++ +
d. 1 1 2 2 3 3
1 1 2 2 3 3
d d d
s s s
θ θ θθ θ θ
+ ++ +
9. Two monatomic ideal gases 1 and 2 of molecular masses
1m and
2m respectively are enclosed in separate containers
kept at the same temperature. The ratio of the speed of
sound in gas 1 to the gas 2 is given by:
a. 1
2
m
m b. 2
1
m
m
c. 1
2
m
m d. 2
1
m
m
10. A gas is compressed adiabatically till its temperature is
doubled. The ratio of its find volume to initial volume will
be:
a. 1
2 b. more than
1
2
c. less than 1
2 d. between 1and 2
A B A B
A B B A
Physics 4
11. A black body is at a temperature of 2880 K. The energy of
radiation emitted by this body with wavelength between
499 nm and 500 nm is 1,U between 999 nm and 1000 nm
is 2U and between 1499 nm and 1500 nm is
3.U The
Wien constant, 62.88 10 nm-K.= ×b Then,
a. 1 0=U b.
3 0=U
c. 1 2>U U d.
2 1>U U
12. In moving from A to B along an electric field line, the
electric field does 196.4 10−× J of work on an electron. If
1 2,φ φ are equipotential surfaces, then the potential difference
( )C AV V− is
a. – 4V b. 4V
c. Zero d. 64 V
13. Force acting on a charged particle kept between the plates
of a charged condenser is F. If one of the plates of the
condenser is removed, the force acting on the same
particle becomes:
a. 0 b. 2
F
c. F d. 2F
14. An electric cable of copper has just one wire of radius
9 mm. Its resistance is 50 Ω. This single copper wire of
cable is replaced by 6 different well insulated copper
wires, each of radius 3 mm. The total resistance of the
cable will now be equal to:
a. 7.5 Ω b. 45 Ω
c. 90 Ω d. 270 Ω
15. An infinitely long conductor PQR is bent to form a right
angle as shown. A current I flows through PQR. The
magnetic field due to this current at the point M is H1.
Now another infinitely long straight conductor QS is
connected at Q so that the current is I/2 in QR as well as in
QS, The current in PQ remaining unchanged. The
magnetic field at M is now .2H .The ratio 21 /HH is given
by
a. 1
2 b. 1
c. 2
3 d. 2
16. Double slit interference experiment is carried out with
monochromatic light and interference fringes are observed. If
now monochromatic light is replaced by white light, what
change is expected in interference pattern?
a. no change
b. pattern disappears
c. white and dark fringes are observed throughout the
pattern
d. a few coloured fringes are observed on either side of
central white fringe
17. The electric field of an electromagnetic wave in free space
is given by 7 ˆ10cos (10 ) V/m,E t kx j= + Where, t and x are
in seconds and metres respectively. It can be inferred that
(a) the wavelength λ is 188.4 m
(b) the wave number k is 0.33 rad/m
(c) the wave amplitude is 10 V/m
(d) the wave is propagating along + x direction
Which one of the following pairs of statements is correct?
a. (c) and (d) b. (a) and (b)
c. (b) and (c) d. (a) and (c)
18. Light of wavelength λ strikes a photosensitive surface
and electrons are ejected with kinetic energy E. If the
kinetic energy is to be increased to 2E, the wavelength
must be changed to λ′ where:
a. 2
λλ ′= b. 2λ λ′=
c. 2
λλ λ′< < d. λ λ′=
19. The ionisation potential of hydrogen atom is 13.6 V. The
energy required to remove an electron in the n = 2 state of
the hydrogen atom is
a. 27.2 eV b. 13.6 eV
c. 6.8 eV d. 3.4 eV
20. Hydrogen bomb is based on which of the following
phenomenon?
a. Nuclear fission
b. Nuclear fusion
c. Radioactive decay
d. None of these
90o
90o
M
Q P – ∞
S + ∞
– ∞
R
I
5 Mock Test-1
SECTION 2 (Numeric Value Question)
21. A cube has a side of length 21.2 10−× m. Calculate its
volume:
a. 6 31.7 10 m−× b. 6 31.73 10 m−×
c. 6 31.70 10 m−× d. 6 31.732 10 m−×
22. A body of mass 2 kg is released at the top of a smooth
inclined plane having inclination 30 .° It takes 3 seconds to
reach the bottom. If the angle of inclination is doubled
keeping same height (i.e., made 60° ), what will be the
time taken?
a. 3 sec b. 3 sec
c. 3 3 sec d. 1.5 sec
23. There are two bodies of masses 100 kg and 10000 kg
separated by a distance 1m. At what distance from the
smaller body, the intensity of gravitational field will be
zero:
a. 1
9m b.
1
10m
c. 1
11 m d.
10
11m
24. Due to a small magnet intensity at a distance x in the end
on position is 9 gauss. What will be the intensity at a
distance 2
x on broad side on position?
a. 9 gauss b. 4 gauss
c. 36 gauss d. 4.5 gauss
25. When both source and observe approach each other with a
velocity equal to the half the velocity of sound the change
in frequency of sound as detected by the listener is:
a. 0 b. 25 %
c. 50% d. 150%
Space for rough work
Physics 6
JEE ADVANCE PAPER-I
Time 3 Hours. Max. Marks 180 (60 for Physics)
Read the Instructions Carefully
Question Paper Format and Marking Scheme:
1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections.
2. Section 1 contains 6 multiple choice questions with one or more than one correct option.
Marking Scheme: +4 if only (all) the correct option(s) is (are) chosen, +3 if all the four options are correct but only three
options are chosen, +2 if three or more options are correct but only two options are chosen, both of the options must be
correct, +1 if two or more options are correct buy only one option is chosen and it must be correct, -2 (in all other cases).
3. Section 2 contains 8 Numerical value answer type questions. The answer to each question is a single digit integer ranging
from 0 to 9 (both inclusive).
Marking Scheme: +3 if only correct numerical value is given.
4. Section 3 contains 2 Paragraph based questions (2 paragraphs, each having 2 MCQs with one correct answer only)
Marking Scheme: +3 if only the correct option is selected, -1 in all other cases.
Note:
Possible response Section 1 Section 2 Section 3
Not attempted 0 0 0
Partial Correct +1 for each correct option selected 0 0
Correct +4 +3 3
incorrect -2 0 -1
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. A transparent thin film of uniform thickness and refractive
index 1 1.4=n is coated on the convex spherical surface of
radius R at one end of a long solid glass cylinder of
refractive index 2 1.5,=n as shown in the figure. Rays of
light parallel to the axis of the cylinder traversing through
the film from air to glass get focused at distance 1f from
the film, while rays of light traversing from glass to air get
focused at distance 2f from the film. Then
a. 1| | 3=f R b.
1| | 2.8=f R
c. 2| | 2=f R d.
2| | 1.4=f R
2. A conductor (shown in the figure) carrying constant
current I is kept in the x-y plane in a uniform magnetic
field .B
If F is the magnitude of the total magnetic force
acting on the conductor, then the correct statement(s)
is(are)
a. If B
is along ˆ, ( )z F L R∝ +
b. If B
is along ˆ, 0x F =
c. If B
is along ˆ, ( )y F L R∝ +
d. If B
is along ˆ, 0z F =
3. Planck’s constant h, speed of light c and gravitational
constant G are used to form a unit of length L and a unit of
mass M. Then the correct option(s) is(are)
a. M c∝ b. M G∝
c. L h∝ d. L G∝
4. Two independent harmonic oscillators of equal mass are
oscillating about the origin with angular frequencies 1ω
and 2ω and have total energies
1E and 2,E respectively.
The variations of their momenta p with positions x are
/6π /4πL R R L
R R y
x
n1
n2 Air
7 Mock Test-1
shown in the figures. If 2an
b= and ,
an
R= then the correct
equation(s) is(are)
a. 1 1 2 2E Eω ω= b. 22
1
nωω
=
c. 2
1 2nω ω = d. 1 2
1 2
E E
ω ω=
5. Inner and outer radii of a spool are r and R respectively.
A thread is wound over its inner surface and spool is
placed over a rough horizontal surface. Thread is pulled
by a force F as shown in figure. In case of pure rolling,
which of the following statements are false?
a. Thread unwinds, spool rotates anticlockwise and
friction acts leftwards
b. Thread winds, spool rotates clockwise and friction acts
leftwards
c. Thread winds, spool moves to the right and friction acts
rightwards
d. Thread winds, spool moves to the right and friction
does not come into existence
6. Two ideal batteries of emf 1V and
2V and three resistances
1 2,R R and 3R are connected as shown in the figure. The
current in resistance R2 would be zero if
a. 1 2=V V and 1 2 3= =R R R
b. 1 2=V V and
1 2 32= =R R R
c. 1 22=V V and
1 2 32 2= =R R R
d. 1 22 =V V and
1 2 32 = =R R R
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. A small filament is at the centre of a hollow glass sphere
of inner and outer radii 8 cm and 9 cm respectively. The
refractive index of glass is 1.50. Calculate the position of
the image of the filament when viewed from outside the
sphere.
8. An infinitely long uniform line charge distribution of
charge per unit length λ lies parallel to the y-axis in the
y-z plane at3
z2a= (see figure). If the magnitude of the
flux of the electric field through the rectangular surface
ABCD lying in the x-y plane with its center at the origin is
0
L
n
λε
0(ε = permittivity of free space), then the value of n
is
9. For an atom of an ion having single electron, the
wavelength observed1 2λ = are units and
3 3λ = units
figure. The value of missing wavelength 2λ is
10. Consider an elliptically shaped rail PQ in the vertical
plane with 3 mOP = and 4 m.OQ = A block of mass
1 kg is pulled along the rail from P to Q with a force of
18 N, which is always parallel to line PQ (see the given
figure). Assuming no frictional losses, the kinetic energy
1λ
n3 orbit
n2 orbit
n1 orbit
3λ
2λ
C
z
D L
O
B A
x
a
3
2a
y
Glass
O B Air
OA = 8 cm
OB = 9 cm
m = 1.50
2V
1V
1R
2R
3R
r
R α
F
Energy = E1 P
a
b x x
R
P Energy = E2
A
Physics 8
of the block when it’s reaches Q in ( 10)×n joules. The
value of n is (take acceleration due to gravity210 ms−)
11. Two identical uniform discs roll without slipping on two
different surfaces AB and CD (see figure) starting at A and
C with linear speeds 1v and
2 ,v respectively, and always
remain in contact with the surfaces. If they reach B and D
with the same linear speed and 1 3 / ,v m s= then
2v in m/s
is 2( 10 m / s )g =
12. Two spherical stars A and B emit blackbody radiation. The
radius of A is 400 times that of B and A emits 410 times
the power emitted from B. The ratio ( / )A Bλ λ of their
wavelengths Aλ and
Bλ at which the peaks occur in their
respective radiation curves is
13. The half life of a freshly prepared radioactive sample is 2 h.
If the sample emits radiation of intensity, which is 16 times
the permissible safe level, then the minimum time taken
after which it would be possible to work safely with source
is
14. A Young’s double slit interference arrangement with slits
1S and 2S is immersed in water (refractive index = 4/3) as
shown in the figure. The positions of maxima on the
surface of water are given by 2 2 2 2 2 ,x p m dλ= − where λ
is the wavelength of light in air (refractive index = 1), 2d
is the separation between the slits and m is an integer. The
value of p is
SECTION 3 (Maximum Marks: 12)
2 Paragraph based questions (2 paragraphs, each having
2 MCQs with one correct answer only)
Paragraph for Question No. 15 to 16
Light guidance in an optical fiber can be understood by
considering a structure comprising of thin solid glass cylinder
of refractive index1n surrounded by a medium of lower
refractive index 2 .n
The light guidance in the structure takes
place due to successive total internal reflections at the interface
of the media 1n and
2n as shown in the figure. All rays with
the angle of incidence i less than a particular value mi are
confined in the medium of refractive index 1.n The numerical
aperture (NA) of the structure is defined as sin .mi
15. For two structures namely 1S with
1 45 / 4n = and
2 3/ 2,n = and 2S with
1 8 / 5n = and 2 7 / 5n = and taking
the refractive index of water to be 4/3 and that of air to be
1, the correct option(s) is(are)
a. NA of 1S immersed in water is the same as that of 2S
immersed in a liquid of refractive index 16
3 15
b. NA of 1S immersed in liquid of refractive index
6
15
is the same as that of 2S immersed in water
c. NA of 1S placed in air is the same as that of
2S
immersed in liquid of refractive index 4
.15
d. NA of 1S placed in air is the same as that of
2S placed
in water
16. If two structures of same cross-sectional area, but different
numerical apertures1NA and
2 2 1( )NA NA NA< are joined
longitudinally, the numerical aperture of the combined
structure is
1nθ
2n
1 2n n>
Air Cladding
i Core
xd
2S
d
1S
Air
1 3m / sv =
30 mB
A 2v
27 mD
C
Q
4m
3m P
90°
O
water
9 Mock Test-1
a. 1 2
1 2
NA NA
NA NA+ b. 1 2NA NA+
c. 1NA d. 2NA
Paragraph for Question No. 17 to 18
In a thin rectangular metallic strip a constant current I flows
along the positive x-direction, as shown in the figure. The
length, width and thickness of the strip are ,ℓ w and d,
respectively. A uniform magnetic field B
is applied on the
strip along the positive y-direction. Due to this, the charge
carriers experience a net deflection along the z-direction.
This results in accumulation of charge carriers on the surface
PQRS and appearance of equal and opposite charges on the
face opposite to PQRS. A potential difference along the z-
direction is thus developed. Charge accumulation continues
until the magnetic force is balanced by the electric force. The
current is assumed to be uniformly distributed on the cross-
section of the strip and carried by electrons.
17. Consider two different metallic strips (1 and 2) of the
same material. Their lengths are the same, widths are
1w and 2w and thicknesses are
1d and 2 ,d respectively.
Two points K and M are symmetrically located on the
opposite faces parallel to the x-y plane (see figure). 1V
and 2V are the potential differences between K and M in
strips 1 and 2, respectively. Then, for a given current I
flowing through them in a given magnetic field strength
B, the correct statement(s) is(are)
a. If 1 2w w= and
1 22 ,d d= then 2 12V V=
b. If 1 2w w= and 1 22 ,d d= then 2 1V V=
c. If 1 22w w= and
1 2 ,d d= then 2 12V V=
d. If 1 22w w= and
1 2 ,d d= then 2 1V V=
18. Consider two different metallic strips (1 and 2) of same
dimensions (lengths ,ℓ width w and thickness d) with
carrier densities n1 and n2, respectively. Strip 1 is
placed in magnetic field B1 and strip 2 is placed in
magnetic field B2, both along positive y-directions.
Then V1 and V2 are the potential differences developed
between K and M in strips 1 and 2, respectively.
Assuming that the current I is the same for both the
strips, the correct option(s) is(are)
a. If 1 2B B= and
1 22 ,n n= then 2 12V V=
b. If 1 2B B= and
1 22 ,n n= then 2 1V V=
c. If 1 22B B= and
1 2 ,n n= then 2 10.5V V=
d. If 1 22B B= and
1 2 ,n n= then 2 1V V=
y
x
z
I
Q P
S
I W
K•
M•
R
d
Space for rough work
l
Physics 10
JEE ADVANCE PAPER-II
Time 3 Hours. Max. Marks 180 (60 for Physics)
Read the Instructions Carefully
Question Paper Format and Marking Scheme:
1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections.
2. Section 1 contains 6 MCQs with one or more than one correct answer
Marking Scheme: +4 if only (all) the correct option(s) is (are) chosen, +3 if all the four options are correct but only three
options are chosen, +2 if three or more options are correct but only two options are chosen, both of the options must be
correct, +1 if two or more options are correct buy only one option is chosen and it must be correct, -2 (in all other cases)
3. Section 2 contains 8 Numerical value answer type questions.
Marking Scheme: +3 if only correct numerical value is given
4. Section 3 contains 4 Matching type questions with 4 options.
Marking Scheme: +3 if only the correct option is selected, -1 in all other cases
Note:
Possible response Section 1 Section 2 Section 3
Not attempted 0 0 0
Partial Correct +1 for each correct option selected 0 0
Correct +3 +4 +3
Incorrect -1 -2 0
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. A fission reaction is given by
236 140 94
92 54 38U Xe Sr ,x y→ + + +
where x and y are two
particles. Considering 236
92 U to at rest, the kinetic energies
of the products are denoted by Xe ,K
Sr ,K x (2MeV)K and
y (2MeV),K respectively. Let the binding energies per
nucleon of 236 140
92 54U, Xe and 94
38Sr be 7.5 MeV, 8.5 MeV
and 8.5 MeV respectively. Considering different
conservation laws, the correct option(s) is(are)
a. Sr Xe, , 129 MeV, 86 MeVx n y n K K= = = =
b. Sr Xe, , 129 MeV, 86 MeVx p y e K K
−= = = =
c. Sr Xe, , 129 MeV, 86 MeVx p y n K K= = = =
d. Sr Xe, , 86 MeV, 129 MeVx n y n K K= = = =
2. Two spheres P and Q of equal radii have densities 1ρ and
2 ,ρ respectively. The spheres are connected by a mass
less string and placed in liquids 1L and
2L of densities 1σ
and 2σ and viscosities
1η and 2 ,η respectively. They float
in equilibrium with the sphere P in 1L and sphere Q in 2L
and the string being taut (see figure). If sphere P alone in
2L has terminal velocity
PV and Q alone in 1L has
terminal velocity ,
QV then
a. 1
2
| |
| |
ηη
=
P
Q
V
V b. 2
1
| |
| |
ηη
=
P
Q
V
V
c. 0⋅ >
P QV V d. 0⋅ <
P QV V
3. In terms of potential difference V, electric current I,
permittivity 0 ,ε permeability 0µ and speed of light c, the
dimensionally correct equation(s) is(are)
a. 2 2
0 0µ ε=I V b. 0 0ε µ=I V
c. 0ε=I cV d.
0 0µ ε=cI V
4. Consider a uniform spherical charge distribution of radius
1R centred at the origin O. In this distribution, a spherical
1LP
Q 2L
11 Mock Test-1
cavity of radius R2, centred at P with distance
1 2= = −OP a R R (see figure) is made. If the electric field
inside the cavity at position r is ( ),
E r then the correct
statement(s) is(are)
a. E is uniform, its magnitude is independent of 2R but its
direction depends on r
b. E is uniform, its magnitude depends on
2R and its
direction depends on r
c. E is uniform, its magnitude is independent of a but its
direction depends on a
d. E is uniform and both its magnitude and direction
depend on a
5. In plotting stress versus strain curves for two materials P
and Q, a student by mistake puts strain on the y-axis and
stress on the x-axis as shown in the figure. Then the
correct statement(s) is(are)
a. P has more tensile strength than Q
b. P is more ductile than Q
c. P is more brittle than Q
d. The Young’s modulus of P is more than that of Q
6. A spherical body of radius R consists of a fluid of constant
density and is in equilibrium under its own gravity. If P(r)
is the pressure at r(r < R), then the correct option(s) is(are)
a. ( 0) 0= =P r b. ( 3 / 4) 63
( 2 / 3) 80
==
=
P r R
P r R
c. ( 3 / 5) 16
( 2 / 5) 21
==
=
P r R
P r R d.
( / 2) 20
( / 3) 27
==
=
P r R
P r R
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. An electron in an excited state of 2Li +
ion has angular
momentum 3 / 2 .πh The de Broglie wavelength of the
electron in this state is 0πp a (where
0a is the Bohr
radius). The value of p is
8. A large spherical mass M is fixed at one position and two
identical point masses m are kept on a line passing
through the centre of M (see figure). The point masses are
connected by a rigid mass less rod of length ℓ and this
assembly is free to move along the line connecting them.
All three masses interact only through their mutual
gravitational interaction. When the point mass nearer to M
is at a distance 3= ℓr from M, the tension in the rod is
zero for .288
Mm k
=
The value of k is
9. The energy of a system as a function of time t is given as 2( ) exp( α ),= −E t A t where 1α 0.2 .−= s The measurement
of A has an error of 1.25%. If the error in the
measurement of time is 1.50%, the percentage error in the
value of ( ) at 5=E t t s is
10. The densities of two solid spheres A and B of the same
radii R vary with radial distance r as ( )ρ =
A
rr k
R and
5
( ) ,ρ =
B
rr k
R respectively, where k is a constant. The
moments of inertia of the individual spheres about axes
passing through their centres are AI and ,BI respectively.
If ,10
=B
A
I n
Ithe value of n is
11. Four harmonic waves of equal frequencies and equal
intensities 0I have phase angles 0,
2,
3 3
π πand .π When
they are superposed, the intensity of the resulting wave is
0 .nI The value of n is
12. For a radioactive material, its activity A and rate of change
of its activity R are defined as = −dN
Adt
and ,= −dA
Rdt
where N(t) is the number of nuclei at time t. Two
radioactive sources P (mean lifeτ ) and Q (mean life 2 )τ
have the same activity at t = 0. Their rates of change of
activities at 2τ=t are PR and ,QR respectively. If
,=P
Q
R n
R e then the value of n is
M
r ℓ
m m
Str
ain
Stress
P
Q
1R
O
P R2
a
Physics 12
13. A monochromatic beam of light is incident at 60° on one
face of an equilateral prism of refractive index n and
emerges from the opposite face making an angle
( )nθ with the normal (see the figure). For 3=n the
value of θ is 60° and .d
mdn
θ= The value of m is
14. The figure shows a portion of an electric circuit. Resistors
are known and are indicated on the diagram and the
voltmeters are identical. If the voltmeters V1 and V2 read
7.5 V and 5.0 V respectively, find the reading of the volt-
meter V3.
SECTION 3 (Maximum Marks: 12)
Matching type questions with 4 options
15. Match the statement of Column with those in
Column II:
Column I Column II
(A) In any Bohr orbit of
the hydrogen atom,
the ratio of kinetic
energy to potential
energy of the electron
is
1. 1
2−
(B) The ratio of the
kinetic energy to the
total energy of an
electron in a Bohr
orbit is
2. 2
(C) In the lowest energy
level of hydrogen
atom, the electron has
the angular
momentum
3. 2
h
π
(D) Ratio of the
wavelengths of first
line of Lyman series
and first line of
Balmer series is
4. 5 : 27
5. 0
a. A→ 1; B→ 2, 4; C→3; D→ 4
b. A→3; B→1, 2; C→1; D→ 2
c. A→4; B→3, 4; C→2; D→ 1
d. A→2; B→2, 3; C→4; D→ 3
16. You are given many resistances, capacitors and inductors.
These are connected to a variable DC voltage source
(the first two circuits) or an AC voltage source of 50 Hz
frequency (the next three circuits) in different ways as
shown in Column II. When a current I (steady state for DC
or r.m.s. for AC) flows through the circuit, the
corresponding voltage V1 and V2. (indicated in circuits) are
related as shown in Column I.
Column I Column II
(A) 10,I V≠ is
proportional to I
1.
(B) 2 10,I V V≠ > 2.
(C) 1 20,V V V= = 3.
(D) 20,I V≠ is
proportional to I
4.
5.
a. A→3, 4; B→2, 3, 4; C→1, 2; D→2, 3, 4
b. A→1, 2; B→1, 2, 3; C→3, 4; D→1, 3 ,4
c. A→2, 4; B→2, 3, 4; C→2, 3; D→1, 2, 3
d. A→1, 3; B→1, 3, 4; C→1, 3; D→2, 3, 4
V1 V2
1 kΩ
V
3µF
V1 V2
6 mH
V
3µF
V1 V2
6 mH
V
2 Ω
V1 V2
6 mH
V
2 Ω
V1 V2
6 mH
V
3 µF
R1 = 9
R2 = 14
R3 = 24 V3
V2
V1
θ 60°
13 Mock Test-1
17. The gravitational field intensity E
of earth at any point is
defined as the gravitational force per unit mass at that
point. If varies from place to place. The variation is shown
in column II with form of position r
v s the E
in the form
of graphs. The variation of r
is given in column I.
Choose the correct form of graphs for the corresponding
variations of .r
Column I Column II
(A) Position r
of body
measured from
surface earth
upward
1.
(B) Position r
of body
measured from
surface of earth
along diameter to
opposite point on
surface of earth
2.
(C) Position r
measured
from center of earth
to any point
3.
(D) Position r
measured
from center of
hollow sphere to
any point
4.
a. A→1; B→2; C→3; D→4
b. A→4; B→3; C→2; D→1
c. A→4; B→1; C→3; D→2
d. A→3; B→2; C→1; D→4
18. A person in a lift is holding a water jar, which has a small
hole at the lower end of its side. When the lift is at rest, the
water jet coming out of the hole hits the floor of the lift at a
distance d of 1.2 m from the person. In the following, state
of the lift’s motion is given in Column I and the distance
where the water jet hits the floor of the lift is given in
Column II. Match the statements from Column I with those
in Column II and select the correct using the code given
below the lists.
Column I Column II
(A) Lift is accelerating
vertically up.
1. d = 1.2 m
(B) Lift is accelerating
vertically down with
an acceleration less
than the gravitational
acceleration.
2. d > 1.2 m
(C) Lift is moving
vertically up with
constant speed.
3. d < 1.2 m
(D) Lift is falling freely. 4. No water leaks out of
the jar
a. A→1; B→1; C→1; D→4
b. A→4; B→3; C→2; D→1
c. A→4; B→1; C→3; D→2
d. A→3; B→2; C→1; D→4
O R
E
rE
O
E
r
O R
E
rE
O R
E
rE
Space for rough work
Physics 14
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
c a c b c b c b b b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
d b b a c d d c d b
21. 22. 23. 24. 25.
a b c c c
1. (c) Displacement vector ˆˆ ˆr xi yj zk= ∆ + ∆ + ∆
ˆˆ ˆ ˆ ˆ(3 2) (4 3) (5 5)i j k i j= − + − + − = +
2. (a) Tangential force (Ft) of the bead
will be given by the normal reaction
(N), while centripetal force (Fc) is
provided by friction (Fr). The bead
starts sliding when the centripetal
force is just equal to the limiting
friction. Therefore, tF ma m L Nα= = =
∴ Limiting value of frictionmax( )rf N m Lµ µ α= = . . . (i)
Angular velocity at time t is tω α=
∴ Centripetal force at time t will be
2 2 2 cF mL mL tω α= = . . . (ii)
Equation (i) and (ii), we get tµα
=
For ( )max
, . .c rt F f i eµα
> > , the bead starts sliding.
In the figure Ft is perpendicular to the paper inwards.
3. (c) Power F v Fv= = i
( ) volume
ddm
F v vdt dt
ρ × = =
( )volumedv
dtρ
=
( ) 2v Av Avρ ρ= =
∴ Power 3P Avρ= or 3P v∝
4. (b) Let ρ be the density of lead. Then 34
3M Rπ ρ= =
= mass of total sphere
3
1
4
3 2
Rm π ρ = =
mass of removed part
8
M =
2
7
8 8
M Mm M= − = = mass of remaining sphere
Choosing the centre of big sphere as the origin,
1 1 2 2
1 2
CM
m x m xX
m m
+=
+
( ) ( ) ( ) 2/8 / 2 7 /8
0M R M x
M
× + ×=
Solving, we get 214
Rx
−=
5. (c) In air effg g=
In water, Buoyant force
Masseffg = g –
4
w
b
d gg g
d= − =
Now 2 ,eff
tt
gπ= Hence 02t t=
6. (b) Soap solution has lower surface tension as compared
to pure water so h is less for soap solution.
7. (c) 1 2F F=
⇒ 1 2Y A Y A1 2α ∆θ = α ∆θ ⇒ 1 2
2 1
3
2
Y r
Y r= =
8. (b) Let 0°C be the reference temperature for zero heat,
then initial heat energy = final heat energy
1 1 1 2 2 2 3 3 3m s m s m sθ θ θ+ + ( )1 1 2 2 3 3m s m s m s= + +
⇒ θ 1 1 1 2 2 2 3 3 3
1 1 2 2 3 3
m s m s m s
m s m s m s
θ θ θθ
+ +=
+ +
1 1 1 2 2 2 3 3 3
1 1 2 2 3 3
Vd s Vd s Vd s
Vd s Vd s Vd s
θ θ θ+ +=
+ +1 1 1 2 2 2 3 3 3
1 1 2 2 3 3
d s d s d s
d s d s d s
θ θ θ+ +=
+ +
9. (b) Speed of sound in a gas is given by1
,RT
v vM M
γ= ∝
∴ 1 2 2
2 1 1
v M m
v M m= =
Here, γ 5
3
p
V
C
Cγ = = for both the gases
monoatomic
5
3γ =
10. (b)
1
1 1 2 11 1 2 2
1 2
V TT V T V
V T
γ
γ γ
−
− − = ⇒ =
or
1 1
1 1 1
2 2 22
V T T
V T T
γ γ− −
= = ( )1
1 1
22γ−= >
11. (d) Wien’s displacement law is mT bλ = (b = Wien’s
constant)
∴ 62.88 10 nm-K
2880Km
b
Tλ
×= =
y
x
60o
F sin 60o
F c
os
60
o
F
A L
cF
tF
is inwardstF
15 Mock Test-1
∴ λ = 1000 nm
Energy distribution with wavelength will be as follows:
From the fig. it is clear that 1
2
U
U (In fact U2
is maximum)
12. (b) Work done by the field
( ) ( )B A C Ae V V e V V= − = − (∵ VB = VC)
⇒ 19
19
6.4 10( ) 4
1.6 10C A
WV V V
e
−
−
×− = = =
× 4V
13. (b) In the electric field between plates of parallel plate
capacitor 0
Eσε
=
∴ 0
qF qE
σε
= = . . . (i)
When one plate is removed, the electric field becomes
02
Eσε
′ =
∴ 02
F qEσε
′ ′= = . . . (ii)
1
2 2
F FF
F
′′= ⇒ =
14. (a) Resistance of 9 mm cable 5= Ω
As 1
RA
∝ or2
1;R
r∝ resistance of 3 mm
Cable 9 5 45× = Ω
In second case 6 wires are connected in parallel, so total
resistance of cables 45
7 56
= = ⋅ Ω
15. (c) Magnetic field at any point lying on the current
carrying straight conductor is zero. Here
H1 = Magnetic field at M due to current in PQ.
H2 = Magnetic field at M due to QR + magnetic field at M
due to QS + magnetic field at M due to PQ
11 1
30
2 2
HH H= + + = ⇒ 1
2
2
3
H
H=
16. (d) If monochromatic light is replaced by white light, a
few coloured fringes are seen on either side of a central
white fringe.
17. (d) Electric field of an electromagnetic wave in free space
is given by 7 ˆ10cos(10 )E t kx j= +
which is acting along
y-direction. As E is varying with x and t, hence
propagation of electromagnetic wave takes place along –x
axis. Thus statement (d) is wrong. Comparing the relation,
710cos(10 )E t kx= + with standard equation of
electromagnetic wave
0 0
2 2 2cos ( ) cosE E t x E t x
π πυ πυ
λ λ λ = + = +
We have, 0 10 / .E V m= Thus statement (c) is correct.
7210
πυλ
= or 8
72 (3 10 )10
πλ
× ×=
or 22
60 60 188.47
λ π= = × ≈ m
Thus, statement (a) is correct.
⇒ 2 2 1
0.033 /60 30
k rad mπ πλ π
= = = = rad / m
Thus, statement (b) is wrong.
18. (c)hc
E Wλ
= = − . . . (i)
2hc
E Wλ
= = −′
. . . (ii)
Dividing (ii) by (i)2
2 2
hcW
hc hcW W
hcW
λλ λ
λ
−′= = − = −
′−
⇒ 2hc hc
Wλ λ
= −′
⇒ 1 2 W
hcλ λ= −
′
⇒ 1
2 W
hc
λ
λ
′ =−
⇒ 2
λλ′ > but less than λ
19. (d) Energy required to remove electron in the n = 2
state2
13.63.4
(2)eV= + = + eV
20. (b) Hydrogen bomb is based on nuclear fusion.
21. (a) 3 2 3(1.2 10 m)V l
−= = × 6 31.728 10 m−= ×
∵ Length (l) has two significant figures, the volume (V) will
also have two significant figures. Therefore, the correct
answer is 6 31.7 10 mV lV −= = ×
22. (b) sin ,a g lθ= = sin
h
θ
⇒ 21sin
sin 2
hg t= θ
θ⇒ t ∝
1
sin θ
⇒ 2 1
1 2
sin
sin
t
t=
θ
θ2 1
1 2
sin 1/ 2 1
sin 3 / 2 3= = = 1
2
3/ 3
3 3
tt = = = s
499
5
99
999
1
000
1499
1
500
( ) ( )A BW q dV e V V= − = − −
Physics 16
23. (c)
( )
1 2
221
GM GM
x x=
−
⇒ ( )22
100 10, 000
1x x=
−
or ( )
2
2
1
100 1
x
x=
−⇒
21 1
10 1 11
xx
x= ⇒ =
−m
24. (c) In C.G.S. 3
29axial
MB
x= = . . . (i)
equaterial 3 3
8
2
M MB
xx= =
. . . (ii)
From equation (i) and (ii) equaterial 36B = gauss.
25. (c) rvn n nr
′ = +/ 2 3
2
vn n n
v= + =
% change in frequency
100% 1 100%n n n
n n
′ ′− = × = − ×
31 100% 50%
2
= − × =
JEE Advance Paper -I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
c a,b,c a,c,d b,d a,c,d a,b,d 9 6 6 5
11. 12. 13. 14. 15. 16. 17. 18.
7 2 8 5 a,c d a,d a,c
1. (c) For air to glass 1
1.5 1.4 1 1.5 1.4− −= +
f R R
∴∴∴∴ 1 3=f R
For glass to air. 2
1 1.4 1.5 1 1.4− −= +
− −f R R
∴∴∴∴ 2 2=f R
2. (a, b, c) ˆ2 ( )[ ]F I L R i B= + ×
3. (a, c, d) 2 1 1 1 3 2[ ], [ ], [ ]h ML T c LT G M L T
− − − −≡ ≡ ≡
⇒ 3
,hc hG
M LG c
∝ ∝
4. (b, d) For first oscillator 2 2
1 1
1
2E m aω=
and 1p mv m a bω= = =
⇒ 1
1a
b mω= . . .(i)
For second oscillator
2 2
2 2 2
1, and 1
2E m R mω ω= =
. . .(ii)
⇒ 22
1
an
b
ωω
= =
⇒ 1 2
2 2 2 2
1 2
E E
a aω ω=
⇒ 1 2
1 2
E E
ω ω=
5. (a, c, d) Since, the spool rolls over the horizontal surface,
therefore, instantaneous axis of rotation passes through the
point of contact of spool with the horizontal surface.
About the instantaneous axis of rotation, moment
produced by F is clockwise. Therefore, the spool rotates
clockwise. In that case acceleration will be rightward and
thread will wind. If rotational motion of spool is
considered about its own axis then resultant moment on it
must be clockwise. But moment produced by the force F
is anticlockwise and its magnitude is equal to F.r,
Hence, moment produced by the friction (about its own
axis) must be clockwise and its magnitude must be greater
than F.r. It is possible only when friction acts leftwards.
Therefore, option (b) is correct.
6. (a, b, d) 1 1 21
1 3
( )+=
+
R V VV
R R⇒
1 3 2 1=V R V R
⇒ 3 1 22
1 3
( )+=
+
R V VV
R R⇒
2 1 2 3=V R V R
7. (9) For refraction at the first surface, 8 cm,u = −
1 1 28 cm, 1, 1.5R µ µ= − = =
⇒ 2 1 2 1
1v u R
µ µ µ µ−− =
⇒ 1.5 1 0.5
' 8 8v+ =
−⇒ ' 8 cmv = −
It means due to the first surface the image is formed at the
centre.
For the second surface 1 29 cm, 1.5, 9 cmu Rµ= − = = −
⇒ 2 1 2 1
2
m m m m
v u R
−− =
⇒ 1 1.9 1 1.5
5 9v
−+ =
−⇒ 9 cmv = −
Thus, the final image is formed at the centre of the sphere.
8. (6) From the figure θ = 60°
So, No. of rectangular surfaces used to form a360
660
°= =
°m
θ3
2a
Line change
Rectangular
surface a
2(L + R)
1 r1 P r2
Planet 2
17 Mock Test-1
9. (6) As is clear from figure. 1 2 3E E E= +
1 2 3
hc hc hc
λ λ λ= +
∴∴∴∴ 2 1 3
1 1 1 1 1 1
2 3 6λ λ λ= − = − = = 6 units
10. (5) Using work energy theorem + = ∆mg FW W KE
− + = ∆mgh Fd KE 1 10 4 18(5)− × × + = ∆KE
⇒ 50∆ =KE ∴∴∴∴ 5=n
11. (7) Kinetic energy of a pure rolling disc having velocity of
centre of mas 2 2
2 2
2
1 1 3
2 2 2 4
mR vv mv mv
R
= + =
So, 2 2
2
3 3(3) (30) ( ) (27)
4 4m mg m v mg+ = +
∴∴∴∴ 2 7 m / sv =
12. (2) 410
A B
dQ dQ
dt dt
=
2 4 4 2 4(400 ) 10 ( )A BR T R T=
So, 2 and 2A B
A B
B A
TT T
T
λλ
= = =
13. (8) Here, 2h, ?T t= =
To work safely with the sample, its activity must be
reduced to 1
.16
From
4
0
1 1 1
2 16 2
nN
N
= = =
∴∴∴∴ n = 4
⇒ 2 4 8ht n T= = × =
14. (3) For maxima, 2 2 2 24,
3d x d x mλ+ − + = m is an
integer
So, 2 2 2 29x m dλ= − ⇒ 3=p
15. (a, c) cθ ≥
⇒ 90°− ≥r c
⇒ sin(90 )° − ≥r c
⇒ cos sin≥r c
Using sin
sin
i
m
ni
r n= and 2
3
sin =n
n
We get, 2 2
2 1 2
2sin
−=m
m
n ni
n
Putting values, we get, correct options as a & c
16. (d) For total internal reflection to take place in both
structures, the numerical aperture should be the least one
for the combined structure and hence, correct option is d.
17. (a, d) 1 2
=I I
⇒ 1 1 2 2=neA v neA v
⇒ 1 1 1 2 2 2=d w v d w v
Now, potential difference developed across MK, =V Bvw
⇒ 1 1 1 2
2 2 2 1
= =V v w d
V v w d and hence correct choice is a & d.
18. (a, c) As 1 2=I I
⇒ 1 1 1 1 2 2 2 2=n w d v n w d v
Now, 2 2 2 2 2 2 1 1 1 2 1
1 2 1 1 1 1 2 2 2 1 2
= = =
V B v w B w n wd B n
V B v w B w n w d B n
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a a,d a,c d a,b b,c 2 7 4 6
11. 12. 13. 14. 15. 16. 17. 18.
3 2 2 3 a a d a
1. (a) Q value of reaction
(140 94) 8.5 236 7.5 219= + × − × = MeV
So, Total kinetic energy of Xe and Sr 219 2 2 215= − − = MeV
So, by conservation of momentum, energy, mass and
charge, only option a. is correct.
2. (a, d) From the given conditions, 1 1 2 2
ρ σ σ ρ< < <
From equilibrium, 1 2 1 2σ σ ρ ρ+ < = +
1
2
2
9
ρ ση
2 −=
pV g and 2 1
1
2
9
ρ ση
−=
QV g
So, 1
1
| |
| |
P
Q
V
V
ηη
= and . 0P QV V
<
3. (a, c) BI c VI≡ℓ
⇒ 2
0I c VIµ ≡
⇒ 0Ic Vµ =
⇒ 2 2 2 2
0µ =I c V
⇒ 2 2
0 0I Vµ ε=
⇒ 0cV Iε =
4. (d) 1 2
03
ρε
=
E CC
C1 ⇒ Centre of sphere and
C2 ⇒ centre of cavity.
5. (a, b) stress
strain=Y
⇒ 1 strain
stress=
Y
1nmn
i
r
θ
n2
ρ
Physics 18
⇒ 1 1
P QY Y> ⇒ P QY Y<
6. (b, c)2
2( ) 1
= −
rP r K
R
7. (2)3
2 2π π= =nh h
mvr
de-Broglie Wavelength h
mvλ =
2
00
(3)2 22
3 3 Li
ara
z
π ππ= = =
8. (7) For m closer to M; 2
2 29− =
ℓ ℓ
GMm Gmma . . .(i)
and for the other m: 2
2 216+ =
ℓ ℓ
Gm GMmma . . .(ii)
From both the equations, 7=k
9. (4) 2( )
−= atE T A e
⇒ 2 2α − −= − +at atdE A e dt AdAe
Putting the values for maximum error,
⇒ 4
100=
dE
E ⇒ % error = 3.
10. (6) 2 224
3ρ π= ∫I r r dr
⇒ 2 2( )( )( )∝ ∫AI r r r dr
⇒ 5 2 2( )( )( )∝ ∫BI r r r dr
∴
6
10=B
A
I
I
11. (3) First and fourth wave interfere destructively. So from
the interference of 2nd
and 3rd
wave only,
⇒ 0 0 0 0 0
22 cos 3
3 3
π π = + + − =
netI I I I I I ⇒ 3=n
12. (2) 1 1
;2
P Qλ λτ τ
= = ⇒ 0
0
( ) p
Q
t
PP
t
Q Q
A eR
R A e
λ
λ
λ
λ
−
−=
At 2 ;τ=t 2P
Q
R
R e=
13. (2) Snell’s Law on 1st surface: 1
3sin
2= n r
1
3sin
2=r
n . . .(i)
⇒ 2
1 2
3 4 3cos 1
4 2
−= − =
nr
n n
1 2
60+ = °r r . . .(ii)
Snell’s Law on 2nd
surface:2sin sinn r = θ
Using equation (i) and (ii) 1sin(60 ) sinθ° − =n r
⇒ 1 1
3 1cos sin sin
2 2n r r
− = θ
⇒ ( )234 3 1 cos
4
d dn
dn dn
θ− − = θ
For 60θ = ° and 3=n ⇒ 2d
dn
θ=
14. (3) ( )
( ) ( )1 2 2 1
3
1 3 1 2 3 2
3.0VVV R R
VV R R V R R
−= =
− − −
15. (a) A→ 1; B→ 2, 4; C→ 3; D→ 4
16. (a) A→ 3, 4; B→ 2, 3, 4; C→ 1, 2; D→ 2, 3, 4
17. (d) Gravitational field intensity at any point outside due to
earth is given by 2 2
( )
GM GME g
R h r= = =
+
where h is height from surface or 2
1E
r∝
but – ve in sign.
Graph is parabolic.
Inside earth g decreases are r increases form surface of
earth.
1x g
E g g xR R
= + = −
2
1E
r− ∝
or E x∝ is straight line decreasing to 0 at centre of earth and
then increases in magnitude but – ve as directed towards
the centre always.
At any point outside 2
1E
r∝ (parabolic graph)
Inside earth E r∝ (linear graph)
In hollow sphere inside 0E = graph along r-axis; outside
as if whole mass lies at centre 2
1E
r∝ decreasing
parabolic graph after sudden rise at surface of earth.
18. (a) A→1; B→1; C→1; D→4
In A, B, C no horizontal velocity is imparted to falling
water, so d remains same.
In D, since its free fall, 0effa =
∴ Liquid won’t fall with respect to lift.
19 Mock Test-2
JEE-MAIN: PHYSICS MOCK TEST-2
SECTION 1 (Multiple Choice Question)
1. In the relation Z
kp eα
−θ
α=
β p is pressure, Z is distance, k
is Boltzmann constant and θ is the temperature. The
dimensional formula of α will be:
a. 2[ ]MLT− b. 2[ ]ML T
c. 0 1[ ]ML T− d. 0 2 1[ ]M L T
−
2. A small block is shot into each of the four tracks as shown
below. Each of the tracks rises to the same height. The
speed with which the block enters the track is the same in
all cases. At the highest point of the track, the normal
reaction is maximum in
a. b.
c. d.
3. A particle, which is constrained to move along x-axis, is
subjected to a force in the same direction which varies
with the distance x of the particle from the origin as 3( ) .F x kx ax= − + Here, k and a are positive constants.
For 0,x ≥ the functional form of the potential energy
U(x) of the particle is
a. b.
c. d.
4. A carpenter has constructed a toy as
shown in the adjoining figure. If the
density of the material of the sphere
is 12 times that of cone, the position
of the center of mass of the toy is
given by.
a. at a distance of 2R from O
b. at a distance of 3R from O
c. at a distance of 4R from O
d. at a distance of 5R from O
5. The total energy of a particle executing simple harmonic
motion is:
a. ∝ x b. ∝ x2
c. independent x d. ∝ x1/2
6. Two satellites A and B go around a planet P in circular
orbits having radii 4R and R respectively. If the speed of
satellite A is 3v, the speed of satellite B will be:
a. 12 v b. 6 v
c. 4
3v d.
3
2v
7. The following diagram shows three soap bubbles A, B and
C prepared by blowing the capillary tube fitted with stop
cocks, S1, S2 and S3. With stop cock S closed and stop
cocks S1, S2 and S3 opened.
a. B will start collapsing with volumes of A and C
increasing
b. C will start collapsing with volumes of A and B
increasing
c. C and A both will start collapsing with the volume of B
increasing
d. Volumes of A, B and C will becomes equal at
equilibrium
8. A steel wire of length 20 cm and uniform cross-section
1 mm2 is tied rigidly at both the ends. The temperature of
the wire is altered from 40°C to 20°C. What is the
magnitude of force developed in the wire? (Coefficient of
linear expansion for steel, α = 1.1 × 10–5
/°C and Y for
steel is 11 22.0 10 N/m× ).
a. 62.2 10× N b. 16 N
c. 8 N d. 44 N
9. The intensity level of two waves of same frequency in a
given medium are 20 dB and 60 dB. Then the ratio of
their amplitudes is:
a. 1 : 4 b. 1 : 16
c. 1 : 410 d. 1 : 100
10. The specific heat of a substance varies as (2t2 + t) ×10
–3
cal/g °C. What is the amount of heat required to raise the
temperature of 100 g of substance through 20°C to 40°C.
S3 S1
S S2
B A
C
U(x)
x
U(x)
x
U(x)
x
U(x)
x
v v
v v
O2 2R
2R
O1
O
2R
Physics 20
a. 37.9 kilocal b. 3.79 kilocal
c. 379 kilocal d. 82 cal
11. A real gas behaves like an ideal gas if its
a. pressure and temperature are both high
b. pressure and temperature are both low
c. pressure is high and temperature is low
d. pressure is low and temperature is high
12. The plates of a parallel plate capacitor of capacity of 50µF
are charged by a battery to a potential of 100 volt. The
battery remains connected the plates are separated from
each other so that the distance between them is doubled.
How much is the energy spent by battery in doing so?
a. 25 × 10–2
J b. –12.5 × 10–2
J
c. – 25 × 10–2
J d. 12.5 × 10
–2 J
13. The magnetic moment produced in a substance of 1gm is
6 × 10–7
ampere-metre2. If its density is 5 gm/cm
3, then
the intensity of magnetisation in A/m will be
a. 8.3 × 106 b. 3.0
c. 1.2 × 10–7
d. 3 × 10–6
14. An ionized gas contains both positive and negative ions. If
it is subjected simultaneously to an electric field along the
+x direction and a magnetic field along the +z direction,
then
a. Positive ions deflect towards +y direction and negative
ions towards –y direction
b. All ions deflect towards +y direction
c. All ions deflect towards –y direction
d. Positive ions deflect towards –y direction and negative
ions towards +y direction
15. A hundred turns of insulated copper wire are wrapped
around an iron cylinder of area 1 × 10–3
m2 are connected
to a resistor. The total resistance in the circuit is 10 ohms.
If the longitudinal magnetic induction in the iron changes
from 1 Wb m–2
, in one direction to 1 Wb 2m− in the
opposite direction, how much charge flows through the
circuit?
a. 2 × 10–2
C b. 2 × 10–3
C
c. 2 × 10–4
C d. 2 × 10–5
C
16. A resistor R, an inductor L and a capacitor C are
connected in series to an oscillator of frequency n. If the
resonant frequency is nr, then the current lags behind
voltage, when
a. n = 0 b. n < nr
c. n = nr d. n > nr
17. The intensity ratio at a point of observation due to two
coherent waves is 100 : 1. The ratio between their
amplitudes is:
a. 1 : 1 b. 1 : 10
c. 1 : 100 d. 10 : 1
18. An electromagnetic wave propagating along north has its
electric field vector upwards. Its magnetic field vector
points towards
a. north b. east
c. west d. downwards
19. The ionisation energy of 10 times ionised sodium atom is
a. 13.6 eV b. 13.6 11× eV
c. 13.6
11 eV d. 213.6 (11)× eV
20. The example of nuclear fusion is
a. Formation of barium and krypton from uranium
b. Formation of helium from hydrogen
c. Formation of plutonium 235 from uranium 235
d. Formation of water from hydrogen and oxygen
SECTION 2 (Numeric Value Question)
21. A force of 5 N acts on a particle along a direction making
an angle of 60° with vertical. Its vertical component be:
a. 10 N b. 3 N
c. 4 N d. 2.5 N
22. A neutral water molecule (H2O) in it's vapor state has an
electric dipole moment of magnitude 6.4 ×10–30
C–m.
How far apart are the molecules centres of positive and
negative charge?
a. 4 m b. 4 mm
c. 4 µm d. 4 pm
23. A cell of e.m.f. 1.5 V having a finite internal resistance is
connected to a load resistance of 2 .Ω . For maximum
power transfer, the internal resistance of the cell in ohms
should be:
a. 4 b. 0.5
c. 2 d. None of these
24. Two plane mirrors are at 45° to each other. If an object is
placed between them, then the number of images will be
a. 5 b. 9
c. 7 d. 8
25. A photon of energy 8 eV is incident on a metal surface of
Threshold frequency 1.6×1015
Hz. The kinetic energy of the
photoelectrons emitted (in eV) 34(Take 6 10 J s)h −= × −
a. 1.6 b. 6
c. 2 d. 1.2
21 Mock Test-2
JEE ADVANCE PAPER-I
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. A piece of wire is bent in the shape of a parabola 2=y kx
(y-axis vertical) with a bead of mass m on it. The bead can
slide on the wire without friction. It stays at the lowest
point of the parabola when the wire is at rest. The wire is
now accelerated parallel to the x-axis with a constant
acceleration a. The distance of the new equilibrium
position of the bead, where the bead can stays at rest with
respect to the wire, from the y-axis is
a. a
gk b.
2
a
gk c.
2a
gk d.
4
a
gk
2. A block of mass m is placed on a surface with a vertical
cross-section given by
3
.6
xy = If the coefficient of friction
is 0.5, the maximum height above the ground at which the
block can be placed without slipping is:
a. 1
m3
b. 1
m2
c. 1
m6
d. 2
m3
3. A ball moves over a fixed track as shown in the figure.
From A to B the ball rolls without slipping. Surface BC is
frictionless. KA, KB and KC are kinetic energies of the ball
at A, B and C, respectively. Then
a. ;> >A C B Ch h K K b. ;> >A C C Ah h K K
c. ;= =A C B Ah h K K d. ;< >A C B Ch h K K
4. A small block of mass of 0.1 kg lies on a fixed inclined
plane PQ which makes an angle θ with the horizontal. A
horizontal force of 1 N acts on the block through its centre
of mass as shown in the figure. The block remains
stationary if (take g = 10 m/s2).
a. 45θ = °
b. 45θ > ° and a frictional force acts on the block towards P.
c. 45θ < ° and a frictional force acts on the block towards
d. a frictional force acts on the block towards
5. In the figure, a ladder of mass m is shown leaning against
a wall. It is in static equilibrium making an angle θwith
the horizontal floor. The coefficient of friction between
the wall and the ladder is µ1 and that between the floor
and the ladder is µ2. The normal reaction of the wall on
the ladder is N1 and that of the floor is N2.
If the ladder is
about to slip, then
a. 1 20, 0= ≠µ µ
and
2 tan2
mgN θ =
b. 1 20, 0≠ =µ µ
and
1tan
2
mgN θ =
c. 1 20, 0≠ ≠µ µ and
1
1 21
mgN =
+ µ µ
d. 1 20, 0= ≠µ µ
and 1 tan
2
mgN θ =
6. Two small particles of equal masses start moving in
opposite directions from a point A in a horizontal circular
orbit. Their tangential velocities are v and 2v
respectively, as shown in the figure. Between collisions,
the particles move with constant speeds. After making
how many elastic collisions, other than that at A, these
two particles will again reach the point A?
a. 4 b. 3
c. 2 d. 1
A 2v v
1µ
2µ
θ
O
Q
1 N
P θ
hc hA
A
B
C
µ1
µ2
Physics 22
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. The densities of two solid spheres A and B of the same
radii R vary with radial distance r as ( )ρ =
A
rr k
R and
5
( ) ,ρ =
B
rr k
R respectively, where k is a constant.
The moments of inertia of the individual spheres about
axes passing through their centres are IA and IB,
respectively. If ,10
=B
A
I n
Ithe value of n is
8. Gravitational acceleration on the surface of a planet is
6,
11g where g is the gravitational acceleration on the
surface of the earth. The average mass density of the
planet is 1
3times that of the earth. If the escape speed on
the surface of the earth is taken to be 11 kms–1
, the escape
speed on the surface of the planet in kms–1
will be
9. A bullet is fired vertically upwards with velocity v from
the surface of a spherical planet. When it reaches its
maximum height, its acceleration due to the planet’s
gravity is th1/ 4 of its value at the surface of the planet. If
the escape velocity from the planet is ,escv v N= then the
value of N is (ignore energy loss due to atmosphere)
10. The ends Q and R of two thin wires, PQ and RS, are
soldered (joined) together. Initially each of the wires has a
length of 1m at 10 C.° Now the end P is maintained at
10 C,° while the end S is heated and maintained at
400 C.° The system is thermally insulated from its
surroundings. If the thermal conductivity of wire PQ is
twice that of the wire RS and the coefficient of linear
thermal expansion of PQ is 5 11.2 10 ,K− −× the change in
length of the wire PQ is:
11. Two soap bubbles A and B are kept in a closed chamber
where the air is maintained at pressure 8 N/m2. The radii
of bubbles A and B are 2 cm and 4 cm, respectively.
Surface tension of the soap-water used to make bubbles is
0.04 N/m. Find the ratio ,B
A
n
n where An and Bn are the
number of moles of air in bubbles A and B, respectively.
[Neglect the effect of gravity.]
12. A cylindrical vessel of height 500 mm has an orifice
(small hole) at its bottom. The orifice is initially closed
and water is filled in it up to height H. Now the top is
completely sealed with a cap and the orifice at the bottom
is opened. Some water comes out from the orifice and the
water level in the vessel becomes steady with height of
water column being 200 mm. Find the fall in height
(in mm) of water level due to opening of the orifice.
[Take atmospheric pressure 5 21.0 10 N / m ,= × density of
water = 1000 kg/m3 and g = 10 m/s
2. Neglect any effect of
surface tension.]
13. Consider two solid spheres P and Q each of density 8 gm
3cm−and diameters 1 cm and 0.5 cm, respectively. Sphere
P is dropped into a liquid of density 0.8 gm 3cm− and
viscosity 3η = poiseulles. Sphere Q is dropped into a
liquid of density 1.6 gm 3cm− and viscosity 2η =
poiseulles. The ratio of the terminal velocities of P and Q
is:
14. A 0.1 kg mass is suspended from a wire of negligible
mass. The length of the wire is 1m and its cross-sectional
area is 7 24.9 10 m .−× If the mass is pulled a little in the
vertically downward direction and released, it performs
simple harmonic motion of angular frequency 140 rad 1.s− If the Young’s modulus of the material of the wire is
9 110 Nm ,n −× the value of n is
SECTION 3 (Maximum Marks: 12)
Paragraph based questions (2 paragraphs, each having 2
MCQs with one correct answer only)
Paragraph for Question No. 15 to 16
A small block of mass 1 kg is released from rest at the top of a
rough track. The track is circular arc of radius 40 m. The block
slides along the track without toppling and a frictional force
acts on it in the direction opposite to the instantaneous velocity.
The work done in overcoming the friction upto the point Q, as
shown in the figure, below, is 150 J. (Take the acceleration due
to gravity, 210 m / s ).g −=
R
P R
Q
30°
23 Mock Test-2
15. The speed of the block when it reaches the point Q is
a. 15 ms− b. 110 ms−
c. 110 3 ms −
d. 120 ms−
16. The magnitude of the normal reaction that acts on the
block at the point Q is
a. 7.5 N
b. 8.6 N
c. 11.5 N
d. 22.5 N
Paragraph for Question No. 17 to 18
Two waves 1 cos(0.5 100 )y A x tπ π= −
2 cosy A=
(0.4 92 )x tπ π− are travelling in a pipe placed along x-axis.
17. Find the number of times intensity is maximum in time
interval of 1 sec
a. 4 b. 6
c. 8 d. 10
18. Find wave velocity of louder sound
a. 100 m/s b. 192 m/s
c. 200 m/s d. 96 m/s
Space for rough work
Physics 24
JEE ADVANCE PAPER-II
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. A person trying to lose weight by burning fat lifts a mass
of 10 kg upto a height of 1 m 1000 times. Assume that the
potential energy lost each time he lowers the mass is
dissipated. How much fat will he use up considering the
work done only when the weight is lifted up? Fat supplies 73.8 10 J× of energy per kg which is converted to
mechanical energy with a 20% efficiency rate (Take 2
9.8 ms ) :g−
=
a. –32.45 10 kg× b. –3
6.45 10 kg×
c. –39.89 10 kg× d. –3
12.89 10 kg×
2. If the resultant of all the external forces acting on a system
of particles is zero, then from an inertial frame, one can
surely say that
a. linear momentum of the system does not change in
time
b. kinetic energy of the system does not change in time
c. angular momentum of the system does not change in
time
d. potential energy of the system does not change in time
3. A point mass of 1 kg collides elastically with a stationary
point mass of 5 kg. After their collision, the 1 kg mass
reverses its direction and moves with a speed of 2
1ms .−Which of the following statement(s) is (are) correct
for the system of these two masses?
a. Total momentum of the system is 3 kg 1ms−
b. Momentum of 5 kg mass after collision is 4 kg 1ms−
c. Kinetic energy of the centre of mass is 0.75 J
d. Total kinetic energy of the system is 4 J
4. A solid sphere of mass M, radius R and having moment of
inertia about an axis passing through the centre of mass as
I, is recast into a disc of thickness t, whose moment of
inertia about an axis passing through its edge and
perpendicular to its plane remains I. Then, radius of the
disc will be
a. 2
15
R b.
2
15R
c. 4
15
R d.
4
R
5. A ball of mass (m) 0.5 kg is attached to the end of a string
having length (L) 0.5 m. The ball is rotated on a
horizontal circular path about vertical axis. The maximum
tension that the string can bear is 324 N. The maximum
possible value of angular velocity of ball (in rad/s)
is
a. 9 b. 18
c. 27 d. 36
6. A small mass m is attached to a massless string whose
other end is fixed at P as shown in the figure. The mass is
undergoing circular motion in the x-y plane with centre at
O and constant angular speed ω . If the angular
momentum of the system, calculated about O and P are
denoted by
OL and P
L
respectively, then.
a.
OL and
PL do not vary with time
b.
OL varies with time while
PL remains constant.
c.
OL remains constant while
PL varies with time
d.
OL and
PL both vary with time.
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. A body sends waves 500 mm long through medium A and
0.25 m long in medium B. If velocity of waves in medium
A is 16 m/s, what is the velocity (in m/s) of waves in
medium B?
8. A light pointer fixed to one prong of a tuning fork touches
gently a smoked vertical plate. The fork is set vibrating
and the plate is allowed to fall freely. Two complete
oscillations are traced when the plate falls through 40 cm.
What is the frequency (in Hz) of the tuning fork?
9. A water tank is 20 m deep. If the water barometer reads
10 m at that place, then what is the pressure at the bottom
of the tank in atmosphere?
m
P
z
O
L
m
25 Mock Test-2
10. There is a soap bubble of radius 42.4 10
−× m in air cylinder
which is originally at the pressure of 5 210 N/m . The air in
the cylinder is now compressed isothermally until the
radius of the bubble is halved. The pressure of air in the
cylinder now becomes 5 210 N/m .n× The surface tension
of soap film is –10.08Nm . Find the integer value of n.
11. A length of wire carries a steady current I. It is bent first
to form a circular plane coil of one turn. The same length
is now best more sharply to give double loop of smaller
radius. If the same current I is passed, the ratio of the
magnitude of magnetic field at the centre with its first
value is.
12. A current 1 amp is flowing in the sides of an equilateral
triangle of side 24 5 10 m.−⋅ × the magnetic field at the
centroid of the triangle in the unit of 5(10 T)− is.
13. There are two infinite long parallel straight current
carrying wires, A and B separated by a distance r (Fig.)
The current in each wire is I. The ratio of magnitude of
magnetic field at points P and Q when points P and Q lie
in the plane of wires is
14. A length of wire carries a steady current I. It is bent first
to form a circular plane coil of one turn. The same length
is now bent more sharply to give double loop of smaller
radius. If the same current I is passed, the ratio of the
magnitude of magnetic field at the centre with its first
value is
SECTION 3 (Maximum Marks: 12)
Matching type questions with 4 options
15. Work is defined as dot product of force and displacement
.W dW F dS= = ⋅∫ ∫
It is a scalar quantity. The total work
done will depend on the displacement and the force,
which may be constant or variable.
Thus in different situations of variable force applied in
column I, the final expression for work done can be
expressed as in column II.
Column I Column II
(A) Force constant in magnitude acts
at constant angle θ with direction
of motion
1. 21cos
2kx θ
(B) Force constant in magnitude acts
at angle θ which varies as θ kxθ = 2. sinF
Kθ θ
(C) Force varies with distance x as
F kx= but angle θ is constant
3. sinFx θ
(D) Force is constant in magnitude
but changes in direction with
changing angle and always acts
along radius of circular path
4. zero
a. A→1, B→2, C→3, D→4 b. A→3, B→2, C→1, D→4
c. A→4, B→3, C→2, D→1 d. A→2, B→1, C→4, D→3
16. Four charges 1 2 3, ,Q Q Q and 4Q of same magnitude are fixed
along the x-axis at 2 , ,x a a a= − − + and 2a+ respectively.
A positive charge q is placed on the positive y-axis at a
distance 0b > . Four options of the signs of these charges
are given in column I. the direction of the forces on the
charge q is given in column II. Match column I with
column II and select the correct answer using the code
given below the lists.
Column I Column II
(A) 1 2 3 4, , ,Q Q Q Q all positive 1. x+
(B) 1 2,Q Q positive;
3 4,Q Q negative 2. x−
(C) 1 4,Q Q positive; 2 3,Q Q negative 3. y+
(D) 1 3,Q Q positive;
2 4,Q Q negative 4. y−
q+
1
( 2 ,0)
Q
a−
(0, )b
2
( ,0)
Q
a−3
( ,0)
Q
a+4
( 2 ,0)
Q
a+
P
O
a
2a
r
r
P
I
Q
A B
I
Physics 26
a. A→3, B→1, C→4, D→2
b. A→4, B→2, C→3, D→1
c. A→3, B→1, C→2, D→4
d. A→4, B→2, C→1, D→3
17. The vibration of body can be under various types of
forces. The vibration are classified mentioned in column I
under the conditions mentioned in column II. Match the
type of vibrations in column I with conditions in
column II.
Column I Column II
(A) Free vibrations 1. A body vibrating in
viscous medium
(B) Forced vibrations 2. A body vibrating under
its natural restoring force
(C) Resonant vibrations 3. A body vibrating under
the influence of another
vibrating body
(D) Damped vibrations 4. A body vibrating with
its natural frequency
under the influence of
another vibrating body
of same frequency
a. A→1, B→2, C→3, D→4
b. A→4, B→3, C→2, D→1
c. A→2, B→3, C→4, D→1
d. A→3, B→4, C→1, D→2
18. To determine specific heat of different substances we use
different types of calorimeters. Can you match the type of
colorimeter used named in column I to the substance
whose specific heat is determined by corresponding
calorimeter in column II.
Column I Column II
(A) Regnault’s
calorimeter
1. To determine specific
heat of solids at very
low temperatures
(B) Joly’s differential
calorimeter
2. To determine specific
heat of solids or gases
at constant pressures
(C) Calender and
Barmer’s calorimeter
3. Used to measure specific
heat at constant volume
(D) Nernst’s vacuum
calorimeter
4. Specific heat of liquids
and gases at constant
pressure
a. A→1, B→2, C→3, D→4
b. A→2, B→3, C→4, D→1
c. A→3, B→4, C→1, D→2
d. A→4, B→3, C→2, D→1
Space for rough work
27 Mock Test-2
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a a d c c b c d d a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
d a b c a d d b d c
21. 22. 23. 24. 25.
d d c c c
1. (a) 1 3 2
2 2[GM] [M L T M][V] [L T ]
[ ] [L]r
− −−= = = ⇒ [ ]
Z
kθ α =
Further [ ]p α= β
∴ [ ]Z
k
p p
α θβ = =
Dimensions of k θ are that to energy.
Hence, [ ]2 2
2[K ] [ML T ][MLT ]
[ ] [L]z
−−θ
α = = =
2. (a) Since, the block rises to the same heights in all the
four cases, from conservation of energy, speed of the block
at highest point will be same in all four cases. Say it is v0
Equation of motion will be
2
0mv
N mgR
+ =
or
2
0mvN mg
R= −
R (the radius of curvature) in first case is minimum.
Therefore, normal reaction N will be maximum in first
case.
Note: In the question it should be mentioned that all the
four tracks are frictionless. Otherwise, v0 will be different
in different tracks.
3. (d) dU
Fdx
= −
∴ . dU F dx= − or 3
0 ( ) ( )
x
U x kx ax dx= − − +∫
2 4
( )2 4
kx axU x = −
( ) 0U x = at x = 0 and
2kx
a=
( )U x = Negative for
2kx
a>
From the given function we can see that
F = 0 at x = 0 i.e., slope of U – x graph is zero at x = 0.
4. (c) If the density of cone be ρ, then its mass will be
( ) ( )2 3
1
1 162 4
3 3m R R Rπ ρ π ρ= =
And its centre of mass will be at a height 4
4 4
h RR= =
From O on the line of symmetry, i.e., 1y R=
Similarly the mass of the sphere
( )3 3
2 1
412 16 3
3m R R mπ ρ π ρ= = =
And its centre of mass will be at its centre 2 ,O i.e.,
2 4 5y R R R= + = (from O).
Now treating the sphere and cone as point masses with
their masses concentrated at their centres of mass
respectively and taking the line of symmetry as y-axis
with origin at O, the centre of mass of the toy is given by
1 1 2 2
1 2
CM
m y m yY
m m
+=
+1 1
1 1
3 54
3
m R m RR
m m
× + ×= =
+
i.e., centre of mass of the toy is at a distance 4R from O on
the line of symmetry, i.e., at the apex of the cone.
5. (c) The total energy of a particle in simple harmonic
motion is constant.
6. (b) Orbital speed of a satellite 0
GMv
r=
⇒ 4
2B A
A B
v r R
v r R= = =
⇒ ( )2 2 3 6B Av v v v= = × =
7. (c) Excess pressure inside soap bubble is inversely
proportional to the radius of bubble, i.e., ∆P ∝ 1/r′. This
means that bubbles A and C posses greater pressure inside
it than B. So the air will move from A and C, towards B.
2R
2R
C
O2 m2
y
O2
O1 O1 m1
O O
4R
v0
N + mg
Physics 28
8. (d) F Y A= ∆α∆θ
( ) ( )11 62.0 10 1 10−= × × × ( )( )51.1 10 40 20 44 N−× × − =
9. (d) 22 1 10
1
10logI
L LI
− =
⇒ 210
1
60 20 10logI
I− =
24 42 2
2
1 1
10 10I a
I a= ⇒ =
⇒ 21 2
1
100 : 1 : 100a
a aa
= ⇒ =
10. (a) dQ mc dt=
2
1
t
t
Q mc dt= ∫
( )2
1
40
2 3
20
100 2 10
t
t
t t dt
=−
=
= × + ×∫40
3 21
20
210
3 2
t t− = +
( ) ( ) ( ) ( )3 3 21 2 110 40 20 40 20
3 2
− = − + −
33.79 10= × cal = 3.79 kilocal
11. (d) A real gas behaves like an ideal gas at low pressure
and high temperature.
12. (a) When separation between the plates is doubled the
capacitance becomes one half i.e., 25C F′ = µF
Energy spent by battery qV=
( ) 2C V V C V′ ′= =
( )26 225 10 100 25 10− −= × × = × J
13. (b) mass/density
M MI
V= = ,
Given mass = 1gm = 310− kg
And density3
3 3 3
2 3 3
5 105 / 5 10 /
(10 )
kggm cm kg m
m
−
−
×= = = ×
Hence 7 3
3
6 10 5 103
10I
−
−
× × ×= =
14. (c) As the electric field is switched on, positive ion will
start to move along positive x-direction and negative ion
along negative x-direction. Current associated with motion
of both types of ions is along positive x-direction.
According to Fleming's left hand rule force on both types
of ions will be along negative y-direction.
15. (a) 3
2100 1 10 22 10
10
d nAdBdQ C
R R
φ −−× × ×
= = = = × C
16. (d) The current will lag behind the voltage when reactance
of inductance is more than the reactance of condenser.
Thus, 1
LC
ωω
> or LC
1>ω
or LC
nπ2
1> or rnn >
where, nr = resonant frequency.
17. (d) 2I a∝ ⇒ 1 1
2 2
100 10
1 1
a I
a I= = =
18. (b) The direction of propagation of electromagnetic wave
is given by the direction of ( ).E B×
Here, the em wave is
propagating along north. The electric field vector is acting
upwards, so the magnetic field vector will point towards
east.
19. (d) (Eion)Na2 2( ) (11) 13.6ion HZ E eV= = eV
20. (c) Fast neutrons can escape from the reaction. So as to
proceed the chain reaction. Slow neutrons are best.
21. (d) The component of force in vertical direction
1
cos cos60 5 2.5 N2
F F= θ = ° = × =
22. (d) There are 10 electrons and 10 protons in a neutral
water molecule.
So it's dipole moment is ( )2 10 (2 )p q l e l= =
Hence length of the dipole i.e,. distance between centres
of positive and negative charges is
20
12
19
6.4 102 4 10 m 4
10 10 1.6 10
pl pm
e
−−
−
×= = = × =
× ×4 pm
23. (c) For maximum power int 2extR R= = Ω
24. (c) 360
1 745
n°
= − =°
25. (c)34 15
0 6 10 1 6 10W hv −= = × × ⋅ × joule
34 15
19
6 10 1 6 10
1 6 10
−
−
× × ⋅ ×= =
⋅ ×eV = 6 eV
∴ 8 6 2kE hv W eV eV eV= − = − = 8 eV –6 eV = 2eV
y
x
60o
F sin 60o
F c
os
60
o
F
29 Mock Test-2
JEE Advance Paper -I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
b c a,b,c a,c c,d c 6 3 2 7
11. 12. 13. 14. 15. 16. 17. 18.
6 6 3 4 b a a c
1. (b) tana
gθ =
tan 2dy
kxdx
θ = =
⇒ 2
ax
gk=
2. (c) sin cossmg mg=θ µ θ [when partied is just balanced]
⇒ tan su=θ ⇒
2
tan2
dy x
dx
θ = =
∴
2
0.52
x= ⇒ 1x =
⇒
1
6y =
3. (a, b, c) = +A A AE mgh K
⇒ =B BE K
⇒ = +C C CE mgh K
Using conservation of energy = =A B CE E E
⇒ >B CK K
⇒ >B AK K
⇒ ( ) ( ) 0− + − =A C A CMg h h K K ⇒−
− = C AA C
K Kh h
Mg
4. (a, c) 0 ( 45 )f = θ = °
So, for block to be at rest cos sinF mgθ = θ
LHS cos 1 cos45F θ = × °
1
2=
RHS 1 1
sin 0.1 102 2
mg θ = × × =
LHS = RHS)
If 45 ; sin cosmg Fθ > ° θ > θ
∴ Friction acts towards Q.
If 45θ < ° F cos sinmgθ > θ
∴ Friction acts towards P.
5. (c, d) Condition of translational equilibrium 1 2 2µ=N N
⇒ 2 1 1µ= =N N Mg , Solving 2
1 21 µ µ=
+mg
N
⇒ 2
1
1 21
µµ µ
=+
mgN Applying torque equation about corner (left)
point on the floor
⇒ 1 1cos sin cos
2mg N N= +ℓ
ℓ ℓ1θ θ µ θ
Solving 1 2
2
1tan
2
−µ µθ =
µ
6. (c) Velocity will exchange after each collision
7. (6) 2 22
43
ρ π= ∫I r r dr
⇒ 2 2( )( )( )∝ ∫AI r r r dr
⇒ 5 2 2( )( )( )∝ ∫BI r r r dr
∴ 6
10=B
A
I
I
8. (3)' 6 ' 2
;11 3
g
g
ρρ
= =
Hence, ' 3 6
22
R
R=
⇒ 2
esc
2
sec
' ' ' 3
V 11
V R
R
ρρ
∝ = ⇒ esc' 3km/s.v =
9. (2) At height R from the surface of planet acceleration due
to planet’s gravity is 1
4th in comparison to the value at the
surface
So, 2 21 1and 0
2 2esc
GMm GMm GMnmv mv
R R R R− + = − − + =
+
∴ 2escv v=
2 /3πV
st1 collision
A 2V V
V 2V
2V
nd2 collision
2 /3π
2 /3π
F
θmg
F
Fcosθ
MgsinθMg cosθ
fs N
y
N cosθ
ma
N
θ N
θ
sin θ
x
ma
θ θ
Physics 30
10. (1)
1( 10)d dxℓ = θ −α d∆ = ∫ℓ ℓ
∵ 10 130
1x
θ − = ⇒ 10 130xθ = +
1
1
0
(130 )x dxα∆ = ∫ℓ 2
1130
2
xα∆ =ℓ
5 1
130 1.2 102
−∆ = × × ×ℓ = 0.78 mm 1≃
11. (6) 2
0
416N / m
A
A
TP P
R= + =
⇒ 2
0
412N / m
B
B
TP P
R= + =
⇒
3
6
= =
B B B
A A A
n P R
n P R
12. (6) 3 2
0 98 10 N / mP P ghρ= − = ×
⇒ 0 0 =PV PV
⇒ 5 310 [ (500 )] 98 10 [ (500 200)]− = × −A H A
⇒ 206 mmH =
Level fall = 206 – 200 = 6 mm
13. (3) 06 L grv V Vgπη ρ ρ+ =
6( )
6 ( )
Q QP P P L P
Q P P Q Q L Q
rV V V g
V r V V
πηρ ρπη ρ ρ−
= ×−
3
3
.(8 0.8)
. (8 1.6)
Q QP
P P Q
rr
r r
η
η−
= ×−
2
7.2
6.4
QP
Q P
r
r
η
η
= × ×
7.2 24 3
6.4 3= × × =
14. (4) YA
mLω =
15. (b) Using work energy theorem
21sin 30
2fmg R W mv° + =
⇒ 2
200 1502
v− =
⇒ 10 m/sv =
16. (a)2
cos60mv
N mgR
− ° =
⇒ 5
5 7.5 N2
N = + =
17. (a) 1
1 2 4f f s−− =
18. (c) 1 2 200 m / sv v= = m/s
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d a a,c a d c 8 7 3 8
11. 12. 13. 14. 15. 16. 17. 18.
4 4 8 4 b a c a
1. (d) Let m mass of fat is used.
7 1
(3.8 10 ) 10(9.8)(1)(1000)5
m × =
3
9.8 5
3.8 10m
×=
× 312.89 10−= × kg
2. (a) Linear momentum remains constant if net external
force on the system of particles is zero.
3. (a, c) By conservation of linear momentum
5 2u v= − . . .(i)
By Newton’s experimental law of collision
2u v= + . . .(ii)
Using (i) and (ii) we have 1m/sv = and 3 m/su =
Kinetic energy of the centre of mass
2
system cm
10.75J
2m v= =
4. (a) 2 22 3
5 2=MR Mr ⇒
2
15=
Rr
5. (d) 2
max maxm r Tω =
⇒ max
max
324
0.5 0.5
T
mrω = =
×1296 36 rad/s= =
5 kg
before collision
1 kg
u
after collision
1 kgV
5 kg
H 500 mm
500 mm
200 mm
RBR
4cm
BA
AR
2cm
28 N / m
P 1m ,Q R 1m S
10 C° 140 C°400 C°
2,k α12 ,k α
31 Mock Test-2
6. (c) 2 ˆ( )OL m r kω=
Direction of PL
varies with time as its direction will be
perpendicular to string i.e. changing with time.
OL =
Constant , PL
= Variable
7. (8) Here, 1 500mm 0.5mλ = =
2 1 20.25 m; 16 m/s, ?λ υ υ= = =
As frequency of body is fixed, say n, therefore,
1 1 2 2;n nυ λ υ λ= =
⇒ 2 2
1 1
0.25 1
0.5 2
υ λυ λ
= = =
⇒ 12
2
168m/s
2
υυ
υ= = =
8. (7) Here, . 2, 40cm, ?m h n= = =
Time taken by the plate to fall down,
2 2 40 2
sec980 7
ht
g
×= = =
Frequency of fork, 2
7Hz.2 / 7
mn
t= = =
9. (3) The pressur at the bottom,0P P h gρ= +
Where, 0p = atmospheric pressure, therefore,
0
1h g
PP
ρ= + atmosphere
20
1 310
g
g
ρρ
= + = atmosphere.
10. (8) Here, 5 2 410 N/m , 2.4 10 m; 0.08N/mp R S
−= = × =
Initial pressure inside the soap bubble in air cylinder,
1
4.S
P PR
= +
Let 1V be the initial volume of the soap bubble;
3
1
4
3V Rπ=
After compression, volume of the soap bubble;
3
12
4
3 2 8
R VV π = =
If is the air pressure after compression in the cylinder,
then pressure inside the bubble is
2
4 8
( / 2)
S SP P P
R R
′ ′= + = +
Using Boyle’s law, we have, 1 1 2 2P V PV=
or 11
4 8
8
S S VP V P
R R
′+ = +
or 4 1 8
9
S SP P
R R
′+ = +
or 5
4
24 24 0.088 8 10
2.4 10
SP P
R −
+′ = + = × +×
5 58.08 10 10n= × = ×
∴ 0.08 8.n = ≈
11. (4) When wires is taken in the form of one turn circular
coil, then length, 2l rπ= or , 12
lr n
π= = Magnetic field
induction at the centre of circular coil due to current I is
0 0 02 2 1
4 4 ( / 2 )
InI IB
r l l
µ µ µ ππ ππ π π
× ×= = =
When wire is taken in the form of double loop, then
12 2l rπ= ×
or 1
4
Ir
π= and 2n =
∴ Magnetic field induction at the centre of the circular coil.
01
1
2
4
n IB
r
µ ππ
× ×= 0
0
2 24
4 ( / 4 )
I I
l l
µ π πµ
π π× ×
= = ×
∴ 1 4B
B=
12. (4) Refer Fig.
The magnetic field induction at the centriod O due to
current I through one side BC of the triangle will be
0
1 1 2(sin sin )
4
IB
r
µ= θ + θ
π
It will be acting perpendicular to the plane of triangle
upwards. Total magnetic field induction at O due to
current through all the three sides of the triangle will be
0
1 1 2
33 [sin sin ]
4
IB B
µ= = θ + θ
π π
Here, 1 2
1 , 60I A= θ = ° = θ and
/ 2
tan 60 3
BD ar OD= = =
°
A
C
I
D B I
O
I
a
60°r
60°
OL
PLθ
Physics32
2
2
4 5 10.
3 2 3
am
−⋅ ×= =
∴ 7
2
13 10
(4 5 10 / 2 3)B −
−= × ×
⋅ ×[sin 60 sin 60 ]× ° + °
On solving, 54 10 T.B −= ×
13. (8) Magnetic field at P due to currents in two wires will be
acting perpendicular to the plane of wires, upwards and is
given by.
Magnetic field at Q due to current in A is perpendicular to
the plane of wire upwards and due to current in B is
perpendicular to the plane of wire downwards and is given
by
8.
14. (4) When wire is taken in the form of one turn circular
coil, then length
or
Magnetic field induction at the centre of circular coil due
to current I is
When wire is taken in the form of double loop, then
or and n = 2
Magnetic field induction at the centre of the circular coil,
⇒ = 4.
15. (b) A→3, B→2, C→1, D→4
Displacements while angle θ between and is
constant
θ henced
Kdx=
θ or
ddx
k=
θ
cos sin
d F
K K=
θθ θ
21cos cos
2Kxθ = θ
at every point
Qθ = 90°
16. (a) A→3; B→1; C→4; D→2
17. (c) A→2, B→3, C→4, D→1
When a body is disturbed from its mean position and left
to vibrate under restoring force, e.g., A tunning force
struck with rubber pad, vibrations are called free or
natural vibrations. When a body is mode to vibrate with
another vibrating body placed nearby, e.g., stem of a
vibrating tunning fork when pressed on top of sonometer
makes the sonometer wire to vibrate having forced
vibrations. When a body is forced to vibrate by another
vibrating body the vibrations are forced on other body.
But if frequency of forced vibrations is equal to the
natural frequency of forced body then vibrations of forced
body are called resonant vibrations. When a body is
vibrating under the action of viscous force, gradually the
amplitude of vibration decreases because energy of
vibrating body is dissipated as work done against viscous
force in the form of heat etc. and body gradually stops
vibrating. Such vibratious are called damped vibrations.
18. (a) A→1, B→2, C→3, D→4
The particular names of calorimeters are after the names
of scientists who designed them for specific heat
measurements in different conditions.
( ) ( )0 0 022 2
4 / 2 4 / 2P
II IB
r r r
µ µ µπ π π
= + =
0 0 02 2
4 2 4 4Q
I I IB
r r r
µ µ µπ π π
= + =
∴( )( )
0
0
2 /
/ 4
P
Q
I rB
B I r
µ π
µ π= =
2l rπ=
, 12
lr n
π= =
( )0 0 02 2 1
4 4 / 2
InI IB
r l l
µ µ µ ππ ππ π π
× ×= = =
12 2l rπ= ×1
4
lr
π=
∴
( )0 0
1 0
1
2 2 24
4 4 / 4
n I I IB
r l l
µ µπ π πµ
π π π× × × ×
= = = ×
1B
B
cosW F S Fx θ= ⋅ =rr
∴ S x=r r
Fr
xr
,kxθ =
∴ dW F dS= ⋅∫ ∫uurr
F dx F= ⋅ = ⋅ ⋅ =∫ ∫uurr
W dW F dx= = ⋅∫ ∫uurr
Kx dx= ⋅∫uur uur
K xdx Kx= =∫F x⊥r r
∴ 0dW F dx= ⋅ =uurr
cos90 0Fdx= °=
33Mock Test-3
JEE-MAIN: PHYSICS MOCK TEST-3
SECTION 1 (Multiple Choice Question)
1. If ˆˆ ˆ2 4 5A i j k= + −ur
the direction of cosines of the vector
Aur
are:
a. 45
5and
45
4,
45
2 − b.
45
3and
45
2,
45
1
c. 45
4and0,
45
4 d.
45
5and
45
2,
45
3
2. A particle is projected vertically upwards and it is at a
height h after 2 seconds and again after 10 seconds. The
height h is:
a. 196 m b. 98 m
c. 9.8 m d. 19.8 m
3. An insect crawls up a hemispherical
surface very slowly (see the figure).
The coefficient of friction between
the surface and the insect is 1/3. If
the line joining the centre of the hemispherical surface to
the insect makes an angle α is given
a. cot 3α = b. tan 3α =
c. sec 3α = d. cosec 3α =
4. An ideal spring with spring constant k is hung from the
ceiling and a block of mass M is attached to its lower end.
The mass is released with the spring initially unscratched.
Then the maximum extension in the spring is
a. b.
c. d.
5. A particle executes simple harmonic motion between
and The time taken for it to go from
0 to A/2 is and to go from A/2 to A is Then:
a. b.
c. d.
6. A large number of liquid drops each of radius r coalesce
to form a single drop of radius R. The energy released in
the process is converted into the kinetic energy of the big
drop so formed. The speed of the big is (Given surface
tension of liquid is T, density of liquid is ρ)
a. b.
c. d.
7. A solid sphere of radius R made of material of bulk
modulus K is surrounded by a liquid in a cylindrical
container. A massless piston of area A floats on the
surface of the liquid when a mass m is placed on the
piston to compress the liquid, the fractional change in the
radius of the sphere, is:
a. b.
c. d.
8. The wavelength of light observed on the earth, from a
moving star is found to decreases by 0.05%. Relative to
the earth, the star is:
a. moving away with a velocity of 5
1.5 10× m/s
b. coming closer with a velocity of 51.5 10× m/s
c. moving away with a velocity of41.5 10× m/s
d. coming closer with a velocity of 4
1.5 10× m/s
9. An ideal gas expands isothermally from a volume to
and then compressed to original volume
adiabatically. Initial pressure is and final pressure is
The total work done is W. Then,
a. b.
c. d.
10. At NTP one mole of diatomic gas is compressed
adiabatically to half of its volume ( ).=γ 1.40 The work
done on the gas will be
a. 1280 J b. 1610 J
c. 1792 J d. 2025 J
11. Two identical conducting rods are first connected
independently to two vessels, one containing water at
100°C and the other containing ice at 0ºC. In the second
case, the rods are joined end to end connected to the same
vessels. Let and gram per second be the rate of
melting of ice in the two cases respectively. The ratio
is
a. b.
c. d.
4Mg
k
2Mg
k
Mg
k 2
Mg
k
x A= − .x A= +
1T 2.T
1 2T T< 1 2T T>
1 2T T= 1 22T T=
1 1T
r Rρ −
2 1 1T
r Rρ −
4 1 1T
r Rρ −
6 1 1T
r Rρ −
R
R
δ
mg
AK 3
mg
AK
mg
A 3
mg
AK
1V
2V 1V
1p
3.p
3 1, 0> >p p W 3 1, 0< <p p W
3 1, 0p p W> < 3 1, 0= =p p W
( )0.42 1.40 :=
1q 2q
1 2/q q
1
2
2
1
4
1
1
4
α
Physics34
12. The energy density (energy per unit volume) in an electric
field caused by a point charge falls off with the distance
from the point charge as:
a. b.
c. d.
13. The needle of a deflection galvanometer shows a
deflection of 60° due to a short bar magnet at a certain
distance in position. If the distance is doubled, the
deflection is
a. b.
c. d.
14 If power factor is in a series RL circuit .
ac mains is used then L is
a. 3
henryπ
b. henryπ
c. henry2
π d. None of these
15. The amplitude ratio of two superposing waves is 2 : 1.
The ratio of maximum and minimum intensities is:
a. 1 : 1 b. 2 : 1
c. 4 : 1 d. 9 : 1
16. The electric and the magnetic field, associated with an
e.m. wave, propagating along the + z-axis, can be
represented by
a. b.
c. d.
17. If the wavelength of the first line of the Balmer series of
hydrogen is 6561 Å, the wavelength of the second line of
the series should be
a. 13122 Å b. 3280 Å c. 4860 Å d. 2187 Å
18. In the nuclear reaction X,++
β what does
X stand for
a. An electron b. A proton
c. A neutron d. A neutrino
19. The symbol given in figure represents
a. NPN transistor
b. PNP transistor
c. Forward biased PN junction diode
d. Reverse biased NP junction diode
SECTION 2 (Numeric Value Question)
20. For television broadcasting, the frequency employed is
normally
a. 30–300 MHz b. 30–300 GHz
c. 30–300 kHz d. 30–300 Hz
21. The potential difference V and current i flowing through
an appliance in an ac circuit are given by
and power dissipated in
the appliance is
a. 0 W b. 10 W
c. 5 W d. 2.5 W
22. A wire has a mass (03. ± 0.003) g, radius (0.5 ± 0.005) mm
and length (6 ± 0.06) cm. The maximum percentage error
in the measurement of its density is:
a. 1 b. 2
c. 3 d. 4
23. An aircraft with a wing-span of 40 m flies with a speed of
1080 km h–1 in the eastward direction at a constant
altitude in the northern hemisphere, where the vertical
component of earth's magnetic field is 48.3 10 T.−× Then
the e.m.f. that develops between the tips of the wings is:
a. 0.5 V b. 0.35 V
c. 1 V d. 2.1 V
24. If an observer is walking away from the plane mirror with
3 m/sec. Then the velocity of the image with respect to
observer will be
a. b.
c. d.
25. The eccentricity of earth’s orbit is 0.0167. The ratio of its
maximum speed in its orbit to its minimum speed is:
a. 2.507 b. 1.033
c. 8.324 d. 1.000
1/ r 21/ r
31/ r41/ r
Atan
1 3sin
8−
1 3cos
8−
1 3tan
8−
1 3cot
8−
1
2100R = Ω
0 0ˆ ˆ,E E i B B j = =
r r
0 0ˆ ˆ,E E k B B i = =
r r
0 0ˆ ˆ,E E j B B i = =
r r
0 0ˆˆ,E E j B B k = =
r r
BC +→ 115
116
5cos volt, 5sin ampV t i tω ω= =
6 m/sec 6 m/sec−
12 m/sec 3 m/sec
E C
B
Space for Rough Work
35 Mock Test-3
JEE ADVANCE PAPER-I
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. In a large building, there are 15 bulbs of 40 W, 5 bulbs of
100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage
of the electric mains is 220 V. The minimum capacity of
the main fuse of the building will be:
a. 12 A b. 14 A
c. 8 A d. 10 A
2. When 5V potential difference is applied across a wire of
length 0.1 m, the drift speed of electrons is 4 12.5 10 ms .− −× If the electron density in the wire is
28 38 10 m ,−× the resistivity of the material is close to:
a. 81.6 10 m−× Ω b. 71.6 10 m−× Ω
c. 61.6 10 m−× Ω d. 51.6 10 m−× Ω 3. A galvanometer having a coil resistance of 100 Ω gives a
full scale deflection, when a current of 1 mA is passed
through it. The value of the resistance, which can convert
this galvanometer into ammeter giving a full scale
deflection for a current of 10 A, is:
a. 0.01 Ω b. 2 Ω
c. 0.1 Ω d. 3 Ω 4. The temperature dependence of resistances of Cu and
undoped Si in the temperature range 300–400 K, is best
described by:
a. Linear increase for Cu, linear increase for Si
b. Linear increase for Cu, exponential increase for Si
c. Linear increase for Cu, exponential decrease for Si
d. Linear decrease for Cu, linear decrease for Si
5. Two ideal batteries of emf 1V and 2V and three resistances
1 2,R R and are 3R connected as shown in the figure. The
current in resistance 2R would be
zero if
a. 1 2=V V and 1 2 3= =R R R
b. 1 2=V V and 1 2 32= =R R R
c. 1 22=V V and 1 2 32 2= =R R R d. 1 22 =V V and 1 2 32 = =R R R
6. Heater of an electric kettle is made of a wire of length L
and diameter d. It takes 4 minutes to raise the temperature
of 0.5 kg water by 40 K. This heater is replaced by a new
heater having two wires of the same material, each of
length L and diameter 2d. The way these wires are
connected is given in the options. How much time in
minutes will it take to raise the temperature of the same
amount of water by 40 K?
a. 4 if wires are in parallel
b. 2 if wires are in series
c. 1 if wires are in series
d. 0.5 if wires are in parallel
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. A current 1 amp is flowing in the sides of an equilateral
triangle of side 24.5 10 m−× . The magnetic field at the
centroid of the triangle in the units of ( )510 T− is
8. If the maximum values of signal and carrier waves are 4 V
and 5V respectively, the percentage of amplitude
modulation is a × 10%. What is the value of a ?
9. A signal wave of frequency 4.5 kHz is modulated with a
carrier wave of frequency 3.45 MHz. The bandwidth of
FM wave is kHz is
10. What is the maximum usuable frequency (in MHz) for
E-layer of atmosphere having critical frequency 4 MHz,
when the angle of incidence is 60°?
11. A T.V. Tower has a height 100 m. In order to triple its
coverage range, the height of tower to be increased is 210 m.a × What is the integer value of a?
12. A microwave telephone link operating at the central
frequency of 10 GHz has been established. If 2% of this is
available for microwave communication channel and each
telephone is allotted a bandwidth of 8 kHz, the number of
telephone channels which can be simultaneously granted
is 2.5 10 .a× What is the integer value of a ?
13. A uniform circular disc of mass 1.5 kg and radius is
initially at rest on a horizontal frictionless surface. Three forces of equal magnitude 0.5F N= are applied
simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the
angular speed of the disc in –1rad s is
F
F
FX
Z Y
O
2V
1V1R
2R
3R
Physics 36
14. A horizontal circular platform of radius 0.5 m and mass
0.45 kg is free to rotate about its axis. Two mass less
spring toyguns, each carrying a steel ball of mass 0.05 kg
are attached to the platform at a distance 0.25 m from the
centre on its either sides along its diameter (see figure).
Each gun simultaneously fires the balls horizontally and
perpendicular to the diameter in opposite directions. After
leaving the platform, the balls have
horizontal speed of 19 ms− with
respect to the ground. The
rotational speed of the platform in –1rad s after the balls leave the platform is
SECTION 3 (Maximum Marks: 12)
Paragraph based questions (2 paragraphs, each having 2
MCQs with one correct answer only)
Paragraph for Question No. 15 to 16
In a thin rectangular metallic strip a constant current I flows
along the positive x-direction, as shown in the figure. The
length, width and thickness of the strip are l, w and d,
respectively. A uniform magnetic field B
is applied on the
strip along the positive y-direction. Due to this, the charge
carriers experience a net deflection along the z-direction. This
results in accumulation of charge carriers on the surface PQRS
and appearance of equal and opposite charges on the face
opposite to PQRS. A potential difference along the z-direction
is thus developed. Charge accumulation continues until the
magnetic force is balanced by the electric force. The current is
assumed to be uniformly distributed on the cross-section of the
strip and carried by electrons.
15. Consider two different metallic strips (1 and 2) of the
same material. Their lengths are the same, widths are w1
and w2 and thicknesses are d1 and d2, respectively. Two
points K and M are symmetrically located on the opposite
faces parallel to the x-y plane (see figure). V1 and V2 are
the potential differences between K and M in strips 1 and
2, respectively. Then, for a given current I flowing
through them in a given magnetic field strength B, the
correct statement(s) is(are)
a. If w1 = w2 and d1 = 2d2, then V2 = 2V1
b. If w1 = w2 and d1 = 2d2, then V2 = V1
c. If w1 = 2w2 and d1 = d2, then V2 = 2V1
d. If 1 22w w= and 1 2 ,d d= then 2 1V V=
16. Consider two different metallic strips (1 and 2) of same
dimensions (lengths ,ℓ width w and thickness d) with
carrier densities n1 and n2, respectively. Strip 1 is placed
in magnetic field B1 and strip 2 is placed in magnetic field
B2, both along positive y-directions. Then V1 and V2 are
the potential differences developed between K and M in
strips 1 and 2, respectively. Assuming that the current I is
the same for both the strips, the correct option(s) is(are)
a. If 1 2B B= and 1 22 ,n n= then 2 12V V=
b. If 1 2B B= and 1 22 ,n n= then 2 1V V=
c. If 1 22B B= and 1 2 ,n n= then 2 10.5V V=
d. If 1 22B B= and 1 2 ,n n= then 2 1V V=
Paragraph for Question No. 17 to 18
The capacitor of capacitance C can be charged (with the help of
a resistance R) by a voltage
source V, by closing switch 1S
while keeping switch 2S open.
The capacitor can be
connected in series with an
inductor ‘L’ by closing switch
2S and opening 1S
17. Initially, the capacitor was uncharged. Now, switch S1 is
closed and S2 is kept open. If time constant of this circuit
is τ, then
a. after time interval τ, charge on the capacitor is CV/2
b. after time interval 2 ,τ charge on the capacitor is 2(1 )CV e−−
c. the work done by the voltage source will be half of the
heat dissipated when the capacitor is fully charged
d. after time interval 2 ,τ charge on the capacitor is 1(1 )CV e−−
18. After the capacitor gets fully charged, S1 is opened and S2 is closed so that the inductor is connected in series with
the capacitor. Then
a. at 0,t = energy stored in the circuit is purely in the
form of magnetic energy
b. at any time 0,t > current in the circuit is in the same
direction
c. at 0,t > there is no exchange of energy between the
inductor and capacitor
d. at any time 0,t > instantaneous current in the circuit
may C
VL
y
x
z
I
Q P
S
I W
K•
M•R
d
V
C
R S1
S2 L
l
37 Mock Test-3
JEE ADVANCE PAPER-II
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. The figure shows certain wire segments joined together to
form a coplanar loop. The loop is placed in a
perpendicular magnetic field in the direction going into
the plane of the figure. The magnitude of the field
increases with time. 1I and 2I are the currents in the
segments ab and .cd Then,
a. 1 2>I I
b. 1 2<I I
c. 1I is in the direction ba and 2I is in the direction cd
d. 1I is in the direction ab and 2I is in the direction dc
2. A thin flexible wire of length L is connected to two
adjacent fixed points and carries a current I in the
clockwise direction, as shown in the figure. When the
system is put in a uniform magnetic field of strength B
going into the plane of the paper, the wire takes the shape
of a circle. The tension in the wire is
a. IBL b. IBL
π c.
2
IBL
π d.
4
IBL
π
3. Along insulated copper wire is closely wound as a spiral
of 'N' turns. The spiral has inner radius 'a' and outer radius
'b'. The spiral lies in the X-Y plane and a steady current 'I'
flows through the wire. The Z component of the magnetic
field at the center of the spiral is
a. 0 bIn
2( )
NI
b a a
−
µ b. 0 +
In2( ) –
NI b a
b a b a
−
µ
c. 0 In2
NI b
b a
µ d. 0 In
2
NI b+a
b b – a
µ
4. A loop carrying current I lies in the x-y plane as shown
in the figure. The unit vector k is coming out of the
plane of the paper. The magnetic moment of the current
loop is
a. 2 ˆa Ik b. 2 ˆ12
π +
a Ik
c. 2 ˆ12
π − +
a Ik d. 2 ˆ(2 1)π + a Ik
5. An infinitely long hollow conducting cylinder with inner
radius R/2 and outer radius R carries a uniform current
density along its length. The magnitude of the magnetic
field, | |B as a function of the radial distance r from the
axis is best represented by.
a. b.
c. d.
6. A circular loop of radius 0.3 cm lies parallel to a much
bigger circular loop of radius cm.20 The centre of the
small loop is on the axis of the bigger loop. The distance
between their centres is .cm15 If a current of 2.0 A flows
through the smaller loop, then the flux linked with bigger
loop is
a. weber101.9 11−× b. weber106 11−×
c. weber103.3 11−× d. weber106.6 9−×
r
B
R/2 R r
B
R/2 R
r
B
R/2 R r
B
R/2 R
y
a x
a
I
X
Y
I a b
c d
a b
Physics 38
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. The isotope 125 B having a mass 12.014 u undergoes
β - decay to 12 126 6C. C has an excited state of the nucleus
12 *6( C ) at 4.041 MeV above its grounds state. If
125 B decays to 12 *
6 C , the maximum kinetic energy of the
β - particle in units of MeV is:
(1 931.5MeV / ,u c= c2, where c is the speed of light in
vacuum)
8. A binary star consists of two stars A (mass 2.2Ms) and B
(mass 11Ms), where Ms is the mass of the sun. They are
separated by distance d and are rotating about their centre
of mass, which is stationary. The ratio of the total angular
momentum of the binary star to the angular momentum of
star B about the centre of mass is
9. Two homogeneous spheres A and B of masses m and 2m
having radii 2a and a respectively are placed in contact.
The ratio of distance of c.m. from first sphere to the
distance of c.m. from second sphere is:
10. A non-uniform thin rod of length L is placed along X-axis
so that one of its ends is at the origin. The linear mass
density of rod is 0xλ λ= ⋅ The centre of mass of rod
divides the length of the rod in the ratio:
11. A sphere of mass 5 kg and diameter 2 m rotates about a
tangent. What is its moment of inertia?
12. A uniform rod of length 1 m and mass 0.5 kg rotates at
angular speed of 6 rad/sec about one of its ends. What is
the KE of the rod?
13. A particle performing uniform circular motion has angular
momentum L. When its angular velocity is doubled and
KE is also doubled, the new angular momentum becomes
x times. What is x?
14. The fundamental frequency of a closed organ pipe is equal
to first overtone frequency of an open organ pipe. If the
length of the open pipe is 36 cm, what is the length
(in cm) of the closed pipe?
SECTION 3 (Maximum Marks: 12)
Matching type questions with 4 options
15. Match Column I with Column II and select the correct
answer using the codes given below the Columns:
Column I Column II
(A) Boltzmann Constant 1. 2 1[ML T ]−
(B) Coefficient of viscosity 2. 1 1[ML T ]− −
(C) Plank Constant 3. 3 1[MLT K ]− −
(D) Thermal conductivity 4. 2 2 1[ML T K ]− −
a. A-4; B-1, 4; C- 4; D-2
b. A-1; B- 1,4; C-4; D-4
c. A-4; B-1,4; C-4; D-2
d. A-1; B- 1,4; C-2; D-4
16. Column I give a list of possible set of parameters
measured in some experiments. The variations of the
parameters in the form of graphs are shown in Column II.
Match the set of parameters given in Column I with the
graph given in Column II.
Column I Column II
(A) Potential energy of a
simple pendulum (y axis)
as a function of
displacement (x axis)
1.
(B) Displacement(y axis) as a
function of time (x axis)
for a one dimensional
motion at zero or constant
acceleration when the
body is moving along the
positive x-direction
2.
(C) Range of a projectile
(y-axis) as a function of its
velocity (x-axis) when
projected at a fixed angle
3.
(D) The square of the time
period (y-axis) of a simple
pendulum as a function of
its length (x -axis)
5.
a. A-4; B- 1,4; C- 4; D- 2
b. A-1; B- 1,4; C- 4; D- 4
c. A-2; B- 1,4; C- 4; D- 2
d. A-1; B- 1,4; C- 2; D- 4
17. Motion is defined as rate of change of position. In the
motions described in column I, match the facts about that
motion in column II for the changes in position vector.
y
x O
y
x O
y
x O
y
x O
39 Mock Test-3
Column I Column II
(A) Magnitude only 1. A coin dropped from
roof of house
(B) Direction only 2. A coin thrown at any
angle with horizontal
(C) Both in magnitude and
direction
3. A coin held in your
hand
(D) Remains invariant 4. A coin in rotated in
circular path with
variable speed
a. A-1; B-2; C-3; D-4
b. A-4; B-2; C-3; D-1
c. A-1; B-4; C-2; D-3
d. A-3; B-2; C-4; D-1
20. Match the statement of Column with those in Column II:
Column I Column II
(A) In any Bohr orbit of the
hydrogen atom, the ratio 1.
1
2−
of kinetic energy to
potential energy of the
electron is
(B) The ratio of the kinetic
energy to the total energy
of an electron in a Bohr
orbit is
2. 2
(C) In the lowest energy level
of hydrogen atom, the
electron has the angular
momentum
3. 2
h
π
(D) Ratio of the wavelengths
of first line of Lyman
series and first line of
Balmer series is
4. 5 : 27
a. A-1; B-3,4; C-4; D-2,3
b. A-1; B-3,4; C-4; D-3
c. A-2; B-3,4; C-1; D-3
d. A-1; B-2,4; C-3; D-4
Space for Rough Work
Physics 40
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a b a b b d b b c c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c d c a d a c d a a
21. 22. 23. 24. 25.
a d c a b
1. (a) kjiA ˆ5ˆ4ˆ2 −+=
∴ 45)5()4()2(|| 222 =−++=A
∴ 2
cos45
α = , 4
cos45
β = , 5
cos45
−γ =
2. (b) 21
2h ut gt= −
⇒ ( )212 10
2h u g= × − × ( )21
10 102
u g= × − ×
⇒ 2 2 10 50u g u g− = − ⇒ 6u g=
∴ 2 2 2 6 2 10h u g g g g= − = × − = = 98 m
3. (a) Equilibrium of insect gives
cos N mg= α . . . (i)
sin N mg= αµ . . . (ii)
From equation (i) and (ii) we get 1
cot 3α = =µ
4. (b) Let x be the maximum
extension of the spring. From
conservation of mechanical energy
decrease in gravitational potential
energy
= increase in elastic potential energy
∴ 21
2Mg x kx=
or
2Mgx
k=
5. (b) For block P, friction will provide the necessary
restoring force.
∴ 2
maxf m Aω= with 2
2
k k
m m mω = =
+
Hence, max2 2
k kAf m A
m
= =
6. (d) Work done, W T A= ∆
or 2 2 2 2[(4 ) 4 ] .4 [ ]nW T R R T nr Rπ π π= − = −
Where 3 3 24 4 1and
3 3 2n r R W mπ π υ× = =
∴ 2 2 21.4 ( )
2m T nr Rυ π= −
or 3 2 2 21 4.4 ( )
2 3R T nr Rπ ρυ π× = −
or 2 2
3 3
6T nr R
R Rυ
ρ
= −
2
3
6 1 6 1 1T nr T
nr R r Rρ ρ = − = −
7. (b) /
/ /
P mg A
V V V V
∆=
∆ ∆V mg
V AK
∆= . . .(i)
But for a sphere 4
3V = π 2R
Differentiating 4
( )3
V =δ π δ23R R
⇒
4)
334
3
V=
V R=
π(3 δδ δ
π
2
3
R RR
R
∴ From (i) 3
V mg R mg=
R AK R AK⇒ =
3δ δ
8. (b) 0.05
100λ λ∆ = , As
v
cλ λ∆ =
⇒ 0.05
100
v
cλ λ= or 45 10v c−= ×
or 4 8 55 10 3 10 1.5 10v −= × × × = × m/s
As λ decreases, the star is approaching the observer.
9. (c) Slope of adiabatic process
at a given state (p, V, T) is
more than the slope of
isothermal process. The
corresponding p-V graph for
the two processes is as shown
in figure. In the figure, AB is isothermal and BC is
adiabatic.
ABW = positive (as volume is increasing) and
BCW = negative (as volume is decreasing) plus,
,BC ABW W> as area under p-V graph gives the work done.
Hence, 0AB BCW W W+ = <
From the graph itself, it is clear that 3 1.p p>
Hence, the correct option is
Note: At point B, slope of adiabatic (process BC) is
greater than the slope of isothermal (process AB).
N
cosmg αmg
sinmg α
αNµ
α
x
k
v = 0
v = 0
M
M
A
V
p
B
C
41 Mock Test-3
10. (c) 11 1 2 2 2 1
2
VP V P V P P
V
γ
γ γ = ⇒ =
work ( )2 2 1 11
RP V P V
γ= −
−
1
11
2
11
1
VP
V
γ
γ
− = −
−
( )0.405 3110 22.4 10 2 1
1 1.40− = × × × × −
−
[ ]2122.4 10 1.32 1 1792
0.40= × × − = J
11. (c)dQ dm
Ldt dt
=
or Temperature difference
Thermal resistance
dmL
dt
=
or 1
Thermal resistance
dm
dt∝ ⇒
1q
R∝
In the first case rods are in parallel and thermal resistance
is 2
R while in second case rods are in series and thermal
resistance is 2R.
⇒ 1
2
2 4
/ 2 1
q R
q R= =
12. (d) Energy u ∝ E2 and 2
1u
r∝ for a point charge; so
4
1u
r∝
13. (c) For short bar magnet in tan A-position
03
2tan
4
MH
d
µπ
= θ
. . . (i)
When distance is doubled, then new deflection θ ′ is given by
03
2tan
4 (2 )
MH
d
µπ
= ′θ
. . . (ii)
∴ tan
tan
′θθ
tan 1
tan 8θ′
=
⇒ –1tan tan 60 3 3tan tan
8 8 8 8
° ′ ′ = = = ⇒
θθ θ =
14. (a) 1
cos 602⇒ = °φ = φ
3
tan 60 HL
LR
° = ⇒ =ω
π
15. (d) ( )( )
2 2
1 2max2
min 1 2
2 1 9
2 1 1
a aI
I a a
+ + = = = − −
16. (a) In the given question, the electromagnetic wave is
propagating along z+ axis. In e.m. wave, the electric field
( )E
and magnetic field ( )B
are perpendicular to each other
and also perpendicular to the direction of propagation of
wave,
So, 0ˆE E i=
and 0
ˆ.B B j=
17. (c) The wavelength of spectral line in Balmer series is
given by 2 2
1 1 1
2R
nλ
= −
For first line of Balmer series, n = 3
⇒ 2 2
1
1 1 1 5
2 3 36
RR
λ = − =
; For second line n = 4.
⇒ 2 2
2
1 1 1 3
2 4 16
RR
λ = − =
∴ 21
1
20 206561 4860
27 27Å
λλ
λ= ⇒ = × = Å
18. (d) 11 116 5C B→ + β+ + γ because β+ 0
1 e=
19. (a) The base is always thin
20. (a) VHF (Very High Frequency) band having frequency
range 30 MHz to 300 MHz is typically used for TV and
radar transmission.
21. (a) 5cos 5sin2
V t tπ
ω ω = = +
⇒ 2sinI tω=
∴ 2
πφ = ⇒ cos 0
2rms rsmP E Iπ
= = W
22. (d) Density 2
m
r Lρ
π=
∴ 100 2 100m r L
m r L
ρρ
∆ ∆ ∆ ∆ × = + + ×
After substituting the values, we get the maximum
percentage error in density = 4%
23. (c) L = 40 m, v = 1080 km,
h–1 = 300 m 1sec− and B = 48.3 10 T−×
⇒ 48.3 10 40 300 1e Blv −= = × × × = V
24. (a) Relative velocity of image ... trw object
= 3 – (–3) = 6 m/sec
25. (b) Angular momentum of satellite about center of earth
remains constant, i.e., mvr = constant
⇒ 2 1
1 2
1
2
v rv
v r∝ ⇒ =
Speed is maximum at position 2,
∴ ( )( )
max 1
min 2
1
1
av r
v r a
εε
+= =
−1 1 0.0167
1.0331 1 0.0167
εε
+ += = =
− −
3m/sec –3m/sec
I O
Physics 42
JEE Advance Paper -I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a d a c a,b,d b,d 4 8 9 8
11. 12. 13. 14. 15. 16. 17. 18.
8 4 2 4 a,d a,c b d
1. (a)e 1 2
1 1 1
R q R R= +
⇒ 2 2 2
e 1 2
.....R q
V V VP
R R= = + +
15 40 5 100 80 5 1 1000 2500w= × + × + × + × = W ⇒ P VI=
⇒ 2500 220P I= = ⇒ 250 125
22 11I = =
2. (d) d
i
Aneυ = ∴ d
ine
Aυ= ⋅
⇒ V V
iR
Aρ
= =ℓ
∴ Vi
A
ρ =ℓ
⇒ d
V
neρ
υ=
⋅ ℓ
⇒ 4 28 19
5
2.5 10 8 10 1.6 10 0.1ρ − −=
× × × × × ×
4 5
4
2 110 1.6 10 m
8 1.6 10 6.4− −
−= = × = × Ω× ×
3. (a)1
g
g
i GS
i=
−Here 310gi A−= A
⇒ 210 ,G = Ω 10I A= ⇒ 210S −= Ω
4. (c) For conductor (Cu) resistance increases linearly and
for semiconductor resistance decreases Exponentially in
given temperature range.
5. (a, b, d) 1 1 21
1 3
( )+=
+
R V VV
R R
⇒ 1 3 2 1=V R V R
⇒ ⇒ ⇒ ⇒ 3 1 22
1 3
( )+=
+
R V VV
R R ⇒ 2 1 2 3=V R V R
6. (b, d)2 2 2
1 24/ 2 / 8
= =V V V
H t tR R R
⇒ 1 2 min.=t ⇒ 2 0.5 min.=t
7. (4) Refer Fig., The magnetic field induction at the
centroid O due to current I through one side BC of the
triangle will be 01 1 2
4
IB
r
µπ
= +. (sin θ1 + sin θ2)
It will be acting perpendicular to the plane of triangle
upwards. Total magnetic field induction at O due to
current through all the three sides of the triangle will be
01 1 2
33 sin sin
4
IB B
r
µπ
= = +[sin θ1 + sin θ2]
Here, I = 1 A, θ1 = 60° = θ2
And 2/ 2 4.5 10
mtan 60 3 2 3 2 3
BD a ar OD
−×= = = = =
°
∴ ( )
[ ]7
2
13 10 sin 60 sin 60
4.5 10 / 2 3B −
−= × × × ° + °
×
On solving, B = 4 × 10–5 T.
8. (8) Here, Am = 4 V ; Ac = 5 V
Percentage of modulation,
A 4
100 100 80% 10%.A 5
m
c
aµ = × = × = = × ⇒ a = 8
9. (9) Here, 4.5kHz; 3.45MHz 3450kHzs cv v= = =
Bandwidth 2 2 4.5 9.0kHz.sv= = × =
10. (8) Here, 4MHz, 90cv i= = °
Maximum usuable frequency seccv i=
4 sec60 4 2 8MHz= × ° = × =
11. (8) ' 2 ' 3 3 2d h R d hR= = =
or ' 9 9 100 900h h m= = × =
Increase in height of tower 2900 100 800 10a= − = = ×
⇒ 8a =
12. (4) Microwave frequency used in telephone link = 10 GHz
= 10 × 109 Hz = 1010 Hz
Frequency available for microwave communication
2%= of 10 GHz = 10 8210 2 10 Hz
100× = ×
Bandwidth of each telephone channel = 8 kHz
38 10 Hz= ×
Number of microwave telephone channels
8
4
3
2 102.5 10 2.5 10
8 10a×
= = × = ××
∴ a = 4
B I
a D
r 60° 60°
I
I
A
C
O
43 Mock Test-3
F
R
sin 30°R
30°
F
F
a
a
a a
13. (2) Iατ = ⇒ 3 sin 30 Iα° =FR
⇒ 2
2=MR
I
⇒ α 2=
⇒ 0 αtω ω= + ⇒ 2 /ω = rad s
14. (4) Since net torque about centre of rotation is zero, so we
can apply conservation of angular momentum of the
system about centre of disc =i fL L
0 2 ( / 2);ω= +I mv r comparing magnitude
∴ 0.45 0.5 0.5 0.5
0.05 9 22 2
ω× × = × × ×
∴ 4ω =
15. (a, d) 1 2=I I
⇒ 1 1 2 2=neAv neA v ⇒ 1 1 1 2 2 2=d wv d w v
Now, potential difference developed across MK
=V Bvw
⇒ 1 1 1 2
2 2 2 1
= =V v w d
V v w dand hence, correct choice is a & d.
16. (a, c) As 1 2=I I 1 1 1 1 2 2 2 2=n wd v n w d v
Now, 2 2 2 2 2 2 1 1 1 2 1
1 2 1 1 1 1 2 2 2 1 2
= = =
V B v w B w n wd B n
V B v w Bw n w d Bn
∴ Correct options are a & c.
17. (b) /0 (1 )tQ Q e τ−= −
/(1 )tQ CV e τ−= − after time interval 2 .τ
18. (d) 0 cosq Q tω=
⇒ 0 sindq
i Q tdt
ω ω= − =
⇒ kax
Ci C V V
Lω= =
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d c a b d a 9 6 2 2/3
11. 12. 13. 14. 15. 16. 17. 18.
7 3 1 9 c a c d
1. (d) According to Lenz’s Law, current will be in
anticlockwise sense as magnetic field is increasing into
the plane of paper.
2. (c) 2 sin2
dT BiRd
θθ= (for θ small)
⇒ Td BiRdθ θ=
⇒ 2
BiLT BiR
π= =
3. (a)0
2 ( )
b
a
INdr
r b a
µ−∫ 0 0 ln
2( ) 2( )
b
a
IN INdr b
b a r b a a
µ µ = = − − ∫
4. (b) =M IA
⇒
2
224
2
π = × +
a
A a
2
224
π= × +
aa
22 21
2 2
ππ
= + +
aa a
∴ 2 ˆ12
π = +
M a Ik
5. (d) Let current be I.
Current density 22
4π
=
−
IJ
RR
⇒ 2 2
4
3 3
4
ππ= =
I IJ
R R
Field for 02
≤ =
Rr B For
2< ≤
Rr R
Applying Ampere’s Law .2πB r
2 2
2 200 2 2
44.
3 4 3 4
µµ π
π
= − = −
II R Rr r
R R
2
02
2
3 4
µπ
= −
I RB r
R r ⇒ At , 0
2= =R
r B
∴ At 2
=R
r filed is continuous.
From the above expression as r increases B increases.
For ≥r R 0.2π µ=B r I
⇒ 0
2
µπ
=I
Br
1∴ ∝B
r
⇒ At ,=r R 0
2
µπ
=I
BR
From inside expression at =r R
⇒ 2
0 02
2.
3 4 2
µ µπ π
= − =
I IRB R
R R R
This proves the continuity in the graph at .=r R From the
above only correct option is (d).
cos( / 2)T dθ cos( / 2)T dθTT
2T sin( / 2)dθ
Amperian
loop
R/2
R r
Physics 44
B A
C 2m
m
a 2a
6. (a) First we find the mutual inductance of the assembly.
Let current I flow through the larger loop.
The strength of induction at the smaller loop
ddR
IR/322
20
)(2 +=
µ
∴ Flux through smaller loop 2/322
220
)(2 dR
rIR
+=
πµ
∴ Mutual inductance 2/322
220
)(2 dR
rR
+=
µ
∴ Flux linked through coil idR
rR.
)(2 2/322
220
+=
πµ
2/322
627
)1025.2104(2
210910104−−
−−−
×+×
××××××=
ππ
101.9)10(625.15
1810
2
16 11153 −−+ ×=××
×= weber
7. (9) 121265 B C β γ− −→ + +
12 12 * 25 6( )Q m B m C c= − 12 12 2
5 6[ ( )]m B m C m C= − + ∆
12 12 2 25 6( )m B m C C mC= − − ∆ 0.014 931 4.041 9= × − =
8. (6) 2
total 1 12
B 2 2
1L
L m r
m r= +
9. (2) Assuming that masses of the two spheres are
concentrated at their centres A and B and taking x = 0 at A,
Fig., we get
0 2 3
2( 2 )cm
m m ax a AC
m m
× + ×= = =
+
2
2AC a
CB aCB a
= ∴ = = ⋅
10. (2/3) As shown in Fig., mass of small element of length dx
is 0 dm dx x dxλ λ= =
⇒ 0
0 0
0
0
( )L L
cm L
xdm x xdx
xdm
xdx
λ
λ= =
∫ ∫
∫ ∫
3 3
02 2
0
[ / 3] / 3 2
[ / 2] / 2 3
L
cm L
x L Lx
x L= = =
If 1 2
2 2,
3 3 3
L L Lx x L= = − = ⇒ 1
2
2 / 3 1
/ 3 2
x L
x L= = ⋅
11. (7) Here, m = 5 kg; 2
1 ; ?2
r m I= = =
Moment of inertia of the sphere about a tangent to the
sphere, 2 2 27 75 1 7kgm
5 5I mr= = × × = ⋅
12. (3) Here, l = 1 m, m = 0.5 kg, 6 rad/secω =
2 2
2 2 21 1 1 0 5 16 3 J
2 2 3 2 3
mlKE Iω ω
⋅ ×= = = × × = ⋅
13. (1) Here, 1 1 1,L I ω= When 2 12 ,ω ω=
and 2 2
2 2 1 1
1 12
2 2KE I Iω ω = =
2 22 1 1 1 2 1
1 1 1(2 ) 2 ;
2 2 2I I I Iω ω= × =
∴ 2 2 2 1 1 1 1 1
12
2L I I I Lω ω ω= = × = = ⇒ 2
1
1L
xL
= = ⋅
14. (9) Let 1l be the length of closed pipe 2 36cm=l = length
of open pipe
Fundamental frequency of closed pipe 1
14n
l
υ=
Frequency of first overtone of open pipe,
2
1 2
22 36
nl l
υ υ υ= × = =
As 1 2n n=
∴ 1
1
;4 364 36
ll
υ υ= =
⇒ 1
369cm.
4l = =
15. (c) A-4; B-1, 4; C- 4; D-2
A. 2 2 2 2 13[ ] [ ] [ ]
2KE KT ML T K K K ML T K− − −′ ′= ⇒ = ⇒ =
B. 2 1 1 16 [ ] [ ] [ ] [ ]F rv MLT L LT ML Tπη η η− − − −= ⇒ = ⇒
C. 2 2 2 1[ ] [ ][ ]
hE hf ML T h ML T
T− −= ⇒ = ⇒ =
D. 2 2 2( ) [ ] [ ] [ ]
[ ] [ ]
dQ K A T ML T k L K
dt x T L
−′ ′∆= ⇒ =
∆
⇒ 3 1[ ]K MLT K− −′ =
16. (a) A-4; B- 1,4; C- 4; D- 2
17. (c) c. A-1; B-4; C-2; D-3
When coin is dropped if falls is straight line where direction is same but magnitude changes with change in position. When a coin is rotated in circular path direction goes on changing but radius remains constant. In projectile motion both direction and magnitude of r changes. A coin held in your hand has constant position vector and hence it is invariant.
18. (d) A-1, B-2,4, C-3, D-4
I
R
r
L
x dm
dx
45 Mock Test-4
JEE-MAIN: PHYSICS MOCK TEST-4
SECTION 1 (Multiple Choice Question)
1. Which of the following sets have different dimensions?
a. Pressure, Young’s modulus, Stress
b. Emf, Potential difference, Electric potential
c. Heat, Work done, Energy
d. Dipole moment, Electric flux, Electric field
2. The 100 coplanar forces each equal to 10 N act on a body.
Each force makes angle 50
πwith the preceding force.
What is the resultant of the forces?
a. 1000 N b. 500 N
c. 250 N d. Zero
3 A string of negligible mass going over a clamped pulley
of mass m supports a block of mass M as shown in the
figure. The force on the pulley by the clamp is given by
a. 2Mg b. 2mg
c. 2 2( )M m m g+ + d. 2 2( ( ) )M m M g+ +
4. A simple pendulum is oscillating without damping. When
the displacement of the bob is less than maximum, its
acceleration vector a
is correctly shown in
a. b.
c. d.
5. Four holes of radius R are cut form of a thin square plate
of side 4R and mass M. The moment of inertia of the
remaining portion about z-axis is:
a. 2
12MR
π b. 24
3 4MR
π −
c. 28 10
3 16MR
π −
d. 24
3 6MR
π −
6. The radius of earth is 6400 km and g = 10 m/s2. In order
that a body of 5 kg weighs zero at the equator, the angular
speed of earth in radian/sec is:
a. 1
80 b.
1
400
c. 1
800 d.
1
1600
7. The lower end of a capillary tube of radius r is placed
vertically in water. Then with the rise of water in the
capillary, heat evolved is:
a. 2 2 2r h
dgJ
π+ b.
2 2 2
2
r h dg
J
π+
c. 2 2
2
r h dg
J
π− d.
2 2r h dg
J
π−
8. A wire of length L and cross-sectional area A is made of a
material of Young’s modulus Y. If the wire is stretched by
the amount x, the work done is:
a. 2
2
YAx
L b.
2YAx
L
c. 2
YAx
L d. 2
Yax L
9. Which of the following graphs correctly represent the
variation of β =/
= −dV dp
V with p for an ideal gas at
constant temperature?
a. b.
c. d.
10. Two rods of different materials and identical
cross-sectional area are joined face to face at one end and
their free ends are fixed to the rigid walls. If the temperature
of the surroundings is increased by 30°C, the magnitude of
the displacement of the joint of rods is: (length of the
p p
p p
y
x
a
a
a
a
m
M
Physics46
rods unit, ratio of their Young’s
modulii, ; coefficients of linear expansion are α1
and α2)
a. 2 15( )aα − b. 1 210( )aα −
c. 2 110( 2 )aα − d. 1 25(2 )aα −
11 A steel ball of mass kg moving with velocity
50 ms–1
collides with another stationary steel ball of mass
kg. During the collision, their internal
energies change equally. If T1 and T2 are rise in
temperature of balls m1 and m2
respectively, specific heat
of steel = 0.105 and mechanical equivalent of heat J = 4.2
J/cal, then:
a. 3 3
1 27.0 10 C, 1.4 10 C,T T= × ° = × °
b. 3 3
1 21.4 10 C, 7.0 10 C,T T= × ° = × °
c. 1 21.4 C, 7.0 C,T T= ° = °
d. 1 27.0 C, 1.4 C,T T= ° = °
12. Three identical dipoles are arranged as shown below.
What will be the net electric field at P ?
a. b.
c. Zero d.
13. The area of the plates of a parallel plate capacitor is A and
the distance between the plates is 10 mm. There are two
dielectric sheets in it, one of dielectric constant 10 and
thickness 6 mm and the other of dielectric constant 5 and
thickness 4 mm. The capacity of the capacitor is:
a. A b. A
c. A d. A
14. If a magnet is suspended at an angle 30o to the magnetic
meridian, it makes an angle of 45o
with the horizontal. The
real dip is
a. b.
c. d.
15. Two straight long conductors AOB and COD are
perpendicular to each other and carry currents and
The magnitude of the magnetic induction at a point P at a
distance a from the point O in a direction perpendicular to
the plane ACBD is
a. b.
c. d.
16. A circular coil and a bar magnet placed nearby are made
to move in the same direction. The coil covers a distance
of 1 m in 0.5 sec and the magnet a distance of 2 m in
1 sec. The induced e.m.f. produced in the coil:
a. Zero
b. 1 V
c. 0.5 V
d. Cannot be determined from the given information
17. Following figure shows an ac generator connected to a
"box" through a pair of terminals. The box contains
possible R, L, C or their combination, whose elements and
arrangements are not known to us. Measurements outside
the box reveals that 75 sin(sin )e tω= volt, i = 1.5 sin
(ωt + 45°) amp, then, the wrong statement is
a. There must be a capacitor in the box
b. There must be an inductor in the box
c. There must be a resistance in the box
d. The power factor is 0.707
18. The focal length of a concave mirror is f and the distance
from the object to the principle focus is x. The ratio of the
size of the image to the size of the object is
a. b.
c. d.
1 21l l= =
1 2/ 2Y Y =
1 1m =
2 0.200m =
0
1
4k
πε
=
3
.k p
x 3
2kp
x
3
2 kp
x
0
12
35ε 0
2
3ε
0
5000
7ε 01500ε
1tan ( 3 / 2)− 1tan ( 3)−
1tan ( 3 / 2)− 1tan (2 / 3)−
1i 2.i
01 2( )
2i i
a
µπ
+ 01 2( )
2i i
a
µπ
−
2 2 1/ 201 2( )
2I I
a
µπ
+ 0 1 2
1 22 ( )
I I
a I I
µπ +
f x
f
+ f
x
f
x
2
2
f
x
? ~
y
x
A B
C D
x
x
P
+Q – Q
– Q +Q
+Q – Q
p →
p →
p →
x
47 Mock Test-4
D S
r
19. The energy of incident photon is 12.375 eV while the
energy of scattered photon is 9.375 eV. Then the kinetic
energy of recoil electron is
a. 3 eV b. less than 3 eV
c. more than 3 eV d. 21.75 eV
20. In a transistor circuit shown here the base current is
35 µA. The value of the resistor Rb is
a. 123.5 kΩ b. 257 k Ω
c. 380.05 k Ω d. None of these
SECTION 2 (Numeric Value Question)
21. A wooden block of mass 10 g is dropped from the top of a
cliff 100 m high. Simultaneously a bullet of mass 10 g is
fired from the foot of the cliff upwards with a velocity of
100 m/s. The bullet and the wooden block will meet, each
other, after a time:
a. 10 s b. 0.5 s
c. 1 s d. 7 s
22. The displacement of a particle varies according to the
relation 4(cos sin ).x t t= +π π The amplitude of the
particle is:
a. 8 b. – 4
c. 4 d. 4 5
23. A sound wave of wavelength 32 cm enters the tube at S as
shown in the figure. Then the smallest radius r so that a
minimum of sound is heard at detector D is.
a. 7cm b. 14 cm
c.21cm d. 28 cm
24. Three rods made of the same material and having the
same cross-section have been joined as shown in the
figure. Each rod is of the same length. The left and right
ends are kept at 0°C and 90ºC respectively. The
temperature of junction of the three rods will be
a. 45°C b. 60°C
c. 30° C d. 20° C
25. Six resistance of 6 ohm each are connected to form a
hexagon. The resistance between any two adjacent
terminals is:
a. 6 Ω b. 36 Ω
c. 5 Ω d. 1 ohm
0°C
90°C
90°C
E C
B
Rb RL
9V
Space for Rough Work
Physics 48
JEE ADVANCE PAPER-I
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. A glass tube of uniform internal radius r has a valve
separating the two identical ends. Initially, the valve is in
a tightly closed position. End 1 has a hemispherical soap
bubble of radius r. End 2 has sub-hemispherical soap
bubble as shown in figure. Just after opening the valve,
a. air from end 1 flows towards end 2. No change in the
volume of the soap bubbles
b. air from end 1 flows towards end 2. Volume of the soap
bubble at end 1 decreases
c. no changes occurs
d. air from end 2 flows towards end 1. volume of the soap
bubble at end 1 increases
2. A thin uniform cylinder shell, closed at both ends is
partially filled with water. It is floating vertically in water
in half-submerged state. If ρc is the relative density of the
material of the shell with respect to water, then the correct
statement is that the shell is.
a. more than half-filled if ρc is less than 0.5
b. more than half-filled if ρc is less than 1.0
c. half-filled if ρc is more than 0.5
d. less than half- filled if ρc is less than 0.5
3. An open glass tube is immersed in mercury in such a way
that a length of 8 cm extends above the mercury level. The
open end of the tube is then closed and sealed and the tube
is raised vertically up by additional 46 cm. What will be
length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
a. 38 cm b. 6 cm
c. 16 cm d. 22 cm
4. There is a circular tube in a vertical plane. Two liquids
which do not mix and of densities d1 and d2 are filled in
the tube. Each liquid subtends 90° angle at centre. Radius
joining their interface makes an angle α with vertical.
Ratio 1
2
d
d is
a. 1 tan
1 tan
αα
+−
b. 1 sin
1 cos
αα
+−
c. 1 sin
1 sin
αα
+−
d. 1 cos
1 cos
αα
+−
5. On heating water, bubbles being formed at the bottom of
the vessel detaches and rise. Take the bubbles to be
spheres of radius R and making a circular contact of radius
r with the bottom of the vessel. If r << R, and the surface
tension of water is T, value of r just before bubbles
detaches is: (density of water is wρ )
a. 2 wgRT
ρ
b. 2 3 wgRT
ρ
c. 2 2
3wgRT
ρ d. 2
6wgRT
ρ
6. Two solid spheres A and B of equal volumes but of
different densities Ad and
Bd are connected by a string.
They are fully immersed in a fluid of density .Fd They
get arranged into an equilibrium state as shown in the
figure with a tension in the string. The arrangement is
possible only if
a. A Fd d< b.
B Fd d>
c. A Fd d> d. 2A B Fd d d+ =
A
B
R
2r
d
α
d
2 1
49 Mock Test-4
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. A 25 watt bulb, which is producing monochromatic light
of wavelength 6600 Å is used to illuminated a metal
surface. If the surface has 3% efficiency for photoelectric
effect, then the photoelectric current produced in deci
ampere is (use 346.6 10 Js) :h−= ×
8. de-Broglie wavelength associated with an electron
accelerated through a potential difference 4 V is 1.λ When
accelerating voltage is decreased by 3 V, its de-Broglie
wavelength is 2.λ The ratio 1
2
λλ
is
9. What is the energy in eV that should be added to an
electron of energy 2 eV to reduce its de-Broglie
wavelength from 1 nm to 0.5 nm?
10. The angular momentum of electron in 2nd orbit of
hydrogen atom is L. The angular momentum of electron in
4th orbit of hydrogen atom is n L where n = ?
11. For an atom of an ion having single electron, the
wavelength observed 1 2λ = are units and 3 3λ = units
figure. The value of missing wavelength 2λ is
12. How many different wavelength may be observed in the
spectrum from a hydrogen sample if the atoms are excited
to 3rd excited state?
13. Light waves from two coherent sources having intensities
I and 2I cross each other at a point with a phase diff. of
60º. What is the resultant intensity at the point?
14. A pn-junction diode can withstand currents upto 10 mA.
When it is forward biased, the potential drop across it is
1.0 V. Assuming that this potential drop is independent of
the current, find the maximum voltage of the battery used
to forward bias the diode when a resistance of is
connected in series with the diode.
SECTION 3 (Maximum Marks: 12)
Paragraph based questions (2 paragraphs, each having 2
MCQs with one correct answer only)
Paragraph for Question No. 15 to 16
The general motion of a rigid body can be considered to be a
combination of (i) a motion of its centre of mass about an axis,
and (ii) its motion about an instantaneous axis passing through
the centre of mass. These axes need not be stationary. Consider,
for example, a thin uniform disc welded (rigidly fixed)
horizontally at its rim to a massless stick, as shown in the
figure.
When the disc-stick system is rotated about the origin on a
horizontal frictionless plane with angular speed ,ω the motion
at any instant can be taken as a combination of (i) a rotation of
the centre of mass of the disc about the z-axis, and (ii) a
rotating of the disc through an instantaneous vertical axis
passing through its centre of mass (as is seen from the changed
orientation of points P and Q). Both these motions have the
same angular speed ω in this case.
Now consider two similar systems as shown in the figure: Case
(a) the disc with its face vertical and parallel to x-z plane; Case
(b) the disc with its face making an angle of 45° with x-y plane
and its horizontal diameter parallel to x-axis. In both the cases,
the disc is welded at point P, and the systems are rotated with
constant angular speed ω about the z-axis.
15. Which of the following statements regarding the angular
speed about the instantaneous axis (passing through the
centre of mass) is correct?
a. It is 2ω for both the cases
b. It is ω for case a.; and 2
ωfor case b.
c. It is ω for case a.; and ω for case b.
d. It is ω for both the cases
z ω
Q
y
x
P
z ω Q
y
x
P 45°
ω
P Q y P
Q
x
1λ
n3 orbit
n2 orbit
n1 orbit
3λ
2λ
Physics 50
16. Which of the following statements about the instantaneous
axis (passing through the centre of mass) is correct?
a. It is vertical for both the cases a. and b.
b. It is vertical for case a.; and is at 45°to the x-z plane
and lies in the plane of the disc for case b.
c. It is horizontal for case a.; and is at 45° to the x-z plane
and is normal to the plane of the disc for case b.
d. It is vertical for case a.; and is at 45° to the x-z plane
and is normal to the plane of the disc for case b.
Paragraph for Question No. 17 to 18
A frame of reference that is accelerated with respect to an
inertial frame of reference is called a non-inertial frame of
reference. A coordinate system fixed on a circular disc rotating
about a fixed axis with a constant angular velocity ω is an
example of a non-inertial frame of reference.
The relationship between the force rotF
experienced by a
particle of mass m moving on the rotating disc and the force
inF
experienced by the particle in an inertial frame of reference
is, rot in rot2 ( ) ( ) ,F F m v m rω ω ω= + × + × ×
Where rotv
is the velocity of the particle in the rotating frame of
reference and r
is the position vector of the particle with
respect to the centre of the disc.
Now consider a smooth slot along a diameter of a disc of radius
R rotating counter-clockwise with a constant angular speed ω
about its vertical axis through its center. We assign a coordinate
system with the origin at the center of the disc, the x-axis along
the slot, the y-axis perpendicular to the slot and the z-axis along
the rotation axis ˆ( ).kω ω=
A small block of mass m is gently
placed in the slot at ˆ( / 2)r R i=
at 0t = and is constrained to
move only along the slot.
17. The distance r of the block at time t is:
a. cos 22
Rtω b. 2 2( )
4t tR
e eω ω−+
c. cos2
Rtω d. ( )
4t tR
e eω ω−+
18. The net reaction of the disc on the block is :
a. 2 ˆˆsinm R tj mgkω ω −
b. 2 ˆˆcos 6m R tj mgkω ω− −
c. 21 ˆˆ( )2
t tm R e e j mgkω ωω −− +
d. 2 2 21 ˆˆ( )2
t tm R e e j mgkω ωω −− +
R
R/2
m
ω
Space for Rough Work
51 Mock Test-4
JEE ADVANCE PAPER-II
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. One end of a horizontal thick copper wire of length 2L
and radius 2R is welded to an end of another horizontal
thin copper wire of length L and radius R. When the
arrangement is stretched by applying forces at two ends,
the ratio of the elongation in the thin wire to that in the
thick wire is
a. 0.25 b. 0.50
c. 2.00 d. 4.00
2. The pressure that has to be applied to the ends of a steel
wire of length 10 cm to keep its length constant when its
temperature is raised by 100°C is: (For steel Young’s
modulus 2 × 1011 Nm–2 and coefficient of thermal
expansion is 1.1 × 10–5 K–1 )
a. 72.2 10 Pa× b. 62.2 10 Pa×
c. 82.2 10 Pa× d. 92.2 10 Pa×
3. A pendulum made of a uniform wire of cross-sectional
area A has time period T. When an additional mass M is
added to its bob, the time period changes to .MT If the
Young's modulus of the material of the wire is Y then 1/Y
is equal to: (g = gravitational acceleration)
a.
2
1MT A
T Mg
−
b. 2
1MT Mg
T A
−
c. 2
1 MT A
T Mg
−
d.
2
1M
T A
T Mg
−
4. Function 2 2sin cos sin cosω ω ω ω= + +x A t B t C t t
represents SHM.
a. For any value of A, B and C (except C = 0)
b. If A = – B; C = 2B, amplitude | 2 |B=
c. If A = B; C = 0
d. If A = B; C = 2B amplitude = |B|
5. Two coaxial solenoids of different radii carry current I in
the same direction. Let 1F
be the magnetic force on the
inner solenoid due to the outer one and 2F
be the magnetic
force on the outer solenoid due to the inner one. Then:
a. 1 2 0F F= =
b. 1F
is radially inwards and 2F
is radially outwards
c. 1F
is radially inwards and 2 0F =
d. 1F
is radially outwards and 2 0F =
6. Consider a spherical shell of radius R at temperature T.
The black body radiation inside it can be considered as an
ideal gas of photons with internal energy per unit volume
4Uu T
V= ∝ and pressure
1.
3
UP
V
=
If the shell now
undergoes an adiabatic expansion the relation between T
and R is:
a. RT e−∝ b. 3RT e−∝
c. 1
TR
∝ d. 3
1T
R∝
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. An electron in an excited state of Li2+ ion has angular
momentum 3 / 2 .πh The de-Broglie wavelength of the
electron in this state is pπa0 (where a0 is the Bohr radius).
The value of p is
8. Consider a hydrogen atom with its electron in the nth
orbital. An electromagnetic radiation of wavelength 90 nm
is used to ionize the atom. If the kinetic energy of the
ejected electron is 10.4 eV, then the value of n is (hc =
1242 eV nm)
9. A hydrogen atom in its ground state is irradiated by light
of wavelength 970 Å. Taking hc/ e 61.237 10 −= × eV m and
the ground state energy of hydrogen atom as –13.6 eV, the
number of lines present in the emission spectrum is:
10. To determine the half life of a radioactive element, a
student plots a graph of( )
n
dN t
dtℓ versus t.
Here dN(t)/dt
is the rate of radioactive decay at time t. If
the number of radioactive nuclei of this element decreases
by a factor of p after 4.16 years, the value of p is
Years
6
5
4
3
2
1 2 3 4 5 6 7 8
ℓn|dN
(t)/dt|
L θ
I I
Physics 52
11. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 910 s.The mass of an atom of this radioisotope is 2510 kg.−
kg. The mass (in mg) of the radioactive sample is
12. A freshly prepared sample of a radioisotope of half-life
1386s has activity 103 disintegrations per second.
Given that ln 2 = 0.693, the fraction of the initial number
of nuclei (expressed in nearest integer percentage)
that will decay in the first 80s after preparation of the
sample is
13. A nuclear power plant supplying electrical power to a
village uses a radioactive material of half life T years as
the fuel. The amount of fuel at the beginning is such that
the total power requirement of the village is 12.5 % of the
electrical power available from the plant at that time. If
the plant is able to meet the total power needs of the
village for a maximum period of nT years, then the value
of n is
14. For a radioactive material, its activity A and rate of change
of its activity R are defined as = −dN
Adt
and ,= −dA
Rdt
where N(t) is the number of nuclei at time t. Two
radioactive sources P (mean lifeτ ) and Q (mean life 2 )τ
have the same activity at t = 0. Their rates of change of
activities at 2τ=t are PR and ,QR respectively. If
,=P
Q
R n
R e then the value of n is
SECTION 3 (Maximum Marks: 12)
Matching type questions with 4 options
15. There is a definite relation between velocity, mass and
acceleration of body to know about the work done by the
forces applied on body. Column I has some statements
showing some relations which are related with some
statements in Column II. Match the correct options.
Column I Column II
(A) If kinetic energy is K the
momentum P is 0
1. zero
(B) If momentum p is zero the
kinetic energy is 0 2. 2mk
(C) If different masses have
same kinetic energy the
ratio of their momenta is 0
3. 2
1
m
m
(D) If different masses have
same momentum, the ratio
of their kinetic energy is
4. 1
2
m
m
a. A-1, B-2, C-3, D-4 b. A-2, B-1, C-4, D-3
c. A-4, B-3, C-2, D-1 d. A-3, B-2, C-1, D-4
16. Waves are classified on the basis of frequency range.
Column I mentions the type of waves while column II
gives the corresponding frequency range. Can you match
the proper options?
Column I Column II
(A) Infrasonic 1. Objects having velocity
more than velocity of
sound in air at 0°C
(B) Audible 2. Frequency 20 Hz<
(C) Ultrasonics 3. Frequency 20,000 Hz>
(D) Supersonics 4. Frequency lies between
20 Hz to 20, 000 Hz
a. A-4, B-3, C-2, D-1 b. A-2, B-4, C-3, D-1
c. A-4, B-4, C-1, D-2 d. A-3, B-2, C-1, D-4
17. Various electromagnetic waves are given in column I and
various frequency ranges in column II
Column I Column II
(A) Radiowaves 1. 16 211 10 to 3×10 Hz×
(B) γ-rays 2. 9 111 10 to 3×10 Hz×
(C) Microwaves 3. 18 223 10 to 5×10 Hz×
(D) X-rays 4. 5 95 10 to 10 Hz×
a. A-1, B-2, C-3, D-4 b. A-2, B-3, C-4, D-1
c. A-3, B-4, C-1, D-2 d. A-4, B-3, C-2, D-1
18. In each of the following questions, match column I and
column II, and select the correct match out of four given
choices.
Column I Column II
(A) Transducer 1. Range of frequencies over
which communication
system works
(B) Attenuation 2. A device that has input in
electrical form or provides
output in electrical form
(C) Range 3. Loss of strength of a
signal during propagation.
(D) Bandwidth 4. Argest distance between
transmitter and receiver.
a. A-,2 B-4, C-1, D-3 b. A-4, B-3, C-1, D-2
c. A-1, B-2, C-3, D-4 d. A-2, B-3, C-4, D-1
53 Mock Test-4
ANSWERS & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d d d c c c b a a c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c c c a c a b b a b
21. 22. 23. 24. 25.
c d b b c
1. (d) Dipole moment = (charge) × (distance)
Electric flux = (electric field) × (area)
2. (d) Total angle 100 250
ππ= × =
So, all the force will pass through one point and all forces
will be balanced. i.e. their resultant will be zero.
3. (d) Free body diagram of pulley is shown in figure.
Pulley is in equilibrium under four forces. Three forces as
shown in figure and the fourth, which is equal and
opposite to the resultant of these three forces, is the force
applied by the clamp on the pulley (say F).
Resultant R of these three forces is 2 2( ( ) )R M m M g= + +
Therefore, the force F is equal and opposite to R as shown
in figure.
∴ 2 2( ( ) ) .F M m M g= + +
4. (c) Net acceleration a
of the bob in position
B has two components.
(a) na =
radial acceleration (towards BA)
(b) ta =
tangential acceleration (perpendicular to BA)
Therefore, direction of a
is correctly shown in option (c).
5. (c) M = Mass of the square plate before cutting the holes
Mass of one hole 2
216 16
M Mm R
R
ππ
= =
∴ Moment of inertia of the remaining portion,
4square holeI I I= −
( ) ( )2
2 2 216 16 4 212 2
M mRR R m R
= + − +
2 2810
3MR mR= − 28 10
3 16MR
π = −
6. (c) 2g g Rω′ = − (at equator)
Given 20g g Rω′ = ⇒ =
⇒ 3
10 1
6400 10 800
g
Rω = = =
×rad/s
7. (b) When the tube is placed vertically in water, water rises
through height h given by 2 cosT
hrdg
=θ
Upward force 2 cosr T= ×π θ
Work done by this force in raising water column through
height h is given by (2 cosW rT h=∆ π θ)
2 2(2 cos (2 cos
2cos
rhdgrh T = rh r h dg
= = π θ) π θ) π
θ
However, the increase in potential energy pE∆ of the raised
water column ,2
hmg= where m is the mass of the raised
column of water.
∴ 2m r hdπ=
So, 2 2
2( )2 2
p
gh r h dgE r hd
ππ ∆ = =
Further, 2 2
2p
r h dgW E
π∆ − ∆ =
The part ( )pW E∆ − ∆ is used in doing work against viscous
forces and frictional forces between water and glass
surface and appears as heat.
So, Heat released 2 2
2pW E r h dg
J J
π∆ − ∆=
8. (a) Energy stored, 1
2U = × (force)×(extension)
1
2Fx=
As xFL YAY F
Ax L= ⇒ =
∴ 21 1
2 2xYA YAx
U xL L
= =
9. (a) β =/dV dp
V= − = compressibility of gas
1
Bulk modulus of elasticity= and β =
1
p= under isothermal
na
A
B a
ta
R
F
Mg + mg
Mg
T = Mg
mg T=Mg
Physics54
conditions. Thus, β versus p graph will be a rectangular
hyperbola.
10. (c) Applying,
temperature difference across the rods are 20°C and 10°C
respectively.
∴ Displacement of the joint
11. (c) J m1s ∆θ1
⇒ ∆θ1
2
4Js
v= = = 1.4 °C
1 2m s m s=1 2
∆θ ∆θ
⇒ 1
2
11.4 7 C
0.2
m
m= = × = °
2 1∆θ ∆θ
12. (c) Point P lies at equatorial positions of dipole 1 and 2
and axial position of dipole 3. Hence field at P
Due to dipole 1 (towards left)
Due to dipole 2 (towards left)
Due to dipole 3 (towards right)
So, Net field at P will be zero.
13. (c) The two capacitor C1 and C2
are connected in series
This gives
14. (a) Let the real dip be φ, then
For apparent dip,
tancos
V
H
B
B′ =φ
β
or φ or φ
15. (c) At P :
16. (a) Speed of the magnet m/s
Speed of the coil m/s
Relative speed between coil and magnet is zero, so there
is no induced emf in the coil.
17. (b) Since voltage is lagging behind the current, so there
must be no inductor in the box.
18. (b) where ∴
19. (a) Kinetic energy of recoil electron,
12.375 9.375 3eV= − =
20. (b) kΩ
21. (c)
⇒
⇒ ⇒ = 1s
( )1 2KAQ
l
θ θ−=
2 2 1 1t tα α= −
( )2 1 2 110 20 10 2α α α α= − = −
2
1 1 1
1 1
2 2m v J m s⋅ = ∆
( )2
3
50
4 4 4.2 10 0.105∆ = =
× × ×
1 3
.k pE
x=
2 2
.k pE
x=
3 3
.(2 )k pE
x=
1 0 2 01 2
1 2
,K A K A
C Cd d
ε ε= =
1 2
1 2 1 0 2 0
1 1 1,
d d
C C C K A K Aε ε= + =
1 2
0 1 2
1 d d
A K Kε
= +
3 3
0
1 6 10 4 10
10 5Aε
− − × ×= +
3
0
1 14 10
10Aε
−×=
0
5000
7C Aε=
tan V
H
B
Bφ =
2
cos cos30 3
V V V
H H H
B B B
B B Bβ= = =
°
2tan 45 .tan
3° = 1 3
tan2
φ − =
2 2
1 2netB B B= +
2 2
0 01 22 2
4 4
i i
a a
µ µπ π
= +
2 2 1/ 20
1 2( )2
i ia
µπ
= +
1
22 /
1v m s= =
2
12 /
0.5v m s= =
;I f
O f u=
−u f x= + I f
O x= −
kE hv hv′= −
6
9257
35 10b b b b
V i R R k−
= ⇒ = = Ω×
1 2 100s s+ =
2 21 1100
2 2gt ut gt m+ − =
100ut m=100 100
100t s
u= = =
S N
v1
v2
i2
D
O
B
C
A
i1
a
B1
B2
P
P
+Q – Q
– Q +Q
+Q – Q
E1 E2 E3
1
3
2
10 mm
6 mm 4 mm
55Mock Test-4
22. (d) 4(cos sin )x t t= +π π
Standard equation of displacement is
Comparing the given equation with standard equation
23. (b) Path difference 162
32
2)2( ===−
λπ rr
⇒ 16
14cm2
rπ
= =−
24. (b) Let θ be the temperature of the junction (say B).
Thermal resistance of all the three rods is equal. Rate of
heat flow through AB + Rate of heat flow through CB =
Rate of heat flow through BD
∴ 90 90 0
R R R
°− °− −+ =
θ θ θ, Here, R =Thermal resistance
∴ 3 180 ,= °θ or 60 C= °θ
25. (c) Resistance and are in parallel; so
equivalent resistance
JEE Advance Paper -I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
b a a a c a,b,d 4 2 6 2
11. 12. 13. 14. 15. 16. 17. 18.
6 6 4 2 d a d c
1. (b) P1 = pressure just inside the bubble at the end
0
42
TP
R= +
2. (a) If water is half filled shell displaced( )ρ= ∆M g V g
⇒ shell( )2
∆ =
VV
⇒ liwuid shellshell
2
ρ=
VM
⇒
liquidshell
shell 2
ρ =
M
V
⇒ liquid
shell2
ρσ =
⇒
1. .
2=R d
1If .
2<R d , Then the cylinder should be more than half
filled so that it is half submerged and floating.
3. (a) 0(54 ) 8P h P′ − = (Boyel’s Law)
⇒ 0 8
54
PP
h′ =
− ⇒ 0P gh Pρ′ + =
⇒ 00
8
54
Pgh P
hρ+ =
−
⇒ 0
46
54
hgh P
hρ
− = − 0( (76))P gρ=
⇒ 46
(76)54
hh
h
− = −
⇒ 254 76 46 76h h h− = × −
⇒ 2 130 76 46 0h h− + × = ⇒ ( 38)( 92) 0h h− − =
⇒ ⇒ 38cmh=
4. (a) Pressure at 0 level is same from both sides.
1(1 sin )d α−
1 2(1 cos ) [cos sin ]d dα α α= − + +
⇒ 1
2
sin cos
cos sin
d
d
α αα α
+=
−
5. (c) 2 2
3
wgR
T
ρ, The downward force on the bubble due to
surface tension 2 . sinr T= π θ
22 Tr
R
π=
1 14 2 cos sin
2 2t tπ π
+
4 2 sin cos cos sin2 2
t tπ π
π π = + 4 2 sin
4t
ππ = +
( )sinx a tω φ= +
4 2a =
( )6 5 30× = Ω 6 Ω
30 65
30 6R
×= = Ω
+
R
r
θ
θ δT
ααα
d
0
(54-h)
54 cm
h
P’ P0
0°C
90°C
90°C
A
B
C
D
2l
2Y θ Y θ1 θ2
l
Not possible
38,92h =↓
Physics 56
The upward buoyant force exceeds, the surface tension
force then the bubble detaches.
∴ 2
34 2
3w
TrR g
R
ππ ρ = ⇒ 2 2
3wgr RT
ρ=
6. (a, b, d)
7. (4) No. of photons emitted per second by 25 watt source
will be
10
34 8 19
25 (6600 10 ) 25
/ (6.6 10 ) (3 10 ) 3 10
E En
hc hc
λλ
−
− −
× ×= = = =
× × × ×
Current, I = 3% of number of photoelectrons emitted per
sec × charge of electron
19
19
3 3 251.6 10 0.4A 4 deciampere
100 100 3 10ne
−−= × × × = =
×
8. (2) de-Broglie wavelength, 4 1
or2 V Vmq
λ λ= ∝
∴ 2 1
1 2
V 4V2
V 4V 3V
λλ
= = =−
9. (6) As 2 K
h
mEλ =
or 2
2 2
1or
2K K
hE E
mλ λ= ∝
∴ 1
2
2 2
2
1
0.5 1
1 4K
K
E
E
λλ
= = =
or 1 1
4 4 2 8K KE E eV eV= = × == 8 eV
Increase in energy 2 1
8 2 6eVK KE E= − = − =
10. (2) 22
hL
π =
and ' 42
hL
π =
∴ ' 2 ,L L n L= = where n = 2
11. (6) As is clear from figure. 1 2 3E E E= +
1 2 3
hc hc hc
λ λ λ= +
∴ 2 1 3
1 1 1 1 1 1
2 3 6λ λ λ= − = − = = 6 units
12. (6) 3rd excited state means n = 4. No. of different
wavelength observed ( 1) 4(4 1)
62 2
n nN
− −= = =
13. (4) Here, 1I I=
2 2 , 60I I φ= = °
Amplitude R of resultant wave is
2 2 2 cosR a b ab φ= + +
2 2 2 2 cosR a b ab φ= + + As intensity ∝ (amplitude)2
∴ Resultant intensity, 1 2 1 22 cosRI I I I I φ= + +
2 2 2 cos60I I I I= + + × × °
3 2 (3 2) 4.414I I I I= + = + =
14. (2) See figure
Voltage drop across diode is
dV V= 1.0V
Voltage drop across R is RV RI=
3100 10 10−= × × = 1.0V
max 1.0 1.0 2.0d RV V V V= + = + == 2.0V
15. (d) It is ω for both case.
As angular velocity of a rigid body about any point should
be same. = +
PC COv v v
⇒ 'ω= + +
PO PC COv r v
ˆˆ ˆ ˆ( ) .4522 2
d dLi k j L iω ω ω
− = − + × + − + °
⇒ ˆpr j=ℓ
⇒ ˆˆcos45 sin 452 2c
d dr j k
= + ° + °
ℓ
⇒ ˆˆcos45 sin 452 2
= − ° − °
PC
d dr j k
⇒ ˆcos452
CO
dv iω
= − + °
ℓ
⇒ ˆ ˆ( ) '2 2 2 2
ωω= − + ×d d
i i j ⇒ ˆ'ω ω=
k
16. (a) 2 2ω= +ℓV r TOPView
2 sin
2
VW
r
θ =
2 2
2 2
ω +=
+
ℓ
ℓr
r r
rω=
kω
P
ˆ( dcos 45 )iℓ− + °ω
iℓ−ω
d cos 45°
V
I
VR
R
Vd
A
B
2 Fd Vg
( )A Bd d Vg+
57 Mock Test-4
Here there is a relv of P′ and Q′ (not for P, Q) (vrel)P’Q’ will
be in horizontal plane
⇒ ω must be along vertical. Same situation will be there for
P as mentioned ω should be same for both cases.
17. (d)2
mr maω =
⇒ 2a rω= ⇒ 2vdv
rdr
ω=
⇒ 2
0 / 2
v r
R
vdv rdrω=∫ ∫
⇒2
2
4
Rv rω= −
2
/ 2 02
4
v t
R
drdt
Rr
ω=
−∫ ∫ . . .(i)
Assume: sec2
Rr = θ
⇒ sec tan2
Rdr d= θ θ θ
⇒ 2
02
sec tan2
tan4
tR
d
dtR
θ θ θ=
θ∫ ∫ω
⇒ 2 22 4r r R
t nR R
ω −
= +
ℓ
⇒ [ ]4
t tRr e eω ω−= +
18. (c) rot in rotˆ ˆ ˆˆ ˆ2 ( ) ( )F F m v i k m k ri kω ω ω= + × + × ×
2 2in rot
ˆ ˆ ˆ2 ( )mr i F mv j m riω ω ω= + − +
inˆ2 rF mv jω=
. . .(i)
4
t tRr e eω ω− = +
4
t t
r
dr Rv e e
dt
ω ωω ω − = = −
inˆ2
4t tR
F m e e jω ωωω− = −
Also reaction is due to disc surface then
2
ˆˆ2
t t
reaction
mRF e e j mgk
ω ωω − = − +
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
c c a a,b,c a a,b 2 2 6 8
11. 12. 13. 14. 15. 16. 17. 18.
1 4 3 2 b b d d
1. (c)
⇒ 2 2
1 2
4,
2
π π= =
R x R yk k
L L
⇒ 1 2= =F k x k y ⇒ 1
2
2= =ky
x k
2. (c) F
YA
∆=
ℓ
ℓ⇒ Tα
∆= ∆ℓ
ℓ
⇒ F
P Y TA
α= = ∆
11 5 8 22 10 1.1 10 100 2.2 10 /−= × × × × = ×
Pa
3. (a) 12 ,Tg
π=ℓ
22MTg
π=ℓ
2 2 14 ,T
gπ= ⋅ℓ
2 2 22MT
gπ=ℓ
⇒ 2
22
1
MT xT
=ℓ
ℓ
⇒ 2
2 2 12
1 1
1 1MT
T
−− = − =
ℓ ℓ ℓ
ℓ ℓ
⇒ 1
2 1( )
Mg
Aγ = ⋅
−ℓ
ℓ ℓ
⇒
2
2 1
1
( )11MTA A
Mg Mg Tγ
− = ⋅ = −
ℓ ℓ
ℓ
4. (a, b, c) (1 cos2 ) (1 cos 2 ) sin 22 2 2
ω ω ω= − + + +A B C
x t t t
For 0, 0= =A B sin 22
ω=C
x t
For = −A B and 2=C B cos 2 sin 2ω ω= +x B t B t
Amplitude | 2 |= B
For ; 0= =A B C =x A
Hence this is not correct option
For , 2= =A B C B sin 2ω= +x B B t
It is also represents SHM.
5. (a) Solenoid consists of circular current loops which are
placed in uniform field, so magnetic fore on both
solenoids 0.=
F F
2L, 2R L, R
2 20 rω +ℓ
2 2r+ℓ
2r+ℓ
1p
θ θ
θ
ˆ( )kω ˆ( )y j
rv
i
r
Physics 58
⇒ 1 2 0F F= =
6. (c) 41
3
UP T
V
= ∝
For an ideal gas PV nRT=
∴ nR T
PV
′= ( R′ -molar gas constant)
∴ RT e−∝ ⇒ 3nRT
V
′∝
∴ 31T
V∝ ⇒
3
3
14
3
T
Rπ
∝
∴ 3
3
1T
R∝ ⇒
1T
R∝
7. (2) 3
2 2π π= =nh h
mvr de-Broglie wavelength
2
00
(3)2 22
3 3
π πλ π= = = =
Li
ah ra
mv z
8. (2) photon ionize atom kinetic energyE E E= +
⇒ 2
1242 13.610.4
90 n= + . From this, 2n =
9. (6) [Here we are assuming that in original paper atom
⇒ atoms] 2 2
1 1 1237013.6
1 970n
− =
⇒ 2 16n = ⇒ 4n =
No. of lines 42C=
4 36
2
×= =
10. (8) 0tN N e λ−=
⇒ 0| / | ( )n dN dt n N tλ λ−ℓ ℓ
From graph, 1
2λ = per year
⇒ 1/ 2
0.6931.386
1/ 2t = = year
⇒ 1/ 24.16 yrs =3t ∴ 8p =
11. (1) A Nλ= ⇒A
N Aτλ
= =
So 25 10 9 610 10 10 10 kg 1 mgM mN mAτ − −= = = × × = =
12. (4) –(1 ) 1 1 (1 )t tf e e t t
λ λ λ λ−= − = − ≈ − − =
⇒ 0.04f = . Hence % decay 4%≈
13. (3) 0
1
2
t
TA
A
=
Where, 0A is the initial activity of the radioactive material
and A is the activity at t.
So, 12.5 1
100 2
t
T =
∴ 3 .t T=
14. (2) λP1 1
;2
λ λτ τ
= =p Q
⇒ 0
0
( ) 2, At 2
P
Q
t
PP P
t
Q QQ
A eR Rt
R R eA e
−
−= = =
λ
λ
λτ;
λ
15. (b) A-2, B-1, C-4, D-3
Kinetic energy 21
2K mv=
or,
2
2
pK
m= or 2p mK=
If 0,K = 0p = as above
For different mass 1m and 2m if 1 2E E=
2 21 2
1 22 2
p p
m m= or 1 1
2 2
p m
p m=
If 1 2p p= for different masses.2
1 1 2 2
2 1 2 1
2
2
E p m m
E m p m= × =
16. (b) A-2, B-4, C-3, D-1
Waves having frequency less than minimum audible
frequency are infrasonics.
Waves of frequency between 20Hz to 20,000Hz are called
audible waves. Waves having frequency > 20,000Hz
are ultrasonic waves. Supersonic objects are those which
have velocity more than velocity of sound.
17. (d) A-4, B-3, C-2, D-1
The frequency ranges of various waves are as under :
Radiowaves; 55 10× to 910 ; Hz;
γ-rays; 183 10× to 225 10× Hz
Microwaves; 91 10× to 113 10 ;× Hz;
X-rays; 161 10× to 213 10 .× Hz.
18. (d) A-2, B-3, C-4, D-1 A transducer is a device that has input in electrical form or
provides output in electrical form. Attenuation is loss of strength of a signal during propagation. Range is largest distance between transmitter and receiver. Bandwidth is range of frequencies over which communication system works.
F1
B = 0
B2
x x x x x x x x x x x x
x x x x x x x x x x x x
59Mock Test-5
JEE-MAIN: PHYSICS MOCK TEST-5
SECTION 1 (Multiple Choice Question)
1. Frequency is the function of density )(ρ , length )(a and
surface tension )(T . Then its value is
a. 1/ 2 3/ 2k a Tρ / b. 3/ 2 3/ 2
/k a Tρ
c. 1/ 2 3/ 2 3/ 4
/k a Tρ d. 1/ 2 1/ 2 3/ 2
/k a Tρ
2. With respect to a rectangular cartesian coordinate system,
three vectors are expressed as: ˆ ˆ ˆ ˆ4 , 3 2a i j b i j= − = − +rr
and ˆc k= −r
where ˆˆ ˆ, ,i j k are unit vectors, along the X, Y
and Z-axis respectively. The unit vectors r along the
direction of sum of these vector is:
a. 1 ˆˆ ˆˆ ( )3
r i j k= + − b. 1 ˆˆ ˆˆ ( )2
r i j k= + −
c. 1 ˆˆ ˆˆ ( )3
r i j k= − + d. 1 ˆˆ ˆˆ ( )2
r i j k= + +
3. If and represent the work done in moving a
particle from A to B along three different paths 1, 2 and 3
respectively (as shown) in the gravitational field of a
point mass m. Find the correct relation between
and
a. W1 > W2 > W3 b. W1 = W2 = W3
c. W1 < W2 < W3 d. W2 > W1 > W3
4. The moment of inertia of a uniform rod of length 2l and
mass m about an axis passing through its center and
inclined at an angle is:
a. b.
c. d.
5. A mass M is suspended from a spring of negligible mass.
The spring is pulled a little and then released, so that the
mass executes SHM of time period T. If the mass is
increased by m, the time period becomes 5T/3. The ratio
of m/M is
a. b. c. d.
6. The gravitational potential difference between the surface
of a planet and a point 10 m above is 4.0 J/kg. The
gravitational field in this region, assumed uniform is
a. 0.025 N/kg b. 0.40 N/kg
c. 40 N/kg d. 4.0 N/kg
7. 1000 drops of water all of same size join together to form
a single drop and the energy released raises the
temperature of the drop. Given the T is the surface tension
of water. r the radius of each small drop, the density of
liquid, J the mechanical equivalent of heat. What is the
rise in temperature?
a. b.
c. d. None of these
8. A simple pendulum is made by attaching a 1 kg bob to
5 m long copper wire. Its period is T. now if 1 kg bob is
replaced by 10 kg bob, the period of oscillations:
a. remains T
b. becomes greater than T
c. becomes less than T
d. any of above answer depends on locality
9. On which of the given scales of temperature, the temperature
is inverse negative?
a. Celsius b. Fahrenheit
c. Reaumur d. Kelvin
10. An ideal gas is initially at temperature T and volume V. Its
volume is increased by due to an increase in
temperature pressure remaining constant. The quantity
varies with temperature is:
a. b.
c. d.
1 2,W W 3W
1 2,W W
3.W
xx′
α
22sin
3
mlα
22sin
12
mlα
22cos
6
mlα
22cos
2
mlα
5
3
3
5
16
9
28
9
ρ
/T Jr 10 /T Jr
100 /T Jr
∆V
,∆T
/δ = ∆ ∆V V T
T T
T T
T T
T T
A B α
C
x′
x
A
B
1 2 3
Physics 60
11. A lead ball moving with velocity V strikes a wall and
stops. If 50T of its energy is converted into heat, then
what will be the increases in temperature (specific heat of
lead is s)?
a. 22V
Js b.
2
4
V
Js
c. 2V s
J d.
2
2
V s
J
12. The plots of intensity versus wavelength for three black
bodies at temperatures 1 2,T T and 3T respectively are as
shown. Their temperatures are such that
a. 1 2 3> >T T T b.
1 3 2> >T T T
c. 2 3 1> >T T T d.
3 2 1> >T T T
13. The electric field in a region surrounding the origin is
uniform and along the x-axis. A small circle is drawn with
the centre at the origin cutting the axes at points A, B, C, D
having co-ordinates (a, 0), (0, a), (– a, 0), (0, – a);
respectively as shown in figure then potential in minimum
at the point
a. A b. B
c. C d. D
14. A spherical capacitor consists of two concentric spherical
conductors of inner one of radius1R maintained at
potential1V and the outer one of radius
2R at potential 2.V
The potential centre ( )2 1R x R> > is:
a. ( )1 21
2 1
V Vx R
R R
−−
− b.
( ) ( )( )
1 1 2 2 2 1
2 1
V R R x V R x R
R R x
− + −
−
c. 21
1 1
V xV
R R+
− d. 1 2
2 1
V Vx
R R
++
15. Two identical short bar magnets, each having magnetic
moment M, are placed a distance of 2d apart with axes
perpendicular to each other in a horizontal plane. The
magnetic induction at a point midway between them is
a. 0
3( 2)
4
M
d
µπ
b. 0
3( 3)
4
M
d
µπ
c. 0
3
2 M
d
µπ
d. 0
3( 5)
4
M
d
µπ
16. The figure shows four wire loops, with edge lengths of
either L or 2L. All four loops will move through a region
of uniform magnetic field B
(directed out of the page) at
the same constant velocity. Rank the four loops according
to the maximum magnitude of the e.m.f. induced as they
move through the field, greatest first:
a. ( ) ( )c d a be e e e= < = b. ( ) ( )c d a be e e e= > =
c. c d b ae e e e> > >
d.
c d b ae e e e< < <
17. A resistance is connected to an AC source. The average
power dissipation in the resistance is
a. 50% of the peak power
b. 25% of the peak power
c. 41.4% of the peak power
d. equal to peak power
18. When a point source of monochromatic light is at a
distance of 0.2 m from a photoelectric cell, the cut-off
voltage and the saturation current are 0.6 V and 18 mA
respectively. If the same source is placed 0.6 m away from
the photoelectric cell, then
a. the stopping potential will be 0.2 V
b. the stopping potential will be 0.6 V
c. the saturation current will be 6 mA
d. the saturation current will be 18 mA
19. A beam of fast moving alpha particles were directed
towards a thin film of gold. The parts BA ′′, and C ′ of the
transmitted and reflected beams corresponding to the
incident parts A, B and C of the beam, are shown in the
adjoining diagram. The number of alpha particles in
a. B′ will be minimum and in C′ maximum
b. A′ will be maximum and in B′ minimum
c. A′ will be minimum and in B′ maximum
d. C′ will be minimum and in B′ maximum
B′ B A
C C′
A′
a b
c d
• • •
• • •
• • •
• • •
•
•
•
•
•
•
•
•
D
C
B E →
T1
T3
T2
I
λ
61 Mock Test-5
20. Energy released in the fission of a single 235
92 U nucleus
is 200 MeV. The fission rate of a 235
92 U fuelled reactor
operating at a power level of 5W is
a. 1101056.1 −+× s b. 1111056.1 −+× s
c. 1161056.1 −+× s d. 1171056.1 −+× s
SECTION 2 (Numeric Value Question)
21. A train of 150 m in length is going towards north direction
at a speed of 10 m/s. A parrot flies at a speed of 5 m/s
towards south direction parallel to the railway track. Then
the time taken by the parrot to cross the train is equal to:
a. 12 sec b. 8 sec
c. 10 sec d. 15 sec
22. What is the maximum value of the force F such that the
block shown in the arrangement, does not move?
a. 20 N b. 10 N
c. 12 N d. 15 N
23. When an astronaut in a rocket reaches near the moon, he
sends a radiowave of frequency 5000 MHz towards the
moon. He finds that the reflected wave from the moon has
a frequency 86 kHz more than the real frequency. The
velocity of the rocket with respect to the moon is:
a. 1.29 km/sec b. 2.58 km/sec
c. 3.87 km/sec d. 5.16 km/sec
24. A battery of 10 cells each of e.m.f. E = 1.5 V, and internal
resistance 0.5Ω has 1 cell wrongly connected. It is being
charged by 220V power supply with an external resistance
of 47 Ω in series. The potential difference across the
battery is:
a. 32 V b. 8 V
c. 180 V d. 188 V
25. A cell is connected between the points A and C of a
circular conductor ABCD of centre O with angle
A 60 .OC = ° If
1B and 2B are the magnitudes of the
magnetic fields at O due to the currents in ABC and ADC
respectively, the ratio 1
2
B
Bis
a. 0.2 b. 6
c. 1 d. 5
1A i2
300o B
C A
D
60o
i1
O
60°
F 3m kg=1
2 3µ =
Space for Rough Work
Physics 62
JEE ADVANCE PAPER-I
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. Given,
1R 1= Ω
1C 2 Fµ= µF
2R 2= Ω
2C 4 F= µF
(I) (II) (III)
The time constants (in µS) for the circuits I, II, III are
respectively
a. 18, 8 / 9, 4 b. 18, 4, 8 / 9
c. 4, 8 / 9,18 d. 8 / 9,18, 4
2. A circuit is connected as shown in the figure with the
switch S open. When the switch is closed the total amount
of charge that flows from Y to X is
a. 0 b. 54 µC
c. 27 µC d. 81 µC
3. A parallel plate capacitor C with plates of unit area and
separation d is filled with a liquid of dielectric constant
2.K = The level of liquid is 3
dinitially. Suppose the
liquid level decreases at a constant speed ,V the time
constant as a function of time t is
a. 06
5 3
R
d Vt
ε+
b. 0
2 2 2
(15 9 )
2 3 9
d Vt R
d dVt V t
ε+− −
c. 06
5 3
R
d Vt
ε−
d. 0
2 2 2
(15 9 )
2 3 9
d Vt R
d dVt V t
ε−+ −
4. A 2µF
capacitor is charged as shown in figure. The
percentage of its stored energy dissipated after the switch
S is turned to position 2 is
a. 0% b. 20% c. 75% d. 80%
5. In the given circuit, a charge of +80 µF is given to the
upper plate of the 4µF
capacitor. Then in the steady state,
the charge on the upper plate of the 3µF capacitor is.
a. +32 µC b. +40 µC
c. +48 µC d. +80 µC
6. Two capacitors C1 and C2
are charged to 120 V and 200 V
respectively. It is found that by connecting them together
the potential on each one can be made zero. Then
a. 21 35 CC = b. 21 53 CC =
c. 053 21 =+ CC d. 21 49 CC =
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. The magnetic flux of φ (in weber) in a closed circuit of
resistance 3 ohm varies with time t (in second) according
to the equation 22 10 3.t tφ = − + What will be the
magnetic of induced current at t = 0.25 s ?
8. An a.c. generator gives an output voltage of E = 170 sin
56⋅52t. What is the frequency of alternating voltage
produced?
9. A condenser of capacitance 0.144 pH is used in a
transmitter to transmit at wavelength λ. If inductance of
2
1
πmH is used for resonance, what is the value of λ?
10. In the series RLC circuit as shown in Fig., what would be
the ammeter reading?
4 Fµ80 C+ µ
2 Fµ 3 Fµ
1 2
s
2µF 8µF V
R
C
d 3
d
3µF 6µF X
Y
9V
3Ω 6Ω
S
V
R2
R2
C1
C2 R1 R2 C1 C2
V
C1 C2
V R1
R2 + –
+ –
63 Mock Test-5
11. An a.c. source of frequency 1000 Hz is connected to a coil
of 200
πmH and negligible resistance. If effective current
through the coil is 7.5 mA, what is the voltage across the
coil?
12. Two alternating currents are given by ( )1 0 sinI I tω φ= −
and ( )2 0 cosI I tω φ= + . What is the ratio of virtual values
of the two currents?
13. A silver sphere of radius 1 cm and work function 4.7 eV is
suspended from an insulating thread in free-space. It is
under continuous illumination of 20 nm wavelength light.
As photoelectrons are emitted, the sphere gets charged
and acquires a potential. The maximum number of photo-
electrons emitted from the sphere is A 10ZA× (where 1 <
A < 10). The value of Z is
14. When a piece of metal is illuminated by monochromatic
light of wavelength λ then the stopping potential for
photoelectric current is 3V0. When the same surface is
illustrated by light of wavelength 2λ, then stopping
potential becomes the V0 value of Threshold wavelength
for photoelectric current is aλ, where the value of a is
SECTION 3 (Maximum Marks: 12)
Paragraph based questions (2 paragraphs, each having 2
MCQs with one correct answer only)
Paragraph for Question No. 15 to 16
A fixed thermally conducting cylinder
has a radius R and height 0.L The
cylinder is open at its bottom and has a
small hole at its top. A piston of mass
M is held at a distance L from the top
surface, as shown in the figure. The
atmospheric pressure is 0.P
15. The piston is now pulled out slowly and held at a distance
2L from the top. The pressure in the cylinder between its
top and the piston will then be
a. 0P b. 0
2
P
c. 0
22
P Mg
Rπ+ d. 0
22
P Mg
Rπ−
16. While the piston is at a distance 2L from the top, the hole
at the top is sealed. The piston is then released, to a
position where it can stay in equilibrium. In this condition,
the distance of the piston from the top is
a. 2
0
2
0
2(2 )
P RL
R P Mg
ππ
+ b.
2
0
2
0
(2 )P R Mg
LR P
ππ
−
c. 2
0
2
0
(2 )P R Mg
LR P
ππ
+
d. 2
0
2
0
(2 )P R
LR P Mg
ππ
−
Paragraph for Question No. 17 to 18
The nuclear charge (Ze) is non-uniformly distributed within a
nucleus of radius R. The charge density ( )rρ [charge per unit
volume] is dependent only on the radial distance r from the
centre of the nucleus as shown in figure. The electric field is
only along the radial direction.
17. The electric field at r R= is
a. independent of a
b. directly proportional to a
c. directly proportional to 2a
d. inversely proportional to a
18. For 0,a = the value of d (maximum value of ρ as shown
in the figure) is
a. 3
3
4
Ze
Rπ b.
3
3Ze
Rπ
c. 3
4
3
Ze
Rπ d.
33
Ze
Rπ
o a R r
d
( )ρ r
200 V 200 V
100 V, 50 Hz
R = 500 Ω L C
~
A
L
L0
2
Piston
Space for Rough Work
Physics 64
JEE ADVANCE PAPER-II
SECTION 1 (Maximum Marks: 24)
MCQs with one or more than one correct answer
1. A parallel plate capacitor has a dielectric slab of dielectric
constant K between its plates that covers 1/3 of the area of
its plates, as shown in the figure. The total capacitance of
the capacitor is C while that of the portion with dielectric
in between is C1. When the capacitor is charged, the plate
area covered by the dielectric gets charge Q1 and the rest
of the area gets charge Q2. The electric field in the
dielectric is 1E and that in the other portion is E2.
Choose
the correct option/options, ignoring edge effects.
a. 1
2
1=E
E b. 1
2
1=
E
E K
c. 1
2
3=
Q
Q K d.
1
2 +=
C K
C K
2. A parallel plate capacitor having plates of area S and plate
separation d, has capacitance 1C in air. When two
dielectrics of different relative permittivities (ε1 = 2 and
ε2 = 4) are introduced between the two plates as shown in
the figure, the capacitance becomes2 .C The ratio 2
1
C
Cis
a. 6/5 b. 5/3 c. 7/5 d. 7/3
3. Consider a cylindrical element as shown in the figure.
Current flowing the through element is I and resistivity of
material of the cylinder is .ρ Choose the correct option out
the following
a. Power loss in first half is four times the power loss in
second half
b. Voltage drop in first half is twice of voltage drop in
second half
c. Current density in both halves are equal
d. Electric field in both halves is equal
4. Figure shows three resistor configurations R1, R2, and R3
connected to 3V battery. If the power dissipated by the
configuration R1, R2, and R3 is P1, P2, and P3, respectively,
then
a. 1 2 3P P P> > b. 1 3 2P P P> >
c. 2 1 3P P P> > d. 3 2 1P P P> >
5. Incandescent bulbs are designed by keeping in mind that
the resistance of their filament increases with the increase
in temperature. If at room temperature, 100 W, 60 W and
40 W bulbs have filament resistances R100, R60 and R40,
respectively, the relation between these resistances is
a. 100 40 60
1 1 1
R R R= + b.
100 40 60R R R= +
c. 100 60 40R R R> >
d.
100 60 40
1 1 1
R R R> >
6. To verify Ohm’s law, a student is provided with a test
resistor RT, a high resistance R1, a small resistance R2, two
identical galvanometers G1 and G2, and a variable voltage
source V. The correct circuit to carry out the experiment is
a. b.
c.
d.
SECTION 2 (Maximum Marks: 24)
Numerical value answer type questions
7. Image of an object approaching a convex mirror of radius
of curvature 20 m along its optical axis is observed to
move from 25
3 m to
50
7 m in 30 seconds. What is the
speed of the object in km per hour?
G2
R2
RT
R1
V
G1
G2
R1
RT
R2
V
G1
G1
G2 R1
RT R2
V
G1
G2 R2
RT R1
V
R1 R2 R3
1
1
1
1
1
3V
1
1
1
1
1 3V 1Ω
1Ω
1Ω
1Ω
1Ω
3V
4r I
./ 2ℓ ./ 2ℓ
2r
A B C
d/2
d
S/2
S/2
ε1
ε2
2E–
+
–
+ –
+
65 Mock Test-5
8. Water (with refractive index = 4/3) in a tank is 18 cm
deep. Oil of refractive index 7
4lies on water making a
convex surface of radius of curvature ‘R = 6 cm’ as shown.
Consider oil to act as a thin lens. An object ‘S’ is placed
24 cm above water surface. The location of its image is at
‘x’ cm above the bottom of the tank. Then ‘x’ is
9. Consider a concave mirror and a convex lens (refractive
index = 1.5) of focal length 10 cm each, separated by a
distance of 50 cm in air (refractive index = 1) as shown in
the figure. An object is placed at a distance of 15 cm from
the mirror. Its erect image formed by this combination has
magnification 1.M When the set-up is kept in a medium of
refractive index 7/6, the magnification becomes 2.M The
magnitude 2
1
M
Mis
10. A monochromatic beam of light is incident at 60° on one
face of an equilateral prism of refractive index n and
emerges from the opposite face making an angle θ (n)
with the normal (see the figure). For 3=n the value of
θ is 60° and .d
mdn
θ= The value of m is
11. An α-particle and a proton are accelerated from rest by a
potential difference of 100 V. After this, their de Broglie
wavelengths are αλ and pλ respectively. The ratio ,
p
α
λ
λto
the nearest integer, is
12. A silver sphere of radius 1 cm and work function 4.7 eV is
suspended from an insulating thread in free space. It is
under continuous illumination of 200 nm wavelength
light. As photoelectrons are emitted, the sphere gets
charged and acquires a potential. The maximum number
of photoelectrons emitted from the sphere is A 10 z×
(where 1 < A < 10). The value of 'Z' is
13. A proton is fired from very far away towards a nucleus
with charge Q = 120 e, where e is the electronic charge. It
makes a closest approach of 10 fm to the nucleus. The de
Broglie wavelength (in units of fm) of the proton at its
start is: (take the proton mass,
27(5 / 3) 10 kg;pm −× 15/ 4.2 10h e −= × 9
0
1. / ; 9 10
4J s C
πε= ×
15/ ;1 10 )m F fm m
−=
14. The work functions of silver and sodium are 4.6 eV and
2.3 eV, respectively. The ratio of the slope of the stopping
potential versus frequency plot for silver to that of sodium
is
SECTION 3 (Maximum Marks: 12)
Matching type questions with 4 options
15. Some laws / processes are given in Column I. Match these
with the physical phenomena given in Column II.
Column I Column II
(A) Intensity of light
received by lens
1. Radius of aperture (R)
(B) Angular magnification 2. Dispersion of lens
(C) Length of telescope 3. Focal length ,a ef f
(D) Sharpness of image 4. Spherical aberration
a. A-1; B-3; C-3; D-1,2,3
b. A-1,2,3; B-4; C-2,3, D-1
c. A-1; B-3; C-1,2,3; D-2
d. A-2,3; B-1; C-4; D-3
16. Study of properties of matter helps the student to know
some phenomenon having important applications of the
principles learnt by him in his daily life. Can you match
some of the phenomenon you observe with the principles
you have studied.
Column I Column II
(A) A cricketer spins his ball
while bowling to change
the direction and
momentum of ball is
1. Surface tension
θ 60°
15 cm
50 cm
R 6cm=S
1.0µ =
7 / 4µ =
4 / 3µ =
Physics 66
according to principle of
(B) A painter askes you for
primers coating on walls
before panting to
remove sewage of water
through bricks in walls
2. Gravitational pull
(C) Formation of stars in
universe
3. Bernoulli’s pull
(D) Ships are asked to
spread oil on sea surface
during his tides
4. Capillarity
a. A-1; B-2; C-3; D-4 b. A-4; B-3; C-2; D-1
c. A-3; B-2; C-1; D-4 d. A-3; B-4; C-2; D-1
17. Match the statement of Column I with those in Column II:
Column I Column II
(A) Magnetic field intensity is
defined as
1. The spin motion
of electron
(B) Which of the following,
the most suitable material
for making permanent
magnet is
2. Steel
(C) In the case of bar magnet,
lines of magnetic
induction
3. Run continuously
through the bar
and outside
(D) A sensitive magnetic
instrument can be shielded
very effectively from
outside magnetic fields by
placing it inside a box of
4. Soft iron of high
permeability
a. A-4; B-3; C-2; D-1 b. A-2; B-4; C-3; D-1
c. A-1; B-2; C-3; D-4 d. A-4; B-1; C-3; D-2
18. Column I defines the type of modulus or coefficient of
elasticity. Column II gives the type of corresponding
modulus. Match the definition with proper type of
elasticity.
Column I Column II
(A) Ratio of longitudinal
or tensile stress to
longitudinal strain
1. Modulus of Rigidly
(B) Ratio of normal or
hydrostatic stress to
volumetric strain
2. Poisson’s ratio
(C) Ratio of lateral strain
to longitudinal strain
3. Bulk modulus
(D) Ratio of tangential
stress to shear strain
4. Young’s modulus
a. A-1; B-2; C-3; D-4 b. A-4; B-3; C-2; D-1
c. A-3; B-2; C-1; D-4 d. A-2; B-1; C-4; D-3
Space for Rough Work
67 Mock Test-5
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d a b a c b d b d c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b b a b d b a b b b
21. 22. 23. 24. 25.
c a d a c
1. (d) Velocity gradient 1
1[ ][ ]
[ ]
v LTT
x L
−−= = =
Potential gradient2 3 1
3 1[ ][ ]
[ ]
V ML T AMLT A
x L
− −− −= = =
Energy gradient 2 2
2[ ][ ]
[ ]
E ML TMLT
x L
−= = =
Pressure gradient1 2
2 2[ ][ ]
[ ]
P ML TML T
x L
− −− −= = =
2. (a) ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ4 3 2r a b c i j i j k i j k= + + = − − + − = + −
2 2 2
ˆˆ ˆˆ
| | 1 1 ( 1)
r i j kr
r
+ −= =
+ + −
ˆˆ ˆ
3
i j k+ −=
3. (b) Gravitational field is a conservative force field. In a
conservative force field work done is path independent
∴ 1 2 3
W W W= =
4. (a) The desired moment of inertia is
x l
x l
I dI
=+
=−
= ∫
( )2
2 2sin sin2 3
l
l
m mldx x
lα α
+
−
= = ∫
5. (c) 1 1,m M T T= =
⇒ 2 2
5,
3
Tm M m T= + =
⇒ 11 1
2 22
2 /
2 /
m KT m M
T m M mm K
π
π= = =
+
or,
2
2
1
5 /3 25
9
M m T T
M T T
+ = = =
⇒ 16
9
m
M=
6. (b) 4
0.410
g
g
Vf
r
∆= = =
∆ N/kg
7. (d) 3 34 4
3 3R n rπ π= ×
1/ 3( ) 10R n r r= =
Decrease in surface area 2 2(4 ) 4A n r Rπ π∆ = −
2 21000(4 ) 4 (10 )r rπ π= −
2 24 [1000 100] 900 4r rπ π= − = ×
Energy released A T= ∆ × =2900 4 T
calr
J
×=
π
Let θ be the rise in temperature;’ then
2
34 900 4 T1000 1
3
rr d
J
×× × ×
ππ θ =
∴ 2.7T
dJr
=θ
8. (b) For simple pendulum 2l
Tg
π= when weight
suspended on elastic wire is increased, it length increases,
so time period increases.
9. (d) Zero Kelvin –237 C= ° (absolute temperature). As no
matter can attain this temperature, hence temperature can
never be negative on Kelvin scale.
10. (c) For an ideal gas : pV = nRT
For p = constant p V nT T∆ = ∆
∴ V nR nR V
nRTT p T
V
∆= = =
∆
∴ 1V
V T T
∆=
∆ or
1
Tδ =
Therefore, δ is inversely proportional to temperature T.
i.e., when T increases, δ decreases and vice-versa.
Hence, Tδ − graph will be a rectangular hyperbola as
shown in the above figure.
11. (b) 21 1
2 2mv J ms⋅ = ∆
∆θ
⇒ 2
4
v
Js=∆θ
12. (b) Wien’s displacement law for a perfectly black body is
mTλ = constant = Wien’s constant b
Here, mλ is the minimum wavelength corresponding to
maximum intensity I.
• x
dx
α x sin α
T
Physics 68
or 1
mT
λ ∝
From the figure ( ) ( ) ( )1 3 2m m mλ λ λ< <
Therefore, 1 3 2T T T> >
13. (a) In the direction of electric field, potential decreases.
14. (b) Let 1q and 2q be charges on inner and outer spherical
conductors.
Then 1 21
0 1 2
1
4
q qV
R Rπε
= +
. . .(i)
Also 1 22
0 2
1
4
q qV
Rπε+
= . . .(ii)
Then potential at point P, 1 2
0 2
1
4
q qV
x Rπε
= +
. . .(iii)
Solving (i) and (ii), 1q and 2q may be found. If these are
substituted in (iii), we get
( ) ( )
( )1 1 2 2 2 1
2 1
V R R x V R x RV
R R x
− + −=
−
15. (d) At point P net magnetic field 2 2
1 2netB B B= +
where 01 3
2.
4
MB
d
µπ
= and 02 3
.4
MB
d
µπ
=
⇒
0
3
5.
4net
MB
d
µπ
=
16. (b) Emf induces across the length of the wire which cuts
the magnetic field.
(Length of c = Length d) > (Length of a = b).
So, ( )c de e= ( )a be e> =
17. (a) Peak power, 0 0 0P V I=
Average power dissipation, 0 0 cos2
av
V IP φ=
For a resistive circuit 0φ =
⇒ cos 1φ =
⇒ 0 0 0 50%2 2
av
V I PP = = = of
0P
18. (b) By changing distance the intensity change but
frequency remains same; so stopping potential remains
same.
19. (b) Because atom is hollow and whole mass of atom is
concentrated in a small centre called nucleus.
20. (b) 200Fission
Energy=
200 MeV
196106.110200
−×××= J
Fission rate 5
200 MeV=
111056.1 ×= fission/sec.
21. (c) Relative velocity of parrot and train
10 5 15 / = + = m/s
∴ Time 150
1015
s
v= = = 10 s
22. (a) Free body diagram (FBD) of the block (shown by a
dot) is shown in figure. For vertical equilibrium of the
block
sin 60N mg F= + ° 3 32
Fg= +
For no motion, force of friction or cos 60f F≥ °
⇒ cos 60N F≥ °µ or 1 3
32 22 3
F Fg
+ ≥
or 2
Fg ≥
or 2 or 20NF g≤
Therefore, maximum value of F is 20 N.
23. (d) Here change in frequency takes place in two steps
∴ 2v
v vc
∆ =
⇒ 3 6
86 10 2 5000 10v
c× = × ×
⇒
3
10
86 10
10v
×=
7 886 10 3 10c −= × × × m/s
3
2.58 10= × m/s
24. (a) Current in circuit
( )220 9 1 1 5
47 0 5 10i
− − × ⋅=
+ ⋅ ×
220 124 A
52
−= =
∴ Potential difference across battery which is being charged
V E ir= +
( ) ( )8 1 5 4 0 5 10 32= × ⋅ + × ⋅ × = V
Vertical
Horizontal
mg + F sin 60°
N
f F cos 60°
P
B2
B1
d
S
N
1
2
N
S
d
69 Mock Test-5
25. (c) 0
4
iB B i
r= ⇒
µ θ∝ θ
π (but 1 2 2
2 1 1
i l
i l= =
1
2θ
θ)
⇒
1 1
2 2 2
.B i
B i=
1θ
θ
So, 1
2 2 1
B
B= ×
1 2θ θ
θ θ
⇒1 2B B=
JEE Advance Paper-I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d c a d c b 3 9 7 2
11. 12. 13. 14. 15. 16. 17. 18.
3 1 7 4 a d a b
1. (d) 1 8 / 9τ µ= µ S
⇒ 2 18τ µ= µ S ⇒
3 4τ µ= µ S
2. (c) 27 Cµ
Initial charge distribution (when switch S is open)
Final charge distribution (when switch S is closed)
3. (a)
0 0
equivalent0 0
2.2
3 32
2
3 3
d dvt vt
C
d dvt vt
ε ε
ε ε
− +=
+− +
∴ equivalentC Rτ =
4. (d) 21(2)
2iU V= ×
⇒ 21 (2) (8)
2 10lossU V
×∆ = × ×
⇒ 4
80%5
losslost
i
U
Uη
∆= = =
5. (c) Charge on capacitor of 3 Fµ
⇒ 3
80 48 C2 3
Q = = µ+
6. (b) Potential 0= on connecting them together i.e., 0=Q
i.e. 2211 VCVC = [capacitance is positive but they are
connected with opposite polarity]
⇒ 21 200120 CC =
⇒ 21 53 CC =
7. (3) Here, 3R = Ω
⇒⇒⇒⇒ 22 10 3t tφ = − +
? 0.25I t= = g
⇒⇒⇒⇒ 2(2 10 3)d d
e t tdt dt
φ= = − − +
⇒⇒⇒⇒ e = – 4 t + 10
At 0.25 , 4 0.25 10 9 voltt s e= = − × + =
⇒⇒⇒⇒ 9
3 A3
eI
R= = =
8. (9) Here, 170sin 56.52E t= Compare with the standard form of equation of alternating
emf
0 sinE E tω=
⇒ 56.52ω =
⇒ 2 56.52vπ =
⇒ 56.52 56.52
92 2 3.14
vπ
= = =×
Hz.
9. (7) Here, 120.144pF 0.144 10C F−= = × F
⇒ 3
2 2
1 10mH HL
π π
−
= =
⇒ 1
2 LCυ
π=
⇒ .2v
LCλ υ πυ
= =
3
8 12
2
103 10 2 0.144 10π
π
−−= × × × ×
⇒ 8
8 1.2 103 10 2λ π
π
−×= × × ×
3µF 6µF X
Y
9V 3Ω 6Ω
S
+9µC +18µC
+27µC
+9µC +36µC
3µF 6µF X
Y
9V
3Ω 6Ω
S
+18µC +18µC
1A
1A
i2
300o 1
2
60o
i1
O
+ –
Physics70
The correct answer is 7 m.
10. (2) As
∴ The circuit is series resonance circuit Z = R
⇒
11. (3) Here, v = 1000 Hz,
As resistance of coil is negligible,
∴
12. (1)
Phase difference has no impact on virtual value of current.
13. (7) Here, eV,
Let be the stopping potential. According to Einstein’s
photoelectric equation (using hc = 1240 eV nm), we have
1240eV
eV 4.7eV200nm
s
hc− = −
0φ
λ
6.2eV 4.7eV 1.5eV 1.5eV= − = =
∴
The sphere will stop emitting photoelectrons, when the
potential on its surface becomes 1.5 V. Let n be the no. of
photoelectrons emitted from sphere.
The charge on sphere,
∴
As per questions,
14. (4) 03eV
and 0eV
or 02eV or 0eV
hc
If is the stopping wavelength, then 0eV
or
15. (a) Po
16. (d) 2 2
0( )Mg P R P Rπ π+ =
⇒ 2 2
0(2 ) ( )P L R P x Rπ π=
1 1 2 2(PV PV= for isothermal process)
⇒ 2
0
2
0
(2 )P R
x LR P Mg
ππ
= −
17. (a) independent of a
18. (b) 2
0
( )42
Rd
q R x x dx Zeπ= − =∫
⇒ 3
3Zed
Rπ=
JEE Advance Paper-II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a,d d b c d c 3 2 7 2
11. 12. 13. 14. 15. 16. 17. 18.
3 7 7 1 a d c b
1. (a, d) As E = V/d
1 2/ 1=E E (both parts have common potential difference)
Assume 0C be the capacitance without dielectric for
whole capacitor.
⇒ 0 02
3 3+ =
C Ck C
⇒ 1
2++
C k
C k⇒ 1
2
.2
+Q k
Q
2. (d)0
010
442
/ 2
ε ε= =
SS
Cd d
⇒ 0 020 30
2,
ε ε=
S SC C
d d
⇒ '
10 1010
1 1 1
C CC= +
0
11
2 2
d
Sε = +
⇒ 10
' 04
3
ε=
SC
d
⇒ ' 02 30 10
7
3
ε= + =
SC C C
d
⇒ 2
1
7
3=
C
C
7.2 mλ =
200VL CV V= =
1002A
50
vE E
IZ R
υυ = = = =
200 0.2mH = HL
π π=
37.5mA 7.5 10 AIυ−= = ×
?Eυ =
( ) ( )2L
E I X I vLυ υ υ π= =
3 0.27.5 10 2 1000 3Vπ
π− = × × × =
1
2
0
0
/ 21
/ 2
I I
I I
υ
υ
= =
2
01 cm 10 m; 4.7 ,r eVφ−= = = 200nm.λ =
sV
1.5sV =
, 1.5Vsq ne V= =
270
19 9
4 (1.5) (10 ) 11.05 10
1.6 10 (9 10 )
sV rn
e
πε −
−
× × ×= = × = ×
× ×710 1.05 10 or 7ZA Z× = × =
0 03hc
eV φλ
= − 0 02
hceV φ
λ= −
0
12 1
2 2
hc hfeV
λ λ = − = 0
4hc
eVλ=
0λ 0 0/eV hc λ=
0
0
4hc
eVλ λ= =
10C 20C
30C
71 Mock Test-5
3. (b) 1 1
2 2
4
1= =
R A
R A
⇒ 2
1 1
2
2 2
4
1= =
P I R
P I R
⇒ 1 1
2 2
4
1= =
V IR
V IR⇒ 1
2
1
4=
J
J
4. (c)2V
PR
=
⇒ 1 21 ,R R= Ω 31/ 2 , 2R= Ω = Ω
∴ 2 1 3P P P> >
5. (d) Power 1/ R∝
6. (c) 1G is acting as voltmeter and 2G is acting as ammeter.
7. (3) For 1
50,
7v m= 1 25mu = −
⇒ 2 2
25, 50m
3v m u= = −
Speed of object 25 18
330 5
= × = kmph.
8. (2)1
71
7 1 4
4 24 6V
−− =
−
⇒ 1 21 cmV =
⇒ 2
4 / 3 7 / 40
21V− =
⇒ 2 16 cmV =
⇒ 18 16 2 cmx = − =
9. (7) Image by mirror is formed at 30 cm from mirror at its
right and finally by the combination it is formed at 20 cm
on right of the lens. So in air medium, magnification by
lens is unity. In second medium, 7
,6
µ = focal length of
the lens is given by,
1 2
1 2
1 11 (1.5 1)101 1.5 1 1
17 / 6
R R
f R R
− −
= − −
⇒ 35
2f cm= cm
So, in second medium, final image is formed at 140 cm to the
right of the lens. Second medium does not change the
magnification by mirror.
So 2 2
1 1
2
1
7m
m
M MM
M M M= =ℓ
ℓ
10. (2) Snell’s Law on 1st surface: 1
3sin
2= n r
1
3sin
2=r
n . . .(i)
⇒
2
1 2
3 4 3cos 1
4 2
−= − =
nr
n n
1 2 60+ = °r r . . .(ii)
Snell’s Law on 2nd
surface: 2sin sinn r = θ
Using equation (i) and (ii) 1sin(60 ) sinn r° − = θ
1 1
3 1cos sin sin
2 2n r r
− = θ
( )234 3 1 cos
4
d dn
dn dn
θ− − = θ
For 60θ = ° and 3=n
⇒ 2d
dn
θ=
11. (3) 21
2mv qV=
⇒ h
mvλ =
⇒ 8 3.= ≃λ
12. (7) Stopping potential Whc
λ= −
6.2 eV 4.7 eV= − 1.5 eV=
⇒ 1.5Kq
Vr
= =
⇒ 2
7
9 19
1.5 101.05 10
9 10 1.6 10n
−
−
×= = ×
× × ×
⇒ 7Z =
13. (7) 2
2
0
1 1 (120)
2 4 (10 )
emv
fmπε∞ =
[By Conservation of energy]
Assuming the nucleus to be considerably massive, we can
disregard its motion.
∴ Let momentum of proton be p mv∞=
∴ 2 2
15
0
1 (120)
2 4 (10 10 )p
p e
m πε −=
×
∴ 2
9 27
14
5 1209 10 2 10
3 10
ep −
−
×= × × × × ×
Physics 72
⇒ 9 27 14 230 120 10p e− += × × ×
∴ 4 23600 10p e−= × ×
∴ 260 10p e
−= × ×
∴ h
pλ =
560 10
h
e=
× ×
1515
2
42 10 4210 m
660 10
−−
−
×= = ×
×
∴ 7fm=λ
14. (1) Slope of graph is h/e = constant ⇒ 1
15. (a) A→1, B→3, C→3, D→1,2,3
A, C and D in case of concave mirror convex lens image
can be real, virtual, diminished magnified or of same size.
B is case of convex mirror image is always virtual (for
real object)
16. (d) A-3; B-4; C-2; D-1
According to Bernoulli’s theorem the spinning of ball
along with horizontal throw will change the velocity of
streamlines of air above and below the ball which will
change the pressure on ball above and below it which
cause change in its direction of motion. This is called
magnus effect.
Primers contain materials which make obtuse angle of
contact due to their surface tension and do not allow
capillary rise of water in pours of bricks.
Interstellar dust particles attract each other by
gravitational pull to build a star.
Oil has smaller surface tension than water and is lighter
than water so it spreads move over the surface of water to
decrease height of tides.
17. (c) A-1; B-2; C-3; D-4
18. (b) A-4; B-3; C-2; D-1
Longitudinal stress
Longitudinal strain
F lY
A l= × = =
∆Young’s modulus
Normal stress
Volumetric strain
F VK
A V= × = =
∆ Bulk modulus
= Volume elasticity
Poisson’s ratio Lateral strain
Longitudinal strain=
⇒
DD lD
l D l
l
σ
∆∆
= = ×∆ ∆
73Mock Test-1
Mock Test 1 “JEE-Main”
Do not open this Test Booklet until you are asked to do so.
Read carefully the Instructions on the Back Cover of this Test Booklet.
Important Instructions:
1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly
prohibited.
2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer
Sheet and fill in the particulars carefully.
3. The test is of 3 hours duration.
4. The Test Booklet consists of 90 questions. The maximum marks are 360.
5. There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 30
questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.
6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. 1/4 (one
fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will
be made if no response is indicated for an item in the answer sheet.
7. There is only one correct response for each question. Filling up more than one response in each question will be treated as
wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.
8. Use Blue/Black Ball Point Pen only for writing particulars/ marking responses on Side-1 and Side-2 of the Answer Sheet.
Use of pencil is strictly prohibited.
9. No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any
electronic device, etc., except the Admit Card inside the examination hall/room.
10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the
bottom of each page and at the end of the booklet.
11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall.
However, the candidates are allowed to take away this Test Booklet with them.
12. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial
number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate
should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet.
13. Do not fold or make any stray marks on the Answer Sheet.
Name of the Candidate (in Capital letters):
Roll Number : in figures
in words
Examination Centre Number:
Name of Examination Centre (in Capital letters):
Candidate’s Signature: Invigilator’s Signature:
A Test Booklet code
Chemistry74
Read the following instructions carefully:
1. The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with
Blue/Black Ball Point Pen.
2. For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.
3. The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test
Booklet/Answer Sheet.
4. Out of the four options given for each question, only one option is the correct answer.
5. For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted from
the total score. No deduction from the total score, however, will be made if no response is indicated for an item
in the Answer Sheet.
6. Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in
Test Booklet Code and Answer Sheet Code), another set will be provided.
7. The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All
calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself,
marked ‘Space for Rough Work’. This space is given at the bottom of each page and at the end of the booklet.
8. On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the
Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
9. Each candidate must show on demand his/her Admit Card to the Invigilator.
10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.
11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the
Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance
Sheet a second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair
means case. The candidates are also required to put their left hand THUMB impression in the space
provided in the Attendance Sheet.
12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited
13. The candidates are governed by all Rules and Regulations of the JAB/Board with regard to their conduct in the
Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board.
14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager,
mobile phone, electronic device or any other material except the Admit Card inside the examination
hall/room.
75Mock Test-1
JEE-MAIN: CHEMISTRY MOCK TEST-1
1. In a galvanic cell, the salt bridge a. does not participate chemically in the cell reaction. b. stops the diffusion of ions from one electrode to another. c. is necessary for the occurrence of the cell reaction d. ensures mixing of the two electrolytic solutions
2. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately
a. 1.86 Å b. 3.22 Å c. 5.72 Å d. 0.93 Å
3. Which of the following is the energy of a possible excited state of hydrogen?
a. 13.6 eV b. 6.8 eV c. 3.4 eV d. 6.8 eV
4. The number of atoms in 100 g of an fcc crystal with
density –3=10.0 g cm and cell edge equal to 200 pm is
equal to a. 245 10 b. 255 10 c. 236 10 d. 252 10
5. The following reaction is performed at 298 K.
2 22NO(g) O (g) 2NO (g)
The standard free energy of formation of NO (g) is 86.6
kJ/mol at 298 K. What is the standard free energy of
formation of 2NO (g) at 298 K? 12p(K 1.6 10 )
a. 12R(298) n (1.6 10 ) 86600 b. 1286600 R(298) n (1.6 10 )
c. 12n (1.6 10 )
86600R(298)
d. 120.5[2 86600 R(298) n (1.6 10 )]
6. The vapour pressure of acetone at 20 C is 185 torr. When
1.2 g of a non-volatile substance was dissolved in 100 g of
acetone at 20 C, its vapour pressure was 183 torr. The
molar mass (g mol–1) of the substance is:
a. 32 b. 64 c. 128 d. 488
7. The standard Gibbs energy change at 300 K for the
reaction 2A B C is 2494.2 J. At a given time, the
composition of the reaction mixture is 12[A] , [B] 2 and
12[C] . The reaction proceeds in the:
[ R 8.314 J / K / mol, e 2718 ]
a. forward direction because cQ K
b. reverse direction because cQ K
c. forward direction because cQ K d. reverse direction because cQ K
8. Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is:
a. 0 g b. 63.5 g c. 2 g d. 127 g
9. What is the half life period of a radioactive substance if 87.5% of any given amount of the substance disintegrates in 40 minutes?
a. 160 min b. 10 min c. 20 min d. 13 min 20 sec
10. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :
a. 18 mg b. 36 mg c. 42 mg d. 54 mg
11. The ionic radii (in Å) of 3 2N , O and F are respectively:
a. 1.36, 1.40 and 1.71 b. 1.36, 1.71 and 1.40 c. 1.71, 1.40 and 1.36 d. 1.71, 1.36 and 1.40
12. In the context of the Hall-Heroult process for the extraction of A , which of the following statements is
false?
a. CO and 2CO are produced in this process
b. 2 3A O is mixed with 2CaF which lowers the melting
point of the mixture and brings conductivity
c. 3A is reduced at the cathode to form A
d. 3 6Na A F serves as the electrolyte
13. From the following statements regarding 2 2H O , choose
the incorrect statement a. It can act only as an oxidising agent b. It decomposes on exposure to light c. It has to be stored in plastic or wax lined glass bottles in
dark d. It has to be kept away from dust 14. Which one of the following alkaline earth metal sulphates
has its hydration enthalpy greater than its lattice enthalpy?
a. 4CaSO b. 4BeSO
c. 4BaSO d. 4SrSO
Chemistry76
15. Which among the following is the most reactive?
a. 2Cl b. 2Br c. 2I d. ICl
16. Which one has the highest boiling point? a. He b. Ne c. Kr d. Xe
17. The number of geometric isomers that can exist for square
planar [Pt(Cl)(py) 3 2(NH )(NH OH)] is (py = pyridine):
a. 2 b. 3 c. 4 d. 6
18. The color of 4KMnO is due to
a. M L charge transfer transition
b. d d transition
c. L M charge transfer transition
d. * transition
19. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is: (Atomic mass
Ag 108, Br 80)
a. 24 b. 36 c. 48 d. 60
20. In the reaction shown below, the major product(s) formed is/are
a.
b.
c.
d.
21. Which compound would give 5-keto-2-methyl hexanal
upon ozonolysis?
a.
b.
c.
d.
22. The synthesis of alkyl fluorides is best accomplished by :
a. Free radical fluorination
b. Sandmeyer’s reaction
c. Finkelstein reaction
d. Swarts reaction
23. In the following sequence of reactions:
Toluene 2KMnO 2 2
4
SOCl H / Pd
BaSOA B C, the
product C is:
a. 6 5C H COOH b. 6 5 3C H CH
c. 6 5 2C H CH OH d. 6 5C H CHO
24. In the reaction
2NaNO / HCl
0 5 C CuCN/ KCN2D E N The product E
is
a.
b.
c.
d.
25. Which polymer is used in the manufacture of paints and
lacquers?
a. Bakelite b. Glyptal
c. Polypropene d. Poly vinyl chloride
CHH3
CN
CH3
CH3
COOH
CH3
NH3
CH3
3CH
3H C
3CH
3CH
3CH
3CH
3CH
3CH
3 3N H CH COO
3CH
O O
H|N 2+ H O
H|N
3CH
O
3CH
O O
H|N 2+ H O
3CH
O O
H|N
2NH
H|N
O
2NH
O
3CH
3+ CH COOH
77Mock Test-1
26. Which of the vitamins given below is water soluble? a. Vitamin C b. Vitamin D c. Vitamin E d. Vitamin K 27. Which of the following compounds is not colored yellow? a. 2 6Zn [Fe(CN) ] b. 3 2 6K [Co(NO ) ]
c. 4 3 3 10 4(NH ) [As(Mo O ) ] d. 4BaCrO 28. Match the statement of Column with those in Column II:
Column I Column II
(A)
1. Pseudo first order
(B)
2. Zero order
(C) 3. Second order
(D) 4. First order
a. A1; B3, 4; C2; D3 b. A2; B4; C3; D1 c. A1; B3, 2; C2; D4 d. A4; B1; C3; D2
Assertion and Reason
Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below: a. If both assertion and reason are true and the reason is the
correct explanation of the assertion. b. If both assertion and reason are true but reason is not the
correct explanation of the assertion. c. If assertion is true but reason is false.
d. If the assertion and reason both are false.
e. If assertion is false but reason is true
29. Assertion: Enthalpy and entropy of any elementary
substance in the standard state are taken as zero.
Reason: At zero degree absolute, the constituent particles
become completely motionless.
30. Assertion: Molecularity has no meaning for a complex
reaction.
Reason: The overall molecularity of a complex reaction is
equal to the molecularity of the slowest step.
H12 22 11 2C H O H O
6 12 6C H O 6 12 6C H OHOH
3 2 5 H or OHCH COOC H
3 2 5CH COOH C H OHhv
2 2H Cl 2HCl
3 3CH Cl OH CH OH Cl
Space for rough work
Chemistry78
JEE ADVANCE PAPER-I
Time 3 Hours. Max. Marks 264 (88 for Chemistry)
Read The Instructions Carefully
Question Paper Format and Marking Scheme:
1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections.
2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
Marking Scheme: +4 for correct answer and 0 in all other cases.
3. Section 2 contains 10 multiple choice questions with one or more than one correct option.
Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases.
4. Section 3 contains 2 “match the following” type questions and you will have to match entries in Column I with the entries
in Column II.
Marking Scheme: for each entry in Column I, +2 for correct answer, 0 if not attempted and – 1 in all other cases.
NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and
marking scheme too.
SECTION 1 (Maximum Marks: 32)
This section contains EIGHT questions.
The answer to each question is a SINGLE DIGIT INTEGER
ranging from 0 to 9, both inclusive.
For each question, darken the bubble corresponding to the correct
integer in the ORS.
Marking scheme:
+4 If the bubble corresponding to the answer is darkened.
0 In all other cases.
1. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as
a strong electrolyte) is 0.0558 C, the number of
chloride(s) in the coordination sphere of the complex is
[ fK of water = 1.86 K kg mol–1]
2. The total number of stereoisomers that can exist for M is
3. The number of resonance structures for N is
4. The total number of lone pairs of electrons in 2 3N O is
5. For the octahedral complexes of 3Fe in SCN
(thiocyanato-S) and in CN ligand environments, the
difference between the spin-only magnetic moments in
Bohr magnetons (When approximated to the nearest
integer) is [Atomic number of Fe = 26]
6. Among the triatomic molecules/ions, 2 3 2BeCl , N , N O,
2 3 2 2 3NO ,O ,SCl , ICl , I and 2XeF , the total number of
linear molecule(s)/ion(s) where the hybridization of the
central atom does not have contribution from the d-
orbital(s) is
[Atomic number: S = 16, Cl = 17, I = 53 and Xe = 54]
7. Not considering the electronic spin, the degeneracy of the
second excited state (n = 3) of H atom is 9, while the
degeneracy of the second excited state of H is
8. All the energy released from the reaction X Y. 0rG
1193 kJ mol is used for oxidizing M as 3M M 02e ,E 0.25V. Under standard conditions, the
number of moles of M oxidized when one mole of X is
converted to Y is –1[F 96500 C mol ]
OH NaOH N
3CH
OM
3H C
3H C
79Mock Test-1
SECTION 2 (Maximum Marks: 40)
This section contains TEN questions.
Each question has FOUR options (a), (b), (c) and (d). ONE OR
MORE THAN ONE of these four option(s) is (are) correct.
For each question, darken the bubble(s) corresponding to all the
correct option(s) in the ORS.
Marking scheme:
+4 If only the bubble(s) corresponding to all the correct
option(s) is(are) darkened.
0 If none of the bubbles is darkened
–2 In all other cases
9. For the reaction: 3 2 4I ClO H SO 4 2HSO I
The correct statement(s) in the balanced equation is/are
a. Stoichiometric coefficient of 4HSO is 6
b. Iodide is oxidised c. Sulphur is reduced
d. 2H O is one of the products.
10. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are)
a. b.
c. d.
11. The major product of the following reaction is
a.
b.
c.
d.
12. In the following reaction, the major product is
a.
b.
c.
d.
13. The correct statement(s) for orthoboric acid is/are a. It behaves as a weak acid in water due to self ionisation b. Acidity of its aqueous solution increases upon addition
of ethylene glycol c. It has a three dimensional structure due to hydrogen
bonding. d. It is a weak electrolyte in water
14. The major product of the reaction is
a.
b.
c.
d.
15. The reactivity of compound Z with different halogens under appropriate conditions is given below:
The observed pattern of electrophilic substitution can be
explained by a. the steric effect of the halogen b. the steric effect of the tert-butyl group c. the electronic effect of the phenolic group d. the electronic effect of the tert-butyl group
16. An ideal gas in thermally insulated vessel at internal
pressure 1P , volume 1V and absolute temperature
1T expands irreversibly against zero external presssure,
as shown in the diagram. The final internal pressure,
volume and absolute temperature of the gas are 2 2P , V
and 2T , respectively. For this expansion.
extP 0
1 2 2P ,V ,T
extP 0
1 2 2P ,V ,T
Thermal insulation
Z 3 3C(CH )
OH2X
2 2mono halo substituted derivative when X I
2 2di halo substituted derivative when X Br
2 2tri halo substituted derivative when X Cl
3H C
3CH
2NH
OH
3H C
3CH
2CO H
OH
3H C
3CH
2CO H
OH
3H C 2NH
3CH OH
3H C
3CH
2CO H
2NH
2NaNO , aqueous HCl0 C
Br3H C
3CH
Br3H C
3CH
Br3H C
3CH3CH
Br3H C
3CH
2CH
3H C
3CH1 equivalent HBr
O3CH
O
3CH
O3CH
O 3CH
O
O3CH 2i. KOH , H O
ii. H , heat
HBr 3CH
3H C
Br H
3CH3H C
3CH
BrH 3CH2H C
Br H
3CH3H C
Chemistry80
a. q 0 b. 2 1T T c. 2 2 1 1P V P V d. 2 2 1 1P V P V
17. 3Fe is reduced to 2Fe by using
a. 2 2H O in presence of NaOH
b. 2 2Na O in water
c. 2 2H O in presence of 2 4H SO d. 2 2Na O in presence of 2 4H SO
18. The % yield of ammonia as a function of time in the
reaction 2 2 3N (g) 3H (g) 2NH (g), H 0 at 1(P,T )
is given below:
If this reaction is conducted at 2(P,T ), with 2 1T T , the
% yield of ammonia as a function of time is represented by
a. b.
c. d.
SECTION 3 (Maximum Marks: 16)
This section contains TWO questions.
Each question contains two columns, Column I and Column II
Column I has four entries (A), (B), (C) and (D)
Column II has five entries (1), (2), (3), (4) and (5)
Match the entries in Column I with the entries in Column II
One or more entries in Column I may match with one or more
entries in Column II
The ORS contains a 4 5 matrix whose layout will be similar to
the one shown below:
(A) (1) (2) (3) (4) (5)
(B) (1) (2) (3) (4) (5)
(C) (1) (2) (3) (4) (5)
(D) (1) (2) (3) (4) (5)
For each entry in Column I, darken the bubbles of all the
matching entries. For example, if entry (A) in Column I, matches
with entries (2), (3) and (4), then darken these three bubbles in the
ORS. Similarly, for entries (B), (C) and (D).
Marking scheme:
For each entry in Column I
+2 If only the bubble(s) corresponding to all the correct match(es)
is(are) darkened
0 If none of the bubbles is darkened
–1 In all other cases
19. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from Column -I with an appropriate structure from Column –II is:
Column I Column II
(A) Pathway A 1.
(B) Pathway B 2.
(C) Pathway C 3.
(D) Pathway D 4.
a. →1; B→3; C→4; D→2
b. →2; B→4; C→4; D→1
c. →1; B→1; C→2; D→4
d. →3.; B→2; C→1; D→3
20. Match the orbital overlap figures shown in Column -I with the description given in Column -II and select the correct
OO
RR
O
(Peroxyester)
1
PCO
R R O
1
QCO
R R O
1 1
R2CO CO
RCO R O
S2RCO R O
R X carbonyl compound
R X carbonyl compound
1COR R O
% Yield
time
1T
2T
% Yield
time
1T
2T
% Yield
time
1T
2T
% Yield1T
time
1T
time
3CH
O
6 5 2C H CH
OO
3CH
O
6 5C H
OO
3CH
O
6 5 2C H CH
OO
3CH
2 6 5CH C H
3CH
O
6 5C HO
O3CH
6 5C H
81Mock Test-1
answer using the code given below the lists.
2 2 2 2(en H NCH CH NH ; atomic numbers :
Ti 22, Cr 24, Co 27; Pt 78)
Column I Column II
(A)
1. p-d antibonding
(B)
2. d-d bonding
(C)
3. p-d bonding
(D)
4. d-d antibonding
a. →2; B→3; C→1; D→4
b. →2; B→4; C→4; D→1
c. →1; B→1; C→2; D→4
d. →3.; B→2; C→1; D→3
Space for rough work
Chemistry82
JEE ADVANCE PAPER-II
Time 3 Hours. Max. Marks 240 (80 for Chemistry)
Read The Instructions Carefully
Question Paper Format and Marking Scheme:
1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections.
2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
Marking Scheme: +4 for correct answer and 0 in all other cases.
3. Section 2 contains 8 multiple choice questions with one or more than one correct option.
Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases.
4. Section 3 contains 2 “paragraph” type questions. Each paragraph describes an experiment, a situation or a problem. Two
multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.
Marking Scheme: +4 for correct answer, 0 if not attempted and – 2 in all other cases.
NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and
marking scheme too.
SECTION 1 (Maximum Marks: 32)
This section contains EIGHT questions.
The answer to each question is a SINGLE DIGIT INTEGER
ranging from 0 to 9, both inclusive.
For each question, darken the bubble corresponding to the correct
integer in the ORS.
Marking scheme:
+4 If the bubble corresponding to the answer is darkened.
0 In all other cases.
1. In dilute aqueous 2 4H SO , the complex diaquodioxalato-
ferrate (II) is oxidized by 4
MnO .− For this reaction, the
ratio of the rate of change of [H ]+ to the rate of change of
4[MnO ]− is
2. The number of hydroxyl group(s) in Q is
3. Among the following, the number of reaction(s) that
produce(s) benzaldehyde is
I
II
III
IV
4. The crystal of a solid is square packing of identical
spheres in each layer and spheres of one layer are placed
just above the voids made by spheres in previous layer.
The packing efficiency of such crystal (in %) is
[ 3.15, 2 1.4]π = =
5. Among the complex ions, 2 2 2[Co(NH - CH - CH -NH2)2
2Cl ] ,+ 3–
2 2 4 2[CrCl (C O ) ] , 2 4 2[Fe(H O) (OH) ] ,+ [Fe(NH3)2]
4(CN) ] ,−
2 2 2 2 2[Co(NH CH CH NH )− − −2
3(NH )Cl]+ and
2
3 4 2[Co(NH ) (H O)Cl] ,+ the number of complex ion(s) that
show(s) cis-trans isomerism is
2CO Me
2
DIBAL-H
Toluene, 78 C H O− °→
COCl2
4
H
Pd BaSO−→
2CHCl2H O
100 C°→
3
CO, HCl
Anhydrous AlCl / CuCl→
4aqueousdiluteKMnO (excess)H
heat 0 CP Q
+
°→ →
3H C
H
HO
3CH
83Mock Test-1
6. Three moles of 2 6B H are completely reacted with
methanol. The number of moles of boron containing
product formed is
7. The molar conductivity of a solution of a weak acid HX
(0.01 M) is 10 times smaller than the molar conductivity
of a solution of a weak acid HY (0.10 M). If 0 0
X Y,
the difference in their apK values, apK (HX) apK (HY),
is (consider degree of ionisation of both acids to be << 1)
8. A closed vessel with rigid walls contains 1 mol of 23892 U
and 1 mol of air at 298 K. Considering complete decay of 238
92 U to 20682 Pb, the ratio of the final pressure to the initial
pressure of the system at 298 K is
SECTION 2 (Maximum Marks: 32)
This section contains EIGHT questions.
Each question has FOUR options (a), (b), (c) and (d). ONE OR
MORE THAN ONE of these four option(s) is (are) correct.
For each question, darken the bubble(s) corresponding to all the
correct option(s) in the ORS.
Marking scheme:
+4 If only the bubble(s) corresponding to all the correct option(s)
is(are) darkened.
0 If none of the bubbles is darkened
–2 In all other cases
9. Which of the following compounds will exhibit
geometrical isomerism?
a. 1-Phenyl-2-butene
b. 3-Phenyl-1-butene
c. 2-Phenyl-1-butene
d. 1,1-Diphenyl-2-propane
10. In the following reactions, the product S is
a.
b.
c. d.
11. The major product U in the following reactions is
a.
b.
c.
d.
12. In the following reactions, the major product W is
a.
b.
c.
d.
13. The correct statement(s) regarding, (i) HClO, (ii) 2HClO ,
(iii) 3HClO and (iv) 4HClO , is (are)
a. The number of Cl = O bonds in (ii) and (iii) together is two
b. The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
c. The hybridisation of Cl in (iv) is 3sp
d. Amongst (i) to (iv), the strongest acid is (i)
N N
OH
N NOH
N N
OH
N N OH
2NHOH
NaOH 2NaNO , HCl
0 CV W
OH
HCH2
CH2
O
H|
O
O
H3C CH3 O
HCH3
O
H|
O
2 3 2CH CH CH , H radical initiator, O
high pressure, heat T U
3H C
NN
3H C
3H C NN 3H C
3H C3 3
2
i. O NHii. Zn,H O
R S
Chemistry84
14. The pair(s) of ions where BOTH the ions are precipitated
upon passing 2H S gas in presence of dilute HCl, is(are)
a. 2 2Ba , Zn b. 3 3Bi , Fe
c. 2 2Cu , Pb
d. 2 3Hg , Bi
15. Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are
a. 3 3CH SiCl and 3 4Si(CH )
b. 3 2 2(CH ) SiCl and 3 3(CH ) SiCl
c. 3 2 2(CH ) SiCl and 3 3 3(CH ) SiCl d. 4SiCl and 3 3(CH ) SiCl
16. When 2O is adsorbed on a metallic surface, electron
transfer occurs from the metal to 2O . The TRUE
statement(s) regarding this adsorption is(are)
a. 2O is physisorbed
b. heat is released
c. occupancy of *2p of 2O is increased
d. bond length of 2O is increased
SECTION 3 (Maximum Marks: 16)
This section contains TWO questions. Based on each paragraph, there will be TWO questions Each question has FOUR options (A), (B), (C) and (D). ONE OR
MORE THAN ONE of these four option(s) is(are) correct. For each question, darken the bubble(s) corresponding to all the
correct option(s) in the ORS. Marking scheme: +4 If only the bubble(s) corresponding to all the correct option(s)
is(are) darkened. 0 If none of the bubbles is darkened –2 In all other cases
Paragraph for Question Nos. 17 to 18
The hydrogen-like species 2+Li is in a spherically symmetric
state 1S with one radial node. Upon absorbing light the ion
undergoes transition to a state 2S . The state 2S has one radial
node and its energy is equal to the ground state energy of the hydrogen atom.
17. Energy of the state 1S in units of the hydrogen atom
ground state energy is : a. 0.75 b. 1.50 c. 2.25 d. 4.50 18. The orbital angular momentum quantum number of the
state 2S is :
a. 0 b. 1 c. 2 d. 3
Paragraph for Question No. 19 to 20 In the following reactions
19. Compound X is
a.
b.
c. d.
20. The major compound Y is
a.
b.
c. d.
CH3
CH3
CH3
CH2
CH3CH3
CHOOH
CH3
OH
CH3
O
2i. EtMgBr, H O8 8 ii. H , heat
C H O Y
2
4 2 4
H O
HgSO , H SO
4 2 2
2 2 2 2
Pd BaSO i. B H8 6 8 8H ii. H O . NaOH, H O
C H C H X
Space for rough work
85Mock Test-1
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a a c a d b b b d a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c d a b d d b c a a
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
b d d c b a a a c b
1. (a)
2. (a) For BCC unit 6 cell, 3a 4r=
3 3
r a 4.294 4
= = × Å 1.85= Å
3. (c) Energy in 1st excited state 3.4 eV= −
4. (a) eff3 30
A
Z Mw
a 10 Nρ
−
×=
× × [For fcc, effZ 4= unit cell]
∴ 3 30
A
eff
a 10 NM
Z
ρ −× × ×
=w
–3 3 30 3 23(10.0 g cm (200pm) 10 cm 6 10 atoms)
4
−× × × ×
=
Thus, 112gmol− contains AN= atoms 236 10= × atoms
∴ 100 g contains 23
246 10100 5 10 atoms
12
×= × = ×
5. (d) n2NO NORx
G 2 G 2 G∆ = ∆ − ∆
2
1NO NO Rx2G G G∆ = ∆ + ∆
1
NO e p2G ( RT ln K )= ∆ + −
120.5[2 86600 R(298) n (1.6 10 )]= × − ×l
6. (b)
01
solute01 2
nP PX
P n n
−= =
+
185 183 2 1.2 / M
1.2 100185 185M 58
−= =
+
⇒ M 64=
7. (b) e cG RT ln K∆ = −
e c2494.2 8.314 300 ln K= × ⇒ 1cK e−=
1c
1K e 0.36
2.718−
= = =
12
2 2
2(B)(C)Q 4
[A] [1/ 2]
×= = =
cQ K ,> i.e. backward reaction.
8. (b) 2Cu 2e Cu (s)+ −+ →
2 mol 1 mol 63.5=
9. (d) min
10. (a) Amount of acetic acid adsorbed
33(0.06 0.042) 50 10 60
16 10 18 mg3
−−− × × ×
= = × =
11. (c) Ionic Radii order: 3 2N O F− − −> >
12. (d)
13. (a) It can act as an oxidising as well as reducing agent.
14. (b) 4BaSO is least soluble, 4BeSO is most soluble.
15. (d) The interhalogen compounds are generally more reactive
than halogens (except F2).
16. (d) Xe has the highest boiling point.
17. (b) No. of Geometrical isomers of
3 2[Pt(Cl)(py)(NH )(NH OH)] 3+ =
18. (c)
19. (a) % of Bromine 80 141 mg
100 24%188 250 mg
= × × =
20. (a) Only amines undergo acetylation and not acid amides.
21. (b)
t 40=
2.303 ak log
t a x=
−
2.303 ak log
40 a 0.875 a=
−
2.303 alog
40 0.125 a=
2.303log8 0.051 min
40= =
1/ 2
0.693 0.693t 13.58 min
k 0.051= = =
3CH
3CH
3CH
3CH
CHOO
3
2
O
Zn / H O→
5-keto-2-methyl hexanal
C
NH
2NH
C3CH
3CH COOH+
O
O
C
2NH
2NH
C3H C O
COH
OO
Chemistry86
22. (d) R I AgF R F AgI (Swarts Reaction)
23. (d)
24. (c)
25. (b) Glyptal is used in the manufacture of paints and lacquers.
26. (a) Vitamin ‘B’ and ‘C’ are water soluble.
27. (a) 2 6Zn [Fe(CN) ] is bluish white ppt.
28. (a) A1; B3, 4; C2; D3
(A) Inversion of cane sugar is pseudo first order reaction. (B) Hydrolysis of ester in the presence of acid is first order
while in the presence of base is second order reaction. (C) Photochemical reactions are of zero order. (D) SN2 reactions are of second order.
29. (c) Enthalpy is zero but entropy is not zero. Vibrational motion exists even at absolute zero.
30. (b) Molecularity of a reaction can be defined only for an elementary reaction because complex reaction does not take place in one single step and it is almost impossible for all the total molecules of the reactants to be in a state of encounter simultaneously.
JEE Advance Paper -I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
1 2 9 8 4 4 3 4 a,b,d b,d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
a d b,d c a,b,c a,b,c a,b b a a
1. (1) f fT iK m
0.0558 i 1.86 0.01 i 3
3 5 2Complex is [Co(NH ) Cl]Cl .
2. (2) Bridging does not allow the other 2 variants to exist.
Total no. of stereoisomers of M = 2
3. (9)
N is
4. (8)
5. (4) 3 36 6[Fe(SCN) ] and [Fe(CN) ]
In both the cases the electronic configuration of 3Fe will
be 2 2 6 2 6 51s , 2s , 2p ,3s ,3p ,3d .
Since SCN is a weak field ligand and CN is a strong field
ligand, the pairing will occur only in case of 36[Fe(CN) ] .
Case – 1 n(n 2) 5(5 2) 35 5.91 BM Case – 2 n(n 2) 1(1 2) 3 1.73 BM
Difference in spin only magnetic moment 5.91 1.73 4.18 4
6. (4) 2 3 2 2 3 2 2 3 2BeCl , N , N O, NO ,O ,SCl , ICl , I , XeF
5Case-1 3d (high spin) (no pairing)
Case-2(low spin)(pairing)
+:O N–– N
:O
: :O O N O :O N
Total no. of lone pairs = 8
–O O O
1 2 3
O O
4 5 6
–O
OO
7 8 9
–O
–O
HONaOH N
3CH
O
3H C
3H C
N Cl
CH3
NH3
CH3
CuCN / KCN2N 2NaNO / HCl
0 5 C
CN
CH3
3CH
Toluene
COOH
Benzoic acid
COCl
Benzoyl chloride
CHO
Benzaldehyde
2
4
H / PdBaSO
2SOCl2KMnO
87Mock Test-1
2BeC sp linear
3N sp linear
2N O sp linear
2NO sp linear
23O sp linear
3
2SCl sp linear
33I sp d linear
3
2ICl sp d linear
32XeF sp d linear
So, among the following only four (4) has linear shape and no
d-orbital is involved in hybridization.
7. (3) Single electron species don’t follow the (n ) rule but
multi electron species do.
Ground state of 2H 1s
First excited state of 1 1H 1s ,2s Second excited state of 1 0 1H 1s , 2s , 2p
8. (4) X Y 0rG 193KJ / mol
3M M 2e 0E 0.25V
0G for the this reaction is
0 0G nFE 2 ( 0.25) 96500 48250 J / mol
48.25 KJ/mol
So the number of moles of M oxidized using
X Y will be
193
4 moles48.25
9. (a, b, d) 3 2 46I ClO 6H SO
4 2 2Cl 6HSO 3I 3H O
10. (b, d)
(1)
(2)
(3)
(4)
11. (a)
12. (d)
13. (b, d) (a) 3 3H BO is a weak monobasic Lewis acid.
3 3 4H BO H OH B(OH) H . . .(i)
(b) Equilibrium (i) is shifted in forward direction by the
addition of syn-diols like ethylene glycol which forms a
stable complex with 4B(OH) .
3
3
H C
H C2C CH — CH — Br:Br
2C CH — HC
3
3
H C
H C
3
3 2
|— C — CH CH
CH
H C
2
3
H Br2
CH|
H C C — CH CH
O+
2H ,H O/Δ
3CH
O3CH
OH
O:OH
3CH2CH
O O
3CHCH
O
2H O
2H /Pt2 2 3H C CH —C —CH —CH
BrH3 2 2 3H C —CH —C—CH —CH
Optically inactiveBrH
2H /Pt2 3H C C —C — CH
BrH2H C
Optically inactiveBrH
H
2H C
3 3H C — C — C — CH
2H /Pt3 2 3H C CH —C — CH — CH
BrH
Optically inactiveBrH
3 2 3H C — CH — C — CH — CH
3H C — CH CH — C3CH
BrH3CH
BrHOptically active
2H /Pt3 2 2H C — CH — CH — C
(3 degenerate orbitals)
x y zP P P
Chemistry88
(c) It has a planar sheet like structure due to hydrogen
bonding. (d) H3BO3 is a weak electrolyte in water.
14. (c)
N2NH N
23 3NaNO / HCl
2 2
3 3
H HCH—CH —CH CH—CH —HC
H H
OH :OH| |C O C OC C
C C
2
3
2 H O
3
H CCH—CH —CH—COOH
H C 3
2
3
H CCH—CH —HC
H COH
OH
C O==
15. (a, b, c)
16. (a, b, c) Since container is thermally insulated so, q = 0, and it is a case of free expansion therefore W = 0 and
E 0. So, 1 2T T . Also, 1 1 2 2P V P V .
17. (a, b) 32 22Fe H O 2OH 2
2 22Fe 2H O O
2 2 2Na O H O 2 2H O NaOH
18. (b) 2 2N (g) 2H (g)
2 2 3N (g) 2H (g) 2NH (g); H 0
Increasing the temperature lowers equilibrium yield of ammonia. However, at higher temperature the initial rate of
forward reaction would be greater than at lower temperature that is why the percentage yield of NH3 too would be more initially.
19. (a) →1; B→3; C→4; D→2
(A)
(B)
(C)
3 3Ph CH — CO — Ph CH
(D)
26 5 3 6 5 2C H CO CH O C H CO
20. (a)→2; B→3; C→1; D→4
(A)
(B)
(C)
(D)
d-d antibonding
p-d antibonding
p-d bonding
d-d bonding
O
6 5C H
OO 3CH
26 5 3 3
3
O|
C H — CO CH C — CH|
CH
3CHO
6 5C H
OO
3CH
6 5C H
2 3 3Ph — C H CH — CO — CH
6 5 2 2 2 3
3
O|
C H — C H CO Ph — CH — C — CH|CH
3CHO
5 6 2H C CH
OO
3CH
2 6 5CH C H
6 5 2 2 2C H CH CO CH O
3CH
O
5 6 2H C CH
OO
OH
OH
OHBr
Br
ClCl
3CMe
I
3CMe
3CMe
Cl
3CMe
OH
2I
Rxn(i)
2Br
Rxn(ii)
2Cl
Rxn(iii)
O O
OO
24H O
(stable complex)
B
–
OO O
O
OO
OO H
H
HH
HH
HH
B
89Mock Test-1
JEE Advance Paper-II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
8 4 4 75 5 6 3 9 a a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b a b,c c,d b b,c,d a b c d
1. (8) 2 22 4 2 4[Fe(C O )(H O)] MnO 8H
2 32 2Mn Fe 4CO 6H O
So, the ratio of rate of change of
[H ] to that of rate of change of 4[MnO ] is 8.
2. (4)
3. (4)
I
II
III
IV
4. (75)
Z 1 [1
th8
is shared in first layer and
1
2is shared in second layer
1
Z 4 0.5 18
2 2 2(2 ) r h x 2h rfi =
2 2 2Area 4 [ (2 ) 2 ]r a r a r= = fi =
Vol 24 2 r r 34 2 r
3
3
41
3D 0.75 75%4 2 3 2
r
r
5. (5) 2 2[Co(en) Cl ] will show cis-trans isomerism
32 2 4 2[CrCl (C O ) ] will show cis-trans isomerism
2 4 2[Fe(H O) (OH) ] will show cis-trans isomerism
4 3 2[Fe(CN) (NH ) ] will show cis-trans isomerism
22 3[Co(en) (NH )Cl] will show cis-trans isomerism
23 4 2[Co(NH ) (H O)Cl] will not show cis-trans
isomerism (although it will show geometrical isomerism)
6. (6) 2 6 3 2B H 6MeOH 2B(OMe) 6H
1 mole of 2 6B H reacts with 6 moles of MeOH to give 2
moles of 3B(OMe)
3 moles of 2 6B H will react with 18 moles of MeOH to
give 6 moles of 3B(OMe) .
7. (3) HX H X
[H ][X ]Ka
[HX]
HY H Y
[H ][Y ]Ka
[HY]
m for 1mHX m for
2mHY
1 2m m
1
10 2Ka C
1
2
m1 1 0
m1
Ka C
2
2
m2 2 0
m2
Ka C
2
1 1 1
2 2 2
Ka C m
Ka C m
2
0.01 10.001
0.1 10
1 2pKa pKa 3
8. (9) In conversion of 23892 U to 206
82 Pb, 8 - particles and 6β particles are ejected.
2r
h
x
2CO Me CHO
2
BIBAL HToluene, 78 C H O
COCl CHO2
4
HPd BaSO
2CHCl CHO2H O
100 C
CHO
3
CO, HClAnhydrous AlCl / CuCl
H
+
H
HO
HOHO
OH
OH
(Q)
(P)4aqueous dilute KMnO
(excess) 0 C
Chemistry90
The number of gaseous moles initially 1mol
The number of gaseous moles finally 1 8 mol; (1 mol
from air and 8 mol of 42 He ) So the ratio 9 /1 9
9. (a)
10. (a)
11. (b)
12. (a)
13. (b, c) H — O — Cl
(I) . .
. .H — O — Cl O
(II) . .
H — O — Cl O||O
(III)
O||
H — O — Cl O||O
(IV)
14. (c, d) 2 2 2 3Cu , Pb , Hg , Bi give ppt. with 2H S in presence
of dilute HCl.
15. (b)
16. (b, c, d) * Adsorption of 2O on metal surface is
exothermic.
* During electron transfer from metal to 2O electron
occupies *2p orbital of 2O .
* Due to electron transfer to 2O the bond order of 2O
decreases hence bond length increases.
For Question Nos. 17 to 18
17. (c) 1
21H
S 1H2
E 3 9E 2.25 E .
42
3 2Me SiCl, H O
3 3
n3 3
CH CH| |
Si — O — Si — O — Si ——O — Si| |
CH CH
MeMeMe
MeMe
Me
3 3
n3 3
CH CH| |
H — O — Si — O — Si ——O — H| |
CH CH
2
3
H O
3
CH|
Cl — Si — Cl|
CH
3
3
CH|
HO — Si — OH|
CH
N N Ph OH
(W)
3NaNO ,HCl β Napthol/NaOH0 C
(V)
2NH–
2N Cl
T
CH 3 3H C CH
3
3
CH|
C — CH|
OO
H U
2O
3 3H C — CH — CH
O||C — H3H C
2 2CH — CH — NH||OH
. .
3H C
(S)
N
OH
3H C NH
OH
22H O
O
3H C 2NH
2CHCH
OH
O||C — H3H C
2 3CH — CH — NH||O
3NH
3H C 3
2
(i) O(ii) Zn.H O
3H C
R
O||C — H
2CH — C — H||O
3
H
CHC C2Ph CH
H
trans
3CH
HC C2Ph CH
H
cis
91Mock Test-1
18. (b)
2S is 3p l = 1
For Question Nos. 19 to 20
8 6C H double bond equivalent
6
8 1 62
3
H / heat3 3
OH CH| |
Ph — C — CH Ph — C CH — CH| (Y)Et
19. (c) 20. (d)
2 2CH CH OH
2
(i) EtMgBr(ii) H O
(X)
3
O||C — CH
C CH
2CH CH
4 2 4 2HgSO , H SO , H O
2 8
2 2 2
(1) B H(2) H O , NaOH, H O
Chemistry92
JEE-MAIN: CHEMISTRY MOCK TEST-2
1. 2 g of oxygen contain same number of atoms as contained by
a. 0.5 g hydrogen b. 4.0 g sulphur c. 7.0 g nitrogen d. 2.3 g sodium
2. The axial angles in triclinic crystal system are
a. 90 b. 90 , 90
c. 90 d. 90
3. A sample of drinking water was found to be severely
contaminated with chloroform supposed to be a
carcinogen. The level of contamination was 15 ppm (by mass). Determine the molality of chloroform in the water sample.
a. b.
c. d.
4. The temperature at which hydrogen molecules will have the same root mean square velocity as oxygen molecules
have at 27 C is
a. 25 C b. 7.93 C
c. 248 C d. 127 C
5. An electron makes a transition from orbit n = 4 to the orbit n = 2 of a hydrogen atom. The wave number of the emitted radiations (R = Rydberg's constant) will be
a. 16
3 R b.
2 R
16 c.
3 R
16 d.
4 R
16
6. Select the compound in which chlorine is assigned the oxidation number +5
a. 4HClO b. 2HClO c. 3HClO d. HCl
7. The molar conductivity is maximum for the solution of concentration
a. 0.001 M b. 0.005 M c. 0.002 M d. 0.004 M
8. The number and type of bonds between two carbon atoms in calcium carbide are
a. One sigma, one pi b. One sigma, two pi c.Two sigma, one pi d. Two singma, two pi
9. The standard state gibbs free energy change for the given
isomerization reaction cis-2-pentene trans-2-pentene
is at If more trans-2-pentene is added
to the reaction vessel, then
a. More cis-2-pentene is formed b. Equilibrium is shifted in the forward direction
c. Equilibrium remains unaffected d. Additional trans-2-pentene is formed
10. The dissociation constant of is
The hydrolysis constant for 0.1 M sodium acetate is
a. b.
c. d.
11. The half-life of a first order reaction having rate constant
is
a. 12.1 h b. 9.7 h c. 11.3 h d. 1.8 h
12. Which characteristic is true in respect of colloidal particle a. They always have two phases
b. They are only in liquid state c. They can't be electrolysed
d. They are only hydrophilic
13. The heat of transition of graphite into diamond
would be, where
a. b.
c. d. None of these
14. All the nuclei from the initial element to the final element
constitute a series which is called a. g-series b. b-series
c. b-g series d. Disintegration series
15. The following compound will undergo electrophilic substitution more readily than benzene
a. Nitrobenzene b. Benzoic acid
c. Benzaldehyde d. Phenol
16. Cis and trans 2-butene are a. Conformational isomers
b. Optical isomers
c. Position isomers d. Geometrical isomers
17. Electrolysis of a concentrated solution of sodium fumarate
gives
a. Ethylene b. Ethane c. Acetylene d. Vinyl alcohol
3CHCl ,
4 12.12 10 mol kg 4 11.26 10 mol kg 4 10.12 10 mol kg 4 15.36 10 mol kg
–3.67kJ / mol 400K.
3CH COOH 51.8 10 .
45.56 10 105.56 1051.8 10 91.8 10
5 1K 1.7 10 s
t( H )
2 2C(graphite) O (g) CO (g); H x kJ
2 2C(diamond) C (g) CO (g) ; H y kJ 1(x y) kJ mol 1(x y) kJ mol1(y x) kJ mol
93Mock Test-2
18. Identify and in the following sequence
a.
b.
c.
d.
19. Consider the following alcohols
(i) 1-Phenyl-1-propanol
(ii) 3-Phenyl-1-propanol
(iii) 1-Phenyl-2-propanol
The correct sequence of increasing order of reactivity of
these alcohol in their reaction with HBr is
a. (i), (ii), (iii) b. (ii), (i), (iii)
c. (i), (iii), (ii) d. (ii), (iii), (i)
20. Reaction between diethyl cadmium and acetyl chloride
leads to the formation of
a. dimethyl ketone b. ethylmethyl ketone
c. diethyl ketone d. acetaldehyde
21. Which of the following is basic?
a. b.
c. d.
22. The compound X is:
a. b.
c. d.
23. Which of the following is a synthetic polymer
a. Rubber b. Perspex
c. Protein d. Cellulose
24. The correct statement in respect of protein haemoglobin is that it
a. Acts as an oxygen carrier in the blood b. Forms antibodies and offers resistance to diseases c. Functions as a catalyst for biological reactions d. Maintains blood sugar level
25. Which one of the following is known as broad spectrum antibiotics
a. Streptomycine b. Ampicillin c. Chloramphenicol d. Penicillin G
26. The element or elements whose position is anomalous in the periodic table is
a. Halogens b. and Ni
c. Inert gases d. Hydrogen
27. Flux added in the extraction of iron is a. Silica b. Felspar c. Limestone d. Flint
28. The correct order of the increasing ionic character is
a.
b.
c.
d.
29. Nesseler's reagent is
a. 2 4K HgI b. 2 4K HgI KOH
c. 2 2K HgI KOH d. 2 4K HgI Hg
30. The primary valence of the metal ion in the co-ordination
compound 2 4K Ni CN is
a. Four b. Zero c. Two d. Six
X YX
2 5C H Br Yproduct 3 7C H 2NH
4X KCN,Y LiAlH
3X KCN,Y H O
3 3X CH Cl,Y AlCl / HCl
3 2 2X CH NH ,Y HNO
3 2CH CH OH 2 2H O
2 2HOCH CH OH 3CH COOH
2 5Na, C H OH3CH CN X.
3 2 2CH CH NO 3 2CH CH COOH
6 5 3 2C H N(CH ) 6 5 2C H CONH
Fe, Co
2 2 2 2BeCl MgCl CaCl BaCl
2 2 2 2BeCl MgCl BaCl CaCl
2 2 2 2BeCl BaCl MgCl CaCl
2 2 2 2BaCl CaCl MgCl BeCl
Space for rough work
Chemistry94
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. At 400 K, the root mean square (rms) speed of a gas X
(molecular weight = 40) is equal to the most probable speed
of gas Y at 60 K. The molecular weight of the gas Y is
2. Based on VSEPR theory, the number of 90 degree
F−Br−F angles in is
3. 20% surface sites have adsorbed On heating gas
evolved from sites and were collected at 0.001 atm and
298 K in a container of volume is 2.46 cm3. Density of
surface sites is and surface area is 1000
cm2, find out the no. of surface sites occupied per
molecule of
4. The total number of contributing structure showing
hyperconjugation (involving C–H bonds) for the
following carbocation is
5. The volume (in mL) of 0.1 M required for
complete precipitation of chloride ions present in 30 mL
of 0.01 M solution of as silver chloride
is close to
6. Among and
the total number of black coloured sulphides is
7. 29.2% (w/w) HCl stock solution has a density of 1.25 g
The molecular weight of HCl is 36.5 g The
volume (mL) of stock solution required to prepare a 200
mL solution of 0.4 M HCl is
8. The maximum number of isomers (including
stereoisomers) that are possible on monochlorination of
the following compound is
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately
a. 1.86 Å b. 3.22 Å c. 5.72 Å d. 0.93 Å
10. 18 g glucose is added to 178.2 g water. The vapor
pressure of water (in torr) for this aqueous solution is : a. 7.6 b. 76.0 c. 752.4 d. 759.0
11. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the van der Waals constant a (atm. liter2 mol2) is.
a. 1.0 b. 4.5 c. 1.5 d. 3.0
12. A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant
temperature of As it does so, it absorbs 208 J of
heat. The values of q and w for the process will be
a. b.
c. d.
13. The first ionisation potential of Na is 5.1 eV. The value of
electron gain enthalpy of will be
a. b.
c. d.
14. In a galvanic cell, the salt bridge a. does not participate chemically in the cell reaction. b. stops the diffusion of ions from one electrode to another. c. is necessary for the occurrence of the cell reaction d. ensures mixing of the two electrolytic solutions
15. Stability of the species of and increase in
the order of
a. b.
c. d.
5BrF
2N . 2N
14 26.023 10 / cm
2N .
3AgNO
2 5 2[Cr(H O) Cl]Cl ,
2 2 3PbS, CuS, HgS, MnS, Ag S, NiS,CoS, Bi S
2SnS ,
1mL . 1mol .
6 I2 6(C H P )
37.0 C.
(R 8.314 J / mol K) (ln 7.5 2.01)
q 208 J, w 208 J q 208 J, w 208 J
q 208 J, w 208 J q 208 J, w 208 J
Na
–2.55 eV 5.1eV
10.2 eV 2.55 eV
2 2Li , Li 2Li
2 2 2Li Li Li 2 2 2Li Li Li
2 2 2Li Li Li 2 2 2Li Li Li
(Graph not to scale)
0 2.0 3.0
20.121.623.124.6
PV
(li
ter-
atm
mol
–1)
11(mol liter )
V
3 2CH CH 2 3CH CH
C
H
3CH
3 2 3H C CH CH
95Mock Test-2
16. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?
a. 0.1 L b. 0.9 L c. 2.0 L d. 9.0 L
17. NiCl2 exhibits temperature dependent
magnetic behavior (paramagnetic / diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively.
a. tetrahedral and tetrahedral b. square planar and square planar c. tetrahedral and square planar d. square planar and tetrahedral
18. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is
expected to have the highest value?
a. b.
c. d.
SECTION 3 Contains 2 Match The Following Type Questions
You will have to match entries in Column I with the entries in Column II.
19. Match gases under specified conditions listed in Column-I with their properties/laws in Column-II.
Column I Column II
(A) Hydrogen gas
273K)
1. Compressibility
factor
(B) Hydrogen gas (P 0,
2. Attractive forces are dominant
(C) 3.
(D) Real gas with very large
molar volume
4.
a. A 2; B 3; C 2; D 1, 4
b. A 1, 3; B 1; C 2; D 4
c. A 3; B 4; C 3, 4; D 1, 2
d. A 1, 2; B 3; C 1, 2; D 1, 4
20. The standard reduction potential data at 25ºC is given
below:
Match of the redox pair in Column I with the values
given in Column II.
Column I Column II
(A) 1. – 0.18 V
(B) 2. – 0.4 V
(C) 3. – 0.04 V
(D) 4. – 0.83 V
a. A 2; B 3; C 1; D 4
b. A 1; B 2; C 3; D 4
c. A 3; B 4 ; C 1; D 2
d. A 2; B 3; C 1; D 4
2 5 2 6 5 2P(C H ) (C H )
3 2
0
M / ME
Cr(Z 24) Mn(Z 25)
Fe(Z 26) Co(Z 27)
(P 200 atm,
T 1
T 273K)
2CO (P 1 atm, T 273K) P nRT
P(V nb) nRT
3 2E (Fe , Fe ) 0.77 V; 2E (Fe , Fe) 0.44V 2E (Cu , Cu) 0.34V; E (Cu , Cu) 0.52V
2 2E [O (g) 4H 4e 2H O] 1.23V;
2 2E [O (g) 2H O 4e 4OH ] 0.40V
3E (Cr , Cr) 0.74V; 2E (Cr , Cr) 0.91 V
E
3E (Fe , Fe)
2E (4H O 4 4OH ) 2E (Cu Cu 2Cu )
3 2E (Cr , Cr )
Space for rough work
Chemistry96
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. The maximum number of electrons that can have principal
quantum number, n = 3, and spin quantum number,
is
2. The dissociation constant of a substituted benzoic acid at 25ºC is 1.0 × 10–4. The pH of a 0.01M solution of its
sodium salt is
3. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular
formula is
4. To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol. of
an unknown compound (vapour pressure 0.68 atm. At 0ºC) are introduced. Considering the ideal gas behaviour,
the total volume (in litre) of the gases at 0ºC is close to
5. The coordination number of Al in the crystalline state of
is
6. Among the following, the number of compounds than can
react with to give is
7. If the freezing point of a 0.01 molal aqueous solution of a
cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is –0.0558ºC, the number of
chloride(s) in the coordination sphere of the complex is [Kf of water = 1.86 K kg mol–1]
8. A student performs a titration with different burettes and
finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titer value is
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. The correct statement(s) regarding defects in solids is(are)
a. Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion
b. Frenkel defect is a dislocation defect c. Trapping of an electron in the lattice leads to the
formation of F-centre
d. Schottky defects have no effect on the physical properties of solids
10. Mixture(s) showing positive deviation from Raoult’s law at 35ºC is(are)
a. carbon, tetrahedral + methanol b. carbon disulphide + acetone c. benzene + toluene d. phenol + aniline
11. For a gaseous state, if most probable speed is denoted by
C*, average speed by and mean square speed by C,
then for a large number of molecules the ratio of these speeds are
a.
b.
c.
d.
12. The species having bond order different from that in CO is a. NO– b. NO+ c. CN– d. N2
13. Which of the following is the wrong statement?
a. ONCF and are not isoelectronic
b. molecules is bent
c. Ozone is violet–black in solid state d. Ozone is diamagnetic gas
14. The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of
three different aqueous solutions of KCl, 3CH OH and
3 2 11 3CH (CH ) OSO Na at room temperature. The correct
assignment of the sketches is
a. I : KCl II : 3CH OH III : 3 2 11 3CH (CH ) OSO Na
b. I : 3 2 11 3CH (CH ) OSO Na II : 3CH OH III : KCl
c. I : KCl II : 3 2 11 3CH (CH ) OSO Na III : 3CH OH
d. I : 3CH OH II : KCl III: 3 2 11 3CH (CH ) OSO Na
15. Using the data provided, calculate the multiple bond
energy (kJ mol–1) of CC bond in C2H2. That energy is
(take the bond energy of a C—H bond as 350 kJ mol–1)
s
1m ,
2
5 10C H
3AlCl
5PCl 3POCl
2 2 2 2 2 4 4 10O , CO , SO , H O, H SO , P O
C
C* : C : C 1.225 :1.128 :1
C*: C : C 1.128 :1.1225 :1
C* : C : C 1:1.228 :1.225
C* : C : C 1:1.225 :1.128
ONO
3O
Concentration
Sur
face
tens
ion II
Concentration
Sur
face
tens
ion
I
Concentration
Sur
face
tens
ion III
97Mock Test-2
22C(s) + H (g) ;
a. 1165 b. 837 c. 865 d. 815
16. A solution of (–)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of
due to the formation of
a. carbanion b. carbene c. carbocation d. free radical
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem.
Two multiple choice questions will be asked based on this paragraph.
One or more than one option can be correct.
Paragraph for Question No. 17 to 18
The electrochemical cell shown below is a concentration cell.
(saturated solution of sparingly soluble salt,
M. The emf of the cell depends
on the difference in concentration of ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.
17. The value of for the given cell is
(take 1 F = 96500 C
a. –5.7 b. 5.7 c. 11.4 d. –11.4
18. The solubility product of MX2 at 298 K
based on the information available for the given concentration cell is (take 2.303 × R × 298 / F = 0.059 V)
a. b.
c. d.
Paragraph for Question Nos. 19 to 20
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately
are present in a few grams of any chemical
compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass:
1 Faraday coulombs)
19. The total number of moles of chlorine gas evolved is a. 0.5 b. 1.0 c. 2.0 d. 3.0
20. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is
a. 200 b. 225 c. 400 d. 446
2 2C H (g) 1H 225 kJ mol
2C(s) 2C(g)1H 1410 kJ mol
2H (g) 2H(g)1H 330 kJ mol
5SbCl ,
2M|M 2 3
2MX ) || M (0.001 mol dm ) |
2M
1G (kJ mol )1mol )
3 9sp(K ;mol dm )
151 10 154 10121 10 124 10
236.023 10 )
Na 23, Hg 200, 96500
Space for rough work
Chemistry98
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
b c b c c c a b a b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c a b d d d c a d b
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
a a b a c d c a b c
1. (b) Number of atoms in 2 g of oxygen
232 6 10
16
× ×=
(1) 23 236 10 0.5 3 10× × = ×
(2)
23226 10 4
7.5 1032
× ×= ×
(3)
23236 10 7
1.5 1028
× ×= ×
(4)
23226 10 2.3
6 1023
× ×= ×
∴
The correct answer is (b).
2. (c) The axial angles in triclinic crystal system are different
and none is perpendicular to any of the others i.e.,
90 .α β γ≠ ≠ ≠ °
3. (b) Level of contamination = 15 ppm
= 15 parts in parts Mass % of
Thus, the water sample contains
(by mass) of
Amount of
Mass of water (solvent) =
So, molality of in solution
4. (c) r.m.s. 2
3RT(H )
2=
r.m.s. 2
3R 400(O )
32
×=
( T 273 127 400 K= + = )
3 RT 3R 300
2 32
×=
⇒ T 400
2 32=
∴ 400 2
T 25 K32
×= = or T 25 273 248 C= − = − °
5. (c) Wave number2 2
1 2
1 1 1 1 1 3RR R
4 16 16n nλ
= − = − =
6. (c)
⇒ ;
7. (a) Since molar conductance
8. (b) and
9. (a) Equilibrium shifts backward by Le-chatelier’s
principle.
10. (b)
11. (c)
12. (a) Dispersion medium and dispersed phase are phases of
colloid.
13. (b) Graphite diamond
14. (d) Definition of disintegration series.
15. (d) Phenol will undergo electrophilic substitution more
readily than benzene.
16. (d) Cis and trans 2-butene are geometrical isomers.
17. (c)
18. (a)
19. (d) HBr reacts with alcohols through the formation of
carbocation. The stable is the carbocation formed, more is
the reactivity of alcohol with HBr. The carbocation
formed are
(1) (2) (3)
610
3CHCl
3
6
15100 1.5 10
10
−= × = ×
31.5 10 %
−× 3CHCl
3 1
15gCHCl 0.126mol
119.5g mol−= =
6 310 g 10 kg=
3CHCl
4 1
3
0.126mol1.26 10 molkg
10 kg
− −= = ×
3HClO
∗
1 2 3 0+ − × =x 6 1 5= − = +x
1
Molarity∝
σ1 π2
1410w
b 5
a
K 1 10K 5.56 10
K 1.8 10
−−
−
×= = = ×
×
5 1K 1.7 10 s
− −= ×
5
1/ 2
0.693 0.693t 10 11.32h
K 1.7= = × =
→ 1
tH ( ) kJ mol .−∆ = −x y
CHCOONa||CHCOONa
Electrolysis
22H O+ →
Acetylene
CH|||CH
2 22CO 2KOH H+ + +
4LiAlH (Y)KCN(X)
2 5 2 5C H Br C H CN→ →
2 5 2 2 3 7 2C H CH NH (C H NH )
4X KCN,Y LiAlH= =
2 3CH CH CH+− −2 2 2CH CH CH+
2 3CH CH CH+
99Mock Test-2
20. (b)
21. (a) Ethanol is the weakest acid among these, hence it is
most basic.
22. (a)
23. (b) Perspex is a synthesised polymer.
24. (a) Four ions of each haemoglobin can bind with 4
molecules of and it is carried as oxyhaemoglobin.
25. (c) Chloramphenicol is broad spectrum antibiotic used in the treatment of typhoid, dysentry, acute fever.
26. (d) It shows similarities with both alkali metals as well as halogens.
27. (c) Impurities of 2SiO is present in iron ore so basic flux
3CaCO is added.
2 3Flux Impurity Slag
CaO SiO CaSiO
28. (a)
As we go down the group I.E. decreases. Hence ionic character increases.
29. (b) 2 2 4
Nessler's reagent
2KI HgI K Hgl KOH
30. (c) Primary valencies are also known as oxidation state.
2 4K [ Ni(CN) ], 2 x 4 0 x 2
JEE Advance Paper -I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
4 0 2 6 6 6 8 8 a c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c a b a b d c d d c
1. (4)
,
2. (0)
All four planar bonds (F−Br−F) will reduce from 90° to
84.8° after bp repulsion.
So among the following only four (4) has linear shape and no d-orbital is involved in hybridization.
3. (2)
By ideal gas,
Now molecules of
Now total surface sites available
Surface site used to adsorb
Sites occupied per molecule of
4. (6)
5. (6) Number of ionisable in is 2
Millimoles of
Millimoles of required
6. (6) Black coloured sulphides
* Bi2S3 in its crystalline form is dark brown but
precipitate obtained is black in colour.
7. (8) Molarity
Cd2 3
2 3
CH CH
CH CH32CH COCl
3 2 3 22CH COCH CH CaCl
2 5Na, C H OH3 3 2 2CH CN 4H CH CH NH (X)
2Fe
2O
4 2 4 8Hb 4O Hb O
2 2 2 2BeCl MgCl CaCl BaCl
(Xgas )(400K) (Ygas )(60K )rms mpV V
M.W.(X gas) 40;M.W.(Ygas) x
1 1
1 2
3RT 2RT
M M
400 3 2 60
40 x
12030
x x 4
p
2
2NP 0.001atm, T 298 K, V 2.46 cm
PV nRT
2
37
N
PV 0.001 2.46 10n 1.0 10
RT 0.0821 298
23 7 162N 6.023 10 1 10 6.023 10
14 176.023 10 1000 6.023 10
17 162
20N 6.023 10 12.04 10
100
16
2 16
12.04 10N 2
6.02 10
6 H atoms are there
Cl 2 5 2[Cr(H O) Cl]Cl
Cl 30 0.01 2 0.6
Ag 0.6
0.6 0.1 V
V 6 ml
2PbS,CuS, HgS, Ag S, NiS,CoS
2 3Bi S
solnsol
w w sol sol w
W dn W W 100 d1000
V M v M W W M 100
w
W% d 10
WM
BrF F
F FF
Chemistry100
⇒
8. (8)
Total = 2 + 4 + 1 + 1 = 8
9. (a) For BCC 6 unit cell,
Å Å
10. (c) Moles of glucose =
Moles of water =
⇒
torr
= 752.4 torr
11. (c)
fi Slope = – a
Slope
12. (a) (as it absorb heat)
13. (b) E.A = Ionisation potential
EA of
14. (a)
15. (b) B.O. of , B.O. of
Hence stability order
16. (d) ⇒
17. (c)
18. (d) Factual
19. (d) A → 1, 2 ; B→ 3 ; C→ 1, 2 ; D → 1, 4
(A) at high pressure and low temperature.
Equation reduces to
29.2 1.25 1010M
36.5
× ×= = 1 1 2 2M V M V=
⇒ 110 V 0.4 200× = ×
0.4 200V 8ml
10
×= =
3a 4r=
3 3r a 4.29
4 4= = × 1.85=
180.1
180=
178.29.9
18=
⇒ Totaln 10=
⇒P 0.1
P 10
∆=
°
P 0.01P∆ = °
0.01 760 7.6= × =
SP 760 7.6= −
b 0,T 300K, n 1= = =
2
2
anP (V nb) RT
V
+ − =
2
aP (V) RT
V
+ =
⇒a
PV RTV
+ =
1PV a RT
V
= − × +
y mx C= +
2 1
2 1
y y 20.1 21.61.5
x x 3 2
− −= =
− −
q 208 J,= +
2Re w 10
1
vw 2.303 nRT log
v
= −
10
3752.303 (0.04 8.314 (310) log 208 J
50
= − × × × = −
∴ Na 5.1eV+ =−
2Li 0.5+ = 2Li 0.5− =
2 2 2Li Li Li− += < <
0.1 1 (1 v) 0.01× = + ×0.1
1 v0.01
+ =
⇒ 1 v 10+ = ⇒ v 10 1 9L= − =
22 2 5 2 6 5 2[NiCl P(C H ) (C H ) ]+
2 8Ni (28) : [Ar]4s 3d+ °
mPVZ
RT=
2
2
anP (V nb) nRT
V
+ − =
P(V nb) nRT.− =
Hybridisation is dsp2 (diamagnetic)
Square planar
Hybridisation is sp3 (paramagnetic)
tetrahedral
3d8 In strong ligand
In weak ligand 3d8
3 2 2 3CH CH C CH CH 1−
2CH Cl
H
*
3 2 2 3CH CH C CH CH 1−
3CH
H
3 2 3CH CH C CHCH Two Enantiomeric pairs 4− =
3CH
H
*
*
3CH
Cl
Br
Br
Cl
3CH
101Mock Test-2
(B) For hydrogen gas value of Z = 1 at P = 0 and it increases continuously on increasing pressure.
(C) CO2 molecules have larger attractive forces, under normal conditions.
(D) at very large molar volume Z 1.
20. (c) A 3; B 4 ; C 1 ; D 2
(A)
(B) . . .(i)
. . .(ii)
So . . .(iii)
For IIIrd reduction
(C)
(D)
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
9 8 7 7 6 5 1 3 b, c a, b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c a a d d c d b b d
1. (9) Number of orbital for is
Number of electrons for and
2. (8) ) = 1 × 10–4
(1–h 1)
3. (7) Cyclic
For 3rd structure 2 cis-trans and 1 optical isomer are
possible. Total 7 isomers.
4. (7) Let unknown is X.
Now
=7
5. (6) Coordination number of Al is 6. It exists in ccp lattice with 6 coordinate layer structure.
6. (5)
7. (1) f fT = iK m
8. (3)
9. (b, c)
10. (a, b)
(a) 4 3CCl CH OH Positive deviation from Raoult’s
law
(b) 2CS
Positive deviation from Raoult’s law
(c) 6 6 7 8C H C H Ideal solution
(d)
Negative deviation from Raoult’s law.
11. (c)
1 : 1.128 : 1.225
12. (a) (16 electron system)
Bond order =2.
mPVZ ,
RT
3 3 2 2
o o o
Fe / Fe Fe / Fe Fe / FeG G G
3 3 2 2
o o o
(Fe / Fe) (Fe / Fe ) Fe / Fe3 FE 1 FE ( 2 FE )
3
o
Fe / FeE 0.04 V
2 2O (g) 2H O 4e 4OH E 0.40 V
2 22H O O (g) 4H 4e E 1.23 V
24H O 4H 4OH E 0.40 1.23 0.83 V.
2 2
o o o
(Cu / Cu) Cu / Cu Cu / CuG 1 FE ( 1 F E )
2 2
o o o
Cu / Cu Cu / Cu Cu / Cu2 FE 1 FE ( 1 F E )
2
o
Cu / CuE 0.18 V.
3 2 3 2
o o o
Cr / Cr Cr / Cr Cr / CrG G G
3 2 3 2
o o o
Cr / Cr Cr / Cr Cr / Cr1 F E 3 F E ( 2 F E )
3 2
o
Cr / CrE 0.4 V.
n 3 2n 9
n 3 s
1m 9
2
a 6 5K (C H COOH
6 5pH of 0.01MC H COONa
16 5 2 6 5
0.01h0.01(1 h) 0.01h
C H COO H C H COOH OH
2w
ha
K 0.01hK
K 1 h
14 2 2
4
10 10 h
10 1 h
4 6[OH ] 0.01h 0.01 10 10 8[H ] 10 pH 8
5 10C H
He total xp p P (1 0.68) atm 0.32 atm
He He
RTp n
V
He
RT 0.10 0.082 273v
p 0.32
0.0558 i 1.86 0.01 i 3
3 5 2Complex is [Co(NH ) Cl]Cl
2RT 8RT 3RTC*: C : C
M M M
82 : 3
3.14
NO
+
OH 2NH
3CH 3CH
O
C
Chemistry102
and N2 are isoelectronic with CO therefore all
have same bond order (=3).
13. (a) ONCF and ONO– are isoelectronic in nature.
14. (d) Impurities affect surface tension appreciably. It is observed that impurities which tend to concentrate on surface of liquids, compared to its bulk lower the surface tension.
Substances like detergents, soaps 3 2 11 3[CH (CH ) SO Na ]
decreases the surface tension sharply.
Those like alcohol (e.g., 3 2 5–CH OH, C H OH) lower the
surface tension slightly. This can also be related to the fact
that 3CH OH has smaller dielectric constant. Dielectric
constant is directly proportional to surface tension. So, on
adding 3CH OH in water, overall dielectric constant
decreases and surface tension decreases. Inorganic impurities present in bulk of a liquid such as
KCl tend to increase the surface tension of water.
15. (d)
BE(H2) + BE(H2) + Hsub (C) – BE (C – H) × 2 +
BE(C C ) Hrxn
x= 815
16. (c) Racemises slowly due to formation of intermediate carbocation.
For Question Nos. 17 to 18
17. (d)
18. (b)
5 3 154(10 ) 4 10- -= = ¥
For Question Nos. 19 to 20
19. (b)
At anode:
Moles of in 500 ml.
Therefore 1 mole of Cl2 evolves.
20. (d) Na—Hg (amalgam) formed = 2 moles at cathode.
NO , CN
2 2 22C(s) H (g) C H (g)(H C C H)
330 1410 [350 2 x] 225
cellG nFE
2 96500 0.059 111.387 kJ mol
11.4 kJ
10 3
0.0591 CE log
2 10
10 3
0.0591 C0.059 log
2 10
23
C10
10
2 5C [M ] 10 M.
2 2 3spK [M ][X ] 4s
NaCl Na Cl
22Cl Cl
Cl 2
103Mock Test-3
JEE-MAIN: CHEMISTRY MOCK TEST-3
1. The volume of 1.0 g of hydrogen in litres at N.T.P. is a. 2.24 b. 22.4 c. 1.12 d. 11.2
2. Which set of characteristics of ZnS crystal is correct? a. Coordination number (4: 4) : ccp; Zn2+ ion in the alternate
tetrahedral voids b. Coordination number (6: 6); hcp; Zn2+ ion in all tetrahedral
voids c. Coordination number (6: 4) : hcp; Zn2+ ion in all octahedral
voids d. Coordination number (4: 4); ccp; Zn2+
ion in all tetrahedral voids
3. PtCl4.6H2O can exist as a hydrated complex. 1 molal aqueous solution has depression in freezing point of 3.72º
Assume 100% ionisation and
then complex is
a. b.
c. d.
4. A weather balloon filled with hydrogen gas at 1 atm and 27ºC has volume equal to 12000 litres. On ascending, it reaches a place where temperature is –23ºC and pressure 0.5 atm. The volume of the balloon is
a. 24000 L b. 12000 L c. 10000 L d. 20000 L
5. In Bohr model of the hydrogen atom, the lowest orbit corresponds to
a. Infinite energy b. The maximum energy c. The minimum energy d. Zero energy
6. When 4KMnO is reduced with oxalic acid in acidic
solution, the oxidation number of M n changes from a. 7 to 4 b. 6 to 4 c. 7 to 2 d. 4 to 2
7. The unit of molar conductivity is
a. –1cm–2 mol–1 b. cm–2 mol–1
c. –1cm2 mol–1
d. cm2 mol–1
8. In a double bond connecting two atoms, there is a sharing of a. 2 electrons b. 1 electron c. 4 electrons d. All electrons
9. In a reversible reaction, the catalyst a. Increases the activation energy of the backward reaction b. Increases the activation energy of the forward reaction
c. Decreases the activation energy of both, forward and backward reaction
d. Decreases the activation energy of forward reaction
10. Ammonium cyanide is salt of NH4OH(Kb = 1.8 × 10–5)
and HCN (Kb = 4.0 × 10–10). The hydrolysis constant of
0.1 M at 25ºC is
a. 1.4 b.
c. d.
11. For the reaction A + B C it is found that doubling the
concentration of A increases the rate by 4 times, and
doubling the concentration of B doubles the reaction rate.
What is the overall order of the reaction?
a. 4 b. 3/2
c. 3 d. 1
12. The coagulation power of an electrolyte for arsenious
sulphide decreases in the order
a. + +3 +2Na , Al , Ba b. 3 24 4PO , SO , Cl- - -
c. +3 +2 +Al , Ba , Na d. None of these
13. Correct relationship between heat of fusion heat
of vaporisation and heat of sublimation is
a.
b.
c.
d.
14. The number of neutrons in the parent nucleus which gives
N14 on - emission is
a. 7 b. 14 c. 6 d. 8
15. Which represents nucleophilic aromatic substitution
reaction?
a. Reaction of benzene with in sunlight
b. Benzyl bromide hydrolysis
c. Reaction of NaOH with dinitrofluorobenzene
d. Sulphonation of benzene
16. Which one of the following is the chiral molecule?
a. b.
c. d.
17. n-Propyl chloride and benzene react in the presence of
anhydrous to form
a. ethyl benzene b. methyl benzene
c. n-propyl benzene d. iso-propyl benzene
1f 2K (H O) 1.86 mol kg,
2 6 4[Pt(H O) ]Cl 2 4 2 2 2[Pt(H O) Cl ]Cl 2H O
2 3 3 2[Pt(H O) Cl ],Cl.3H O 2 4 4 2[Pt(H O )Cl ]4H O
4NH CN
157.2 1017.2 10 61.4 10
fus( H ),
vap( H ) sub( H )
fus vap subH H H
vap fus subH H H
sub vap fusH H H
sub vap fusH H H
2Cl
3CH Cl 2 2CH Cl
3CHBr CHClBrI
3AlCl
Chemistry104
18. 1-chlorobutane reacts with alcoholic KOH to form
a. 1-butene b. 2-butane
c. 1-butanol d. 2-butanol
19. An alcohol having molecular formula C5H11OH on
dehydration gives an alkene, which on oxidation yield a
mixture of ketone and an acid. The alcohol is
a. b.
c. d.
20. m-chlorobenzaldehyde on reaction with conc. KOH at
room temperature gives
a. potassium m-chlorobenzate and m-chlorobenzyl alcohol
b. m-hydroxy benzaldehyde and m-chlorobenzyle alcohol
c. m-chlorobenzyl alcohol and m-hydroxy benzyle alcohol
d. potassium m-chlorobenzoate and m-hydroxy bezaldehyde
21. Acetamide is treated separately with the following
reagents. Which one of these would give methyl amine?
a. PCl5 b.
c. Sodalime d. Hot conc.
22. On treating aniline with nitrous acid and HCl at 0–5ºC
gives
a. An alcohol
b. Diazonium salt
c. Nitro aniline
d. Aniline hydrogen chloride
23. ‘Rayon’ is
a. Natural silk b. Artificial silk
c. Natural plastic or rubber d. Synthetic plastic
24. The waxes are long chain compounds of fatty acids, which
belong to the class of
a. Esters b. Ethers
c. Alcohols d. Acetic acid
25. Which of the following is a local anaesthetic?
a. Diazepam b. Procaine
c. Mescaline d. None of these
26. The electronic configuration of the element which is just
above the element with atomic number 43 in the same
periodic group is
a.
b.
c.
d.
27. Complex is formed in the extraction of
a. Na b. Cu c. Ag d. Fe
28. when heated gives
a. Magnesium oxychloride b. Magnesium dichloride
c. Magnesium oxide d. Magnesium chloride
29. Acidified potassium dichromate on reacting with a
sulphite is reduced to
a. 2 2CrO Cl b. 24CrO
c. 3Cr d. 2Cr
30. In the extraction of which of the following, complex ion
forms?
a. Cu b. Ag c. Fe d. Na
3 2 2 3CH CH CH(OH)CH CH 3 2 2 3CH CHCH CH CH|OH
3 2 3(CH ) CHCH(OH)CH 3 2 2(CH ) CCH OH
2NaOH Br
2 4H SO
2 2 6 2 6 5 21s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 10 2 51s 2s 2p 3s 3p 3d 4s 4p
2 2 6 2 6 6 11s 2s 2p 3s 3p 3d 4s
2 2 6 2 6 10 1 61s 2s 2p 3s 3p 3d 4s 4p
2 2MgCl .6H O
Space for rough work
105Mock Test-3
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions
The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
1. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity
of a solution of a weak acid HY (0.10 M). If 0 0
X Y,
the difference in their apK values, apK (HX) apK (HY),
is (consider degree of ionization of both acids to be << 1)
2. The total number of and particles emitted in the
nuclear reaction is
3. The number of hydroxyl group(s) in Q is
4. The total number of basic groups in the following form of lysine is
5. In 1 L saturated solution of AgCl [Ksp(AgCl)=1.6×10–10],
0.1 mol of CuCl is added. The
resultant concentration of in the solution is
The value of “x” is
6. EDTA4– is ethylenediaminetetraacetate ion. The total number of N – Co – O bond angles in [Co(EDTA)1–] complex ion is
7. Reaction of Br2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromated with evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is
8. A decapeptide (Mol. Wt. 796) on complete hydrolysis gives glycine (Mol. Wt.75), alanine and phenylanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes
occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are
a. 1 1
,2 8
b. 1
1,4
c. 1 1
,2 2
d. 1 1
,4 8
10. A monatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals to 1. What is the molar heat capacity of the gas?
a. b. c. d. 0
11. The given graph represents the variation of Z
(compressibility factor ) versus P, for three real
gases A, B and C. Identify the only incorrect statement.
a. For the gas A, a = 0 and its dependence on P is linear at all pressure
b. For the gas B, b = 0 and its dependence on P is linear at all pressure
c. For the gas C, which is typical real gas for which
neither a nor By knowing the minima and the
point of intersection, with Z = 1, a and b can be calculated
d. At high pressure, the slope is positive for all real gases
12. For the reaction: 3 2 4I ClO H SO 4 2Cl HSO I
The correct statement(s) in the balanced equation is/are
a. Stoichiometric coefficient of is 6
b. Iodide is oxidised c. Sulphur is reduced d. H2O is one of the products.
13. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation of such a
reaction will be and
a. b.
c. d.
238 21492 82U Pb
6sp[K (CuCl) 1.0 10 ]
Ag
x1.6 10 .
4R
2
3R
2
5R
2
PV
nRT
b 0.
4HSO
1(R 8.314 JK log 2 0.301)153.6 kJ mol 148.6 kJ mol
158.6 kJ mol 160.5 kJ mol
1
A
B
C
P (atm)
Z
0
AIdeal gasBC
3 2 2 2 2H N CH CH CH CH
CH C
O
2H N O
4aqueous dilute KMnO (excess)Hheat 0 C
P Q
3H C
H
HO 3CH
Chemistry106
14. The freezing point (in ºC) of a solution containing 0.1 g of
K3[Fe(CN)6](Mol. Wt. 329) in 100 g of water (Kf = 1.86K
kg mol–1
) is
a. b.
c. d.
15. The carboxyl functional group (—COOH) is present in
a. picric acid b. barbituric acid
c. ascorbic acid d. aspirin
16. Which compound would give 5-keto-2-methyl hexanal
upon ozonolysis?
a. b.
c.
d.
17. For the process at T=100ºC and 1
atmosphere pressure, the correct choice is
a.
b.
c.
d.
18. Which of the following exists as covalent crystals in the
solid state?
a. Iodine b. Silicon
c. Sulphur d. Phosphorus
SECTION 3 Contains 2 Match The Following Type Questions
You will have to match entries in Column I with the entries in
Column II.
19. According to Bohr’s theory,
Total energy Kinetic energy
Potential energy Radius of nth orbit
Match the following:
Column I Column II
(A) Vn/Kn = ? 1. 0
(B) If radius of nth orbit
2. –1
(C) Angular momentum in
lowest orbital
3. –2
(D) 4. 1
a. A → 3 ; B → 2 ; C → 1 ; D → 4
b. A → 1 ; B → 2 ; C → 3 ; D → 4
c. A → 4 ; B → 2 ; C → 1 ; D → 3
d. A → 3 ; B → 4 ; C → 1 ; D → 2
20. An aqueous solution of X is added slowly to an aqueous
solution of Y as shown in Column - I. The variation in
conductivity of these reactions in Column - II.
Column I Column II
(A) 1. Conductivity
decreases and
then increases
(B) 2. Conductivity
decreases and
then does not
change much
(C) 3. Conductivity
increases and
then does not
change much
(D) 4. Conductivity
does not
change much
and then
increases
a. A → 1 ; B → 2 ; C → 3 ; D → 4
b. A → 2 ; B → 4 ; C → 3 ; D → 1
c. A → 3 ; B → 4 ; C → 2 ; D → 1
d. A → 3 ; B → 4 ; D → 1 ; D → 2
22.3 10
−− ×
25.7 10
−− ×
35.7 10
−− ×
21.2 10
−− ×
2 2H O( ) H O(g)→l
system surroundingS 0 and S 0∆ > ∆ >
system surroundingS 0 and S 0∆ > ∆ <
system surroundingS 0 and S 0∆ < ∆ >
system surroundingS 0 and S 0∆ < ∆ <
nE = nK =
nV =
nr =
x
nE , x ?∝ =
y
n
1Z , y ?
r∝ =
2 5 3 3X Y
(C H ) N CH COOH+
3X Y
KI(0.1M) AgNO (0.01 M)+
3X Y
CH COOH KOH+
X YNaOH HI+
3CH
3H C
3CH
3CH
3CH
3CH
3CH
3CH
Space for rough work
107Mock Test-3
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions
The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
1. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is
2. When the following aldohexose exists in its D–configuration, the total number of stereoisomers in its pyranose form is
3. The number of resonance structures for N is
4. The concentration of R in the reaction RP was measured as a function of time and the following data is obtained:
[R] (molar) 1.0 0.75 0.40 0.10
t (min). 0.0 0.05 0.12 0.18
The order of the reaction is
5. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is
6. The total number of diprotic acids among the following is
7. The number of neutrons emitted when undergoes
controlled nuclear fission to and is
8. In dilute aqueous 2 4H SO , the complex diaquodioxa-
latoferrate (II) is oxidised by 4MnO . For this reaction,
the ratio of the rate of change of [H ] to the rate of
change of 4[MnO ] is
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. The correct statement(s) for cubic close packed (cep) three dimensional structure is (are)
a. The number of the nearest neighbours of an atom
present in the topmost layer is 12
b. The efficiency of atom packing is 74%
c. The number of octahedral and tetrahedral voids per
atom are 1 and 2, respectively
d. The unit cell edge length is 2 2 times the radius of the
atom
10. The standard Gibbs energy change at 300 K for the
reaction is 2494.2 J. At a given time, the
composition of the reaction mixture is and
The reaction proceeds in the:
[ ]
a. forward direction because
b. reverse direction because
c. forward direction because
d. reverse direction because
11. A compound MpXq has cubic close packing (ccp)
arrangement of X. Its unit cell structure is shown below.
The empirical formula of the compound is
a. MX b.
c. d.
12. Which of the following is the energy of a possible excited
state of hydrogen?
a. b.
c. d.
13. Assuming 2s-2p mixing is not operative, the paramagnetic
species among the following is
a. Be2 b. B2
c. C2 d. N2+
14. In allene(C3H4), the type(s) of hybridization of the carbon
atoms is (are)
a. b.
c. d.
KCN 2 4K SO 4 2 2 4(NH ) C O
NaCl 3 2Zn(NO ) 3FeCl
2 3K CO 4 3NH NO LiCN
2
2
CHO|
CH|
CHOH|
CHOH|
CHOH|
CH OH
3 4H PO 2 4H SO 3 3H PO 2 3H CO 2 2 7H S O
3 3H BO 3 2H PO 2 4H CrO 2 3H SO23592 U
14254 Xe 90
38Sr
2A B C 12[A] , [B] 2
12[C] .
R 8.314 J / K / mol, e 2718
cQ K
cQ K
cQ K
cQ K
2MX
2M X 5 14M X
13.6 eV 6.8 eV
3.4 eV 6.8 eV
3sp and sp 2sp and sp
2only sp 2 3sp and sp
M =X =
Chemistry108
15. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure
I.
II.
II.
The correct order of their boiling point is a. I > II > III b. III > II > I c. II > III > I d. III > I > II
16. (isomeric products)
C5H11Cl (isomeric products)
What are N and M ? a. 6, 6 b. 6, 4 c. 4, 4 d. 3, 3
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.
Paragraph for Question Nos. 17 to 18 X and Y are two volatile liquids with molar weights of 10g mol–1 and 40g mol–1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.
17. The value of d in cm (shown in the figure), as estimated
from Graham’s law, is a. 8 b. 12 c. 16 d. 20
18. The experimental value of d is found to be smaller than the estimate obtained using Graham’s law. This is due to
a. larger mean free path for X as compared to that of Y b. larger mean free path for Y as compared to that of X c. increased collision frequency of Y with the inert gas as
compared to that of X with the inert gas d. increased collision frequency of X with the inert gas as
compared to that of Y with the inert gas
Paragraph for Question Nos. 19 to 20
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7ºC was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of
a strong acid with a strong base is a constant 1(–57.0 kJ mol ),
this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M
acetic acid 5a(K 2.0 10 ) was mixed with 100 mL of 1.0 M
NaOH (under identical conditions to Expt. 1) where a
temperature rise of 5.6 C was measured. (Consider heat
capacity of all solutions as 1 14.2 Jg K and density of all
solutions as 1.0 1g mL )
19. Enthalpy of dissociation (in kJ 1mol ) of acetic acid
obtained from the Expt. 2 is a. 1.0 b. 10.0 c. 24.5 d. 51.4
20. The pH of the solution after Expt. 2 is a. 2.8 b. 4.7 c. 5.0 d. 7.0
2Cl , h Nv
fractional distillation M
L = 24 cm
Cotton wool
Soaked in X
Cotton wool
Soaked in Y
Initial formation of
The product
d
CH3
CH3
H3C
and
and
Space for rough work
109Mock Test-3
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d a c d c c c c c b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c c c d b,c d c a c a
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
b d b a b a c c c b
1. (d) 2 g of hydrogen occupy volume 22.4 L=
1 g of hydrogen occupies volume 22.4 1
11.2 L2
×= =
2. (a) ZnS has zinc blende type structure (i.e., ccp structure).
The S2–
ions are present at the corners of the cube and at the
centre of each face. Zinc ions occupy half of the tetrahedral
sites. Each zinc ion is surrounded by four sulphide ions
which are disposed towards the corner of regular tetrahedral.
Similarly, S2–
ions surrounded by four Zn2+
ions.
3. (c)
Hence,α = 1,∴ n = 2
Two species will be produced from single species which
is only possible for Cl3H2O
4. (d) 1
P 1atm,= 1
V 12000 L,=
1T 27 273 300K= + =
2
P 0.5 atm,= 2
V ?=
2T 23 273 250K= − + =
1 1 2 2
1 2
P V P V
T T=
or 1 1 2
2
2 1
P V T 1 12000 250V 20, 000 L
P T 0.5 300
× ×= = =
×
5. (c) In hydrogen atom, the lowest orbit (n = 1) corresponds
to minimum energy (– 13.6 eV).
6. (c) 5
In this reaction oxidation state of Mn change from +7 to +2.
7. (c) Molar conductivity
So, its unit will be Ω–1
cm2
mol–1
8. (c) In a double bond connecting two atoms sharing of 4
electrons take place as in H2C = CH2.
9. (c) Decreases the activation energy of both forward and
backward reaction.
10. (b) Wh
a b
KK
K K=
×
4
4 5
1 101.4
4.0 10 1.8 10
−
− −
×= =
× × ×
11. (c) A + B → C On doubling the concentration of A rate of
reaction increases by four times. Rate ∝[A]2. However on
doubling the concentration of B, rate of reaction increases
two times. Rate ∝[B]
Thus, overall order of reaction = 2 + 1 = 3
12. (c) According to Hardy-Schulze rule.
13. (c) Heats of combustion are always exothermic except
oxidation of N as,
;
;
14. (d)
in no. of neutrons 14 – 6 = 8.
15. (b, c)
16. (d)
A carbon atom which is attached to four different atoms
or groups is called a chiral or asymmetric carbon atom.
such a carbon atom is often marked by an asterisk.
17. (c)
18. (a)
19. (c)
f cal f( T ) K m∆ = ×
f cal( T ) 1.86 1 1.86∆ = × =
f obs
f cal
( T ) 3.72i 2 1 (n 1) .
( T ) 1.86α
∆= = = = + −
∆
∴
2 3 3 2[Pt(H O) Cl ],Cl3H O
COOH|COOH
27
4 2 4 2 4 4 2 22KMnO 3H SO K SO 2MnSO 10CO 8H O++
+ + → + + +
4 2 4 2 4 4 2 22KMnO 3H SO K SO 2MnSO 10CO 8H O+ + → + + +
1
Mρ=
12 2 22
N O N O+ → H ve∆ = +
2 2N O 2NO+ → H ve∆ = +
14 14
6 6 1X Nβ
+→
14
6X
*
H|
I — C — Br|Cl
3 2 2 2CH CH CH CH Cl KOH(alc.)− + →
3 2 2 21-butene
CH CH CH CH KCl H O− = + +
3 3
3
CH — CH — CH — CH||OHCH
Dehydration→
2 2 3CH CH CH
3AlCl
3 3 2CH CH CH Cl+ →
OH
Na
−
+→
DinitrofluorobenzeneNO
NO F
NO
2O OH
+
DinitrophenolDinitro phenol Dinitrofluorobenzene
Chemistry110
Oxidation
3O C — CH
|OH
20. (a)
21. (b)
“Hofmann’s bromamide reaction”
22. (d)
23. (b) ‘Rayon’ is man-made fibre which consists of purified cellulose in the form of long threads. Rayon resembles silk in appearance. Hence called as artificial silk.
24. (a) Waxes are esters of higher fatty acids.
25. (b) The anaesthetics produce temporary insensitibility to the vital function of all type of cells, specially of nervous system and are used during surgical operations.
These are classified as (a) General anasthetic – produces unconsciousness all over
the body e.g. Cyclopropane, chloroform
(b) Local anasthetic – affect only the part of body e.g. Xylocaine, Procain etc.
26. (a) .
27. (c) Hydrometallurgy
2 2 2Ag S 4NaCN 2Na[Ag(CN) ] Na S
2 2 42Na[Ag(CN) ] Zn Na [Zn(CN) ] 2Ag
28. (c)
29. (c) 2 2 3 22 7 3 4 2Cr O 8H 2SO 2Cr 3SO 4H O
30. (b) 2 22
Sodium dicyno argentate
Ag S 4NaCN 2Na Ag CN Na S
22Na[Ag(CN) ] Zn 2 4
Sodium tetracynozincate (ppt)
Na [Zn (CN) ] 2Ag
JEE Advance Paper-I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
3 8 4 2 7 8 5 6 a a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b a, b, d a a d b b b a c
1. (3) HX H X
[H ][X ]Ka
[HX]
HY H Y
[H ][Y ]Ka
[HY]
m for 1mHX m for
2mHY
1 2m m
1
10
2Ka C
1
2
m1 1 0
m1
Ka C
2
2
m2 2 0
m2
Ka C
2
1 1 1
2 2 2
Ka C m
Ka C m
20.01 1
0.0010.1 10
1 2pKa pKa 3
2. (8)
total 8 particles.
3. (4)
3 3
3
CH — C CH — CH|
CH
3
3
CH — C O|
CH
O O|| |
Ph — C — H HO Ph — C — H|
OHI
3 2 2 3 2
O||
CH — C — NH Br NaOH CH NH
2N O,
5 225 Mn 3d 4s
heat2 2 2MgCl .6H O MgO 5H O 2HCl
238 6 214 2 21492 80 82U X Pb
(6 ,2 ),
H
+
H
HO
HOHO
OH
OH
(Q)
(P)4aqueous dilute KMnO
(excess) 0 C
NH2 2N Cl
NH2
O O
HClHONO
Cl
CHOCannizzaro's
reaction
KOH
Cl
COOK
Cl
2CH OH
ææÆ
111Mock Test-3
4. (2)
5. (7) Let the solubility of AgCl is x
and that of CuCl is y
. . .(i)
Similarly
. . .(ii)
On solving (i) and (ii)
6. (8) Total no. of N – Co – O bond angles is 8.
7. (5)
8. (6) Let number of glycine units Mass of decaeptide = 796
Mass of needed = 162 g, Total mass = 958 g
9. (a) In ccp lattice:
Number of O atoms
Number of Octahedral voids
Number of tetrahedral voids
Number of
Number of
Due to charge neutrality
10. (a)
11. (b)
12. (a, b, d)
13. (a)
14. (a) ,
fi
15. (d)
16. (b)
17. (b) At and 1 atmosphere pressure
is at equilibrium. For equilibrium
and
18. (b) Silicon exists as covalent crystal in the solid state.
19. (a) A 3; B 2; C 1; D 4
20. (c) A 3; B 4; C 2; D 1
(A)
Initially conductivity increases due to ion formation after that it becomes practically constant because X alone cannot form ions. Hence (3) is the correct match.
(B)
Number of ions in the solution remains constant until all the AgNO3 precipitated as AgI. Thereafter conductance increases due to increase in number of ions. Hence (4) is the correct match.
1mol litre
xxAgCl Ag Cl 1mol litre
y yCuCl Cu Cl
1spK of AgCl [Ag ] [Cl ]
101.6 10 x(x y)
spK of CuCl [Cu ] [Cl ] 61.6 10 y(x y)
7[Ag ] 1.6 10
x 7
22 3 3 23Br 3CO 5Br BrO 3CO
n
2H O
47958 75 n
100
958 47
n 6100 75
4
4
83Al 4 m
2Mg 8 n
4( 2) 4m( 3) 8n( 2) 0
1 1m and n
2 8
3 2 4ClO 6I 6H SO
2 4 23I Cl 6HSO 3H O
2 1
1 2
T TEa0.3010
2.303R T T
3
Ea 310 3000.3010
2.303 8.314 10 310 300
aE 53.6 kJ mol
33 6 6K [Fe(CN) ] 3K [Fe(CN) ] i 4
f f
m 1000 0.1 1000T K i 1.86 4
M W 329 100
22.3 10 2fT 2.3 10
100 C
2 2H O( ) H O(g)
totalS 0 system surroundingS S 0
system surroundingS 0 and S 0
2 5 3 3 2 5 3 3X Y
(C H ) N CH COOH (C H ) NH CH COO
3 3X Y
KI(0.1 M) A gNO (0.01 M) AgI KNO
3CH
3CH
3CH
3CH
CHOO
3
2
OZn / H O
5-keto-2-methyl hexanal
3
O||
O — C — CH
COOH
(Aspirin)
O
Co N O
O
N O
O O
O
O
and 2— NH are basic groups in lysine.
O||
—C — O
Chemistry112
(C) Initially conductance decreases due to the decrease in the
number of OH ions thereafter it slowly increases due to
the increase in number of H+ ions. Hence (2) is the correct
match.
(D) Initially it decreases due to decrease in H+ ions and then
increases due to the increase in OH ions. Hence (1) is the
correct match.
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
3 8 9 5 6 6 3 9 b,c,d b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b c c b b b c d a b
1. (3) LiCN are basic in nature and their
aqueous solution turns red litmus paper blue.
2. (8)
Total No. of stereoisomers = 24
= 16 which contains 8D –
Configuration and 8 – L Configuration.
3. (9)
4. (5) From two data, (for zero order kinetics)
⇒
5. (6)
6. (6) 3 3 2 3
H PO , H SO
7. (3)
8. (9) 0
0
a2.303K log ,
t a x
Ê ˆ= Á ˜-Ë ¯
9. (b, c, d) (a) For any atom in topmost layer, coordination
number is not 12 since there is no layer above topmost
layer, (b) Fact, (c) Fact, (D) 2 a 4R=
So, a 2 2 R=
10. (b)
i.e. backward reaction.
11. (b)
12. (c) Energy in 1st excited state
13. (c) Assuming that no 2s-2p mixing takes place
(a)
(b)
(diamagnetic)
(c)
(d)
14. (b)
KCN, 2 3K CO ,-
I
x 0.25K 5
t 0.05= = =
II
x 0.60K 5
t 0.12= = =
2 2 2 4 2 2 4(Potassiummanganate)
2MnO 4KOH O 2K MnO 2H O O.S. of Mn 6in K MnO+ + → + = +
2 2 2 4 2 2 42MnO 4KOH O 2K MnO 2H O O.S. of Mn 6in K MnO+ + → + = +
2 4 2 3 2 2 7 2 4 2 3H SO , H CO , H S O , H CrO , H3PO3, H SO
235 142 90 1
92 54 38 0U Xe Sr 3 n→ + +
0
1/80
a2.303K log
1ta
8
=
e cG RT ln K∆ = −
e c2494.2 8.314 300 ln K= ×
⇒1
cK e−=
1
c
1K e 0.36
2.718
−= = =
12
2 2
2(B)(C)Q 4
[A] [1/ 2]
×= = =
cQ K ,>
1 1X :8 6 4
8 2× + × =
1M 4 1 2;
4= × + =
2 4 2M X MX=
3.4 eV= −
2 2 2 22Be 1s , *1s , 2s , * 2s (diamagnetic)σ σ σ σ→
0x0y
2p2 2 2 2 22 z 2p
B 1s , *1s , 2s , *2s , 2p , (diamagnetic)π
πσ σ σ σ σ→
1 0x x1 0y y
2p *2p2 2 2 2 2 02 z z2p *2p
C 1s , *1s , 2s , *2s , 2p , , , *2p (paramagnetic)π π
π πσ σ σ σ σ σ→
1 0x x1 0y y
2p *2p2 2 2 2 2 02 z z2p *2p
C 1s , *1s , 2s , * 2s , 2p , , , *2p (paramagnetic)π π
π πσ σ σ σ σ σ
2 2 2 2 2 02 z zN 1s , *1s , 2s , * 2s , 2p , , , *2p (diamagnetic)
π π
π πσ σ σ σ σ σ→
2 0x x2 0y y
2p *2p2 2 2 2 2 02 z z2p *2p
N 1s , *1s , 2s , * 2s , 2p , , , * 2p (diamagnetic)π π
π πσ σ σ σ σ σ
2 2sp sp sp
H
HC C C== ==
H
H(allene)
–O O O
1 2 3
OO
7 8 9
–O
O O
4 5 6
–O
–O
OH NaOH N→
HO NaOH
N→
N is
H
H
HH HO
O OH
OH
OH2CH
113Mock Test-3
15. (b) III > II > I More the branching in an alkane, lesser will be the surface area, lesser will be the boiling point.
16. (b)
Md, 1 cannot be separated by fractional distillation.
For Question Nos. 17 to 18
17. (c)
18. (d) As the collision frequency increases then molecular speed decreases than that expected.
For Question Nos. 19 to 20
19. (a) 2HCl NaOH NaCl H O
n 100 1 100 m mole 0.1 mole
Energy evolved due to neutralization of HCl and NaOH
0.1 57 5.7 kJ 5700 Joule
Energy used to increase temperature of solution
200 4.2 5.7 4788 Joule Energy used to increase temperature of calorimeter
5700 4788 912 Joule ms. t 912
m.s 5.7 912 ms 160 Joule / C [Calorimeter constant]
Energy evolved by neutralization of 3CH COOH and
NaOH 200 4.2 5.6 160 5.6 5600 Joule So energy used in dissociation of 0.1 mole
3CH COOH 5700 5600 100 Joule Enthalpy of dissociation 1 kJ / mole
20. (b) 3
1 100 1CH COOH
200 2
3
1 100 1CH CONa
200 2
a
[salt]pH pK log
[acid]
1/ 2
pH 5 log2 log1/ 2
pH= 4.7
X
Y
r d 402
r 24 d 10
d 48 2d 3d 48d 16 cm
CH3
CH3
H3C Cl
CH3
CH3 H3C
*
1,d
CH3
CH3 H3C
Cl *
1,d
CH3
CH2Cl H3C
Chemistry114
JEE-MAIN: CHEMISTRY MOCK TEST-4
1. The number of moles of 2 2SO Cl in 13.5 g is
a. 0.1 b. 0.2
c. 0.3 d. 0.4
2. Arrangement of sulphide ions in zinc blende is
a. simple cubic b. hcp
c. bcc d. fcc
3. A pressure cooker reduces cooking is increased
a. Heat is more evenly distributed
b. Boiling point of water inside the cooker is increased
c. The high pressure tenderizes the food
d. All of the above
4. Two flasks of equal volume contains 2SO and 2CO
respectively at 25 C and 2 atm pressure. Which of the
following is equal in them ?
a. masses of the two gas b. number of molecules
c. rates of effusion d. molecular structure
5. The ratio of the kinetic energy to the total energy of an
electron in a Bohr orbit is
a. –1 b. 2
c. 1 : 2 d. None of these
6. Oxygen has oxidation states of +2 in the
a. 2 2H O b. 2CO c. 2H O d. 2OF
7. Given The equivalent
conductance of the electrolytic cell is
a. b.
c. d.
8. In molecule, the atoms are bonded by
a. One , Two b. One , One
c. Two , One d. Three bonds
9. For the reaction the
equilibrium constant changes with
a. Total pressure
b. Catalyst
c. The amounts of and taken
d. Temperature
10. A certain weak acid has a dissociation constant of
The equilibrium constant for its reaction with a
strong base is.
a. b.
c. d.
11. According to Arrhenius theory, the activation energy is a. The energy it should possess so that it can enter into an
effective collision b. The energy which the molecule should possess in order
to undergo reaction c. The energy it has to acquire further so that it can enter
into a effective collison d. The energy gained by the molecules on colliding with
another molecule
12. Which of the following is property of colloid? a. Scattering of light b. They show attraction c. Dialysis d. Emulsion
13. Which of the following is an example of endothermic reaction?
a.
b.
c.
d.
14. The nuclear binding energy for (39.962384 amu) is: (given mass of proton and neutron are 1.007825 amu and 1.008665 amu respectively)
a. b.
c. d. None of these
15. Which is an electrophile?
a. b. c. d.
16. Cyanide and isocyanide are isomers of type a. Positional b. Functional c. Tautomer d. Structural
17. Acetone will be formed by the ozonolysis of a. Butene-1 b. Butene-2 c. Isobutene d. Butyne-2
18. Which of the following reactions gives
?
a. 3Zn /CH OHææææÆ
b.
c.
d.
1/ a 0.5 cm , R 50 ohm, N 1.0. l
1 2 110 ohm cm gm eq 1 2 120 ohm cm gm eq
1 2 1300 ohm cm gm eq 1 2 1100 ohm cm gm eq
2N
2 2H (g) I (g) 2HI(g),
2H 2I
41.0 10 .
41.0 10 101.0 10101.0 10 141.0 10
2 2 2 2 6C H 2H C H ; E 314.0kJ
2 2C O CO ; E 393.5kJ
2 2N O 2NO; E 180.5kJ
2 2 22H O 2H O; E 571.8kJ
Ar
343.81 MeV 0.369096 MeV
931 MeV
3AlCl CN 3NH 3CH OH
2 2H C C C CH
2 2CH Br CBr CH
2 3
o
Aq.K CO2
40 CHC C CH COOH
Zn2 2
HeatCH Br C C CH Br
2 22CH CH CH I
115Mock Test-4
19. Which of the following does not form phenol or phenoxide?
a. 6 5C H Cl b. 6 5C H COOH
c. 6 5 2C H N Cl d. 6 5 3C H SO Na 20. Identify the final product (Z) in the following sequence of
reactions:
a. b.
c. d.
21. When propionic acid is treated with aqueous sodium
bicarbonate, is liberated. The C of comes from?
a. methyl group b. carboxylic acid group c. methylene group d. bicarbonate group
22. Aniline undergoes condensation to form Schiff base on reacting with
a. Acetyl chloride b. Ammonia c. Acetone d. Benzaldehyde
23. The mass average molecular mass and number average molecular mass of a polymer are respectively 40,000 and 30,000. The polydispersity index of polymer will be
a. < 1 b. > 1 c. 1 d. 0
24. Hardening of oils is caused by
a. b. c. d.
25. Which of the following is molecular disease?
a. Allergy b. Cancer
c. German measeles d. Sickel-cell-anaemia
26. Hydrogen can be put in halogen group because
a. It has deuterium and tritium as isotopes
b. It forms hydrides like chlorides
c. It contains one electron only
d. It is light
27. Which of the following metal is extracted by
amalgamation process?
a. Tin b. Silver
c. Copper d. Zinc
28. Which of the following hydroxide is insoluble in water?
a. b.
c. d.
29. The product of oxidation of I ion by 4MnO in alkaline
medium is
a. 2I b. 3IO
c. 4IO d. 3I
30. Potassium ferrocyanide is a
a. Normal salt b. Mixed salt
c. Double salt d. Complex salt
3 2 4H O H SO2 Heat
Me C O HCN X Y Z;
3 2(CH ) C(OH)COOH 2 3CH C(CH )COOH
2 3HOCH CH(CH )COOH 3CH CH CHCOOH
2CO 2CO
2H 2N 2O 2CO
2Be(OH) 2Mg(OH)
2Ca(OH) 2Ba(OH)
Space for rough work
Chemistry116
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions
The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
1. HCl gas is passed into water, yielding a solution of
density and containing 30% HCl by weight.
Calculate the molarity of the solution.
2. A sample contains a mixture of and
HCl is added to 15.0 g of the sample, yielding 11.0 g of
NaCl. What percent of the sample is ?
Reactions are:
Mw of Mw of Mw of
3. Amongst the following, the total number of compounds
soluble in aqueous is
4. The total number of contributing structures showing hyperconjugation (involving C—H bonds) for the following carbocation is.
5. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are)
6.
;
Total number of isomers (including stereoisomers) is 7. Total number of isomers
8. In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is Assuming concentration of solute is Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take
a. 724 b. 740 c. 736 d. 718 10. Two closed bulbs of equal volume (V) containing an ideal
gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then I raised to T2. The final pressure pf is:
11.095g mL
3NaHCO 2 3Na CO .
2 3Na CO
2 3 2 2
3 2 2
Na CO 2HCl 2NaCl CO H O
NaHCO HCl NaCl CO H O
NaCl 58.5, 3NaHCO 84,1
2 3Na CO 106g mol
NaOH
3 2 2 2CH — CH — CH — CH OH
3 2 3
enantiomeric ( )
CH — CH — CH — CH|
OH
3 3
3
OH|
CH — C — CH|CH
3 2
3
CH — CH — CH OH|
CH
2 C.
1bK 0.76K kg mol )
1T
iP , V
1T
iP , V
1T
fP ,V
2T
fP ,V
3
2
1. O 1. NaOH(aq)2. Zn, H O 2. heat
Y
3CH 3CH
H H
meso
3CH
3CHH
H
3CH
3CHH
H
Pair of enantiomers
Mirror
3CH 2 3CH CH 3CH 3CH
Br
Cl
3CH
Cl
3CH
Br
2 3OCH CH
2CH OH
OH
2 3CH CH
2 3CH CH
COOH
2NO
N3H C
3CH
OHN3H C 3CH COOH
117Mock Test-4
a. b.
c. d.
11. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron,
respectively, then the value of h/ (where is wavelength associated with electron wave) is given by:
a. meV b. 2 meV
c. d.
12. Reduction of the metal centre in aqueous permanganate ion involves
a. 3 electrons in neutral medium b. 5 electrons in neutral medium c. 3 electrons in alkaline medium d. 5 electrons in acidic medium
13. For the following electrochemical cell at 298 K,
Given:
The value of x is a. –2 b. –1 c. 1 d. 2
14. Which one of the following molecules is expected diamagnetic behaviour?
a. C2 b. N2
c. O2 d. S2
15. Aqueous solutions of HNO3, KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is (are)
a. and
b. KOH and
c. and
d. and
16. The value of log10K for a reaction A B is
(Given
and 2.303 × 8.314 × 298 = 5705 )
a. 5 b. 10 c. 95 d. 100
17. The number of structural isomers for C6H14 is a. 3 b. 4 c. 5 d. 6
18. Amongst the given options, the compound(s) in which all
the atoms are in one plane in all the possible
conformations (if any), is (are)
a. b.
c. d.
SECTION 3 Contains 2 Matches The Following Type Questions
You will have to match entries in Column I with the entries in
Column II.
19. According to Bohr’s theory,
Total energy Kinetic energy
Potential energy Radius of nth orbit
Match the following:
Column I Column II
(A) 1. 0
(B) If radius on nth orbit
2. –1
(C) Angular momentum in
lowest orbital
3. –2
(D) 4. 1
a. A 3; B 2; C 4; D 1
b. A 1; B 2; C 3; D 4
c. A 3; B 1; C 2; D 4
d. A 3; B 2; C 1; D 4
20. Match the thermodynamic processes given under Column
I with the expression given under Column II:
Column I Column II
(A) Freezing of water at
273 K and 1 atm
1. q = 0
(B) Expansion of 1 mol of
an ideal gas into a
vacuum under
isolated conditions
2. w = 0
(C) Mixing of equal
volumes of two ideal
gases at constant
3.
1 2i
1 2
T Tp
T T
1i
1 2
T2p
T T
2i
1 2
T2p
T T
1 2i
1 2
T T2p
T T
meV 2 meV
4 22Pt(s) | H (g, 1 bar) H (aq, 1 M) || M (aq), M (aq) | Pt(s)
2
cell 4
[M (aq)]E 0.092 V when 10 .
[M (aq)]x
4 2
0
M / M
RTE 0.151 V; 2.303 0.059 V
F
3HNO 3CH COOH
3CH COONa
3HNO 3CH COONa
3CH COOH 3CH COONa
o 1r 298KH 54.07 kJ mol , o 1 1
r 298KS 10 JK mol 1 1R 8.314 JK mol ;
2H C C O 2 2H C C CH
nE nK
nV nr
n nV / K ?
xnE , x ?
yn
1Z , y ?
r
sysS 0
2CHH C C — C
HH H
2H C 2CH
C C
Chemistry118
temperature and
pressure in an isolated
container
(D) Reversible heating of
at 1 atm from
300 K to 600 K,
followed by reversible
4.
cooling to 300 K at 1
atm
5.
a. A 3,5; B 1,2,4 ; C 1,2,4; D 1,2,4,5
b. A 1,2,4; B 1,2,4,5; C 3,4; D 1,2,4
c. A 3,5; B 1,2,4; C 1,2,4,5; D 1,2,4
d. A 1,2,4; B 3,5; C 1, 4,5; D 1,2,4
2H (g)
U 0 G 0
Space for rough work
119Mock Test-4
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. 0.45 g of an acid (mol wt. = 90) required 20 ml of 0.5 N
for complete neutralization. Basicity of acid is
2. The co-ordination number of copper in cuprammonium
sulphate is
3. The co-ordination number of cobalt in the complex
is
4. The primary valence of the metal ion in the co-ordination
compoun is
5. The oxidation number of Cr in is
6. The number of equivalent Cr O bonds in 24CrO is.
7. The number of the following reagents that produce ppt.
with 4ZnSO solution is.
2 3 2 4 2 3 3NaOH, N CO , NaCl, Na HPO , Na S, CH CO Na
8. The change in the magnetic moment value when
2
2 4Cu H O
is converted to 2
3 4Cu NH
is.
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. A gas described by van der Waal’s equation
a. behaves similar to an ideal gas in the limit of large
molar volumes
b. behaves similar to an ideal gas in the limit of large
pressures
c. is characterized by van der Waal’s coefficients that are
dependent on the identity of the gas but are independent of
the temperature
d. has the pressure that is lower than the pressure exerted
by the same gas behaving ideally
10. Assuming that Hund’s rule is violated, the bond order and
magnetic nature of the diatomic molecule is
a. 1 and diamagnetic b. 0 and diamagnetic
c. 1 and paramagnetic d. 0 and paramagnetic
11. The kinetic energy of an electron in the second Bohr orbit
of a hydrogen atom is [a0 is Bohr radius].
a. b.
c. d.
12. The equilibrium in aqueous
medium at 25ºC shifts towards the left in the presence of
a. b. c. d.
13. The bond energy (in kcal mol–1) of C – C single bond is
approximately
a. 1 b. 10 c. 100 d. 1000
14. Which is correct statement if
is added at equilibrium condition?
a. The equilibrium will shift to forward direction because
according to IInd law of thermodynamics the entropy must
increases in the direction of spontaneous reaction
b. The condition for equilibrium is
where G is Gibbs free energy per mole of the gaseous
species measured at that partial pressure. The condition of
equilibrium is unaffected by the use of catalyst, which
increases the rate of both the forward and backward
reactions to the same extent
c. The catalyst will increase the rate of forward reaction
by α and that of backward reaction by .
d. Catalyst will not alter the rate of either of the reaction
15. For a first order reaction AP, the temperature (T)
dependent rate constant (k) was found to follow the
equation log k The pre-exponential
factor A and the activation energy Ea, respectively, are
a. and
b. and
c. and
d. and
16. The initial rate of hydrolysis of methyl acetate (1 M) by a
weak acid (HA, 1M) is 1/100th of that of a strong acid
(HX, 1M), at 25ºC. The Ka of HA is
a. b.
c. d.
KOH
2 2 2[Co(en) Br ]Cl
2 4K [Ni CN ]
3 6 3[Cr(NH ) ]Cl
2B
2
2 20
h
4 ma
2
2 20
h
16 ma2
2 20
h
32 ma
2
2 20
h
64 ma
I II2Cu Cu Cu
3NO Cl SCN CN
2 2 3N 3H 2NH 2N
2 2 3N H NHG 3G 2G
1(2000) 6.0.
T
6 11.0 10 s 19.2 kJmol
16.0s 116.6 kJmol
6 11.0 10 s 116.6 kJmol
6 11.0 10 s 138.3kJmol
41 10 51 1061 10 31 10
Chemistry120
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem.
Two multiple choice questions will be asked based on this paragraph.
One or more than one option can be correct.
Paragraph for Question No. 17 to 18
Thermal decomposition of gaseous 2
X to gaseous X at 298 K
takes place according to the following equation:
2X (g) 2X(g)
The standard reaction Gibbs energy, rG∆ ° of this reaction is
positive. At the start of the reaction, there is one mole of 2
X
and no X. As the reaction proceeds, the number of moles of X
formed is given by .β Thus
equilibriumβ is the number of moles of
X formed at equilibrium. The reaction is carried out at a
constant total pressure of 2 bar. Consider the gases to behave
ideally. (Given: R = 0.083 L bar 1 1K mol )− −
17. The equilibrium constant pK for this reaction at 298 K, in
terms of equilibrium ,β is
a.
2
equilibrium
equilibrium
8
2
β
β− b.
2
equilibrium
2
equilibrium
8
4
β
β−
c.
2
equilibrium
equilibrium
4
2
β
β− d.
2
equilibrium
2
equilibrium
4
4
β
β−
18. The incorrect statement among the following, for this
reaction, is
a. Decrease in the total pressure will result in formation of
more moles of gaseous X
b. At the start of the reaction, dissociation of gaseous 2
X
takes place spontaneously
c. equlibrium 0.7β =
d. C
K 1<
Paragraph for Question No. 19 to 20
Tollen’s reagent is used for the detection of aldehyde when a
solution of is added to glucose with then
gluconic acid is formed
Gluconic acid
[Use 2.303 and at
19.
Find ln K of this reaction.
a. 66.13 b. 58.38
c. 28.30 d. 46.29
20. When ammonia is added to the solution, pH is raised to
11. Which half-cell reaction is affected by pH and by how
much?
a. will increase by a factor of 0.65 from
b. will decrease by a factor of 0.65 from
c. will increase by a factor of 0.65 from
d. will decrease by a factor of 0.65 from
3AgNO
4NH OH
Ag e Ag ;+ −+ → o
redE 0.8 V=
6 12 6 2C H O H O+ →
6 12 7(C H O ) 2H 2e ;+ −+ + o
oxdE 0.05 V=−
3 2 3Ag(NH ) e Ag(s) 2NH ;+ −+ → + o
oxdE 0.337 V=
RT0.0592
F× =
F38.92
RT= 298 K
6 12 6 2 6 12 72Ag C H O H O 2Ag (s) C H O 2H− ++ + → + +
oxdE
o
oxdE
oxdE
o
oxdE
redEo
redE
redE
o
redE
Space for rough work
121Mock Test-4
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a d b b a d a a d c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c a a a a b c c b c
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
d d b a b b b a b d
1. (a) Molecular mass of
2 2SO Cl 32 2 16 35.5 2 135= + × + × =
Moles 13.5
0.1135
= =
2. (d) Arrangement of sulphide ions 2
(S )−
in zinc blende
(ZnS) is fcc while 2Zn +
ions occupy alternate tetrahedral
voids.
3. (b) The temperature at which a liquid boils increases with
increase in pressure.
4. (b) Equal volumes of all gases under similar conditions of
temperature and pressure contain equal number of
molecules.
5. (a) K.E. = – (T.E.)
6. (d) Oxygen have oxidation state in
7. (a)
8. (a)
9. (d) Equilibrium constant changes with temperature,
pressure and the concentration of either reactant or
product.
10. (c)
Clearly, the reverse reaction is the hydrolysis reaction.
11. (c) The definition of activation energy.
12. (a) Scattering of light is a property of colloid.
13. (a) For exothermic reactions
For endothermic reactions
14. (a) Total no of protons
Total no of neutrons
Mass defect
Binding energy = mass defect 931
15. (a) is lewis acid i.e., electron deficient compound.
So it is electrophile.
16. (b) and are functional isomers.
17. (c)
18. (c)
19. (b) Benzoic acid.
20. (b)
21. (d)
22. (d)
23. (b) Average number molecular weight
Average mass molecular weight
Polydispersity index (PDI)
24. (a)
25. (b) “Cancer” is known as molecular disease.
2+ 2OF .
1/ a 0.5 cm , R 50 ohm−= =l
Ra 50p 100
0.5= = =
l
1 2 11000 1 1000 1 1000k 10 ohm cm gm eq
N p N 100 1
− −Λ = × = × = × =
4
aHA : K 10−=
HA NaOH NaA H O+ + 2HA NaOH NaA H O+ +
⇒4
10aRequired 14
h w
K1 10K 10
K K 10
−
−= = = =
p RH H .<
p RH H .>
40
18Ar 18=
22=
[m p m n] 39.962384= × + × −
[1.007825 18 1.008665 22] 39.962384= × + × −
[18.14085 22.19063] 39.962384= + − 0.369=
×
0.369 931 343.62MeV= × =
3AlCl
CyanideR — C N≡≡
IsocyanideR — N C=
r
3 2
3
CH — C CH|CH
==
3
2
O
H O, Zn→
3
3Acetone
CH C O|CH
==
Zn
2 2CH Br C C CH Br∆
− ≡ − →2 2
CH C C CH= = =
2Me C O HCN= + →3
3(X)
OH|
CH — C — CN|
CH
3H O+
→
3
3(Y)
OH|
CH — C — COOH|
CH
2 4H SO→ 2
3(Z)
CH CCOOH|
CH
==
* *
3 2 3 3 2 2 2CH — CH — COOH Na HCO CH CH COONa H O CO+ → + +
* *
3 2 3 3 2 2 2CH — CH — COOH Na H CO CH CH COONa H O CO+ → + +
6 5 2 6 5 6 5 6 5 2Schiff 's base
C H NH O CHC H C H N CHC H H O+ = → − = +
nM 30,000=
wM 40,000=
w
n
M 40,0001.33
30,000M= = =
Ni
2Oil(unsaturated) H Fat (saturated)+ →
N N
π σ
π
Chemistry122
26. (b) Hydrogen, forms hydrides like halides, e.g. HCl.
27. (b) 2 2 2 2Cu Cl Ag S Cu S 2AgCl
2 22AgCl Hg Hg Cl 2Ag
AgCl Hg Ag HgCl
28. (a) The solubility of hydroxides of alkaline earth metals in water increases on moving down the group.
29. (b) 24 3KI MnO K IO Mn
30. (d) In 4 6K Fe(CN) , the species retains its identity in solid
as well as in solution state.
JEE Advance Paper -I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
9 9 4 6 3 5 7 1 a c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
d a, d d c c. d b c b, c d a
1. (9)
2. (9) Let x be the percentage of Then, weight of
Moles of NaCl produced
The NaCl is produced by the reaction of mol of
and mol of Each mol of
produces 2 mol of NaCl.
Solve x :
3. (4) Aromatic alcohols and carboxylic acids form salt with NaOH, will dissolve in aqueous NaOH.
4. (6) These are total to carbon and they all can
participate in hyperconjugation.
5. (3)
6. (5)
7. (7)
8. (1)
9. (a)
10. (c) Initial moles = final moles
11. (d)
12. (a, d) In acidic medium
In neutral medium,
Hence, number of electrons loose in acidic and neutral medium are 5 and 3 respectively.
13. (d)
2
% by weight 10 30 10 1.0959 M
36.5
dM
Mw
2 3Na CO .
3NaHCO (15 )g x
11.0g0.18mol
58.5g
106
x
2 3Na CO(15 )
84
x3NaHCO .
2 3Na CO
2 15
0.188106 84
x x2 313.5g Na CO ,
3NaHCO (15 1.35) 13.6g
2 3 2 3
1.35% Na CO 100 9.0% Na CO
15
6 — H 2sp
L bT k m sb
s solution
Wk
M W
s
2.5 10002 0.76
M 100
s
0.76 2.5 1000M 9.5
100 2
ss
solution
n760 xX
760 n
s
s
solv
solv
W
W 2.5 1836.0
W 9.5 100M
x 760 36 724
i i f f
1 1 2 1
P V P V P V P V
RT RT RT RT
i i f f
1 1 2 1
P P P P
T T T T i
f1 2 1
2P 1 1P
T T T
i 1 2f
1 1 2
2P T TP
T T T
2
f i1 2
TP 2P
T T
K.E. eV
h
2meV h
2meV
24 2MnO 8H 5e Mn 4H O
4 2 2MnO 2H O 3e MnO 4OH
2 2o
cell cell 10 42
0.059 [M ][H ]E E log
2 [M ]pH
x10
0.0590.092 0.151 log 10
2
x 2
H
HH
Three structures Two structures
2 3CH CH3H C
OHCOOH COOH
3H C3CH
OH
N
123Mock Test-4
14. (c) O2 is expected to be diamagnetic in nature but actually
it is paramagnetic.
15. (c, d) In option (c), if HNO3 is present in limiting amount
then this mixture will be a buffer and the mixture given in
option (d), contains a weak acid (CH3COOH) and its salt
with strong base NaOH, i.e. CH3COONa.
16. (b)
Hence (b) is correct.
17. (c)
, Hence (c) is correct
18. (b, c)
19. (d) A → 3; B → 2; C →1; D →4
20. (a) A 3,5; B 1,2,4; C 1,2,4; D 1,2,4,4
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
2 4 6 2 3 4 4 0 a, c, d a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c b,c,d c b d a b c b a
1. (2)
2. (4) In Cuprammonium sulphate co-
ordination no. of is 4.
3. (6)
number of bidentate ligand
number of monodentate ligand
4. (2) Primary valencies are also known as oxidation state.
5. (3)
x – 3 = 0, x + 3, Oxidation number of Cr is = +3
6. (4)
7. (4)
8. (0)
9. (a, c, d)
At low pressure, when the sample occupies a large
volume, the molecules are so far apart for most of the time
that the intermolecular forces play no significant role, and
the gas behaves virtually perfectly.
a and b are characteristic of a gas and are independent of
temperature. The term represents the pressure
exerted by an ideal gas while P represents the pressure
exerted by a real gas.
10. (a)
Bond order (nature is diamagnetic as no
unpaired electron)
11. (c) and ⇒
⇒
⇒ . . .(i)
Expression for
⇒ . . .(ii)
⇒ . . .(iii)
For n = 2 K.E.
12. (c, d) Cu2+
ions will react with CN–
and SCN–
forming
and leading the reaction in
the backward direction.
G H T S 54.07 1000 298 10∆ ° = ∆ ° − ∆ ° = − × − ×1
1057050 J mol 57050 5705log K−= − − = −
10log K 10=
6 14C H
3 2 2 2 2 3H C — CH — CH — CH — CH — CH
3 2 2 3
3
H C — CH — CH — CH — CH|CH
3 2 3
3
H C — CH — CH — CH — CH|
CH
3 3
33
H C — CH — CH — CH||CHCH
3
3 2 3
3
CH|
H C — C — CH — CH|CH
→ → → →
BW 1000
Normality = N =Eq.wt V
×
×
∴0.45 1000
Eq. Wt 450.5 20
×= =
×
∴Molec. Wt 90
Basicity 2Eq. Wt 45
= = =
3 4 4[Cu(NH ) ]SO
Cu
2 2 2[Co(en) Br ]Cl
C.N.of Co 2= ×
1+ × 2 2 1 2 6.= × + × =
2 4K [ Ni(CN) ], 2 x 4 0 x 2+ − = ⇒ = +
x 6 (0) 3 ( 1) 0+ × + × − =
2
2
n aP (V nb) nRT
V
+ − =
2
2
n aP
V
+
2 2 2 2 2x
* *
2 1s 1s 2s 2s 2pB (10) σ σ σ σ π=
6 41
2
−= =
nhmvr
2π=
2 2
2
mv e
r r= 2 2mv r e=
2e 2v
nh
π×= ∴
nh(mvr )
2π=
4 2
2
2 2
1me 4
1 2K.E. mv2 n h
π×= =
2
0 2 2
ha
4 meπ=
22
2
0
hme
4 aπ=
2
2 2 2
0
h 1K.E.
8ma nπ= ×
2
2 2
0
h
32 maπ=
3
4[Cu(CN) ] − 3
4[Cu(SCN) ] −
Chemistry124
also combines with which reacts with Cu to
produce CuCl pushing the reaction in the backward direction.
13. (c)
14. (b)
15. (d) Given, log
Since, log
So,
and Ea
16. (a) Rate in weak acid (rate in strong acid)
For Question Nos. 17 to 18
17. (b) Paragraph-1 2X (g) 2X(g)
Initial mole 1 0
eq.t t (1 ) 2
Given equilibrium2 So, equilibrium
2
Total mole at equilibrium eq(1 ) (12
eq.
eq eq2 total total total
eq. eq eq
1 2 22Px P P P2 2
12
eq eq
x(g) total totaleq eq
2P P P
21
2
So
2
eq.total2
eq.
p2 eq.
totaleq
2P
2(Px)K
(Px ) 2P
(2 )
2 2eq. eq
p total2 2eq. eq
4 8K P
4 4
18. (c) (a) Correct statement. As one decrease in pressure reaction will move in the direction where no. of gaseous molecules increases.
(b) Correct statement
At the start of reaction p pQ K so dissociation of 2X take
place spontaneously. (c) Incorrect statement as
2 2eq
p 2 2eq
8 8 (0.7)K
4 4 (0.7)
1, but
(d) Correct statement.
As pG 0 & G RT n K
pG 1, so K should be less than 1.
So K 1
ng.p cK K (RT) (RT 1)
p
c
KK
RT c pK K
So cK 1
For Question Nos. 19 to 20
19. (b)
20. (a) On increasing concentration of NH3 the concentration of H+
ion decreases. Therefore, Ered increases.
22Cu 2CN Cu(CN)
2 22Cu(CN) 2CuCN (CN) 3
4CuCN 3CN [Cu(CN) ] 2 3
4Cu 4SCN [Cu(SCN) ] 2Cu
2CuCl
2CuCl Cu 2CuCl
2000K 6
T
K log EaA
2.303RT
6 1A 10 sec38.3kJ / mole
1
100
weak acid strong acid
1[H ] [H ]
100
2weak acid
1[H ] M 10 M
100
2C 10
4aK 10
ocell
RTE ln K
nF
1 0.0592(0.8 0.05) ln K
2 2.303
(0.8 0.05) 22.303ln k 58.38
0.0592
125Mock Test-5
JEE-MAIN: CHEMISTRY MOCK TEST-5
1. One mole of potassium dichromate completely oxidises the following number of moles of ferrous sulphate in acidic medium
a. 1 b. 3 c. 5 d. 6
2. In a solid lattice, the cation has left a lattice site and is located at an interstitial position, the lattice defect is
a. Frenkel defect b. Schottky defect c. F-centre defect d. Valency defect
3. At 300 K, the vapour pressure of an ideal solution containing 3 mole of A and 2 mole of B is 600 torr. At the same temperature, if 1.5 mole of A and 0.5 mole of C (non-volatile) are added to this solution the vapour pressure of
solution increases by 30 torr. What is the value of
a. 940 b. 405 c. 90 d. None of these
4. If the root mean square speed of helium is 4.75 m s–1 at
25 C, then its speed will become 9.50 m s–1 at
a. 100 C b. 323 C c. 919 C d. 1192 C
5. Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon?
a. 3s b. 2p
c. 2s d. 1s
6. The product of oxidation of I with 4M nO in alkaline
medium is
a. 3IO b. 2I c. IO d. 4IO
7. The standard reduction electrode potentials of four
elements are
and The element that displaces A from its
compounds is a. B b. C c. D d. None of these
8. The bond angle in carbon tetrachloride is approximately
a.
b.
c.
d.
9. The formation of nitric oxide by contact process
43.200 kcal is favoured by
a. Low temperature and low pressure b. Low temperature and high pressure c. High temperature and high pressure d. High temperature and excess reactants concentration
10. The pH of an aqueous solution is 9.0. If the
solubility product of is what is ?
a. b.
c. d. 0.1
11. If ‘I’ is the intensity of absorbed light and C is the concentration of AB for the photochemical process
the rate of formation of is directly
proportional to a. C b. I
c. d. C.I
12. Gold number is maximum for the lyophilic sol is a. Gelatin b. Haemoglobin c. Sodium oleate d. Potato starch
13. For the allotropic change represented by equation
the enthalpy change is
If 6 g of diamond and 6 g of graphite are
separately burnt to yield carbon dioxide, the heat liberated in the first case is
a. Less than in the second case by
b. More than in the second case by
c. Less than in the second case by
d. More than in the second case by
14. The half-life of if its K or is is
a. b.
c. d.
15. Most stable carbonium ion is
a. b.
c. d.
16. Which of the following compounds will exhibit cis-trans isomerism
a. 2-butene b. 2-butyne c. 2-butanol d. Butanone
oBP ?
A 0.250 V, B 0.136 V, C 0.126 V
D 0.402 V.
90
109
120
180
2 2N O 2NO. H
2Mg OH
2Mg OH 111 10 , 2[ ]Mg
51 10 41.0 1021 10
AB hv AB*, AB*
2I
C(diamond) C(graphite);
H 1.89 kJ.
1.89kJ
1.89kJ
11.34 kJ
0.945 kJ
146C 42.31 10
22 10 yrs 33 10 yrs43.5 10 yrs 34 10 yrs
2 5C H
3 3(CH ) C
6 5 3(C H ) C
6 5 2C H CH
Chemistry 126
17. Acetylene gas is obtained by the electrolysis of a. Sodium fumarate b. Sodium succinate c. Sodium maleate d. Both (a) and (c) 18. The compound added to prevent chloroform to form
phosgene gas is
a.
b.
c.
d.
19. Epoxides are a. Cyclic ethers b. Not ethers c. Aryl-alkyl ethers d. Ethers with another functional group
20. reacts with
a. b.
c. d.
21. In the reaction
The compound X is a. Phthalic anhydride b. Phthalic acid c. o-xylene d. Benzoic acid
22. In the reaction
; C is
a. Pentanal b. Pentanone c. 2-Hexanone d. Hexanal
23. A polymer containing nitrogen is a. Bakelite b. Dacron c. Rubber d. Nylon-66
24. The base present in DNA, but not in RNA is a. Guanine b. Adenine c. Uracil d. Thymine
25. Which of the following is an antidiabatic drug a. Insulin b. Penicillin c. Chloroquine d. Aspirin
26. The correct sequence of elements in decreasing order of first ionisation energy is
a. b.
c. d.
27. In the metallurgical extraction of zinc from ZnO the reducing agent used is
a. Carbon monoxide b. Sulphur dioxide c. Carbon dioxide d. Nitric oxide
28. In the lime (kiln), the reaction
goes to completion because a. Of high temperature
b. CaO is more stable than
c. escapes simultaneously
d. CaO is not dissociated
29. Amalgams are a. Highly coloured alloys b. Always solid c. Alloys which contain mercury as one of the contents d. Alloys which have great resistance to abrasion
30. Generally, a group of atoms can function as a ligand if a. They are positively charged ions b. They are free radicals c. They are either neutral molecules or negatively charged ions d. None of these
2 5C H OH
3CH COOH
3 3CH COCH
3CH OH
NaOH / H
6 5 3C H OCH 3CH OH
3 3
O||
CH — C — CH 2 5C H OH
3NH8 6 4C H O X
Zn, HCl3 2 4CH (CH ) CN A
HONO OA B C
Na Mg Al Mg Na Al
Al Mg Na Mg Al Na
3 2CaCO (s) CO (g)
3CaCO
2CO
Space for rough work
127Mock Test-5
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions
The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
1. The change in the number of unpaired electrons when
2
2 6Fe H O
is changed into 4
6Fe CN
is
2. A compound of mol. wt. 180 is acetylated to give a
compound of mol. wt. 390. The number of amino groups
in the initial compound is?
3. Specific rotations of -anomer of glucose is 112 and
for -anomer is +19 . Specific rotation of equilibrium
mixture is 52.6 . Calculate % composition of -and -
anomers in the equilibrium mixture.
4. Consider all possible isomeric ketones including
stereoisomers of MW = 100. All these isomers are
independently reacted with 4NaBH (Note: Stereoisomers
are also reacted separately). The total number of ketones
that give a racemic product(s) is/are
5. Amongst the following, the total number of compounds
soluble in aqueous NaOH is
6. The half-life period of a radioactive substance is 2 min.
The time taken for 1 g of the substance to reduce to 0.25 g
will be
7. of hypothetical MgCl is and for
is The enthalpy of
disproportionation of MgCl is –49x. Find the value of x.
8. Hydrolysis of an alkyl halide (RX) by dilute alkali
takes place simultaneously by and pathways. A
plot of vs. is a straight line of
the slope equal to and intercept equal to
Calculate the initial rate (mol of
consumption of RX when the reaction is carried out taking
of RX and 0.1 mol of ions.
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. The ionic radii (in Å) of and are
respectively:
a. 1.36, 1.40 and 1.71
b. 1.36, 1.71 and 1.40
c. 1.71, 1.40 and 1.36
d. 1.71, 1.36 and 1.40
10. From the following statements regarding choose
the incorrect statement
a. It can act only as an oxidizing agent
b. It decomposes on exposure to light
c. It has to be stored in plastic or wax lined glass bottles in
dark
d. It has to be kept away from dust
11. was added to an aqueous KCl solution
gradually and the conductivity of the solution was
measured. The plot of conductance versus the volume
of is
a.
b.
c.
d.
f H 1125kJ mol
2MgCl 1642 kJ mol .
[OH ]
2SN 1SN
1 d[R X]
[RX] dt
[OH ]
3 1 12 10 mol L h 2 11 10 h . 1 1L min )
11mol L 1L [OH ]
3 2N , O F
2 2H O ,
3AgNO (aq.)
(^)
3AgNO
volume
(d)
++
++
+ + +
volume
(c)
++ ++ ++++
volume
(b)
+
+ + +
++
++
+ ++
+
volume
(a)
OCH2CH3
CH2OH3
OH N
H3C CH3
NO2 OH
N
H3C CH3
CH2CH3 CH2OH3 COOH
COOH
Chemistry 128
12. The intermolecular interaction that is dependent on the
inverse cube of distance between the molecules is:
a. ion-ion interaction b. ion-dipole interaction
c. London force d. hydrogen bond
13. According to Molecular Orbital Theory
a. 22C is expected to be diamagnetic
b. 22O is expected to have a longer bond length than 2O
c. 2N and 2N have the same bond order
d. 2He has the same energy as two isolated He atoms
14. For the first order reaction
a. the concentration of the reactant decreases exponent-
tially with time
b. the half-life of the reaction decreases with increasing
temperature
c. the half-life of the reaction depends on the initial
concentration of the reactant
d. the reaction proceeds to 99.6% completion in eight
half-life duration
15. In the reaction, the time taken for 75%
reaction of P is twice the time taken for 50% reaction of P.
The concentration of Q varies with reaction time as shown
in the figure. The overall order of the reaction is
a. 2 b. 3 c. 0 d. 1
16. Choose the correct reason(s) for the stability of the
lyophobic colloidal particles.
a. Preferential adsorption of ions on their surface from the
solution
b. Preferential adsorption of solvent on their surface from
the solution
c. Attraction between different particles having opposite
charges on their surface
d. Potential difference between the fixed layer and the
diffused layer of opposite charges around the colloidal
particles
17. The standard enthalpies of formation of
and glucose(s) at are –400 kJ/mol, –300 kJ/mol and
–1300 kJ/mol, respectively. The standard enthalpy of
combustion per gram of glucose at is
a. + 2900 kJ b. – 2900 kJ c. – 16.11 kJ d. +16.11 kJ
18. Which of the following compounds is not colored yellow?
a.
b.
c.
d.
SECTION 3 Contains 2 Match The Following Type Questions
You will have to match entries in Column I with the entries in
Column II.
19. Match each of the diatomic molecules in Column I with its property/properties in Column II.
Column - I Column – II
(A) 1. Paramagnetic
(B) 2. Undergoes oxidation
(C) 3. Undergoes reduction
(D) 4. Bond order ≥ 2
5. Mixing of ‘s’ and ‘p’ orbitals
a. A 5, 3, 1, B 5, 4, C 2, 1, D 4, 1, 2
b. A 1, 3, 5, B 1, 2, 4 C 3, 2, D 1, 2,
c. A 1, 3, B 1, 2, 4 C 1, 2, 3 D 4, 5,
d. A1, 3, 5, B4, 5, C1, 2, D1, 2, 4
[According to MOT]
20. Match the compounds/ions in Column I with their
properties/reactions in Column II.
Column I Column II
(A) 1. gives precipitate with 2, 4-dinitrophenyl-
hydrazine
(B) 2. gives precipitate with
(C) 3. is a nucleophile
(D) 4. is involved in cyano-hydrin formation
a. A 1, 2, 3, B 2, C 1, 3, 4, D 2, 3
b. A 1, 3, 4, B 2, 3 C 1, 2, 3, D 2
c. A 1, 3, B 1, 4, C 1, 3, D 2, 3
d. A 2, 3, B 1, 2, C 1, 3, D 1, 2
2 5 2 22N O (g) 4NO (g) O (g)
P Q R S
2CO (g), H2O( )l
25 C
25 C
2 6Zn [Fe(CN) ]
3 2 6K [Co(NO ) ]
4 3 3 10 4(NH ) [As(Mo O ) ]
4BaCrO
2B
2N
2O
2O
6 5C H CHO
3CH C CH3AgNO
CN
I
Time
[Q]
0[Q]
129Mock Test-5
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. The number of rings formed in is
……………
2. The number of equivalent bonds in is
……………
3. In fructose, the possible optical isomers are……………
4. Oxidation of glucose is one of the most important
reactions in a living cell. What is the number of ATP
molecules generated in cells from one molecule of
glucose?
5. 10 gm of a mixture of hexane and ethanol reacts with
sodium to give 200 ml. of 2H at 27 C and 760 mm
pressure. The percentage of ethanol is……………
6. The polymerisation of propene to linear polypene is
represented by the reaction
n
Where n has large integral value, the average enthalpies of
bond dissociation for and at 298 K
are +590 and respectively. The enthalpy
of polymerization is Find the value of n.
7. In the case of a first order reaction, the time required for
93.75% of reaction to take place is x times that required
for half of the reaction. Find the value of x.
8. In 1 L saturated solution of spAgCl (K of AgCl
0.1mol of CuCl (Ksp is
added. The resultant concentration of in the solution
is Calculate the value of x.
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. Which of the following atoms has the highest first
ionization energy?
a. Rb b. Na
c. K d. Sc
10. Among the electrolytes and
the most effective coagulating agent for
sol is
a. b. c. d.
11. The compound(s) with TWO lone pairs of electrons on the
central atom is(are)
a. 5BrF b. 3ClF
c. 4XeF d. 4SF
12. Solubility product constant of salts of types
and at temperature ‘T’ are
and respectively.
Solubilities (mole of the salts at temperature ‘T’
are in the order
a. b.
c. d.
13. According to the Arrhenius equation
a. a high activation energy usually implies a fast reaction b. rate constant increases with increase in temperature.
This is due to a greater number of collisions whose energy exceeds the activation energy
c. higher the magnitude of activation energy, stronger is
the temperature dependence of the rate constant d. the pre-exponential factor is a measure of the rate at
which collisions occur, irrespective of their energy
14. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is
correct?
a. Both k and 1/n appear in the slope term b. 1/n appears as the intercept
c. Only 1/n appears as the slope d. log (1/n) appears as the intercept.
15. When is adsorbed on a metallic surface, electron
transfer occurs from the metal to The TRUE
statement(s) regarding this adsorption is(are)
a. is physisorbed
b. heat is released
c. occupancy of of is increased
d. bond length of is increased
2Ca EDTA
Cl O 2 7Cl O
3
2
CH|CH CH
3
2 n
CH|
—CH — CH—
(C C) (C C)1331kJ mol ,
1360 kJ mol .
101.6 10 ), 6CuCl 1.0 10 )
Ag
x1.6 10 .
2 4Na SO , 2CaCl , 2 4 3Al (SO )
4NH Cl, 2 3Sb S
2 4Na SO 2CaCl 2 4 3Al (SO ) 4NH Cl
sp(K )
2MX, MX 3M X
8 144.0 10 , 3.2 10 152.7 10 ,3dm )
2 3MX MX M X 3 2M X MX MX
2 3MX M X MX 3 2MX M X MX
2O
2O .
2O
*2p 2O
2O
Chemistry 130
16. Methylene blue, from its aqueous solution, is adsorbed on
activated charcoal at For this process, the correct
statement is
a. The adsorption requires activation at
b. The adsorption is accompanied by a decrease in
enthalpy
c. The adsorption increases with increase of temperature
d. The adsorption is irreversible
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem.
Two multiple choice questions will be asked based on this paragraph.
One or more than one option can be correct.
Paragraph for Question Nos. 17 to 18
Carbon–14 is used to determine the age of organic material.
The procedure is based on the formation of by neutron
capture in the upper atmosphere
is absorbed by living organisms during photosynthesis.
The content is constant in living organism once the plant or
animal dies, the uptake of carbon dioxide by it ceases and the
level of in the dead being, falls due to the decay which
undergoes
The half life period of is 5770 years. The decay constant
can be calculated by using the following formula
The comparison of the activity of the dead
matter with that of the carbon still in circulation enables
measurement of the period of the isolation of the material from
the living cycle. The method however, ceases to be accurate
over periods longer than 30,000 years. The proportion of
to in living matter is
17. Which of the following option is correct?
a. In living organisms, circulation of from atmosphere
is high so the carbon content is constant in organism
b. Carbon dating can be used to find out the age of earth
crust and rocks
c. Radioactive absorption due to cosmic radiation is equal to
the rate of radioactive decay, hence the carbon content
remains constant in living organism
d. Carbon dating cannot be used to determine concen-
tration of in dead beings
18. What should be the age of fossil for meaningful
determination of its age?
a. 6 years
b. 6000 years
c. 60,000 years
d. It can be used to calculate any age
Paragraph for Question Nos. 19 to 20
Rocket propellants consist of rocket engines powered by
propellants. These are used both in space vehicles as well as in
offensive weapons such as missiles. The propellants are
chemical substances which on ignition provide thrust for the
rocket to move forward. These substances are called rocket
propellants. A propellant is a combination of an oxidiser and a
fuel which when ignited undergoes combustion to release large
quantities of hot gases. The passage of hot gases through the
nozzle of the rocket motor provides the necessary thrust for the
rocket to move forward according to Newton's third law of
motion.
19. A biliquid propellant contains
a. Liquid hydrazine
b. A mixture of liquid fuel and a liquid oxidiser
c. A solid rocket fuel
d. A liquid fuel which can also act as an oxidiser
20. A hybrid rocket propellant uses
a. A liquid oxidiser and a solid fuel
b. A composite solid propellant
c. A biliquid propellant
d. A solid, liquid and gas as a propellant
25 C.°
25 C°
14 C
14 1 14 1
7 0 6 1N n C n+ → +14 C
14 C
14 C
14C
14 14
6 7C N β −→ +14 C
( )λ
1/ 2
0.693.
tλ = -β
14 C
12 C121 : 10 .
14 C
14 C
Space for rough work
131Mock Test-5
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d a c c d a c b d d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b d d b c a d a a c
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
a d d d a d a d c c
1. (d) 3
2 7Cr O Cr
6n
−− +→=
; Fe Fe1n
++ +++→=
eq. of 2 2 7
K Cr O = eq. of 4
FeSO 1 6 x 1× = ×
2. (a) When an ion (generally cation due to its small size) is
missing from its normal position and occpy an interstitial
site between the lattice points, the lattice defect obtained
is known as Frenkel defect.
3. (c)
and
4. (c) r.m.s. speed T∝
4.75 298
9.50 T= or
24.75 298
9.50 T
=
or T 1192 K= or T 1192 273 919 C= − = °
5. (d) When electron jumps to lower orbit photons are
emitted while photons are absorbed when electron jumps
to higher orbit, 1s orbital is the lower most, electron in
this orbital can absorb photons but cannot emit.
6. (a)
7. (c) A is displace from D because D have a
8. (b) is hybridised so bond angle will be
approximately
9. (d) High temperature and excess concentration of the
reactant concentration.
10. (d)
11. (b) In photochemical reaction the rate of formation of
product is directly proportional to the intensity of
absorbed light.
12. (d) Gold number shows the protective power of a
lyophilic solution. Lesser the gold number, greater will be
the protecting power of that colloid. Gelatin is one of the
best protective colloid. Among the given colloids, potato
starch has maximum gold number.
13. (d)
or
For combustion of by
14. (b)
15. (c) In the triphenyl methyl carbonium ion the electrons
of all the three benzene rings are delocalised with the
vacant p-orbital of central carbon atom. So, it is resonance
stabilised. It is the most stable of all the carbonium ions
given
The ion is stabilised by hyperconjugation, a
second order resonance.
16. (a)
Cis-trans isomerism shown by compound which have double
or triple bond by which they restrict their rotation, since 2
butyne have no hydrogen on triple bonded carbon.
[It does not show cis-trans]
17. (d)
Sodium salt of maleic acid or fumaric acid
18. (a) Because it float over chloroform and prevent its
oxidation.
19. (a) Cyclic ethers are called epoxies.
o o
A A B BP P x P x= +
o o o o
A B A B
3 2600 P P ;3P 2P 3000
3 2 2 3
= + + = + +
o o o o
A B A B
4.5 2630 P P ;4.5P 2P 4410
4.5 2 0.5 4.5 2 0.5
= + + = + + + + o o
A A1.5P 1410;P 940= = o
BP 90=
2
4 4 3 26MnO I 6OH 6MnO IO 3H O− − − − −+ + → + +
E 0.402 V° = −
4CCl 3sp
109 .°
pH 9= ∴ 9[H ] 10+ −=14
5
9
1 10[OH ] 10
10
−− −
−
×= =
2Mg(OH) Mg 2OH2Mg(OH) Mg 2OH+ −+
2 2
spK [Mg ][OH ]+ −= ⇒ 11 2 5 21 10 [Mg ][10 ]− + −× =
⇒11
2 1
5 2
1 10[Mg ] 10 0.1
(10 )
−+ −
−
×= = =
(graphite) (diamond)C C , H 1.9kJ→ ∆ =
(graphite) 2 2 1C O CO , H H+ → ∆ = −∆
(diamond) 2 2 2C O CO , H H+ → ∆ = −∆
1 2( H ) ( H ) 1.9 kJ− ∆ − − ∆ =
2 1H H 1.9∆ = ∆ +
2 16g, H H∆ > ∆ 1.9 / 2 0.95kJ.=
4
1/ 2 4
0.693 0.693t 0.3 10 yrs
k 2.31 10−= = = ×
×33.0 10 yrs.= ×
π
3
3
3
CH|
CH — C|CH
+
3 32 butyne
CH — C C — CH≡≡
CH — COONa||CH — COONa
→CHCOO||CHCOO
−
−2Na++
CHCOO||CHCOO
−
−→
Acetylene
CH|||CH
22CO 2e−+ +
2 2CH — CH
O
3CH
HC C==
3CH
HCis 2 butene
3CH
HC C==
3
H
CHTrans 2 butene
Chemistry 132
20. (c)
21. (a)
22. (d)
23. (d) Nylon-66- It is a polymer containing nitrogen
24. (d) Thymine is present in DNA while in RNA there is
Uracil. 25. (a) Insulin is an antidiabatic drug. 26. (d) This is due to the presence of fully
filled s-orbital in Mg. 27. (a)
2ZnO CO CO Zn
28. (d) Lime stone – CaCO3, Clay – silica and alumina Gypsum – CaSO4
29. (c) Amalgams are alloys which contain mercury as one of the contents.
30. (c) Ligand must have capacity to donate lone pair of electrons to form coordinate bond.
JEE Advance Paper -I 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
4 5 2 5 4 2 8 5 c a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
d b a, c c b,c,d b c b b a
1. (4)
2. (5) Difference in mass of compound = 390 – 180 = 210
wt. of – group is = 43
Therefore no. of group = .
3. The -anomer of D-glucose has a specific rotation of +112
degrees in water. The β-anomer of D-glucose has a specific rotation of + 19 degrees (18.7 actually, but rounding up to 19).
If the fraction of glucose present as the -anomer is xand
the fraction present as the β-anomer is y, and the rotation of the mixture is +52.6 degree, we have X (+112.2 degree) + y(18.7) = 52.6 degree ……..(i) There is a very little of the open chain form present, so the fraction present as the α-anomer (x) plus the fraction present as the β-anomer (y) should account for all the glucose, i.e., x + y = 1 or y = 1 – x. Putting value of y in equation (i), we get x (+112.2 degree) + (1 – x)(18.7) = 52.6 degree ……..(ii) Solving equation (ii) for (x), we have, x = 0.36 or 36%. Thus (y) must be (1 – 0.36) = 0.64 or 64%.
So, percentage composition of - and -anomers in the
equilibrium mixture is 36% and 64% respectively.
4. (5)
5. (4)
6. (2)
7. (8)
;
;
8. (5) (by pathway)
rate constant of SN2 reaction
This is the equation of a straight line for
vs plot with slope equal to and intercept equal
to
dil NaOH3 32CH CO CH
3 2 3(Diacetone alchol)
3
OH O| ||
CH — C — CH — C — CH|CH
Zn , HCl3 2 4CH (CH ) CN
HONO3 2 4 2 2CH (CH ) CH NH
O3 2 4 2 3 2 4
Hexanal
CH (CH ) CH OH CH (CH ) CHO
2 6 2 4
nNylon-66
H H O| |||
—N — (CH ) — N — C — (CH ) — C —||O
Mg Al Na.
4 2.2H O
3CH CO
2NH210
4.88 543
1/ 2 1/ 2t t1g 0 5g 0 25g 25%t 4min
22MgCl Mg MgCl H ?
2Mg(s) 1/ 2Cl (g) MgCl 11H 125kJ mol
2 2Mg(s) Cl (g) MgCl 12H 642 kJ mol
12 1H H 2 H 642 (2 125) 392 kJ mol
49 392 x 8x
2
d[RX]k [RX][OH ]
dt
2SN
2k
2 1
d[RX]k [RX][OH ] k [RX]
dt
2 1
1 d[RX]k [OH ] k
[RX] dt
1 d[RX]
[RX] dt
[OH ]2k
1k .
N
H3C CH3
OH OH
N
H3C CH3
COOH
2CONH
COOH3NH
COOH
COOHC||O
O||C
133Mock Test-5
From question:
and
Hence,
9. (c) Ionic Radii order:
10. (a) It can act as an oxidising as well as reducing agent.
11. (d)
12. (b) Ion-dipole interaction
13. (a,c) (a) 22C Total no. of electrons = 14 so it is
diamagnetic
(b) 22O Bond order = 3; 2O Bond order = 2
Bond length in 22O is less than bond length in 2O .
(c) Bond order of 2N 2.5 ; Bond order of 2He 1/ 2
Some energy is released during the formation of 2He from
two isolated He atoms.
14. (a, b, d) For first order reaction
Hence concentration of decreases exponentially,
Also, , which is independent of concentration
and decreases with the increase of temperature.
15. (d) Overall order of reaction can be decided by the data
given
It is a first order reaction with respect to P.
From graph [Q] is linearly decreasing with time, i.e., order
of reaction with respect to Q is zero and the rate
expression is
16. (a, d) (a) Preferential adsorption of ions on surface from
the solution
(c) Attraction between particles having same charges on
their surface accounts for the Brownian motion.
(d) Definition of Zeta Potential.
17. (c) Combustion of glucose
18. (a) is bluish white ppt.
19. (d) A 1, 3, 5, B 4, 5, C 1, 2, D 1, 2, 4
[According to MOT]
20. (a) A 1, 2, 3, B 2, C 1, 3, 4, D 2, 3
(Note: Assuming is ammonical)
(a)
(b)
(c) ;
(d)
JEE Advance Paper –II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
5 6 8 38 10 5 4 7 d c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b, c d b,c,d b,c,d b b c b 2 1
1. (5) 2. (6)
3. (8) Fructose has three chiral centres and hence
optical isomers are possible.
4. (38)
5. (10) 22400 ml of 2H is produced by 46 gms of ethanol.
200 ml of 2H is produced by 46 200 23
22400 56
3 1 1 2 12 1k 2 10 mol L hr , k 1 10 hr
[RX] 1.0M [OH ] 0.1M
3 2d[RX]2 10 1 0.1 1 10 1
dt
1 1300 mol L hr 1 15mol L min 3 2N O F
kt0[A] [A] e
2[NO ]
1/ 2
0.693t
K
1/ 2t
99.6
2.303 100t log
K 0.4
99.6 1/ 2
2.303 0.693t (2.4) 8 8 t
K K
75% 50%t 2t
1 0r k[P] [Q] .
6 12 6 2 2 2C H O 6O 6CO 6H O
combustion f 2 f 2 f 6 12 6H (6 H CO 6 H H O) H C H O
(6 400 6 300) ( 1300)
2900 kJ / mol
2900 /180 kJ / g
16.11 kJ / g
2 6Zn [Fe(CN) ]
3AgNO
3NH2
(whiteppt)
PhCHO Ag O PhCOO Ag
KCNPhCHO
CN|
Ph — C — O|H
3ammonical AgNO3 3
(White ppt)
CH C CH CH C C Ag
KCNPhCHO CN|
Ph — C — O|H
3AgNO CN AgCN
3AgNO I AgI
32 8
6 12 6 2 2 2C H O 6O 6CO 6H O 38ATP
2NH NH 2PhCHO O N
NO2
(ppt.)PhHC N NH
O2N
NO2
Volume
++
++
+ + +
Chemistry 134
Percentage of 2 5
23C H OH 100 9.6% 10%
56 10
6. (5) Energy released = Energy due to formation of two
single bonds of propene
polymerisation/mol
polymerisation
7. (4)
Also,
8. (7)
;
The value of x = 7
9. (d) I. P1 = Sc > Na > K > Rb
10. (c) As is a negative sol, so, will be the
most effective coagulant due to higher charge density on
in accordance with Hardy-Schulze rule. Order of effectiveness of cations:
11. (b, c)
One lone pair on central atom (Br) Two lone pair
on central atom (Cl)
Square planar
Two lone pair on central atom (Xe) One lone pair
on central atom (S)
Only, 3 4ClF & XeF contains two lone pair of electrons on
central atom.
12. (d) Solubility of
Solubility of
Solubility of
13. (b, c, d) Ea / RTK A e
2
dk EaK
dT RT
dkEa
dT
A Frequency factor
= No of collisions per unit time per unit volume.
14. (c) According to the Freundlich adsorption isotherm
15. (b, c, d) Adsorption of on metal surface is exothermic.
During electron transfer from metal to electron
occupies orbital of Due to electron transfer to the bond order of
decreases hence bond length increases.
16. (b) Adsorption of methylene blue on activated charcoal is physical adsorption hence it is characterised by decrease in enthalpy.
17. (c) Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organism
18. (b) 6000 years
19. (b) Biliquid Propellant – A double base propellant is a high strength, high modulus gel of cellulose nitrate (gun cotton) in glyceryl trinitrate or a similar solvent.
20. (a) Hybrid Propellant – A hybrid propellant consists of a solid fuel and liquid oxidiser to provide propulsion energy and working substance e.g., Solid acrylic rubber and
liquid .
12 331 662 kJ mol
H 1590 662 72 kJ mol
H 72 n 360 n 5
1/ 21
0.0693t
k
93.751
2.303 100t log
k 100 93.75
1
2.303 100log
k 6.25
4
1
2.303log 2
k 1/ 2
1 1
4 2.303 log 2 4 0.6934t
k k
10sp
(S x)SAgCl Ag Cl (K 1.6 10 )
6sp
x (x S )CuCl Cu Cl (K 1 10 )
10S (S x) 1.6 10 6x(x S ) 10
4S1.6 10
x
4 4 6S (1.6 10 )x. x(x 1.6 10 x) 10
3x 10
7 xS 1.6 10 1.6 10
2 3Sb S 2 4 3Al (SO )
3Al
34Al Ca Na NH
8 4(MX) 4 10 2 10 5
2(MX ) 8 10 4
3(M X) 1 10
3 2MX M X MX .
1/ nxkP
m
x 1log log K log P
m n
2O
2O*2p 2O .
2O 2O
2 4N O
Xe
F
FF
FS
FF
(See-Saw Shape)
FF
Cl F
F
F
(T-Shape)
F
Br
F
F
F
F
(Square Shape)
135Mock Test-1
Mock Test “JEE-Main”
Do not open this Test Booklet until you are asked to do so.
Read carefully the Instructions on the Back Cover of this Test Booklet.
Important Instructions:
1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly
prohibited.
2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer
Sheet and fill in the particulars carefully.
3. The test is of 3 hours duration.
4. The Test Booklet consists of 90 questions. The maximum marks are 360.
5. There are three parts in the question paper A, B, C consisting of, Physics, Chemistry and Mathematics having 30
questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.
6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. 1/4 (one
fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will
be made if no response is indicated for an item in the answer sheet.
7. There is only one correct response for each question. Filling up more than one response in each question will be treated as
wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.
8. Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet.
Use of pencil is strictly prohibited.
9. No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any
electronic device, etc., except the Admit Card inside the examination hall/room.
10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the
bottom of each page and at the end of the booklet.
11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room/Hall.
However, the candidates are allowed to take away this Test Booklet with them.
12. The CODE for this Booklet is A. Make sure that the CODE printed on Side-2 of the Answer Sheet and also tally the serial
number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate
should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet.
13. Do not fold or make any stray marks on the Answer Sheet.
Name of the Candidate (in Capital letters):
Roll Number: in figures
in words
Examination Centre Number:
Name of Examination Centre (in Capital letters):
Candidate’s Signature: Invigilator’s Signature:
A Test Booklet code
Mathematics136
Read the Following Instructions Carefully:
1. The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-1) with
Blue/Black Ball Point Pen.
2. For writing/marking particulars on Side-2 of the Answer Sheet, use Blue/Black Ball Point Pen only.
3. The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test
Booklet/Answer Sheet.
4. Out of the four options given for each question, only one option is the correct answer.
5. For each incorrect response, one-fourth (¼) of the total marks allotted to the question would be deducted from
the total score. No deduction from the total score, however, will be made if no response is indicated for an item
in the Answer Sheet.
6. Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in
Test Booklet Code and Answer Sheet Code), another set will be provided.
7. The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All
calculations/writing work are to be done in the space provided for this purpose in the Test Booklet itself,
marked ‘Space for Rough Work’. This space is given at the bottom of each page and at the end of the booklet.
8. On completion of the test, the candidates must hand over the Answer Sheet to the Invigilator on duty in the
Room/Hall. However, the candidates are allowed to take away this Test Booklet with them.
9. Each candidate must show on demand his/her Admit Card to the Invigilator.
10. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.
11. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the
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12. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited.
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Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the JAB/Board.
14. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
15. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager,
mobile phone, electronic device or any other material except the Admit Card inside the examination
hall/room.
137Mock Test-1
JEE-MAIN: MATHEMATICS MOCK TEST-1
1. The domain of definition of is:
a. b.
c. d.
2. How many roots the equation have
a. One b. Two
c. Infinite d. None
3. If is the cube root of unity, then +
a. 4 b. 0
c. – 4 d. None of these
4. From the following find the correct relation
a. b.
c. d.
5. If the roots of the cubic equation
are in G.P., then
a. b. c. d.
6.
a. b.
c. d.
7.
a. b.
c. d.
8. How many words can be made from the letters of the
word INSURANCE, if all vowels come together.
a. 18270 b. 17280
c. 12780 d. None of these
9. There are four machines and it is known that exactly two
of them are faulty. They are tested, one by one, in a
random order till both the faulty machines are identified.
Then, the probability that only two tests are needed, is
a. b.
c. d.
10.
a. 1/2 b. – ½ c. 1/4 d. 1
11. Find real part of
a. – 1 b. 1
c. 0 d. None of these
12. From a 60 meter high tower angles of depression of the
top and bottom of a house are α and β respectively. If
the height of the house is 60sin ( )
,x
β α−then x=
a. sin sinα β b. cos cosα β
c. sin cosα β d. cos sinα β
13. The function is not defined
at The value which should be assigned to f at
so that it is continuous at is
a. b.
c. d.
14. If ,sin,cos 44 θθ ayax == then ,dy
dx at
3,
4
πθ = is
a. –1 b. 1
c. 2a− d.
2a
15. The minimum value of [(5 )(2 )]/[1 ]x x x+ + + for non-
negative real x is
a. 12 b. 1 c. 9 d. 8
16.
a. b.
c. d.
17. The measurement of the area bounded by the co-ordinate
axes and the curve is
a. 1 b. 2
c. 3 d.
18. The solution of the equation is
a. b.
c. d.
2
2
log ( 3)( )
3 2
xf x
x x
+=
+ +
1, 2R − − − ( 2, )− ∞
1, 2, 3R − − − − ( 3, ) 1, 2− ∞ − − −
2 21
1 1x
x x− = −
− −
ω 2 2(3 5 3 )ω ω+ +2 2(3 3 5 )ω ω+ + + =
( )AB A B′ ′ ′= ( )AB B A′ ′ ′=
1 adj AA
A
− = 1 1 1( )AB A B− − −=
3 2 0ax bx cx d+ + + =
3 3c a b d= 3 3ca bd= 3 3a b c d= 3 3ab cd=
1 2 3 42 3 4 ....
nC C C C nC+ + + + + =
2n . 2nn
1. 2nn − 1. 2nn +
2 4(log ) (log )
12! 4!
e en n+ + + =K
n 1/ n
11( )
2n n
−+1
( )2
n ne e
−+
1
3
1
6
1
2
1
4
3sin sin
10 10
π π =
1cosh (1)−
log(1 ) log(1 )( )
ax bxf x
x
+ − −=
0.x =
0x = 0,x =
a b− a b+
log loga b+ log loga b−
1tan
21
xedx
x
−
=+∫
2log(1 )x c+ +1tanlog xe c−
+1
tan xe c−
+1
1 tantan xe c−− +
logey x=
∞
2x y ydye x e
dx
− −= +
3
3
y x xe e c= + + 2y xe e x c= + +
3y xe e x c= + + xy e c= +
Mathematics138
19. The distance between and is
a. b. 4
c. d. None of these
20. The area of a circle whose centre is (h, k) and radius a is
a. 2 2 2( )h k aπ + − b. 2a hkπ
c. 2aπ d. None of these
21. The locus of the mid-point of the line segment joining the
focus to a moving point on the parabola is
another parabola with directrix
a. b.
c. d.
22. If and then the vector parallel to
is
a. (3, 5) b. (1, 1)
c. (1, 3) d. (8, 5)
23. The acute angle between the line joining the points (2, 1,
–3), (–3,1,7) and a line parallel to
through the point (–1, 0, 4) is
a. b.
c. d.
24. Which of the following is logically equivalent to
a. b.
c. d.
25. The number of solutions of the system of equations
is
a. 3 b. 2 c. 1 d. 0
26. The sum to infinity of the given series
is
a. b.
c. d.
27. The values of A and B such that the function
, is continuous
everywhere are
a. b.
c. d.
28.
a. b.
c. d. None of these
29. The equations of tangents to the circle
2 2 22 4 25 0x y x y+ − − + = which are perpendicular to the
line 5 12 8 0x y+ + = are
a. 12 5 8 0, 12 5 252x y x y− + = − =
b. 12 5 0, 12 5 252x y x y− = − =
c. 12 5 8 0,12 5 252 0x y x y− − = − + =
d. None of these
30. is logically equivalent to
a.
b.
c.
d.
4 3 11+ =x y 8 6 15,+ =x y
7
2
7
10
2 4y ax=
x a= −2
ax = −
0x =2
ax =
(2, 5)a =r
(1, 4),b =r
( )a b+r
r
1 3
3 4 5
x y z− += =
1 7cos
5 10
−
1 1cos
10
−
1 3cos
5 10
−
1 1cos
5 10
−
~ (~ )⇒p q
∧p q ~∧p q
~ ∧p q ~ ~∧p q
2 7, 3 2 1, 4 3 5x y z x y z x y z+ − = − + = + − =
2 3 4
1 1 1 1
2 3 4n n n n− + − +K
1loge
n
n
+
log1
e
n
n
+
1loge
n
n
−
log1
e
n
n
−
2sin ,2
( ) sin ,2 2
cos ,2
x x
f x A x B x
x x
π
π π
π
− ≤ −
= + − < <
≥
0, 1A B= = 1, 1A B= =
1, 1A B= − = 1, 0A B= − =
3 43 5x x dx+ =∫4 3/ 2(3 5 )x c+ + 4 3/ 21
(3 5 )5
x c+ +
4 3/ 21(3 5 )
30x c+ +
~ ∧p q
→p q
→q p
~ ( )→p q
~ ( )→q p
Space for rough work
139Mock Test-1
JEE ADVANCE PAPER-I
Time 3 Hours. Max. Marks 264 (88 for Mathematics)
Read The Instructions Carefully
Question Paper Format and Marking Scheme:
1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections.
2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
Marking Scheme: +4 for correct answer and 0 in all other cases.
3. Section 2 contains 10 multiple choice questions with one or more than one correct option.
Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases.
4. Section 3 contains 2 “match the following” type questions and you will have to match entries in Column I with the entries
in Column II.
Marking Scheme: For each entry in Column I, +2 for correct answer, 0 if not attempted and –1 in all other cases.
NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and
marking scheme too.
SECTION 1 (Maximum Marks: 32)
This section contains EIGHT questions.
The answer to each question is a SINGLE DIGIT INTEGER
ranging from 0 to 9, both inclusive.
For each question, darken the bubble corresponding to the correct
integer in the ORS.
Marking scheme:
+4 If the bubble corresponding to the answer is darkened.
0 In all other cases.
1. If area enclosed between the curves and
and the axis of x is sq unit, then the value of
622
must be
2. The number of distinct solutions of the equation
2 4 4 6 65
cos 2 cos sin cos sin 24
+ + + + =x x x x x
in the
interval [0, 2 ]π is
3. Let the curve C be the mirror image of the parabola 2 4=y x with respect to the line 4 0.+ + =x y If A
and B are the points of intersection of C with the line
5,= −y then the distance between A and B is
4. The minimum number of times a fair coin needs to be
tossed, so that the probability of getting at least two heads
is at least 0.96 is
5. Let n be the number of ways in which 5 boys and 5 girls
can stand in a queue in such a way that all the girls stand
consecutively in the queue. Let m be the number of ways
in which 5 boys and 5 girls can stand in a queue in such a
way that exactly four girls stand consecutively in the
queue. Then the value of m
nis
6. TP and TQ are any two tangents to a parabola and the
tangent at a third point R cuts then in P' and Q', then the
value of must be
7. Let : →f R R be a function defined by
[ ], 2( ) ,
0, 2
x xf x
x
≤=>
where [x] is the greatest integer less
than or equal to x. If 2 2
1
( ),
2 ( 1)−
=+ +∫xf x
I dxf x
then the value
of (4 1)−I is
8. A cylindrical container is to be made from certain solid
material with the following constraints: It has a fixed
inner volume of 3mm ,V has a 2 mm thick solid wall and
is open at the top. The bottom of the container is a solid
circular disc of thickness 2 mm and is of radius equal to
the outer radius of the container.
ln( )y x e= +
1lnx
y
=
λ
TP TQ
TP TQ
′ ′=
Mathematics140
If the volume of the material used to make the container is
minimum when the inner radius of the container is 10
mm, then the value of / 250V π is
SECTION 2 (Maximum Marks: 40)
This section contains TEN questions.
Each question has FOUR options (a), (b), (c) and (d). ONE OR
MORE THAN ONE of these four option(s) is (are) correct.
For each question, darken the bubble(s) corresponding to all the
correct option(s) in the ORS.
Marking scheme:
+4 If only the bubble(s) corresponding to all the correct
option(s) is(are) darkened.
0 If none of the bubbles is darkened
–2 In all other cases
9. Let ∆PQR be a triangle. Let , ,= =uuur uuurrr
a QR b RP and
.=uuurr
c PQ If | | 12, | | 4 3= =rr
a b and . 24,=r rb c then which
of the following is (are) true?
a. 2| |
| | 122
− =r
rca b.
2| || | 30
2− =
rrca
c. | | 48 3× + × =rr r r
a b c a
d. . 72= −rr
a b
10. Let X and Y be two arbitrary, 3 3,× non-zero, skew-
symmetric matrices and Z be an arbitrary 3 3,× non-zero,
symmetric matrix. Then which of the following matrices
is (are) skew symmetric?
a. 3 4 4 3−Y Z Z Y b.
44 44+X Y
c. 4 3 3 4−X Z Z X d.
23 23+X Y
11. Which of the following values of α satisfy the equation
2 2 2
2 2 2
2 2 2
(1 α) (1 2α) (1 3α)
(2 α) (2 2α) (2 3α) 648α ?
(3 α) (3 2α) (3 3α)
+ + +
+ + + = −
+ + +
?
a. 4− b. 9
c. 9− d. 4
12. In 3 ,R consider the planes 1 : 0=P y and 2 : 1.+ =P x z
Let 3P be a plane, different from 1P and 2 ,P which passes
through the intersection of 1P and 2 .P If the distance of
the point (0, 1, 0) from 3P is 1 and the distance of a point
(α, β, γ) from 3P is 2, then which of the following
relations is (are) true?
a. 2α + β + 2γ 2 0+ =
b. 2α – β + 2γ 4 0+ =
c. 2α + β – 2γ – 10 0=
d. 2α – β + 2γ – 8 0=
13. In 3 ,R let L be a straight line passing through the origin.
Suppose that all the points on L are at a constant distance
from the two planes 1 : 2 1 0+ − + =P x y z and
2 : 2 1 0.− + − =P x y z Let M be the locus of the feet of the
perpendiculars drawn from the points on L to the plane
1 .P Which of the following points lie(s) on M?
a. 5 2
0, ,6 3
− −
b. 1 1 1
, ,6 3 6
− −
c. 5 1
,0,6 6
−
d. 1 2
,0,3 3
−
14. Let P and Q be distinct points on the parabola 22=y x
such that a circle with PQ as diameter passes through the
vertex O of the parabola. If P lies in the first quadrant and
the area of the triangle ∆OPQ is 3 2, then which of the
following is (are) the coordinates of P?
a. (4, 2 2)
b. (9,3 2)
c. 1 1
,4 2
d. (1, 2)
15. Let ( )y x be a solution of the differential equation
(1 ) 1.′+ + =x xe y ye If (0) 2,=y then which of the
following statements is (are) true ?
a. ( 4) 0− =y
b. ( 2) 0− =y
c. ( )y x has a critical point in the interval ( 1,0)−
d. ( )y x has no critical point in the interval ( 1,0)−
16. Consider the family of all circles whose centers lie on the
straight line y = x. If this family of circles is represented
by the differential equation 1 0,′′ ′+ + =Py Qy where P, Q
are functions of x, y and 2
2(here , ),′ ′ ′′= =
dy d yy y y
dx dx then
which of the following statements is (are) true?
a. = +P y x
b. = −P y x
c. 21 ( )′ ′+ = − + + +P Q x y y y
d. 2( )′ ′− = + − −P Q x y y y
141Mock Test-1
17. Let : →g R R be a differential function with
(0) 0, (0) 0′= =g g and (1) 0.′ ≠g
Let ( ), 0
( ) | |00,
≠= =
xg x x
f x xx
and | |( ) = xh x e for all
.∈x R Let ( )f ho (x) denote ( ( ))f h x and ( )( )h f xo
denote ( ( )).h f x Then which of the following is (are)
true?
a. f is differentiable at 0x =
b. h is differentiable at 0x =
c. f ho is differentiable at 0x =
d. h fo is differentiable at 0x =
18. Let ( ) sin sin sin6 2
=
π πf x x for all ∈x R and
( ) sin2
=π
g x x for all .∈x R Let (f o g)(x) denote f (g(x))
and (g o f )(x) denote g(f (x)). Then which of the following
is (are) true?
a. Range of f is 1 1
,2 2
−
b. Range of f o g is 1 1
,2 2
−
c. 0
( )lim
( ) 6→=π
x
f x
g x
d. There is an ∈x R such that ( )( ) 1g f x =o
SECTION 3 (Maximum Marks: 16)
This section contains TWO questions.
Each question contains two columns, Column I and Column II
Column I has four entries (A), (B), (C) and (D)
Column II has five entries (1), (2), (3), (4) and (5)
Match the entries in Column I with the entries in Column II
One or more entries in Column I may match with one or more
entries in Column II
The ORS contains a 4 5× matrix whose layout will be similar to
the one shown below:
(A) (1) (2) (3) (4) (5)
(B) (1) (2) (3) (4) (5)
(C) (1) (2) (3) (4) (5)
(D) (1) (2) (3) (4) (5)
For each entry in Column I, darken the bubbles of all the
matching entries. For example, if entry (A) in Column I, matches
with entries (2), (3) and (5), then darken these three bubbles in the
ORS. Similarly, for entries (B), (C) and (D).
Marking scheme:
For each entry in Column I
+2 If only the bubble(s) corresponding to all the correct match(es)
is(are) darkened
0 If none of the bubbles is darkened
–1 In all other cases
19. Match the Column:
Column I Column II
(A) In 2 ,R if the magnitude of
the projection vector of the
vector ˆ ˆα β+i j on ˆ ˆ3 +i j
is 3 and if α | 2 3β,= +
then possible value(s) of |α |
is (are)
1. 1
(B) Let a and b be real numbers
such that the function
2
2
3 2, 1( )
, 1
− − <=+ ≥ax x
f xbx a x
is
differentiable for all .∈x R
Then possible value(s) of a
is (are)
2. 2
(C) Let ω 1≠ be a complex
cube root of unity. If
2 4 3(3 3 2 ) +− +ω ω n
2 4 3(2 3 3 ) ++ + −ω ω n
2 4 3( 3 2 3 ) 0,++ − + + =ω ω n
then possible value(s) of n is
(are)
3. 3
(D) Let the harmonic mean of
two positive real numbers a
and b be 4. If q is a positive
real number such that a, 5,
q, b is an arithmetic
progression, then the
value(s) of | |−q a is (are)
4. 4
5. 5
Mathematics142
20. Match the thermodynamic processes given under Column I
with the expression given under Column II:
Column I Column II
(A) In a triangle let a, b
and c be the lengths of the
sides opposite to the angles
X, Y and Z, respectively. If
and
then
possible values of n for
which is (are)
1. 1
(B) In a triangle let a, b
and c be the lengths of the
sides opposite to the angles
X, Y and Z, respectively. If
then possible value(s) of
is (are)
2. 2
(C) In let
and be the
position vectors of X, Y and
Z with respect of the origin
O, respectively. If the
distance of Z from the
bisector of the acute angle
of with is
then possible value(s) of
is (are)
3. 3
(D) Suppose that denotes
the area of the region
bounded by x = 0, x = 2,
and
where Then the
value(s) of
when and is
(are)
4. 5
5. 6
,∆XYZ
2 2 22( )− =a b c
sin( ),
sin
−=λ
X Y
Z
cos( ) 0=πλn
,∆XYZ
1 cos2+ −X
2cos2 2sin sin ,=Y X Y
a
b
2 ,R ˆ ˆ ˆ ˆ3 , 3+ +i j i j
ˆ ˆ(1 )+ −β βi j
uuur
OXuuur
OY3
,2
| |β
(α)F
2 4=y x
| α 1 | | α 2 | α ,= − + − +y x x x
α 0,1.∈
8(α) 2,
3+F
α 0= α 1,=
Space for rough work
143Mock Test-1
JEE ADVANCE PAPER-II
Time 3 Hours. Max. Marks 240 (80 for Mathematics)
Read The Instructions Carefully
Question Paper Format and Marking Scheme:
1. The question paper has three parts: Physics, Chemistry and Mathematics. Each part has three sections.
2. Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9 (both inclusive).
Marking Scheme: +4 for correct answer and 0 in all other cases.
3. Section 2 contains 8 multiple choice questions with one or more than one correct option.
Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases.
4. Section 3 contains 2 “paragraph” type questions. Each paragraph describes an experiment, a situation or a problem. Two
multiple choice questions will be asked based on this paragraph. One or more than one option can be correct.
Marking Scheme: +4 for correct answer, 0 if not attempted and –2 in all other cases.
NOTE: It’s the mock test as per previous year’s papers but sometimes IIT changes the test paper pattern and
marking scheme too.
SECTION 1 (Maximum Marks: 32)
This section contains EIGHT questions.
The answer to each question is a SINGLE DIGIT INTEGER
ranging from 0 to 9, both inclusive.
For each question, darken the bubble corresponding to the correct
integer in the ORS.
Marking scheme:
+4 If the bubble corresponding to the answer is darkened.
0 In all other cases.
1. Suppose that ,p qr r
and rr
are three non-coplanar vectors
in 3.R Let the components of a vector s
r
along ,p qr r
and
rr
be 4, 3 and 5, respectively. If the components of this
vector sr
along ( ),p q r− + +r r r
( )p q r− +r r r
and ( )p q r− − +r r r
are x, y and z, respectively, then the value of 2x y z+ + is
2. For any integer k, let cos sin ,7 7
k
k ki
π πα
= +
where
1.i = − The value of the expression
12
11
3
4 1 4 21
| |
| |
k kk
k kk
α α
α α
+=
− −=
Σ −
Σ −is
3. Suppose that all the terms of an arithmetic progression
(A.P.) are natural numbers. If the ratio of the sum of the
first seven terms to the sum of the first eleven terms is 6 :
11 and the seventh term lies in between 130 and 140, then
the common difference of this A.P. is
4. The coefficient of 9x in the expansion of 2(1 )(1 )x x+ +
3 100(1 )...(1 )x x+ + is
5. If the normals at the four points
and on the ellipse are concurrent, then
the value of must be
6. Let m and n be two positive integers greater than 1. If
cos( )
0lim
2
n
ma
e e eα
α→
− = −
then the value of m
nis
7. If
21
9 3tan
20
12 9( )
1
x x xe dx
xα
−+ +=
+ ∫ where
1tan x− takes
only principal values, then the value of 3
log |1 |4
e
πα
+
is
8. Let :f R R→ be a continuous odd function, which
vanishes exactly at one point and1
(1) .2
f = Suppose that
1( ) ( )
x
F x f t dt−
= ∫ for all [1 1,2]x∈ − and 1
( ) | ( ( )) |x
G x t f f t dt−
=∫
for all [ 1, 2].x∈ − If1
( ) 1lim ,
( ) 14x
F x
G x→= then the value of
1
2f
is
1 1 2 2 3 3( , ),( , ), ( , )x y x y x y
4 4( , )x y2 2
2 21
x y
a b+ =
1 2 3 4
1 2 3 4
1 1 1 1( )x x x x
x x x x
+ + + × + + +
Mathematics144
SECTION 2 (Maximum Marks: 32)
This section contains EIGHT questions.
Each question has FOUR options (a), (b), (c) and (d). ONE OR
MORE THAN ONE of these four option(s) is (are) correct.
For each question, darken the bubble(s) corresponding to all the
correct option(s) in the ORS.
Marking scheme:
+4 If only the bubble(s) corresponding to all the correct option(s)
is(are) darkened.
0 If none of the bubbles is darkened
–2 In all other cases
9. Let
3
4
192( )
2 sin
xf x
xπ′ =
+for all x R∈ with
10.
2f
=
If
1
1/ 2( ) ,m f x dx M≤ ≤∫ then the possible values of m and M
are
a. 13, 24m M= = b. 1 1
,4 2
m M= =
c. 11, 0M M= − = d. 1, 12M M= =
10. Let S be the set of all non-zero real numbers α such that
the quadratic equation 2 0x xα α− + = has two distinct
real roots 1x and
2x satisfying the inequality 1 2| | 1.x x− <
Which of the following intervals is(are) a subset(s) of S ?
a. 1 1
,2 5
− −
b. 1
, 05
−
c. 1
0,5
d. 1 1
,25
11. If 1 6
3 sin11
α − =
and
1 43 cos ,
9β − =
where the inverse
trigonometric functions take only the principal values,
then the correct option(s) is (are)
a. cos 0β > b. sin 0β <
c. cos( ) 0α β+ > d. cos 0α <
12. Let 1E and
2E be two ellipse whose centers are at the
origin. The major axes of 1E and
2E lie along the x-axis
and the y-axis, respectively. Let S be the circle 2 2( 1) 2.x y+ − = The straight line 3x y+ = touches the
curves S, 1E and
2E at P, Q and R, respectively Suppose
that 2 2
.3
PQ PR= = If 1e and
2e are the eccentricities of
1E and 2 ,E respectively, then the correct expression(s)
is(are)
a. 2 2
1 2
43
40e e+ = b. 1 2
7
2 10e e =
c. 2 2
1 2
5| |
8e e− = d. 1 2
3
4e e =
13. Consider the hyperbola 2 2: 1H x y− = and a circle S with
center 2( , 0).N x Suppose that H and S touch each other at
a point 1 1( , )P x y with
1 1x > and 1 0.y > The common
tangent to H and S at P intersects the x-axis at point M. If
( , )l m is the centroix of the triangle ,PMN∆ then the
correct expression(s) is (are)
a. 2
1 1
11 for 1
3
dlx
dx x= − >
b.
( )1
12
11
for 13 1
xdmx
dx x= >
−
c. 12
1 1
11 for 1
3
dlx
dx x= + >
d. 1
1
1for 0
3
dmy
dy= >
14. The option(s) with the values of a and L that satisfy the
following equation is(are)
46 4
0
1 6 4
0
(sin cos )?
(sin cos )
at at dtL
e at at dt
π
π
+=
+
∫∫
a.
4 12,
1
ea L
e
π
π
−= =
− b.
4 12,
1
ea L
e
π
π
+= =
+
c.
4 14,
1
ea L
e
π
π
−= =
− d.
4 14,
1
ea L
e
π
π
+= =
+
15. Let , :[ 1, 2]f g R− → be continuous functions which are
twice differentiable on the interval (–1, 2). Let the values
of f and g at points –1, 0 and 2 be as given in the
following table:
1x = − 0x = 2x =
( )f x 3 6 0
( )g x 0 1 –1
In each of the intervals (–1, 0) and (0, 2) the function
( 3 )f g ′′− never vanishes. Then the correct statements(s) is
(are)
a. ( ) 3 ( ) 0f x g x′ ′− = has exactly three solutions in
( 1, 0) (0, 2)− ∪
b. ( ) 3 ( ) 0f x g x′ ′− = has exactly one solution in (–1, 0)
145Mock Test-1
c. ( ) 3 ( ) 0f x g x′ ′− = has exactly one solution in (0, 2)
d. ( ) 3 ( ) 0f x g x′ ′− = has exactly two solutions in (–1, 0)
and exactly two solutions in (0, 2)
16. Let 8 6 4 2( ) 7 tan 7 tan 3 tan 3 tanf x x x x x= + − − for all
,2 2
xπ π
∈ −
Then the correct expression(s) is (are)?
a. / 4
0
1( )
12xf x dx
π=∫ b.
/ 4
0( ) 0f x dx
π=∫
c. / 4
0
1( )
6xf x dx
π=∫ d.
/ 4
0( ) 1f x dx
π=∫
SECTION 3 (Maximum Marks: 16)
This section contains FOUR questions.
Each question has FOUR options (a), (b), (c) and (d). ONE OR
MORE THAN ONE of these four option(s) is(are) correct.
For each question, darken the bubble(s) corresponding to all the
correct option(s) in the ORS.
Marking scheme:
+4 If only the bubble(s) corresponding to all the correct option(s)
is(are) darkened.
0 If none of the bubbles is darkened
–2 In all other cases
Paragraph-I
Let :F R R→ be a thrice differentiable function. Suppose that
(1) 0, (3) 4F F= = − and ( ) 0F x′ < for all 1
, 32
x
∈
.
Let
( ) ( )f x xF x= for all .x R∈
17. The correct statement(s) is (are)
a. (1) 0f ′ <
b. (2) 0f <
c. ( ) 0f x′ ≠ for any (1, 3)x∈
d. ( ) 0f x′ = for some (1, 3)x∈
18. If 3
2
1( ) 12x F x dx′ = −∫ and
3
1( ) 40,F x dx′′ =∫ then the
correct expression(s) is(are)
a. 9 (3) (1) 32 0f f′ ′+ − =
b. 3
1( ) 12f x dx =∫
c. 9 (3) (1) 32 0f f′ ′− + =
d. 3
1( ) 12f x dx = −∫
Paragraph-II
Let 1n and
2n be the number of red and black balls,
respectively, in box I. Let 3n and
4n be the number of red and
black balls, respectively, in box II.
19. One of the two boxes, box I and box II, was selected at
random and a ball was drawn randomly out of this box. The
ball was found to be red. If the probability that this red ball
was drawn from box II is 1
,3
then the correct option(s) with
the possible values of 1 2 3, ,n n n and
4n is (are)
a. 1 2 3 43, 3, 5, 15n n n n= = = =
b. 1 2 3 43, 6, 10, 50n n n n= = = =
c. 1 2 3 48, 6, 5, 20n n n n= = = =
d. 1 2 3 46, 12, 5, 20n n n n= = = =
20. A ball is drawn at random from box I and transferred to
box II. If the probability of drawing a red ball from box I,
after this transfer, is 1
,3
then the correct option(s) with
the possible values of 1n and
2n is(are)
a. 1 24, 6n n= = b.
1 22, 3n n= =
c. 1 210, 20n n= = d.
1 23, 6n n= =
Space for rough work
Mathematics146
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d d c a a c c d a c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c d b a c c d a c c
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
c c a d d a c c a d
1. (d) and
2. (d) If multiplying each term by the given
equation reduces to or or
which is not possible as considering Thus
given equation has no roots.
3. (c)
4. (b) It is understandable.
5. (a) Let be the roots of the equation
then Product of the roots
Since is a root of the equation.
6. (c) Trick: Put
Now by alternate (c),
put
7. (c)
8. (d) IUAENSRNC Obviously required number of words
are
9. (b) The probability that only two tests are needed =
probability that the first machine tested is faulty
probability that the second machine tested is faulty
10. (c)
.
11. (c) We know that
12. (d)
tanH d β= and tanH h d α− =
⇒ 60 tan
60 tanh
βα
=−
⇒ 60 tan 60 tan
tanh
α ββ
−− = ⇒
60sin( )
sincos cos
cos
hβ α
βα β
β
−=
⇒ cos sinx α β= .
13. (b) Since limit of a function is as therefore
to be continuous at a function, its value must be at
14. (a) 4siny a θ=
⇒ 34 sin cos
dya
dθ θ
θ=
and 4cosx a θ=
⇒ 34 cos sin
dxa
dθ θ
θ= −
∴ 2
2
2
/ sintan
/ cos
dy dy d
dx dx d
θ θθ
θ θ−
= = = −
∴ 2
3
4
3tan 1.
4
dy
dx πθ
π
=
= − = −
3 0x + > 2 3 2 0x x+ + ≠
1,x ≠ ( 1),x−
( 1) ( 1)x x x− = − 2( 1) 0x − =
1,x = 1.x ≠
2 2 2 2(3 5 3 ) (3 3 5 )ω ω ω ω+ + + + +2 2 2 2 2(3 3 3 2 ) (3 3 3 2 )ω ω ω ω ω ω= + + + + + + +
2 3(1 0, 1)ω ω ω+ + = =2 2 2 2 4(2 ) (2 ) 4 4 4( 1) 4.ω ω ω ω= + = + = − = −
, ,A
A ARR
3 2 0ax bx cx d+ + + =
3A =
d
a= −
⇒1/3
dA
a
= −
A
∴ 3 2 0aA bA cA d+ + + =
⇒2 /3 1/3
0d d d
a b c da a a
− + − + − + =
⇒2 /3 1/3
d db c
a a
=
⇒2
3 3
2
d db c
a a= ⇒ 3 3 .b d c a=
1, 2, 3,....n =
1 21, 2 2 4S S= = + =
1, 2n =0 1
1 21.2 1, 2.2 4S S= = = =
2 4 log log(log ) (log )1
2! 4! 2
e en n
e en n e e
−++ + + =K
1
.2
n n−+=
6 !4 ! 8640
2 !× =
×
2 1 1
4 3 6= × =
3sin sin sin18 .sin 54
10 10
π π= ° °
5 1 5 1 1sin18 .cos36 .
4 4 4
− += ° ° = =
( )1 2cosh log 1x x x− = + −
∴ ( )1 2cosh (1) log 1 1 1 log1 0.− = + − = =
a b+ 0,x →
a b+
0x =
(0) .f a b⇒ = +
H=60m α
β
d
h
147Mock Test-1
15. (c) Given
⇒
⇒
Now
Hence minimum value at
16. (c) Putting we get
17. (d) Area
18. (a)
Now integrating both sides, we get
19. (c) 4 3 11x y+ = and 15
4 32
x y+ =
Therefore, 15
1172
5 10D
−= =
.
20. (c) Since area where
Area
21. (c) Let be the mid-point of the line segment
joining the focus and a general point on the
parabola.
Then
Put these values of x and y in we get
So, locus of is
Its directrix is
22. (c) Hence it is parallel to
23. (a) Direction ratio of the line joining the point
are
⇒
Direction ratio of the line parallel to line
are
Angle between two lines,
⇒
24. (d) Since
25. (d)
Hence, number of solutions is zero.
[(5 )(2 )]( )
[1 ]
x xf x
x
+ +=
+
4 4( ) 1 (5 ) (6 )
1 (1 )f x x x
x x= + + + = + +
+ +
2
4'( ) 1 0;
(1 )f x
x= − =
+2 2 3 0x x+ − =
3,1x = −
3
8( ) ,
(1 )f x
x′′ =
+
( 3) ,f ve′′ − = −
(1)f ve′′ = +
1x =
(5 1)(2 1) 6 3(1) 9.
(1 1) 2f
+ + ×= = =
+
1
2
1tan ,
1t x dt dx
x
−= ⇒ =+
1tan
21
xte
dx e dtx
−
=+∫ ∫
1tan .t xe c e c
−
= + = +
0log .A x dx
∞= ∫
0( log )x x x ∞= − = ∞
2 2( )x y y y xdye x e e e x
dx
− − −= + = +
⇒ 2( )y xe dy x e dx= +
3
.3
y xxe e c= + +
2,rπ=
r a=
⇒ 2 .aπ=
( , )P h k
( ,0)a ( , )Q x y
, 2 , 2 .2 2
x a yh k x h a y k
+= = ⇒ = − =
2 4 ,y ax=
24 4 (2 )k a h a= −
⇒ 2 2 2 24 8 4 2k ah a k ah a= − ⇒ = −
( , )P h k 2 22y ax a= −
⇒ 2 22
ay a x
= −
0.2 2
a ax x− = − ⇒ =
ˆ ˆ ˆ ˆ3 9 3( 3 ).a b i j i j+ = + = +r
r
).3,1(
(2, 1, 3), ( 3, 1, 7)− − 1 1 1( , , )a b c
( 3 2, 1 1, 7 ( 3))− − − − −
⇒ ( 5, 0, 10)−
1 3
3 4 5
x y z− += =
2 2 2( , , )a b c
⇒ (3, 4, 5)
1 2 1 2 1 2
2 2 2 2 2 2
1 1 1 2 2 2
cosa a b b c c
a b c a b cθ
+ +=
+ + + +
( 5 3) (0 4) (10 5)cos
25 0 100 9 16 25θ
− × + × + ×=
+ + + +
35cos
25 10θ =
⇒ 1 7cos
5 10θ − =
qpqp ~)(~ ∧≡⇒
qpqp ~~)(~~ ∧=⇒
2 1 1
1 3 2
1 4 3
−
∆ = −
−
2(9 8) 1( 3 2) 1(4 3) 7 7 0= − − − − − + = − =
y =loge x
(1,0) X
Y
Mathematics148
26. (a)
27. (c) For continuity at all we must have
. . .(i)
and
. . .(ii)
From (i) and (ii),
and .
28. (c) Put then
29. (a) Equation of line perpendicular to
is
Now it is a tangent to the circle, if
Radius of circle
Distance of line from centre of circle
or
Hence equations of tangents are
and
30. (d) .
JEE Advance Paper-I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
4 8 4 8 5 1 0 4 a,c,d c, d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b,c b,d a,b a,d a,c b,c a,d a,b,c c c
1. (4)
Required area sq unit
2. (8)2 4 4 6 65
cos 2 cos sin cos sin 24
x x x x x+ + + + =
⇒ 2 2 25
cos 2 5cos sin 04
x x x− =
⇒ 2tan 2 1, where 2 [0, 4 ]x x π= ∈
Number of solutions = 8
3. (4) Image of 5y = − about the line 4 0x y+ + = is
1x =
⇒ Distance 4AB =
4. (8) Let coin was tossed ‘n’ times
Probability of getting at least two heads 1
12 2n n
n = − +
⇒ 1
1 0.962n
n + − ≥
⇒ 2
251
n
n≥
+⇒ 8n ≥
5. (5) .6! 5!n = (5 girls together arranged along with 5
boys)4 .(7! 2.6!).4!m C5= −
(4 out of 5 girls together arranged with others – number of
cases all 5 girls are together)
5 5 6! 4!
6! 5!
m
n
⋅ ⋅ ⋅=
⋅
2 3 4
1 1 1 1
2 3 4n n n n− + − +K
2 3 41 (1/ ) (1/ ) (1/ )
2 3 4
n n n
n= − + − +K
1 1log 1 log .e e
n
n n
+ = + =
,x R∈
( / 2)lim ( 2sin )
2 xf x
π
π−→ −
− = −
( / 2)lim ( sin )
xA x B
π +→ −= +
⇒ 2 A B= − +
( / 2)lim ( sin )
2 xf A x B
π
π−→
= +
( / 2)lim (cos )
xx
π +→=
⇒ 0 A B= +
1−=A
1=B
4 33 5 20 ,x t x dx dt+ = ⇒ =
3 4 1/ 213 5
20x x dx t dt+ =∫ ∫
3/ 2 4 3/ 22 1 1. (3 5 ) .
3 20 30t c x c= × + = + +
5 12 8 0x y+ + =
12 5 0.x y k− + =
=
12(11) 5(2)121 4 25
144 25
k− ++ − =
+
⇒ 8k = 252.−
12 5 8 0x y− + =
12 5 252x y− =
)(~~ pqqp →=∧
4 (1 1) 4= × × =
∴ 4λ =1 1 1 1 1/ 2
2 4 8 16 1 1/ 2 4λ λ λ λ λ λ λ+ + + + ∞
−∞ = = = =K
K
O
7
6
5
4
3
2
1 –1 –2
–1
–2
–3
1 2 3 4 5
149Mock Test-1
6. (1) Let Parabola be and coordinates of P and Q
on this parabola are and T
is the point of intersection of tangents at and
Coordinates of
Similarly ,
Let
∴
or
Similarly,
or
7. (0)
0 1 2
1 0 1
0 0 1 10
2 0 2 1 2 0 4
x x xI dx dx dx
−
⋅ ⋅ ⋅= + + + =
+ + +∫ ∫ ∫
⇒ 4 1 0I = =
8. (4) Let inner radius be r and inner length be l
2r Vπ =l
Volume of material be M,
2 2( 2) ( 2)M r rπ π= + + −l l
2 3
4 88 0 4
dM V Vr
dr r rπ π= − − + + +
0 when 10dM
rdr
= =
⇒ 1000V π=
⇒ 4250
V
π=
9. (a,c,d) | | | |b c a+ =r
r r
⇒ 2 2 2| | | | 2 | |b c b c a+ = ⋅ =
r r
r r r
⇒ 248 | | 48 144c+ + =r
⇒ | | 4 3c =r
∴2| |
| | 122
ca− =
r
r
Also, | | | |a b c+ =r
r r
⇒ 2 2 2| | | | 2 | |a b a b c+ = ⋅ =
r r
r r r
⇒ 72a b⋅ = −r
r
0a b c+ + =r
r r
⇒ a b c a× = ×r
r r r
⇒ | | 2 | | 48 3a b c a a b× + × = × =r r
r r r r
10. (c,d) 3 4 4 3( )TY Z Z Y− 4 3 3 4( ) ( ) ( ) ( )T T T TZ Y Y Z= −
4 3 3 4Z Y Y Z= − +
⇒ symmetric 44 44 is symmetricX Y+
4 3 3 4 skew symmetricX Z Z X−
23 23 skew symmetric.X Y+
11. (b,c) We get
2 2 2(1 α) (1 2α) (1 3α)
3 2α 3 4α 3 6α
5 2α 5 4α 5 6α
+ + +
+ + +
+ + +
3 3 2 2 2 1648α ( ; )R R R R R R= − → − → −
2 2 2α 2 4α 2 9α 2
3 2α 3 4α 3 6α
2 2 2
− − −
+ + +
1 1 2 3 3 2648α ( ; )R R R R R R= − → − → −
⇒
2 2 22α 5α 9α 3
2α 2α 3 6α 648α
0 0 2
− − − −
− − + = −
12. (b,d) Let the required plane be 1 0x z yλ+ + − =
⇒ 2
| 1 | 11
22
λλ
λ
−= ⇒ = −
+
⇒ 3 2 2 2 0P x y z≡ − + − =
Distance of 3P from ( , , )α β γ is 2
| 2 2 2 |
24 1 4
α β γ− + −=
× +
⇒ 2 2 4 0 and 2 2 8 0α β λ α β λ− + + = − + − =
13. (a,b) Line L will be parallel to the line of intersection of
1P and 2P
Let a, b and c be the direction ratios of line L
⇒ 2 0 and 2 0a b c a b c+ − = − + =
⇒ : : ::1 : 3 : 5a b c − −
Equation of line L is 0 0 0
1 3 5
x y z− − −= =
− −
Again foot of perpendicular from origin to plane 1P is
1 1 1, ,
6 3 6
− −
∴ Equation of project of line L on plane 1P is
1 2 1
6 6 6
1 3 5
x y z
k
+ + −= = =
− −
Clearly points 5 2
0, ,6 3
− −
and 1 1 1
, ,6 3 6
− −
satisfy the
line of projection i.e. M
2 4y ax=2
1 1( , 2 )P at at≡ 2
2 2( , 2 );Q at at≡
1t 2.t
∴ 1 2 1 2 , , ( )T at t a t t≡ +
3 1 3 1 , , ( )P at t a t t′ ≡ +
2 3 2 3 , , ( )Q at t a t t′ ≡ +
: : 1TP TP λ′ =
3 2
1 2
t t
t tλ
−=
−
3 2
1 2
t tTP
TP t t
′ −=
−
1 3
1 2
t tTQ
TQ t t
′ −=
−
1TP TQ
TP TQ
′ ′= =
Mathematics150
14. (a,d) 2( , 2 )P at at ⇒ 2
16 8,
a aQ
tt
−
1
2OPQ OP OQ∆ = ⋅
⇒ 2
2
1 (4) 164 4 3 2
2
aat t
t t+ ⋅ + =
2 3 2 4 0t t− + =
⇒ 2, 2 2t =
Hence, 2
2( , 2 ) ,2
tP at at P t
=
2t = ⇒ (1, 2)P
2 2t = ⇒ (4,2 2)P
15. (a,c)1
1 1
x
x x
dy ye
dx e e+ =
+ +
In(1 )
. . 11
xx
e x
x
e
eI F dx e e
e
+= = = ++∫
⇒ (1 ) 1xy e dx+ = ∫ ⇒1 x
x cy
e
+=
+
(0) 2y =
⇒ 1c =
⇒ 4
1 x
xy
e
+=
+
⇒ ( 4) 0y − =
⇒ 2
(1 ) ( 4)0
(1 )
x x
x
e x ey
e
+ − +′ = =
+
2
(1 ) ( 4)Let ( )
(1 )
x x
x
e x eg x
e
+ − +=
+
2
2 4(0) 0
2g
−= <
2 2
1 3 21 1
( 1) 0 01 1
1 1
e e eg
e e
+ − − − = < = > + +
(0) ( 1) 0.g g⋅ − < Hence g(x) has a root in between ( 1, 0)−
16. (b,c) Let the family of circles be 2 2 α α 0x y x y c+ − − + =
On differentiation 2 2 α α 0x yy y′ ′+ − − =
Again on differentiation and substituting 'α' we get
2 2 2
2 2 2 01
x yyx y yy y
y
′ +′ ′′ ′′+ + − = ′+
⇒ 2( ) (1 ) 1 0y x y y y y′′ ′ ′ ′− + + + + =
17. (a,d) Differentiability of ( )f x at 0x =
0 0
(0) (0 ) 0 ( )LHD (0 ) lim lim 0
f f gf
δ δ
δ δδ δ
−
→ →
− − + − ′ = = =
0 0
(0 ) (0) ( )RHD (0 ) lim lim 0
f f gf
δ δ
δ δδ δ
+
→ →
+ − ′ = = =
⇒ ( )f x is differentiable at 0x =
Differentiabiligy of ( )h x at 0x =
(0 ) 1, ( )h h x+′ = is an even function
Hence non diff. at 0x =
Differentiability of ( ( ))f h x at 0x =
| |( ( )) ( )xf h x g e x R= ∀ ∈
0
( (0)) ( (0 ))LHD ( (0 )) lim
f h f hf h
δ
δδ
−
→
− −′ =
0
(1) ( )lim (1)
g g eg
δ
δ δ→
−′= =
0
( (0 )) ( (0))RHD ( (0 )) lim
f h f hf h
δ
δδ
+
→
+ −′ =
0
( ) (1)lim (1)
g e gg
δ
δ δ→
−′= = Since (1) 0g ′ ≠
⇒ ( ( ))f h x is non diff. at 0x =
Differentiability of ( ( ))h f x at 0x =
( ( )|, 0( ( ))
1, 0
f xe xh f x
x≠==
0
( (0)) ( (0 ))LHD. ( (0 )) lim
h f h fh f
δ
δδ
δ→
− −′ − =
| ( )|
0
1 | ( ) |lim 0
| ( ) |
ge g
g
δ
δ
δδ δ
−
→
− −= ⋅ =
−
18. (a,b,c) Given ( ) sin2
g x x x Rπ
= ∀ ∈
1
( ) sin ( ( ))3
f x g g x =
Also, ( ( ( ))) ,2 2
g g g x x Rπ π ∈ − ∀ ∈
Hence, 1 1
( ) and ( ( )) ,2 2
f x f g x ∈ −
0 0
1 1sin ( ( )) ( ( ))( ) 3 3lim lim
1( ) ( )( ( ))
3
x x
g g x g g xf x
g x g xg g x
→ →
= ⋅
⇒ 0
sin sin2
lim6 6
sin2
x
x
x
ππ π
π→
⋅ =
151Mock Test-1
Range of 1 1
( ( )) sin , sin2 2 2 2
g f xπ π ∈ −
⇒ ( ( )) 1.g f x ≠
19. (c) (A) 3
32
α β+=
3 2 3α β+ =± . . .(i)
Given 2 3α β= + . . .(ii)
From equation (i) and (ii), we get α 2 1or= −
So |α| 1 or 2=
(B) 2
2
13 2,( )
1,
xaxf x
xbx a
<− −=
≥+
For continuity 23 2a b a− − = +
2 3 2a a b+ + = − . . .(i)
For differentiability 6a b− =
6a b= −
2 3 2 0a a− + =
1, 2a =
(C) 2 4 3 2 4 3(3 3 2 ) (2 3 3 )n nω ω ω ω+ +− + + + −
2 4 3( 3 2 3 ) 0nω ω ++ − + + =
2 4 3 2 4 3(3 3 2 ) ( (2 3 3 ))n nω ω ω ω ω+ +− + + + −
2 2 4 3( ( 3 2 3) 0nω ω ω ++ − + + =
⇒ 2 4 3 4 8(3 3 2 ) (1 ) 0n n nω ω ω ω+− + + + + =
⇒ 3 ,n k k N≠ ∈
(D) Let 5a d= −
5q d= +
5 2b d= +
⇒ | | | 2 |q a d− =
Given 2
4ab
a b=
+
⇒ 2ab
a b=
+
(5 )(5 2 ) 2(5 5 2 ) 2(10 )d d d d d− + = − + + = +
225 10 5 2 20 2d d d d+ − − = +
22 3 5 0d d− − =
5
1,2
d d= − =
⇒ | 2 | 2,5d =
20. (c) (A) 2
2 2 (given)2
ca b− =
2
2 2 2 244 (sin sin ) sin ( )
2
RR X Y Z− =
⇒ 22(sin( ) sin( ) sin ( )X Y X Y Z− ⋅ + =
⇒ 22 sin( ) sin( ) sin ( )X Y Z Z⋅ − ⋅ =
⇒ sin( ) 1
sin 2
X Y
Zλ
−= =
⇒ cos 0 for n odd integer.2
nπ = =
(B) 1 cos 2 2 cos 2 2 sin sinX Y X Y+ − =
2 2sin sin sin 2sin 0X X Y Y+ − =
(sin sin )(sin 2sin ) 0X Y X Y− + =
⇒ sin sinX Y=
⇒ sin
1.sin
X a
Y b= =
(C) Here, distance of Z from bisector of
3
and2
OX OY =uuur uuur
⇒
2 21 1 9
2 2 2β β − + − =
⇒ 2, 1β = −
⇒ | | 2,1β =
(D) When α 0=
2
0Area 6 2 x dx= − ∫
8 26
3= −
When α 1=
1 2
0 1Area (3 2 ) ( 1 2 )x x dx x x dx= − − + + −∫ ∫
1 22 2
3/ 2 3/ 2
0 1
4 43
2 3 2 3
x xx x x x= − − + + −
8
5 2.3
= −
JEE Advance Paper-II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
9 4 9 8 4 2 9 7 d a,d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b,c,d a,b a,b,c a,c b,c a,b a,b,c c,d a,b c,d
1. (9) 4 3 5= + +r r r r
s p q r
( ) ( ) ( )= − + + + − + + − − +r r r r r r r r r r
s x p q r y p q r z p q r
( ) ( ) ( )= − + − + − − + + +r r r r
s x y z p x y z q x y z r
⇒ 4− + − =x y z
Mathematics152
⇒ 3− − =x y z
⇒ 5+ + =x y z
On solving we get 9 7
4, ,2 2
= = = −x y z
⇒ 2 9+ + =x y z
2. (4)
127 7
1
3(4 2) 7
1
112
43
1
π π
π
=
−
=
Σ −
= =
Σ −
ki i
k
ii k
k
e e
e e
3. (9) Let seventh term be ‘a’ and common difference be ‘d’
Given 7
11
6
11=
S
S ⇒ 15=a d
Hence, 130 15 140< <d ⇒ 9=d
4. (8)9x can be formed in 8 ways
i.e., 9 1 8 2 7 3 6 4 5 1 2 6 1 3 5 2 3 4, , , , , , ,+ + + + + + + + + +x x x x x x x x and
coefficient in each case is 1
⇒ Coefficient of 9
8 times1 1 1 ......... 1 8= + + + + =x
5. (4) Let point of concurrent is
Equation of normal at is,
It is passes through , then
. . .(i)
But or . . .(ii)
Value of from equation (ii), putting in equation (i),
we get
Arranging above as a fourth degree equation in we get
Above equation being of fourth degree in therefore
roots of the above equation are then
. . .(iii)
. . .(iv)
Multiplying equation (iii) and (iv), we get
6. (2)
( )cos
0lim
2
α
α α→
−= −
n
m
e e e
( )( )
(cos( ) 1)
20
(cos 1)lim
cos( ) 1
α
α
α
α α α
−
→
−
−
nn
n m n
e e 2
2α = −n e
if and only if 2 0− =n m
7. (9)1
21
(9 3tan )
20
12 9
1α
−+ +=
+ ∫ x x x
e dxx
Put 19 3tan−+ =x x t
⇒ 2
39
1
+ = +
dx dtx
⇒ 33
9944
01
ππ
α++
= = −∫ te dt e
⇒ 3
log |1 | 94
πα
+ − =
e
8. (7) 1
1(1) | ( ( )) | 0
−= =∫G t f f t dt ( ) ( )− = −f x f x
Given 1
(1)2
=f
1 1
( ) (1)
( ) (1) 11lim lim( ) (1)( ) | ( (1)) | 14
1
→ →
−−= = =−−
x x
F x F
F x fxG x GG x f f
x
⇒ 1/ 2 1
| (1/ 2) | 14=
f
⇒1
7.2
=
f
9. (D) 3 3
1/ 2 1/ 2
192 192( )
3 2≤ ≤∫ ∫
x x
t dt f x t dt
4 4 316 1 ( ) 24
2− ≤ ≤ −x f x x
1 1 1
4 2
1/ 2 1/ 2 1/ 2
3(16 1) ( ) 24
2
− ≤ ≤ −
∫ ∫ ∫x dx f x dx x dx
1
1/ 2
26 391 ( ) 12
10 10< ≤ ≤ <∫ f x dx
10. (a,d) Here, 2
1 20 ( ) 1< − <x x
⇒ 2
1 2 1 20 ( ) 4 1< + − <x x x x
⇒ 2
10 4 1
α< − <
⇒ 1 1 1 1
, ,2 25 5
α
∈ − − ∪
( , )h k
( , )x y′ ′ 1
2 2/ /
y yx x
x a y b
′′ −−=
′ ′
( , )h k
2 2 2 2 4 2 2 ( ) y a h x b x b k x′ ′ ′ ′− + =
2 2
2 21
x y
a b
′ ′= =
22 2 2
2( )
by a x
a′ ′= −
2y′
22 2 2 2 2 2 4 2 2
2( ) ( )
ba x a h b a x b k x
a′ ′ ′− + − =
⇒2
2 2 4 2 2 2 2 2 2 2 2 4 2 2
2( ) ( ) 2 ( )
ba x a h b a x a hx b a b k x
a′ ′ ′ ′− + − + − =
,x′
⇒ 2 2 2 4 2 2 2 3 2 4 2 2 6 2( ) 2 ( ) ( ) 2 ( ) 0a b x ha a b x x a h a b x a h′ ′ ′ ′− − + − + − − + =K
,x′
1 2 3 4, , ,x x x x
2 2 2 2
1 2 3 4 2 2 2 2 2
2 ( ) 2( )
( ) ( )
ha a b hax x x x
a b a b
−+ + + = − =
− − −
1 2 3
1 2 3 4 1 2 3 4
1 1 1 1 x x x
x x x x x x x x
Σ+ + + =
⋅ ⋅ ⋅ 4 2 2
2 22 2 2
6 2 2
2 2 2
2 ( )
2( )( )
( )
a h a b
a ba b
a h a h
a b
−−− −
= =
− −
1 2 3 4
1 2 3 4
1 1 1 1( ) 4x x x x
x x x x
+ + + × + + + =
153Mock Test-1
11. (b,c,d)3
,2 2
π πα π π β< < < <
⇒ 3 5
2 2
π πα β< + <
⇒ sin 0; cos 0β α< <
⇒ cos( ) 0.α β+ >
12. (a,b) For the given line, point of contact for
2 2
1 2 2: 1+ =
x yE
a b is
2 2
,3 3
a b
and for
2 2
2 2 2: 1+ =
x yE
B A is
2 2
,3 3
B A
Point of contact 3+ =x y of and circle is (1, 2)
Also, general point on 3+ =x y can be taken as
1 , 22 2
±
m
r r where,
2 2
3=r
So, required points are 1 8
,3 3
and 5 4
,3 3
Comparing with points of contact of ellipse,
2 25, 8= =a B
2 24, 1= =b A
∴ 1 2
7
2 10=e e and 2 2
1 2
43
40+ =e e
13. (a,b,d) Tanget at P, 1 1 1− =xx yy intersects x-axis at
1
1, 0
Mx
Slope of normal 1 1
1 1 2
0−= − =
−y y
x x x
⇒ 2 1 12 (2 ,0)= ⇒ ≡x x N x
For centroid 1
1 1
13
,3 3
+= =l
xx y
m
2
1 1
11
3= −
ld
dx x
1 1
21 1 1 1
1 1,
3 3 3 1= = =
−
dy xdm dm
dy dx dx x
14. (a,c) Let 1 6 4
0(sin at cos at)
π+ =∫ e dt A
2
1 6 4(sin at cos at)π
π= + +∫I e dt
Put π= +t x
=dt dx
for 2=a as well as 4=a
Similarly
3
1 6 4 2
2(sin at cos at)
π π
π+ =∫ e dt e A
So,
2 3 43 1
1
π π π π
π
+ + + −= =
−A e A e A e A
LA e
For both 2, 4=a
15. (b,c) Let ( ) ( ) 3 ( )= −H x f x g x
( 1) (0) (2) 3.− = = =H H H
Applying Rolle’s Theorem in the interval [–1, 0]
( ) ( ) 3 ( ) 0′ ′ ′= − =H x f x g x for atleast one ( 1, 0)∈ −c
As ( )′′H x never vanishes in the interval
⇒ Exactly one ( 1,0)∈ −c for which ( ) 0′ =H x
Similarly, apply Rolle’s Theorem in the interval [0, 2]
⇒ ( ) 0′ =H x has exactly one solution in (0, 2)
16. (a,b) 6 2 2( ) (7 tan 3 tan ) (tan 1)= − +f x x x x
/ 4 / 4
6 2 2
0 0( ) (7 tan 3tan )sec
π π= −∫ ∫f x dx x x xdx
⇒ / 4
0( ) 0
π=∫ f x dx
/ 4/ 4 / 4
0 00( ) ( ) ( )
ππ π = − ∫ ∫ ∫ ∫xf x dx x f x dx f x dx dx
/ 4
0
1( )
12
π=∫ xf x dx
17. (a,b,c) (a) ( ) ( ) ( )′ ′= +f x f x xF x
(1) (1) (1)′ ′= +f F F
(1) (1) 0′ ′= <f F
(1) 0′ <f
(b) (2) 2 (2)=F F
( )F x is decreasing and (1) 0=F
Hence (2) 0<F
⇒ (2) 0<f
(c) ( ) ( ) ( )′ ′= +f x F x xF x
( ) 0 (1, 3)< ∀ ∈F x x
( ) 0 (1, 3)′ < ∀ ∈F x x
Hence ( ) 0 (1, 3)′ < ∀ ∈f x x
Mathematics154
18. (c,d)3 3
1 1( ) ( )=∫ ∫f x dx xF x dx
32
32
11
1( ) ( )
2 2
′= −
∫
xF x x F x dx
9 1(3) (1) 6 12
2 2= − + = −F F
3
3 3 2
11
40 [ ( )] 3 ( )′ ′= − ∫x F x x F x dx
40 27 (3) (1) 36′ ′= − +F F . . .(i)
( ) ( ) ( )′ ′= +f x F x xF x (3) (3) 3 (3)′ ′= +f F F
(1) (1) (1)′ ′= +f F F 9 (3) (1) 32 0′ ′− + =f f
19. (a,b) P (Red Ball) (1) ( | ) ( ) ( | )= ⋅ + ⋅P P R I P II P R II
1 ( ) ( | )
( | )3 ( ). ( | ) ( ) ( | )
⋅= =
+ ⋅P II P R II
P II RP I P R I P II P R II
3
3 4
31
1 2 3 4
1
3
+=
++ +
n
n n
nn
n n n n
of the given options (a) and (b) satisfy above condition
20. (c,d) P (Red after Transfer) = P (Red Transfer), P (Red
Transfer in II Case)
+ P (Black Transfer). P (Red Transfer in II Case)
1 1
1 2 1 2
( 1)( )
( 1)
−= +
+ + −
n nP R
n n n n
2 2
1 2 1 2
1.
1 3=
+ + −
n n
n n n n
of the given option 3 and 4 satisfy above condition
155Mock Test-2
JEE-MAIN: MATHEMATICS MOCK TEST-2
1. If is given by then
equals:
a. b.
c. d.
2. Let and be the roots of the equation
The equation whose roots are is
a. b.
c. d.
3. Let then
is equal to
a. 0
b.
c.
d.
4. The value of is
a. –1 b. 0
c. –i d. i
5. The inverse of the matrix is
a. b.
c. d.
6. The sum of the first five terms of the series
will be
a. b.
c. d.
7. If then the
expression has the value
a. 32 b. 63
c. 64 d. None of these
8.
a. b. c. d.
9. is
equal to
a. b.
c. d.
10. There are 10 lamps in a hall. Each one of them can be
switched on independently. The number of ways in which
the hall can be illuminated is.
a. b. 1023
c. d. 10 !
11. A fair coin is tossed repeatedly. If tail appears on first four
tosses, then the probability of head appearing on fifth toss
equals
a. b.
c. d.
12. If then
a. 2 b. 4 c. 8 d. 16
13. If and
,then is equal to
a. b. c. d.
14. Find imaginary part of
a. b.
c. 0 d. None of these
15. An observer on the top of a tree, finds the angle of
depression of a car moving towards the tree to be 30°.
After 3 minutes this angle becomes 60°. After how much
more time, the car will reach the tree
a. 4 min b. 4.5 min c. 1.5 min d. 2 min
: [1, ) [2, )f ∞ → ∞1
( )f x xx
= +
1( )f x−
2 4
2
x x+ −21
x
x+
2 4
2
x x− − 21 4x+ −
α β 2 1 0x x+ + =19 7,α β
2 1 0x x− − = 2 1 0x x− + =2 1 0x x+ − = 2 1 0x x+ + =
22 2cos sin , 1,n i i
n n
π πω = + = −
2
3 3( )x y zω ω+ + 2
3 3( )x y zω ω+ +
2 2 2x y z+ +2 2 2x y z yz zx xy+ + − − −2 2 2x y z yz zx xy+ + + + +
6
1
2 2sin cos
7 7k
k ki
π π
=
−
∑
1 0 0
0 1 0
0 0 1
0 0 1
0 1 0
1 0 0
1 0 0
0 1 0
0 0 1
0 1 0
0 0 1
1 0 0
1 0 0
0 0 1
0 1 0
1 33 4 6 ......
2 4+ + +
939
16
318
16
739
16
913
16
2 6 2 12
1 2 12(1 2 ) 1 .... ,x x a x a x a x+ − = + + + +
2 4 6a a a+ + ....+
12a+
1 2 1 2 3 1 2 3 41
2! 3! 4!
+ + + + + ++ + + + ∞ =K
e 3e / 2e 3 / 2e
1 1 1log 2 log 1 log 1 .... log 1
2 3 1e e e e
n
+ + + + + + + −
log 1e
loge
n
log (1 )e
n+ log (1 )e
n−
210
102
1
2
1
32
31
32
1
5
22
2
tan 60 cosec30sin 45 cos 60 ,
sec 45 cot 30x
° °° ° =
° °x =
sin sin 2 sin3 sinθ θ θ α+ + =
cos cos 2 cos3 cosθ θ θ α+ + = θ
/ 2α α 2α / 6α
1 5 7 9sin
16
i− −
log 2 log 2−
Mathematics156
16. Let If be
continuous for all x, then
a. 7 b. –7
c. d. None of these
17. If ....
,x
x ex ey e
+ ∞++= then dy
dx=
a. 1
y
y− b.
1
1 y−
c.
1
y
y+ d.
1
y
y −
18. One maximum point of sin cosp qx x is
a. 1tan ( / )x p q
−= b. 1tan ( / )x q p
−=
c. 1tan ( / )x p q−= d. 1tan ( / )x q p−=
19.
a. b.
c. d.
20. Area bounded by the parabola and the ordinates
is
a. b.
c. d. None of these
21. The solution of the differential equation is
a. b.
c. d.
22. The solution of the differential equation
is
a. b.
c. d. None of these
23. The vertex of an equilateral triangle is (2,–1) and the
equation of its base in The length of its sides
is
a. b.
c. d.
24. If the equation 2 2( 1) ( 2)
13 4
K x y+ ++ = represents a circle,
then K =
a. 3/4 b. 1 c. 4/3 d. 12
25. The equation of the common tangent to the curves
and is
a. b.
c. d.
26. The vectors and are non-collinear. The value of x for
which the vectors and
are collinear, is
a. 1 b.
c. d. None of these
27. If vectors satisfy the condition
then is equal to
a. 0 b. –1 c. 1 d. 2
28. The angle between the straight lines
and is
a. b. c. d.
29. A variable plane is at a constant distance p from the origin
and meets the axes in A, B and C. The locus of the
centroid of the tetrahedron is
a. b.
c. d. None of these
30. is equal to
a. b.
c. d.
3 2
2
16 20,if 2
( ) .( 2), if 2
x x xx
f x xk x
+ − + ≠= − =
( )f x
k =
7±
2
1
(log )dx
x x=∫
1
logc
x+
1
logc
x− +
log log x c+ log log x c− +
22y x=
1, 4x x= =
4 2.
3sq unit
28 2.
3sq unit
56
3sq. unit
210
dy x
dx x
++ =
11tan
2y x c−= − +
2
log 02
xy x c+ + + =
11tan
2y x c−= +
2
log2
xy x c− − =
cos log(sec tan ) cos logy x x dx x+ = (sec tan )y y dy+2 2sec secx y c+ = sec secx y c+ =
sec secx y c− =
2 1.+ =x y
4 / 15 2 / 15
4 / 3 3 1/ 5
2 8y x= 1xy = −
3 9 2y x= + 2 1y x= +
2 8y x= + 2y x= +
ar
br
( 2)c x a b= − +rr r
(2 1)d x a b= + −r rr
2
1
3
1
,, ca brrr
| | | |,a c b c− = −rr r r
( )2
a bb a c
+− ⋅ −
rrr r r
1 2 3
2 5 4
x y z+ − += =
1
1
x −=
2
2
y + 3
3
z −=
−°45 °30 °60 °90
OABC2 2 2 2
16x y z p− − − −+ + = 2 2 2 1
16x y z p− − − −+ + =
2 2 216x y z
− − −+ + =
~ ( )∨p q
~ ~∨p q ~ ~∧p q
~ ∨p q ~∨p q
Space for rough work
157Mock Test-2
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. The coefficient of in the expansion of
is
2. The minimum value of the sum of real numbers
5 4 3 8a , a , 3a , 1, a− − − and 10a with a 0> is
3. The value of
3
2
1 1 1 16 log 4 4 4 is
3 2 3 2 3 2 3 2
+ − − −
4. The number of distinct solutions of the equation
2 4 45cos 2 cos sin
4x x+ + 6cosx + 6sin 2x x+ = in the
interval is
5. A cylindrical container is to be made from certain solid
material with the following constraints: It has a fixed
inner volume of 3mm ,V has a 2 mm thick solid wall and
is open at the top. The bottom of the container is a solid
circular disc of thickness 2 mm and is of radius equal to
the outer radius of the container.
If the volume of the material used to make the container is
minimum when the inner radius of the container is 10
mm, then the value of is
6. Let the curve C be the mirror image of the parabola
with respect to the line If A
and B are the points of intersection of C with the line
then the distance between A and B is
7. Let be a continuous odd function, which vanishes
exactly at one point and Suppose that
for all and
for all If then the value of is
8. Consider the set of eight vectors
Three non-coplanar vectors
can be chosen from ways. Then p is.
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. Let f be a non-negative function defined on the interval
1[0, ]. If 2
0
1 ( ( ))′−∫x
f t 0
( ) , 0 1,= ≤ ≤∫x
dt f t dt x and
(0) 0,=f then
a. 1 1
2 2
<
f and 1 1
3 3
>
f
b. 1 1
2 2
>
f and 1 1
3 3
>
f
c. 1 1
2 2
<
f and 1 1
3 3
<
f
d. 1 1
2 2
>
f and 1 1
3 3
<
f
10. Let α and β be the roots of 2 6 2 0,x x− − = with .α β> If
n n
na α β= − for 1,n ≥ then the value of 10 8
9
2
2
a a
a
−is
a. 1 b. 2
c. 3 d. 4
11. Let cos sin .θ θ= +z i Then the value of 15
2 1
1
Im( )−
=∑ m
m
z at
2θ = ° is
a. 1
sin 2° b.
1
3 sin 2°
c. 1
2 sin 2° d.
1
4 sin 2°
12. If is a matrix satisfying the equation
where I is identity matrix, then the
ordered pair (a, b) is equal to:
a. b.
c. d.
13. The sum of first 20 terms of the sequence
is
a. b.
c. d.
9x
2 3 100(1 ) (1 ) (1 ).....(1 )x x x x+ + + +
[0, 2 ]π
250πV
24=y x 4 0.+ + =x y
5,= −y
:f R R→
(1) .2
f1
=
1( ) ( )
x
F x f t dt−
= ∫ [1 1, 2]x∈ −1
( ) | ( ( )) |x
G x t f f t dt−
= ∫
[ 1, 2].x∈ −1
( ) 1lim ,
( ) 14x
F x
G x→= 1
2f
ˆˆ ˆ ; , , 1,1.V ai bj ck a b c= + + ∈ −
2pV in
1 2 2
2 1 2
2
= −
A
a b
9 ,=TAA I 3 3×
(2, 1)− ( 2,1)−
(2,1) ( 2, 1)− −
0.7, 0.77, 0.777, .....
207(179 10 )
81
−− 207(99 10 )
9
−−
207(179 10 )
81
−+ 207(99 10 )
9
−+
Mathematics158
14. Coefficient of in the expansion of
is
a. 1051 b. 1106
c. 1113 d. 1120
15. The number of seven digit integers, with sum of the digits
equal to 10 and formed by using the digits 1, 2 and 3 only,
is
a. 55 b. 66
c. 77 d. 88
16. The mean of the data set comprising of 16 observations is
16. If one of the observation valued 16 is deleted and three
new observations values 3, 4 and 5 are added to the data,
then the mean of the resultant data, is
a. 16.8 b. 16.0
c. 15.8 d. 14.0
17. Let ABC be a triangle such that 6
ACBπ
∠ = and let a, b
and c denote the lengths of the sides opposite to A, B and
C respectively. The value (s) of x for which
2 21, 1a x x b x= + + = − and 2 1c x= + is (are)
a. (2 3)− + b. 1 3+
c. 2 3+ d. 4 3
18. Let where and
Then equals:
a.
b.
c. d.
SECTION 3 Contains 2 Match The Following Type Questions
You will have to match entries in Column I with the entries in
Column II.
19. Match the Column:
Column I Column II
(A) In if the magnitude
of the projection vector
of the vector
on is and
if then
possible value(s) of
is (are)
1. 1
(B) Let a and b be real
numbers such that the
function
is
differentiable for all
Then possible
value(s) of a is (are)
2. 2
(C) Let be a
complex cube root of
unity. If
then possible value(s)
of n is (are)
3. 3
(D) Let the harmonic mean
of two positive real
numbers a and b be 4. If
q is a positive real
number such that a, 5,
q, b is an arithmetic
progression, then the
value(s) of is
(are)
4. 4
5. 5
20. Consider the following linear equations
0+ + =ax by cz
0+ + =bx cy az
0+ + =cx ay bz
Match the conditions / expressions in Column I with
statements in Column II
Column I Column II
(A) 0+ + ≠a b c and
2 2 2+ + = + +a b c ab bc ca
1. the equations represent
planes meeting only at
a single point.
(B) 0+ + =a b c and
2 2 2a b c ab bc ca+ + ≠ + +
2. the equations represent
the line .= =x y z
(C) 0+ + ≠a b c and 2 2 2
a b c ab bc ca+ + ≠ + +
3. the equations represent
identical planes.
(D) 0+ + =a b c and 2 2 2+ + = + +a b c ab bc ca
4. the equations represent
the whole of the three
dimensional space.
11x
2 4 3 7 4 12(1 ) (1 ) (1 )+ + +x x
1( ) (sin cos ),
k k
kf x x xk
= + x R∈ 1.k ≥
4 6( ) ( )f x f x−
1
6
1
3
1
4
1
12
2 ,R
ˆ ˆα β+i j
ˆ ˆ3 +i j 3
α | 2 3β,= +
|α |
2
2
3 2, 1( )
, 1
− − <=+ ≥ax x
f xbx a x
.∈x R
ω 1≠
2 4 3(3 3 2 ) +− +ω ω n
2 4 3(2 3 3 ) ++ + −ω ω n
2 4 3( 3 2 3 ) 0,
++ − + + =ω ω n
| |−q a
Space for rough work
159Mock Test-2
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. The number of distinct real roots of
4 3 24 12 1 0− + + − =x x x x is
2. The number of all possible values of ,θ where 0 ,θ π< <
for which the system of equations
( )cos3 ( )sin 3y z xyzθ θ+ =2cos3 2sin3
sin3xy z
θ θθ = +
( )sin ( 2 )cos3 sin3xyz y z yθ θ θ3 = + + have a solution
0 0 0( , , )x y z with 0 0 0,y z ≠ is
3. Let , Rα β ∈ be such that 2
0
sin( )lim 1.
sinx
x x
x x
βα→
=−
Then
6( )α β+ equals
4. The centres of two circles and each of unit radius
are at a distance of 6 units from each other. Let P be the
mid-point of the line segment joining the centres of
and and C be a circle touching circles and
externally. If a common tangent to and C passing
through P is also a common tangent to and C, then the
radius of the circle C is
5. Suppose that and are three non-coplanar vectors in
Let the components of a vector along and be
4, 3 and 5, respectively. If the components of this vector
along and are x, y
and z, respectively, then the value of is
6. Suppose that the foci of the ellipse are
and where and Let and
be two parabolas with a common vertex at (0, 0) and
with foci at and respectively. Let be
a tangent to which passes through and be
a tangent to which passes through The is
the slope of and is the slope of then the value
of is
7. Let f :IR IR be defined as f(x) = The total
number of points at which f attains either a local
maximum or a local minimum is
8. The total number of distinct x R∈ for which
2 3
2 3
2 3
1
2 4 1 8 0
3 9 1 27
x x x
x x x
x x x
+
+ =
+
is
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. If and
where N is the set of natural numbers, then is
equal to:
a. N
b. Y – X c. X d. Y
10. A value of b for which the equations2
2
x bx 1 0x x b 0,
+ − =+ + =
Have one root in common is
a. 2− b. i 3− c. i 5 d. 2
11. The variance of first 50 even natural numbers is:
a.
b. 833 c. 437 d.
12. The expression can be written as
a. b. sec cosec 1A A +
c. d. sec cosecA A+
13. The value of
a. b.
c. d.
14. A man is walking towards a vertical pillar in a straight
path, at a uniform speed. At a certain point A on the path,
he observes that the angle of elevation of the top of the
pillar is After walking for 10 minutes from A in the
same direction, at a point R, he observes that the angle of
elevation of the top of the pillar is Then the time
taken (in minutes) by him, from B to reach the pillar, is:
a. 6 b. 10
c. 20 d. 5
1C 2C
1C
2C 1C 2C
1C
2C
,p qr r
rr
3.R sr
,p qr r
rr
sr
( ),p q r− + +r r r
( )p q r− +r r r
( )p q r− − +r r r
2x y z+ +
2 2
19 5
x y+ =
1( , 0)f 2( , 0)f 1 0f > 2 0.f < 1P
2P
1( , 0)f 2(2 , 0),f 1T
1P 2(2 , 0)f 2T
2P 1( , 0).f 1m
1T 2m 2 ,T
2
22
1m
m
+
→ 2| | | 1 | .x x+ −
4 3 1: nX n n N= − − ∈ 9( 1) : ,Y n n N= − ∈
X Y∪
833
4
437
4
tan cot
1 cot 1 tan
A A
A A+
− −
sin cos 1A A +
tan cotA A+
231
1 1
cot cot 1 2 isn
n k
k−
= =
+
∑ ∑
23
25
25
23
23
24
24
23
30 .°
60 .°
Mathematics160
15. is equal to
a. b. c. 1 d. 2
16. Let the function be given by
Then, is
a. even and is strictly increasing in
b. odd and is strictly decreasing in
c. odd and is strictly increasing in
d. neither even nor odd, but is strictly increasing in
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem.
Two multiple choice questions will be asked based on this paragraph.
One or more than one option can be correct.
Paragraph for Question No. 17 to 18
A box contains 1 white ball, 3 red balls and 2 black balls.
Another box contains 2 white balls, 3 red balls and 4 black
balls. A third box B3 contains 3 white balls, 4 red balls and 5
black balls.
17. If 1 ball is drawn from each of the boxes and
the probability that all 3 drawn balls are of the same
colour is
a. b. c. d.
18. If 2 balls are drawn (without replacement) from a
randomly selected box and one of the balls is white and
the other ball is red, the probability that these 2 balls are
drawn from box is
a. b. c. d.
Paragraph for Question No. 19 to 20
Suppose we define the definite integral using the following
formula for more accurate
result for
When
19. is equal to
a. b.
c. d.
20. If and c is a point such that
and is the point lying on the curve for
which is maximum, then is equal to
a. b.
c. d. 0
0
(1 cos2 ) (3 cos )lim
tan 4x
x x
x x→
− +
1
4−
1
2
: ( , ) ,2 2
π π −∞ ∞ → −
g
1( ) 2 tan ( ) .2
π−= −ug u e
g
(0, )∞
( , )−∞ ∞
( , )−∞ ∞
( , )−∞ ∞
1B
2B
1 2,B B 3 ,B
82
648
90
648
558
648
566
648
2B
116
181
126
181
65
181
55
181
( )2
b
a
d af x dx
−=∫ ( ( ) ( )),f a f b+
( , ) ( ) ( ( ) ( ))2
c ac a b F c f a f c
−∈ = + +
2
b c−
( ( ) ( )).f b f c+ ,2
a bc
+=
( ) ( ( ) ( ) 2 ( )).4
b
a
b af x dx f a f b f c
−= + +∫
/ 2
0
sin x dx
π
∫
(1 2)8
π+ (1 2)
4
π+
8 2
π4 2
π
( ) 0 ( , )f x x a b′′ < ∀ ∈
,a c b< < ( , ( ))c f c
( )F c ( )f c′
( ) ( )f b f a
b a
−−
2( ( ) ( ))f b f a
b a
−−
2 ( ) ( )
2
f b f a
b a
−−
Space for rough work
161Mock Test-2
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
a d c d b a d d b b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
a c a b c a a a b b
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
b d b a d c a d a b
1. (a) Use the identity
replace x by in the given function we get
solve to find
2. (d) Given
But and
Hence the equation will be same.
3. (c)
and
∴
4. (d)
5. (d) It is understandable.
6. (a) Given series is
(in G.P.)
Here then sum of the five terms
7. (d) .
Putting and and adding the results
8. (d)
9. (b) The given series reduces to
10. (b) corresponds to none of the lamps is
being switched on.
11. (a) The event that the fifth toss results in a head is
independent the event that the first four tosses result in tails.
Probability of the required event = ½
12. (c) .
13. (a)
. . .(i)
Now,
( 1( ))f f x x− =
1( ),f x−
1
1( 1( )) 1( )
( )f f x f x
f x−− = − +
⇒1
11( ) ,
( )x f x
f x−= − + 1 .f x−
2 1 0x x+ + =
∴ 1 1 1[ 1 3] ( 1 3), ( 1 3)
2 2 2x i i i= − ± = − + − − 2,ω ω=
19 19α ω ω= = 7 14 2 .β ω ω= =
2 2cos sinn i
n n
π πω = +
⇒ 3
2 2 1 3cos sin
3 3 2 2
ii
π πω ω= + = − + =
2
2
3
2 2 4 4cos sin cos sin
3 3 3 3i i
π π π πω = + = +
21 3.
2 2
iω= − − =
2 2
3 3 3 3( ) ( )x y z x y zω ω ω ω+ + + +2 2( ) ( )x y z x y zω ω ω ω= + + + +
2 2 2 .x y z xy yz zx= + + − − −
6
1
2 2sin cos
7 7k
k ki
π π
=
−
∑
6
1
2 2cos sin
7 7k
k ki i
π π
=
= − +
∑2
6 7
1
i k
k
i e
π
=
= − ∑
2 / 7 4 / 7 6 / 7 8 / 7 10 / 7 12 / 7 i i i i i ii e e e e e eπ π π π π π= − + + + + +
12 / 72 / 7
2 / 7
(1 )
1
ii
i
ei e
e
ππ
π
−= −
−
2 / 7 14 / 7
2 / 7
)
1
i i
i
e ei
e
π π
π
−= −
−
14 / 7( 1)ie π =Q
2 / 7
2 / 7
1
1
i
i
ei i
e
π
π
−= − =
−
1 3 273 4 6 ........ 3 .....
2 4 2 4
9+ + + = + + +
2 3 4 53 3 3 33 .....
2 4 8 16= + + + + +
33, ,
2a r= =
55
5
3 33 1 1 1
2 32( 1)
3 111
2 2
na rS
r
− − − = = =− −
243 32 211 3 633 96 39 .
32 16 16 16
− × = = = =
2 6 2 12
1 2 12(1 2 ) 1 ....x x a x a x a x+ − = + + + +
1x = 1x = −
2 464 2(1 ...)a a= + + +
∴2 4 6 12.... 31.a a a a+ + + + =
( 1)
! 2( )!n
n n nT
n n
Σ += =
1 ( 1) 1 1 2
2 ( 1)! 2 ( 1)! ( 1)!
n n
n n n
+ −= = + − − −
1 1 2 ( 2 ) 3.
2 ( 2)! ( 1)! 2 2
e e e
n n
+= + = = − −
3 4log 2 log log log
2 3 1e e e e
n
n
+ + + + − K
log 2 log 3 log 2 log 4 log 3e e e e e= + − + − +K
log ( ) log ( 1)e en n+ − − log .e n=
102 1 1023,− =
∴
1 1 3.2. . 2 8
42 2.3 4 2
xx x= ⇒ = ⇒ =
sin sin3 sin 2 sinθ θ θ α+ + =
⇒ 2sin 2 cos sin 2 sinθ θ θ α+ =⇒ sin 2 (2 cos 1) sinθ θ α+ =
cos cos3 cos 2 cosθ θ θ α+ + =2cos 2 cos cos 2 cosθ θ θ α+ =
Mathematics162
. . .(ii)
From (i) and (ii),
⇒ ⇒ .
14. (b) Imaginary part of
15. (c)
cot 30 cot 60d h h= ° − ° and time = 3 min.
∴ Speed (cot 30 cot 60 )
3
o oh −= per minute
It will travel distance cot 60h ° in
cot 60 3
1.5(cot 30 cot 60 )
o
o o
h
h
×=
− minute.
16. (a) For continuous
⇒
17. (a) x yy e += ⇒ log ( ) logy x y e= +
⇒ 1
1dy dy
y dx dx= + ⇒ .
1
dy y
dx y=
−
18. (a) Let
Put
Point of maxima
19. (b) Put then
20. (b)
Required area
sq. unit.
21. (b) ⇒
On integrating, we get
22. (d)
⇒
Put and
23. (b)
2 2
2 2 1 1| |
51 2AD
− −= =
+
⇒ tan 60AD
BD° =
⇒ 1/ 5
3BD
=
⇒ 1
15BD =
⇒ 2 2 / 15BC BD= = .
⇒ cos 2 (2cos 1) cosθ θ α+ =
⇒ tan 2 tanθ α= 2θ α= / 2θ α=
1 5 7 9sin
16 16
i−
−
9 9log 1 log(2).
16 16
= − + + = −
2lim ( ) (2)x
f x f k→
= =
3 2
22
16 20lim
( 2)x
x x xk
x→
+ − +=
−2
22
( 4 4) ( 5)lim 7.
( 2)x
x x x
x→
− + += =
−
sin .cosp qy x x=
1 1sin .cos .cos cos .( sin )sinp q q pdyp x x x q x x x
dx
− −= + −
1 1 1 1sin .cos cos .sinp q q pdyp x x q x x
dx
− + − += −
0,dy
dx=
2tan
px
q∴ =
⇒ tanp
xq
= ±
∴ 1tan .p
xq
−=
1log ,x t dx dt
x= ⇒ =
2 2
1 1 1 1.
(log ) logdx dt c c
x x t t x= = − + = − +∫ ∫
' ' 2CDD C ABCD= = ×
41/ 2
1
28 22 2
3x dx= =∫
210
dy x
dx x
++ =
1 0dy x dx
x
+ + =
2
log 0.2
xy x c+ + + =
cos log(sec tan ) cos log(sec tan )y x x dx x y y dy+ = +
sec log(sec tan )y y y dy+∫sec log(sec tan )x x x dx= +∫log(sec tan )x x t+ = log(sec tan )y y z+ =
2 2[log(sec tan )] [log(sec tan )].
2 2
x x y yc
+ += +
60o
A (2, –1)
C D
x + 2y –1 = 0
xy 2=
x =1 x = 4
D C
X B A
D′ O
Y
C′
30° 60°
h
d
163Mock Test-2
24. (a) It represents a circle, if
.
25. (d) Tangent to the curve is
So, it must satisfy
Since, it has equal roots.
z
Hence, equation of common tangent is
26. (c) Since and are
collinear, therefore
or
are linearly independent)
27. (a)
and
∴
Therefore,
28. (d)
29. (a)Plane is , where
or . . .(i)
Now according to equation,
Put the values of x, y, z in (i), we get the locus of the
centroid of the tetrahedron.
30. (a) .
JEE Advance Paper-I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
8 8 4 8 4 4 7 5 c c
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
d d c c c d b d c a
1. (8) can be formed in 8 ways
i.e.,
and coefficient in each case is 1
Coefficient of
2. (8) ( )5 4 3 8 10
1/85 4 3 3 8 103( ) 1
( ) (1) ( )( ) 11 1 3 1 1 1
a a a a aa a a a a
− − −− − −+ + + + +
≥ =+ + + + +
(using AM GM)≥
∴ 5 4 3 8 103 1 8− − −+ + + + + ≥a a a a a
3. (4) 3/ 2
1 1 1 16 log 4 4 4 ...
3 2 3 2 3 2 3 2+ − − −
1 1
put 4 4 ...3 2 3 2
x = − −
2 4
3 2
xx = −
23 2 12 2x x= −
23 2 12 2 0x x+ − =
⇒ 4 2
; 3/ 23
x x= = − not possible
⇒ 3/ 2
1 4 26 log
33 2
+ ×
⇒ 3/ 2
46 log
9
+
⇒ 6 2 4− =
4. (8)
Number of solutions = 8
5. (4) Let inner radius be r and inner length be
Volume of material be M
a b=
⇒1
3 4
k= ⇒
3
4k =
2 8y x= 2 / .y mx m= +
1.xy = −
⇒ 22 21 1 0x mx mx x
m m
+ = − ⇒ + + =
∴ 0D =
⇒ 3
2
44 0 1m m
m− = ⇒ =
⇒ 1m = 2.y x= +
( 2)c x a b= − +rr r
(2 1)d x a b= + −r rr
c dλ=rr
⇒ ( 2) (2 1)x a b x a bλ λ− + = + −r rr r
[( 2) (2 1)] ( 1) 0x x a bλ λ− − + + + =rr
( 2) (2 1) 0, 1 0x xλ λ− − + = + =
( ,a brr
Q
⇒ 2 2 1 0x x− + + =
⇒1
.3
x =
( ) ( )2 2 2
a b a b ab a c b c b a c a b
+ +− ⋅ − = ⋅ − ⋅ − ⋅ + +
r rr r rr r r rr r r r r r
| | | |a c b c− = −rr r r
⇒ 2 2| | | |a c b c− = −rr r r
2a b c+ =rr r
( ) 02
a bb a c
+− ⋅ − =
rrr r r
1 1(2 10 12)cos cos (0)
4 25 16 1 4 9θ − −+ −
= =+ + + +
⇒ 90θ = °
1x y z
a b c+ + =
2
1
(1/ )p
a=
Σ
2 2 2 2
1 1 1 1
a b c p+ + =
, ,4 4 4
a b cx y z= = =
qpqp ~~)(~ ∧≡∨
9x
9 1 8 2 7 3 6 4 5 1 2 6 1 3 5 2 3 4, , , , , , ,+ + + + + + + + + +x x x x x x x x
⇒ 9
8 times1 1 1 ......... 1 8= + + + + =x
2 4 4 6 65cos 2 cos sin cos sin 2
4x x x x x+ + + + =
⇒2 2 25
cos 2 5cos sin 04
x x x− =
⇒2tan 2 1, where 2 [0, 4 ]x x π= ∈
l
2r Vπ =l
2 2( 2) ( 2)M r rπ π= + + −l l
2 3
4 88 0 4
dM V Vr
dr r rπ π= − − + + +
Mathematics164
⇒
⇒
6. (4) Image of
about the line is
7. (7)
⇒
Given
⇒
8. (5) Let be vectors
rest of the vectors are and let us
find the number of ways of selecting co-planar vectors.
Observe that out of any 3 coplanar vectors two will be
collinear (anti parallel)
Number of ways of selecting the anti parallel pair = 4
Number of ways of selecting the third vector = 6
Total = 24
Number of non co-planar selections
Alternate
Required value
9. (c)2
1′ = ± −f f
⇒ ( ) sin=f x x or ( ) sin= −f x x (not possible)
⇒ ( ) sin=f x x
Also, sin 0> ∀ >x x x
10. (c)α is a roots of equation
2 26 2 0; 6 2 0Bα α β− − = − − =
2 6 2 0α α− − =
⇒ 2 2 6α α− =
⇒ 10 10 8 8
10 8
9 9
9
2 ( ) 2( )
2 2( )
a a
a
α β α βα β
− − − −=
−
8 2 8 2
9 9
( 2) ( 2)
2( )
α α β βα β
− − −=
−
8 8
9 9
.(6 ) (6 )
2( )
α α β βα β
−=
−
9 9
9 9
6 ( )3
2 2( )
α βα β
−− =
−
11. (d) sin sin3 ... sin 29θ θ θ= + + +X
2(sin ) 1 cos2 cos2 cos4 ... cos28 cos30θ θ θ θ θ θ= − + − + + −X
⇒ 1 cos30 1
2sin 4sin 2
θθ
−= =
°X
12. (d)
Equation
. . .(i)
and
. . .(ii)
. . .(iii)
Solving
13. (c)
0 when 10dM
rdr
= =
1000V π=
4250
V
π=
5y = −
4 0x y+ + = 1x =
⇒ Distance 4AB =
1
1(1) | ( ( )) | 0
−= =∫G t f f t dt
( ) ( )− = −f x f x
1(1)
2=f
1 1
( ) (1)
( ) (1) 11lim lim( ) (1)( ) | ( (1)) | 14
1
→ →
−−= = =−−
x x
F x F
F x fxG x GG x f f
x
⇒1/ 2 1
| (1/ 2) | 14=
f⇒
17.
2
=
f
(1,1,1), ( 1,1,1), (1, 1,1), ( 1, 1, 1)− − − − −
, , ,a b c dr rr r
, , ,a b c d− − − −r rr r
8 5
3 24 32 2 , 5C p= − = = =
8 6 4
3!
× ×=
∴ 5p =
9=TAA I
1 2 2 1 2
2 1 2 2 1 2 9
2 2 2
− = −
a
I
a b b
⇒2 2
9 0 4 2 9 0 0
0 9 2 2 2 0 9 0
4 2 2 2 2 4 0 0 9
a b
a b
a b a b a b
+ + + − ⇒ + + + − + +
4 2 0+ + =a b
⇒ 2 4a b+ = −
2 2 2 0a b+ − =
⇒ 2 2 2a b− = −
2 24 0a b+ + =
⇒ 2 2 5a b+ =
2, 1a b= − = −
0.7 0.77 0.777 ..... 0.777...7+ + + +
7[0.9 0.99 0.999 ... 0.999...9]
9= + + + +
7[(1 0.1) (1 0.01) (1 0.001...1) ... (1 0.000...1)]
9= − + − + − + + −
2 3 20
7 1 1 1 120 ...
9 10 10 10 10
= − + + + +
2020
20
11
7 1 7 1 10 11020 . 20 .19 10 9 9 10
110
− −= − = −
−
20
20
7 1 7180 1 [179 10 ]
81 10 81
− − − = +
165Mock Test-2
14. (c)
Possibilities are (0, 1, 2); (1, 3, 0); (2, 1, 1); (4, 1, 0).
Required coefficients
15. (c) Coefficient of 10x in 2 3 7( )+ +x x x
Coefficient of 3x in 2 7(1 )+ +x x
Coefficient of 3x in
3 7 7(1 ) (1 )−− −x x 7 3 7
3 7+ += −C 9
3 7= −C
9 8 77 77
6
× ×= − =
16. (d)
17. (b) Using cosine rule for C∠
2 2 2 2 2
2 2
3 ( 1) ( 1) (2 1)
2 2( 1)( 1)
x x x x
x x x
+ + + − − +=
+ + −
⇒
2
2
2 2 13
1
x x
x x
+ −=
+ +
⇒ 2( 3 2) ( 3 2) ( 3 1) 0x x− + − + + =
⇒ (2 3) 3
2( 3 2)x
− ±=
−
⇒ (2 3),1 3 1 3x x= − + + ⇒ = + as ( 0).x >
18. (d)
19. (A)
. . .(i)
Given . . .(ii)
From equation (i) and (ii), we get
So
(B)
For continuity
⇒ . . .(i)
For differentiability
⇒
⇒
⇒
(C)
(D) Let
Given
⇒
⇒
⇒
⇒
⇒
⇒
20.
a b c
b c a
c a b
∆ = 2 2 21( )[( ) ( ) ( ) ]
2a b c a b b c c a= − + + − + − + −
(A) If 0+ + ≠a b c
and 2 2 2+ + = + +a b c ab bc ca
⇒ 0∆ = and 0= = ≠a b c
⇒ the equations represent identical planes.
(B) 0+ + =a b c and 2 2 2+ + ≠ + +a b c ab bc ca
⇒ 0∆ =
⇒ the equations have infinitely many solutions.
( )+ + +ax by a b z
( )+ = +bx cy b c z
⇒ 2 2( ) ( )− = −b ac y b ac z ⇒ =y z
⇒ 0+ + =ax by cy ⇒ =ax ay ⇒ = =x y z
1 2 32 3 4 11x x x+ + =
∴4 7 12 4 7 12 4 7 12 4 7
0 1 2 1 3 0 2 1 1 4 1( ) ( ) ( ) ( 1)C C C C C C C C C C C= × × + × × + × × + × ×
(1 7 66) (4 35 1) (6 7 12) (1 7)= × × + × × + × × + ×
462 140 504 4 113.= + + + =
1616
ixΣ= ⇒ 256ixΣ =
( ) 16 3 4 5 25214
18 18
ixΣ − + + +
= =
1(sin cos )
4
k k
kf x x= + 6 6
6
1( ) (sin cos )
6f x x x= +
4 4
4
1( ) (sin cos )
4f x x x= +
2
6
1 31 sin 2
6 4f K x
= − 2
4
1 sin 2( ) 1
4 2
xf x
= =
2 2
4 6
1 sin 2 1 sin 2 1 1 1( ) ( )
4 8 6 8 4 6 12
x xf x f x
− = − − − = − =
33
2
α β+=
3 2 3α β+ = ±
2 3α β= +
α 2 1or= −
|α| 1 or 2=
2
2
13 2,( )
1,
xaxf x
xbx a
<− −=
≥+2
3 2a b a− − = +2
3 2a a b+ + = −
6a b− =
6a b= −2
3 2 0a a− + =1,2a =
2 4 3 2 4 3 2 4 3(3 3 2 ) (2 3 3 ) ( 3 2 3 ) 0
n n nω ω ω ω ω ω+ + +− + + + − + − + + =2 4 3 2 4 3 2 2 4 3(3 3 2 ) ( (2 3 3 )) ( ( 3 2 3) 0n n nω ω ω ω ω ω ω ω+ + +− + + + − + − + + =
⇒ 2 4 3 4 8(3 3 2 ) (1 ) 0n n nω ω ω ω+− + + + + =
⇒ 3 ,n k k N≠ ∈
5a d= −
5q d= +
5 2b d= +
| | | 2 |q a d− =
24
ab
a b=
+
2ab
a b=
+
(5 )(5 2 ) 2(5 5 2 ) 2(10 )d d d d d− + = − + + = +2
25 10 5 2 20 2d d d d+ − − = +2
2 3 5 0d d− − =
51,
2d d= − =
| 2 | 2,5d =
Mathematics166
(C) 0+ + ≠a b c
and 2 2 2+ + ≠ + +a b c ab bc ca
⇒ 0∆ ≠
⇒ the equation represent planes meeting at only one point.
(D) 0+ + =a b c
and 2 2 2+ + = + +a b c ab bc ca
⇒ 0= = =a b c
⇒ the equation represent whole of the three dimensional space.
JEE Advance Paper -II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
2 3 7 8 9 4 5 2 d b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b b b d d c a d a a
1. (2) Let 4 3 2( ) 4 12 1 0= − + + − =f x x x x x
3 2 3 2( ) 4 12 24 1 4( 3 6 ) 1′ = − + + = − + +f x x x x x x x
2 2( ) 12 24 24 12( 2 2)′′ = − + = − +f x x x x x
( )′′f x has 0 rl roots f(x)has maximum 2 distinct real roots
as (0) 1.= −f
2. (3) ( )cos3 ( )sin 3 0y z xyzθ θ+ − = . . .(i)
sin 3 (2cos3 ) (2sin 3 )xyz z yθ θ θ= + . . .(ii)
∴ ( ) cos3 (2cos3 ) (2sin 3 ) ( 2 )cos3 sin 3y z z y y z y yθ θ θ θ θ+ = + = + +
(cos3 2sin 3 ) cos3zθ θ θ− = and (sin cos3 )y θ θ− = 0
⇒ 0 sin3 cos3 0θ θ⇒ − = ⇒ ⇒ sin 3 cos3θ θ=
∴ 3 / 4nθ π π= +
3. (7)
3 3 5 52
2
3 50 0
......3! 5!sin
lim limsin
.......3! 5!
x x
x xx x
x x
x x x xx x
β ββ
βα
α→ →
− + −
=−
− − − −
3 23
3 50
.....3!
lim 1
( 1) .......3! 5!
x
xx
x xx
ββ
α→
− +
= =− + − +
⇒ 1 0α − =
⇒ 1,α =
Limit 6 1β= =
⇒ 1
6β =
⇒ 1 7
6( ) 6 1 6 76 6
α β + = + = × =
4. (8)
5. (9)
On solving we get
6. (4) The equation of is and is
Tangent to passes through (–4, 0)
⇒
Also tangent to passes through (2, 0)
⇒
7. (5)
So, total no. of local maxima and local minima is = 5
2 2cos
3α =
1sin
3α =
2 2tanα =
R
⇒2 2
8units.tanα
= =R
4 3 5= + +r r r rs p q r
( ) ( ) ( )= − + + + − + + − − +r r r r r r r r r rs x p q r y p q r z p q r
( ) ( ) ( )= − + − + − − + + +r r r rs x y z p x y z q x y z r
⇒ 4− + − =x y z ⇒ 3− − =x y z ⇒ 5+ + =x y z
9 74, ,
2 2= = = −x y z
⇒ 2 9+ + =x y z
1P2y 8x 0− =
2P
2y 16x 0+ =2y 5x 0− =
1
1
20 m ( 4)
m= − +
⇒2
1
12
m=
2y 16x 0+ =
2
2
40 m 2
m= × −
⇒ 2
2m 2=
⇒ 2
22
1
1m 4
m+ =
2 1fx x x= + −
2
2
2
2
if 11
if –1 x<01
if 0 1–x 1
if 1x –1
xx x
x x
xx
xx
≤ − − − ≤− − +
= ≤ <+ +
≥+
1C
A2 2
P αα
3
2C
1
B
C
R α
167Mock Test-2
8. (2)
3
2 3
3
1 1 1
0 2 6 1 0
0 6 24 2
x
x x x
x
+
⋅ − =
−
⇒ 3 3(12 2) 0x x + =
⇒ 6 36 5 0x x+ − =
⇒ 6 3 36 6 5 5 0x x x+ − − =
⇒ 3 3(6 5)( 1) 0x x− + =
⇒ 3 3 5
1,6
x x= − =
⇒
1/35
1,6
x x
= − =
So, two solution
9. (d)
⇒
10. (b) 2x bx 1 0+ − =
2x x b 0+ + = . . .(i)
Common root is (b 1)x 1 b 0− − − =
⇒ b 1
xb 1
+=
− This value of x satisfies equation (i)
⇒
2
2
(b 1) b 1b 0
b 1(b 1)
+ ++ + =
−−
⇒ b 3i, 3i, 0.= −
11. (b) Variance =
12. (b) Exp.
13. (b)
14. (d) tan
for 2y distance time
= 10 min.
So for y dist time = 5 min.
15. (d)
16. (c)
is odd and
increasing.
17. (a) P (required) = P (all are white) + P (all are red) + P
(all are black)
18. (d) Let A : one ball is white and other is red
Both balls are from box
Both balls are from box
4 3 1,n
x n n N= − − ∈
(1 3) 3 1,nx n n N= + − − ∈
⇒ 0,9,54, ...X = 9( 1),y n n N= − ∈
0,9,18, ...y = ⇒ .x y y∪ =
( )2
21
xx
n−∑
⇒22 2 2 2
2 2 4 6 ... 100 2 4 ... 100
50 50σ
+ + + + + + + −
⇒ 23434 2601 833σ = − =
22
2
tan 1 1 1tan
tan 1 tan tan tan 1 tan
AA
A A A A A
= + = − − − −
2tan tan 1
tan cot 1tan
A AA A
A
+ += = + + sec . 1A cosec A= +
231 2
1
cot cot ( 1)n
n n−
=
+ +
∑
231
1
1cot tan
1 ( 1)n
n n
n n
−
=
+ − + + ∑
⇒ 1 23 25cot tan .
325 23
− =
130
3° = =
+x
y z
⇒ 3 = +x y z
⇒ tan 60 3° = =x
y
⇒ 3= = +x y y z
3 = +y y z
⇒ 2 =y z
0
(1 cos2 ) (3 cos )lim
tan 4x
x x
x x→
− +
2
0
(2sin )(3 cos )lim
tan 44
4
x
x x
xx x
x
→
+=
×
2
20
2sin (3 cos ) 2lim (3 1) 2
4 4x
x x
x→
+= = + =
1( ) 2 tan ( )2
ug u eπ−= −
1 1 1 1 12 tan tan cot tan cot− − − − −= − − = −u u u u ue e e e e
( ) ( )− = −g x g x
⇒ ( )g x ( ) 0′ >g x
⇒
1 2 3 3 3 4 2 4 5 5
6 9 12 6 9 12 6 9 12 12= × × + × × + × × ×
6 36 40 82.
648 648 648 648= + + =
1:E
1B
2:E
2B
y B z A
x
60 30
Mathematics168
Both balls are from box
Here, P (required)
19. (a)
20. (a)
⇒
3:E
3B
2 =
EP
A
2
2
1 2 .3
1 2 3
. ( )
A. ( ) . ( ) . ( )
=
+ +
AP P E
E
A AP P E P P E P E
E E E
2 3
1 1
9
2
1 3 2 3 3 4
1 1 1 1 1 1
6 9 12
2 2 2
1 1
3 5561 1 2 181.1 1 1
5 6 113 3 3
××
= = =× × × + +× + × + ×
C C
C
C C C C C C
C C C
/ 2
0
0 02 2sin sin(0) sin 2sin
4 2 2x dx
ππ π
π + + = + +
∫
(1 2)8
π= +
( ) ( ) ( ) ( ) ( )F c b a f c f a f b′ ′= − + −
( ) ( )( ) 0F c f c b a′′ ′′= − <
⇒ ( ) 0F c′ =
( ) ( )( )
f b f af c
b a
−′ =
−
169Mock Test-3
JEE-MAIN: MATHEMATICS MOCK TEST -3
1. Let and Then for all
is equal to:
a. x b. 1 c. d.
2. If then
a. x is an irrational number b.
c. d. None of these
3. If then is equal to
a. i b. – I c. 1 d. – 1
4. The inverse of is
a. b.
c. d.
5. The sum of 100 terms of the series ill be
a. b.
c. d.
6. If n is an integer greater than 1, then
a. b. 0
c. d.
7. The sum of is
a. b.
c. d.
8. If the letters of the word KRISNA are arranged in all
possible ways and these words are written out as in a
dictionary, then the rank of the word KRISNA is
a. 324 b. 341
c. 359 d. None of these
9. Seven white balls and three black balls are randomly
placed in a row. The probability that no two black balls
are placed adjacently, equals
a. b.
c. d.
10. If , then =
a. b.
c. d.
11. Find real part of
a. b.
c. d. None of these
12. A house of height 100 metres subtends a right angle at the
window of an opposite house. If the height of the window
be 64 metres, then the distance between the two houses is
a. 48 m b. 36 m c. 54 m d. 72 m
13. If then
a. b.
c. is continuous at d. None of these
14. If ,y x yx e −= then dy
dx=
a. 2log .[log( )]x ex −
b. 2log .[log( )]x ex
c. 2log .(log )x x d. None of these
15. 20 is divided into two parts so that product of cube of one
quantity and square of the other quantity is maximum. The
parts are
a. 10, 10 b. 16, 4
c. 8, 12 d. 12, 8
16.
a. b.
c. d. None of these
( ) 1 [ ]g x x x= + −
1 0
( ) 0 0.
1 1
x
f x x
x
− <
= = >
, ( )x f g x
( )f x ( )g x
6 6 6 ....to ,x = + + + ∞
2 3x< <
3x =
1 1,z z−+ = 100 100z z−+
2 3
4 2
− −
2 31
4 28
−
3 21
2 48
−
2 31
4 28
3 21
2 48
.9 .09 .009.........+ +100
11
10
−
1001
110
+
1061
110
−
1001
110
+
1( 1)n na C a− − +
2( 2)C a − .... ( 1)n+ + − ( )a n− =
a
2a 2n
2 6 12 20
1! 2! 3! 4!+ + + +K
3
2
ee
2e 3e
1
2
7
15
2
15
1
3
3
2
ππ α< <
1 cos 1 cos
1 cos 1 cos
α αα α
− ++
+ −
2
sinα2
sinα−
1
sinα1
sinα−
1 3cos
2 2
i− +
/3π / 4π
3 1log
2
−
sincos , when 0
( )
2, when 0
xx x
f x x
x
+ ≠= =
0lim ( ) 2x
f x→ +
≠0
lim ( ) 0x
f x→ −
=
( )f x 0x =
4 21tan secx x dx
x=∫
52 tan x c+ 51tan
5x c+
52tan
5x c+
Mathematics170
17. The area bounded by the straight lines and the
curves is
a. b.
c. d.
18. The solution of is
a.
b.
c.
d.
19. The product of the perpendiculars drawn from the points
on the line , is
a. b.
c. d.
20. Area of the circle in which a chord of length 2 makes
an angle 2
πat the centre is
a. 2
π b. 2π
c. π d. 4
π
21. The focal chord to is tangent to
then the possible values of the slope of
this chord, are
a. b.
c. d.
22. If are non-zero vectors such that then
which statement is true
a. b.
c. or d. None of these
23. The point of intersection of the lines
is
a. b.
c. d.
24. is equal to
a. b.
c. d.
25. The value of b and c for which the identity
is satisfied, where
are
a. b.
c. d. None of these.
26. If is the harmonic mean between and then the
value of is
a. 2 b.
c.
d. None of these
27. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6
and cards are to be placed in envelopes so that each
envelope contains exactly one card and no card is placed
in the enveloped bearing the same number and moreover
the card numbered 1 is always placed in envelope
numbered 2. Then, the number of ways it can be done is
a. 264 b. 265
c. 53 d. 67
28. ABCD is a rectangular field. A vertical lamp post of height
12m stands at the corner A. If the angle of elevation of its
top from B is 60° and from C is 45°, then the area of the
field is
a. 48 2 sq. m b. 48 3 sq. m
c. 48 sq. m
d. 12 2 sq. m
29. The area bounded by the curves and
x-axis in the 1st quadrant is:
a. 9 sq unit
b. 27/4 sq unit
c. 36 sq unit
d. 18 sq unit
30. If the line touches the hyperbola
then the point of contact is:
a. b.
c. d.
0, 2x x= =22 , 2xy y x x= = −
4 1
3 log 2−
3 4
log 2 3+
41
log 2−
3 4
log 2 3−
(sin cos )xdye x x
dx= +
(sin cos )xy e x x c= − +
(cos sin )xy e x x c= − +
sinxy e x c= +
cosxy e x c= +
2 2( ,0)± −a b cos sin 1θ θ+ =x y
a b2a 2b
2 2+a b2 2−a b
2 16y x=2 2( 6) 2,x y− + =
1,1− 2,2−
2,1/ 2− 2, 1/ 2−
, ,a b crr r
,a b a c⋅ = ⋅rr r r
b c=r r
( )a b c⊥ −rr r
b c=r r
( )a b c⊥ −rr r
5 7 2 3 3 6,
3 1 1 36 2 4
x y z x y z− − + + − −= = = =
− −
5 1021, ,
3 3(2,10,4)
( 3,3,6)− (5,7, 2)−
~ ( )∧p q
~ ~∨p q ~ ~∧p q
~ ∧p q ~∧p q
( 1) ( ) 8 3f x f x x+ − = +
( ) 2 ,f x bx cx d= + +
2, 1b c= = 4, 1b c= = −
1, 4b c= − =
H p ,q
H H
p q+
pq
p q+
p q
pq
+
, 2 3y x y x= + =
2 6 2x y+ =2 2
2 4,x y− =
( 2, 6)− ( 5,2 6)−
1 1,
2 6
(4, 6)−
171Mock Test-3
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. If z is any complex number satisfying | z 3 2i | 2,− − ≤ then
the minimum value of | 2z 6 5i |− + is
2. Let M be a 3 3× matrix satisfying
0 1 1 1
M 1 2 , M 1 1 ,
0 3 0 1
− = − = −
1 0
and M 1 0 .
1 12
=
Then the sum of the diagonal entries of M is
3. Suppose that all the terms of an arithmetic progression
(A.P.) are natural numbers. If the ratio of the sum of the
first seven terms to the sum of the first eleven terms is
6 : 11 and the seventh term lies in between 130 and 140,
then the common difference of this A.P. is
4. A pack contains n cards numbered from 1 to n. Two
consecutive numbered cards are removed from the pack
and the sum of the numbers on the remaining cards is
1224. If the smaller of the numbers on the removed cards
is k, then
5. Consider a triangle ABC and let a, b and c denote the
lengths of the sides opposite to vertices A, B and C
respectively. Suppose 6, 10a b= = and the area of the
triangle is 15 3. If ∠ACB is obtuse and if r denotes the
radius of the in circle of the triangle, then 2r is equal to
6. Let m and n be two positive integers greater than 1. If
then the value of is
7. Let S be the focus of the parabola y2 = 8x and let PQ be
the common chord of the circle x2 + y
2 – 2x –4y = 0 and
the given parabola. The area of the triangle PQS is.
8. Let be an integer. Take n distinct points on a circle
and join each pair of points by a line segment. Colour the
line segment joining every pair of adjacent points by blue
and the rest by red. If the number of red and blue line
segments are equal, then the value of n is
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. Let A and B be two sets containing four and two elements
respectively. Then the number of subsets of the set
each having at least three elements is:
a. 219 b. 256
c. 275 d. 510
10. The real number for which the equation,
has two distinct real roots in [0, 1]
a. lies between 1 and 2 b. lies between 2 and 3
c. lies between –1 and 0 d. does not exist
11. If is a complex number of unit modulus and argument
then arg equals
a. b.
c. d.
12. If and A adj A = A then is equal
to:
a. –1 b. 5
c. 4 d. 13
13. Three positive numbers form an increasing G.P. If the
middle term in this G.P. is doubled, the new numbers are
in A.P. Then the common ratio of the G.P. is :
a.
b.
c. d.
14. If the number of terms in the expansion of
is 28, then the sum of coefficients
of all the terms in this expansion, is:
a. 64 b. 2187
c. 243 d. 729
k 20 _____− =
cos( )
0lim
2
n
ma
e e eα
α→
− = −
m
n
2n ≥
,×A B
k
32 + 3 0x x k+ =
z
,θ1
1
z
z
+ +
θ−2
πθ−
θ π θ−
5
3 2
a bA
− =
,TA 5a b+
2 3+ 3 2+
2 3− 2 3+
2
2 41 , 0,
n
xx x
− + ≠
Q
P
S
(2,0)
Mathematics172
15. Let be the number of all possible triangles formed by
joining vertices of a n-sides regular polygon. If
then the value of n is
a. 7 b. 5
c. 10 d. 8
16. A computer producing factory has only two plants and
Plant produces 20% and plant produces 80%
of the total computers produced. 7% of computers
produced in the factory turn out to be defective. It is
known that P(computer turns out to be defective given
that it is produced in plant (computer turns out
to be defective given that it is produced in Plant
where P(E) denotes the probability of an event E. A
computer produced in the factory is randomly selected and
it does not turn out to be defective. Then the probability
that it is produced in plant is
a. b.
c. d.
17. A bird is sitting on the top of a vertical pole 20 m high and
is elevation from a point O on the ground is . It flies
off horizontally straight away from the point O. After one
second, the elevation of the bird from O is reduced to
Then the speed (in m/s) of the bird is:
a.
b.
c.
d.
18. Let be differentiable on the interval such that
and for each
Then is
a.
b.
c.
d.
SECTION 3 Contains 2 Match The Following Type Questions
You will have to match entries in Column I with the entries in
Column II.
19. Match the statements given in Column I with the values
given in Column II.
Column I Column II
(A) If ˆ ˆ ˆ ˆa j 3k j 3k= + = − +r
and ˆc 2 3k=r
form a
triangle, then the internal
angle of the triangle
between ar
and br
is
1. 6
π
(B) If
b
2 2
a
(f (x) 3x)dx a b ,− = −∫
then the value of f6
π
is
2. 2
3
π
(C) The value of 2
ln 3
π
5 / 6
7 / 6
sec( x) dxπ∫ is
3. 3
π
(D) The maximum value
1Arg
1 z
−
for
| z | 1, z 1= ≠ f is given by
4. π
5.
2
π
20. Consider the lines given by1 : 3 5 0,L x y+ − =
2 : 3 1 0,L x ky− − = 3 : 5 2L x y+ − 12 0= Match the
Statements /Expressions in Column I with the Statements/
Expressions in Column II
Column I Column II
(A) 1 2 3, ,L L L are
concurrent, if
1. 9k = −
(B) One of 1 2 3, ,L L L is
parallel to at least of
the other two, if
2. 6
5k = −
(C) 1 2 3, ,L L L form a
triangle, if 3.
5
6k =
(D) 1 2 3, ,L L L do not form
a triangle, if
4. 5k =
nT
1 1n nT T+ − =
1T
2.T 1T 2T
1)T 10 P=
2 ),T
2T
36
73
47
79
78
93
75
83
45°
30 .°
40( 2 1)−
40( 3 2)−
20 2
20( 3 1)−
( )f x (0, )∞
(1) 1,f =2 2( ) ( )
lim 1→
−=
−t x
t f x x f t
t x0.>x
( )f x
21 2
3 3+
x
x
21 4
3 3− +
x
x
2
1 2− +
x x
1
x
173Mock Test-3
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. For a point P in the plane, let and be the
distances of the point P from the lines and
respectively. The area of the region R
consisting of all points P lying in the first quadrant of the
plane and satisfying is _______
2. The slope of the tangent to the curve
at the point (1, 3) is
3. Let (x, y, z) be points with integer coordinates satisfying
the system of homogeneous equations: 3 0− − =x y z ,
3 0− + =x z 3 2 0.− + + =x y z
Then the number of such points for which
2 2 2 100+ + ≤x y z is
4. Let 1 sin
f ( ) sin tan ,cos2
θθ
θ−
=
where .4 4
π πθ− < <
Then the value of d
(f ( ))d(tan )
θθ
is
5. Two parallel chords of a circle of radius 2 are at a distance
3 1+ apart. If the chords subtend at the centre, angles of
k
πand
2,
k
πwhere 0,k > then the value of [k] is [Note: [k]
denotes the largest integer less than or equal to k].
6. Let a, b, c be positive integers such that is an integer.
If a, b, c are in geometric progression and the arithmetic
mean of a, b, c is b + 2, then the value of is
7. Let i / 3e ,πω = and a, b, c, x, y, z be non-zero complex
numbers such that
a b c x+ + =
2a b c yω ω+ + =
2a b c z.ω ω+ + =
Then the value of 2 2 2
2 2 2
| x | | y | | z |
| a | | b | | c |
+ ++ +
is
8. If the normals of the parabola drawn at the end
points of its latus rectum are tangents to the circle
then the value of is
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. Let then log is equal to:
a. 2 b. 1 c. d.
10. Let be a non-constant twice differentiable function
defined on such that and
Then
a. vanishes at least twice on [0, 1]
b.
c.
d.
11. The following integral is equal to
a.
b.
c.
d.
12. is equal to
a. b.
c. d.
1 ( )d P 2 ( )d P
0x y− =
0x y+ =
1 22 ( ) ( ) 4,d P d P≤ + ≤
5 2 2 2( ) (1 )y x x x− = +
b
a
214
1
a a
a
+ −
+
24=y x
2 2 2( 3) ( 2) ,− + + =x y r2r
1
2 2
0lim (1 tan ) x
xp x
→ += + p
1
2
1
4
( )f x
( , )−∞ ∞ ( ) (1 )f x f x= −
10.
4f ′ =
( )′′f x
10
2f ′ =
1/ 2
1/ 2
1sin 0
2f x x dx
−
+ =
∫
( )1/ 2 1
sin sin
0 1/ 2
(1 1)t tf t e dt f e dtπ π= −∫ ∫
/ 2
17
/ 4
(2cos )
π
π∫ ec x dx
log(1 2 )
16
0
2( )
+−+∫ u ue e du
log(1 2)
17
0
( )
+−+∫ u ue e du
log(1 2 )
17
0
( )
+−−∫ u ue e du
log(1 2 )
16
0
2( )
+−−∫ u ue e du
2
3 4 2
1
2 2 1
xdx
x x x
−
− +∫
4 2
2
2 2 1x xc
x
− ++
4 2
3
2 2 1x xc
x
− ++
4 22 2 1x x
cx
− ++
4 2
2
2 2 1
2
x xc
x
− ++
Mathematics174
13. If y (x) satisfies the differential equation
' tan 2 secy y x x x− = and y(0) = 0, then.
a. 2
4 8 2yπ π =
b. 2
'4 18
yπ π =
c. 2
3 9yπ π =
d. 2
4 2'
3 3 3 3y
π π π = +
14. Given an isosceles triangle, whose one angle is and
radius of its incircle Then the area of the triangle in
sq. units is
a.
b.
c.
d.
15. The circle passing through the point (–1, 0) and touching
the y-axis at (0, 2) also passes through the point
a.
b.
c.
d. (–4, 0)
16 The normal at a point P on the ellipse 2 2x 4y 16+ = meets
the x-axis at Q. If M is the mid-point of the line segment
PQ, then the locus of M intersects the latus rectums of the
given ellipse at the points
a. 3 5 2
,2 7
± ±
b. 3 5 19
,2 4
± ±
c. 1
2 3,7
± ±
d. 4 3
2 3,7
± ±
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem.
Two multiple choice questions will be asked based on this paragraph.
One or more than one option can be correct.
Tangents are drawn from the point P(3, 4) to the ellipse
touching the ellipse at points A and B.
17. The coordinates of A and B are
a. (3, 0) and (0, 2)
b. and
c. and (0, 2)
d. (3, 0) and
18. The equation of the locus of the point whose distances
from the point P and the line AB are equal, is
a.
b.
c.
d.
Paragraph for Question No. 19 to 20
Let where
and
19. Area of S =
a. b.
c. d.
20.
a.
b.
c.
d.
120°
3.=
7 12 3+
12 7 3−
12 7 3+
4π
3, 0
2
−
5, 2
2
−
3 5,
2 2
−
2 2
19 4
x y+ =
8 2 161,
5 15
−
9 8,
5 5
−
8 2 161,
5 15
−
9 8,
5 5
−
2 29 6 –54 –62 241 0x y xy x y+ − + =2 2+ 9 6 –54 62 –241 0x y xy x y+ + =
2 29 9 –6 –54 –62 – 241 0x y xy x y+ =2 2+ –2 27 31 –120 0x y xy x y+ + =
1 2 3 ,= ∩ ∩S S S S 1 :| | 4,= ∈ <S z C z
2S =1 3
: Im 01 3
z iz C
i
− + ∈ >
−
3 : Re 0.= ∈ >S z C Z
10
3
π 20
3
π
16
3
π 32
3
π
min |1 3 |∈
− − =z S
i z
2 3
2
−
2 3
2
+
3 3
2
−
3 3
2
+
175Mock Test-3
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
b c d a a b d a b b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b a c a d c d c b c
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
a c a a b a c a a d
1. (b)
Since
Hence
2. (c)
⇒
3. (d)
or
For
For
4. (a) Let The matrix of cofactors of the elements of A viz.
transpose of the matrix of cofactors of elements
of
.
5. (a) Series is a G.P. with and
6. (b) L.H.S.
7. (d) Let and let
nth
term of given series
or
Now, sum
8. (a) Words starting from are
Words starting from I are 5 ! 120
Words starting from KA are 4 ! 24
Words starting from KI are 4 ! 24
Words starting from KN are 4 ! 24
Words starting from KRA are 3 ! 6
Words starting from KRIA are 2 ! 2
Words starting from KRIN are 2 ! 2
Words starting from KRISA are 1 ! 1
Words starting from KRISNA are 1 ! 1
Hence rank of the word KRISNA is 324.
9. (b) The number of ways of placing 3 black balls without
any restriction is Since, we have total 10 places of
putting 10 balls in a row. Now the number of ways in
which no two black balls put together is equal to the
number of ways of choosing 3 places marked out of
eight places.
This can be done in ways.
Required probability
10. (b)
11. (b) Expression
where
1 ( ) 0
( ( )) 0 ( ) 0
1 ( ) 0
g x
f g x g x
g x
− <
= = >
( ) 1 0g x ≥ >
( ( )) 1g g x =
6 ,x x= + 20 6 , 0x x x x> ⇒ = + >
26 0, 0x x x− − = >
⇒ 3, 0.x x= >
1z z−+ 1= ⇒2
1 0z z− + =
⇒ z ω= −2ω−
,z ω= −100 100 100 100( ) ( )z z ω ω− −+ = − + −
211ω ω ω
ω= + = + = −
2,z ω= − 100 100 2 100 2 100( ) ( )z z ω ω− −+ = − + −
200
200
1ω
ω= + 2 2
2
1ω ω ω
ω= + = +
1.= −
11 12
21 22
2 ( 4) 2 4
( 3) 2 3 2
c c
c c
− − = = − −
∴ adjA =
2 3
4 2A
=
∴ ( ) | |A adj A A I=
90.9
10a = =
10.1
10r = =
∴100 100
100 100
11
1 9 110 1 .11 10 10
110
rS a
r
− −
= = = − − −
0 1 2 3[ ...( 1) . ]n
na C C C C C= − + − + −
1
1 2 3[ 2 3 .... ( 1) . ]n
nC C C nC−+ − + − + − .0 0 0a= + =
2 6 12 20
1! 2! 3! 4!S = + + + +K
1 2 6 12 20n
S T= + + + + +K
1 12 6 12n n
S T T−= + + + +K
0 2 4 6 8 upto termsn
n T= + + + + −K
2 4 6 8 .......upto termsnT n= + + + +
⇒nT [2 2 ( 1) 2]
2
nn= × + − (2 1) ( 1)n n n n= + − = +
∴
( 1) ( 1)or
! ( 1)!n n
n n n nT T
n n n
+ += =
−
1 2
( 2)! ( 1)!nT
n n= +
− −
1 1
1 12 2 3 .
( 2)! ( 1)!n n
e e en n
∞ ∞
= =
= + = + =− −
∑ ∑
A 5 ! 120=
=
=
=
=
=
=
=
=
=
10
3.C
' '−
W W W W W W W− − − − − − − −8
3C
∴8
3
10
3
8 7 6 7
10 9 8 15
C
C
× ×= = =
× ×
2
1 cos 1 cos 1 cos 1 cos
1 cos 1 cos 1 cos
α α α α
α α α
− + − + ++ =
+ − −
2
sinα=±
2 3, since .
sin 2
ππ α
α
= < < −
Q1cos (cos sin )iθ θ− +
1sin sin log( sin 1 sin ),iθ θ θ−= − + +6
πθ =
Mathematics176
Real part of
12. (a) 64 cot dθ =
Also (100 64) tan dθ− = or 2(64)(36) d= ,
∴ 8 6 48 .d m= × =
13. (c) and
Hence is continuous at
14. (a) y x yx e −=
⇒ logy x x y= −
⇒ 1 log
xy
x=
+
⇒ 2 2log (1 log ) log [log ] .
dyx x x ex
dx
− −= + =
15. (d) Let
and
⇒
Now then
Now
Hence is the point of maxima
16. (c)
Put
then it reduces to
17. (d) Required area =
18. (c) Given equation
⇒
On integrating, we get
19. (b) 2 2 2 2
2 2 2 2 2 2 2 2
cos 0 cos
cos sin cos sin
b a b ab b a b ab
b a b a
θ θ
θ θ θ θ
− + − − − − + +
2 2 2 2 2 2
2 2 2 2
[ ( )cos ]
( cos sin )
b a b a b
b a
θθ θ
− − −=
+
2 2 2 2 2 2
2 2 2 2
[ cos cos ]
cos sin
b a a b
b a
θ θθ θ
− +=
+2 2 2 2 2
2
2 2 2 2
[ sin cos ]
cos sin
b a bb
b a
θ θθ θ+
= =+
20. (c)
Let AB be the chord of length O be centre of the circle
and let OC be the perpendicular from O on AB. Then
In
Area of the circle
21. (a)
Here, the focal chord of is tangent to circle
focus of parabola as ie,
Now, tangents are drawn from to
Since, PA is tangent to circle.
slope of tangent or
Slope of focal chord as tangent to circle
22. (c)
Either or
23. (a) Given lines are,
, (say)
∴ 1 13 1cos sin .
2 2 2 4
i π− − + = =
(0 ) (0 ) 2f f+ = − = (0) 2f =
( )f x 0.x =
20 20x y y x+ = ⇒ = −3 2 3 2. .x y z z x y= ⇒ =
3 2(20 )z x x= −3 5 4400 40z x x x= + −
2 4 31200 5 160dz
x x xdx
= + −
0,dz
dx= 12, 20x =
23 2
22400 16 480 ;
d zx x x
dx= + −
2
2
12x
d zive
dx=
= −
12x =∴ 12, 8.x y= =
4 21tan .secx x dx
x∫
2sectan ,
2
xx t dx dt
x= ⇒ =
4 5 52 22 (tan ) tan .
5 5t dt x c x c= + = +∫
22
0[2 (2 )]x x x dx− −∫
23
2
0
2
log 2 3
x xx
= − +
4 8 14
log 2 3 log 2= − + −
3 4.
log 2 3= −
(sin cos )xdye x x
dx= +
(sin cos )x
dy e x x dx= +
sin .xy e x c= +
2 ,
2 1
2 2AC BC= = =
1, cosec 45 . 2 1
2OBC OB BC∆ = ° = =
∴ 2( )OBπ π= =
2 16y x=2 2( 6) 2.x y− + =
⇒ ( ,0)a (4,0)
(4,0) 2 2( 6) 2.x y− + =
∴ tanθ =2
1,2
AC
AP= = = 1
BC
BP= −
∴ 1.= ±
0 ( ) 0a b a c a b a c a b c⋅ = ⋅ ⇒ ⋅ − ⋅ = ⇒ ⋅ − =r r rr r r r r r r r
⇒ 0 or 0b c a b c− = = ⇒ =r rr r r
( ).a b c⊥ −rr r
1
5 7 2
3 1 1
x y zr
− − += = =
−
y
'y
'x x
A
B
θθ
P
(4,0) 2
C(6,0) 2
Tangent as
focal chord
45°
O
A B C
AB = 2
177Mock Test-3
and , (say)
,
and
On solving, we get
Trick: Check through options.
24. (a) .
25. (b) From the given identity
26. (a) As given
.
27. (c) Plan: A square matrix M is invertible if f det(M) or
Let,
(a) Given that (let)
M is non-invertible.
(b) Given that
(let)
Again
M is non-invertible.
(c) As given
( a and c are non-zero)
M is invertible.
(s)
ac is not equal to square of an integer.
M is invertible.
28. (a)
Let AE is a vertical lamp-post. Given, AE = 12m
tan 45AE
AC° =
AC AE m= =12m
tan 60AE
AB° =
4 33
AEAB = =
2 2 144 48 96 4 6BC AC AB= − = − = =
Area = 4 3 4 6 48 2 . .AB BC sq cm= × = × = sq.cm.
29. (a)
To find the area between the curves,
and x-axis in the Ist quadrant (we can plot the above
condition as);
Area of shaded portion OABO
sq unit
30. (d) The equation of tangent at is
which is same as
∴
and
JEE Advance Paper-I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
5 9 9 5 3 2 4 5 a d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c b d d b c d a c a
1. (5) 5
| 2z 6 5i | 2 z 32
− + = − −
5
2 3 3 i2
≥ − −
(corresponding Pt A) 5
2 52
= =
2
3 3 6
36 2 4
x y zr
+ − −= = =
−
∴ 1 23 5 36 3x r r= + = − −
1 27 3 2y r r= − + = +
1 22 4 6z r r= − = +
5 1021, ,
3 3x y z= = =
qpqp ~~)(~ ∨≡∧
2 2( 1) ( 1) ( ) 8 3b x c x d bx cx d x+ + + + − + + = +
⇒ 2 8 3bx b c x+ + = +⇒ 4, 1b c= = −
2 pqH
p q=
+
∴2 2 2( )
2H H q p p q
p q p q p q p q
++ = + = =
+ + +
| | 0.M ≠
a bM
b c
=
a ba b c
b cα
= ⇒ = = =
⇒ | | 0M Mα αα α
= ⇒ =
⇒
[ ] [ ]b c a b=
⇒ a b c α= = =
| | 0M =
⇒
0| | 0
0
aM M ac
c
= ⇒ = ≠
Q
⇒
2| | 0a b
M M ac bb c
= ⇒ = − ≠
Q
∴
, 2 3y x y x= + =
9 9
0 3
3
2
xx dx dx
− = − ∫ ∫
9 93/ 2 2
0 3
13
3/ 2 2 2
x xx
= − −
2 1 81 927 27 9
3 2 2 2
= ⋅ − − − −
118 (18) 9
2= − =
1 1( , )x y 1 12 4,xx yy− =
2 6 2x y+ =
1 12 4
2 26
x y= − =
⇒ 1 4x =1 6y = −
O
'y
y
'x x
B
A(3,0) (9,0)
y x=
3
2
xy
−=
600
E
A B
C
450
D
Mathematics178
2. (9) Let
a b c
M d e f
g h i
=
0 1
M 1 2 b 1, e 2, h 3
0 3
− = ⇒ = − = =
1 1
M 1 1 a 0, d 3, g 2
0 1
− = ⇒ = = = −
1 0
M 1 1 g h i 12 i 7
1 12
= ⇒ + + = ⇒ =
∴ Sum of diagonal elements = 9.
3. (9) Let seventh term be ‘a’ and common difference be‘d’
Given
Hence,
4. (5) Clearly,
and
5. (3)1
2∆ = ab sin C
⇒ 2 2 15 3 3
sin 1206 10 2
C Cab
∆ ×= = = ⇒ = °
×
⇒ 2 2 2 cosc a b ab C= + −
2 26 10 2 6 10 cos120 14= + − × × × ° =
∴ rs
∆= ⇒
2 225 33.
6 10 14
2
r×
= =+ +
6. (2)
if and only if
7. (4) Solving 2 8y x= and
2 2 2 4 0x y x y+ − − =
Simultaneously, we get (2, 4) and (0, 0)
Focus is (2, 0)
∴ Are 1
2 4 4sq.2
= × × = units.
8. (5) Number of red lines
Number of blue lines = n
Hence,
⇒
⇒
9. (a) Set A has 4 elements
Set B has 2 elements
Number of elements in set
Total number of subsets of
Number of subsets having 0 elements
Number of subsets having 1 element each
∴ Number of subsets having 2 elements each
Number of subsets having at least 3 elements
10. (d)
⇒
Not possible. As condition for two distinct real root is
(where are roots of
11. (c) Let
Now
arg (put
12. (b)
7
11
6
11=
S
S
⇒ 15=a d
130 15 140< <d ⇒ 9=d
1 2 3 2 1224 3 4n n+ + + + − ≤ ≤ + +K K
⇒( 2) ( 1) ( 2)
1224 (3 )2 2
n n nn
− − −≤ ≤ +
⇒ 2 3 2446 0n n− − ≤ 2 2454 0n n+ − ≥
⇒ 49 51n< <
⇒ 50n =
∴ ( 1)(2 1) 1224
2
n nk
+− + =
⇒ 25 20 5k k= ⇒ − =
( )cos
0lim
2
α
α α→
−= −
n
m
e e e
( )( )
(cos( ) 1)
20
(cos 1)lim
cos( ) 1
α
α
α
α α α
−
→
−
−
n n
n m n
e e2
2α = −n e
2 0− =n m
2= −nC n
2 − =n C n n
2 2=n C n
( 1)2
2
−=
n nn
1 4− =n
⇒ 5.=n
∴ ( ) 4 2 8× = × =A B
∴ 8( ) 2 256× = =A B
08 1= =C
18 8= =C
2
8! 8 78 28
2!6! 2
×= = = =C
256 1 8 28 256 37 219= − − − = − =3( ) 2 3f x x x k= + +
2( ) 6 3f x x′ = +
( ) 0f x′ = ⇒ 2 1
2x = −
( ) ( ) 0f fα β =
,α β ( ) 0)f x′ =
z ω=2
2
1 1
1 1
z
z
ω ωω
ω ω+ + −
= = =+ + −
∴ 1arg
1
z
zω θ
+= =
+cos sin )z iθ θ= +
| | TA I AA=
⇒1 0 5 5 3
(10 3 )0 1 3 2 2
a b aa b
b
− + = −
⇒ 2 225 10 3a b a b+ = + &15 2 0 &10 3 13a b a b− = + =
⇒3.15
10 132
+ =a
a ⇒ 65 2 13= ×a
179Mock Test-3
13. (d) Let the numbers be a, ar, ar2 is G.P.
Given a, 2ar, ar2 are in A.P. the
which gives as the G.P. is an increasing G.P.
14. (d) Theoretically the number of terms are 2N + 1 (i.e. odd)
But As the number of terms being odd hence considering
that number clubbing of terms is done hence the solutions
follows:
Number of terms =
Sum of coefficient =
Put
15. (b)
⇒ On solving n = 5
16. (c) Let (computer turns out to be defective given
that it is produced in plate
P (produced in not defective)
17. (d)
Here,
Hence distance covered in one second by the bird is
Thus speed of bird
–1) m/s
18. (a)
Also
Hence
19. (A) a b 1 3 2− = + =rr
| a | b, | b | 2= =rr
2 1
cos2 2 2
θ = =×
2
,3 3
π πθ = but its
2
3
πas its opposite to side of maximum
length.
(B)
b
2 2
a
(f (x) 3x) dx a b− = −∫
b 2 22 2 2 2
a
3 a bf (x)dx (b a ) a b
2 2
− += − + − =∫
⇒ f (x) x.=
(C)
5 / 62
7 / 6ln (sec x tan x)
ln 3
π πππ
+
5 5 7 7
ln sec tan ln sec tan .ln 3 6 6 6 6
π π π π ππ
= + − + =
(D) Let 1 1
u z 11 z u
= ⇒ = −−
1
| z | 1 1 12
= ⇒ − =
⇒
| u 1| | u |− =
∴ locus of u is perpendicular bisector of line segment
joining 0 and 1
⇒ maximum arg u approaches 2
πbut will not attain.
⇒2
5=a ⇒ 5 2=a
⇒ 2 6=a ⇒ 3=b
∴ 5 5+ =a b
2
2 ( 0)2
a arar a
+= ≠
2 3,r = +
2
2 28+ =n C
∴ 6=n
63 3 729= =n
1=x
1
3 3 10n nC C+ − =
x P=
2 ),T
⇒7 1 4
(10 )100 5 5
x x= +
⇒ 7 200 80x x= +
⇒7
280x =
2 /T
( )
( )
P A B
P B
∩=
⇒
4 4 273
5(1 ) 5 280
1 4 1 280 70 4 273(1 10 ) (1 )
5 5 5 280 5 280
x
x x
− =
− − + − +
⇒4 273 2 273 546 78
210 4 273 105 2 273 651 93
× ×= = =
+ × + ×
20AP QB m= =
45 ,POA∠ = ° 30QOB∠ = °
⇒ 20;OA= 20 3OB =
⇒ 20( 3 1)OB OA− = −
20( 3 1)AB = −
20( 3 1) /= −
2 2( ) ( )lim 1→
−=
−t x
t f x x f t
t x
⇒ 2 ( ) 2 ( ) 1 0′ − + =x f x x f x
⇒ 2 1( )
3= +f x cx
x
(1) 1=f
⇒ 2
3=c
22 1( ) .
3 3f x x
x= +
Q
B
P
A O
Mathematics180
20. 3 5 0x y+ − = and 5 2 12 0x y+ − = intersect at (2, 1)
Hence 6 1 0k− − = 5k =
for 1 2,L L to be parallel
1 39
3k
k= ⇒ = −−
for 2 3,L L to be parallel
3 6.
5 2 5
kk
− −= ⇒ =
for 6
5, 9,5
k−
≠ − they will form triangle
for 6
5, 9,5
k k−
= = − they will not form triangle
JEE Advance Paper-II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
6 8 7 1 5 4 3 2 3 All
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
a d a,d c d a d a b c
1. (6)
For
Area of region
sq. units
2. (8)
Now put
and
⇒
⇒ ⇒
3. (7) 3 0− − =x y z 3 0− + =x z
⇒ 0=y and 3=z x
⇒ 2 2 2 2 2 2 2 29 10 100+ + = + = + = ≤x y z x z x x x
⇒ 2
10≤x ⇒ 0, 1, 2, 3= ± ± ±x
There are such seven points.
4. (1) 1 sin
f ( ) sin tan2cos
θθ
θ−
=
1
2
sinsin sin
sin cos 2
θ
θ θ−
= +
2 2 2
sin sin sintan
| cos | cossin (cos sin )
θ θ θθ
θ θθ θ θ= = = =
+ −
⇒ d d(tan )
f ( ) 1d(tan ) d(tan )
θθ
θ θ= =
5 (2) 2cos 2cos 3 12k k
π π+ = +
⇒ 3 1
cos cos2 2k k
π π ++ =
Let 3 1
0,cos cos2 2k
π θθ
+= + =
⇒ 2 3 1
2cos 1 cos2 2 2
θ θ +− + =
⇒ cos2
tθ= ⇒ 2 3 3
2 02
t t+
+ − =
⇒ 1 1 4(3 3) 1 (2 3 1) 2 2 3 3
,4 4 4 2
t− ± + + − ± + − −
= = =
Q 3
[ 1,1],cos2 2
tθ
∈ − =
⇒ 2 6
θ π= ⇒ 3.k =
6. (4)
⇒
only
7. (3) The expression may not attain integral value for all a, b, c
If we consider a = b = c, then x = 3a
2y a(1 ) a(1 i 3)ω ω= + + = +
2z a(1 ) a(1 i 3)ω ω= + + = +
∴ 2 2 2 2 2 2 2| x | | y | | z | 9 | a | 4 | a | 4 | a | 17 | a |+ + = + + =
2=x 2 2=x
( , )α βP
=y x
1 22 ( ) ( ) 4≤ + ≤d p d p
( , ),α β α β>P
⇒ 2 2 2 4 2α≤ ≤
2 2 2α≤ ≤
⇒ 2 2((2 2) ( 2) )= −
8 2 6 .= − =
5 42( ) 5 − −
dyy x x
dx
2 2 21(1 ) ( )(2(1 )(2 ))= + + +x x x x
1, 3= =x y
.=dy
mdx
2(3 1)( 5) 1(4) (1)(4)(2)− − = +m
125
4− =m
5 3 8= + =m 8.= =dy
mdx
(integer)= =b c
a b
2 =b ac
⇒2
=b
ca
23
+ += +
a b cb
3 6+ + = +a b c b
⇒ 2 6− + =a b c
2
2 6− + =b
a ba
⇒2
2
2 61− + =
b b
a aa
26
1 − =
b
a a6=a
181Mock Test-3
8. (2) Equation of normals are and
Distance from on both normals is ‘r’
9 (3) then
⇒
10. (a,b,c,d)
Put
⇒
Hence is an even function or
an odd function.
Also, and for
We have
Also,
(obtained by putting,
Since Also
atleast twice in [0, 1] (Rolle’s Theorem)
11. (a)
Let
and
12. (d)
Let
⇒
13. (a,d) ' tan 2 secy y x x x− =
tan log cosI.F. cos
x dx xe e x∫= = =
∴ cos 2 sec .cosy x x x x dx= ∫
⇒ 2cosy x x c= +
⇒ 2cos ( (0) 0)y x x y⋅ = =Q
⇒ 2 secy x x=
∴ 2 2
4 2and y'
4 3 38 2 3 3yπ π π π π = = +
14. (c) . . .(i)
Also
and and
. . .(ii)
From equation (i) and (ii), we get
15. (d) Circle touching y-axis at (0, 2) is
passes through (–1, 0)
Put
Circle passes through (–4, 0)
3x y+ = 3.x y− =
⇒ (3, 2)−
⇒| 3 2 3 |
2r
− −=
⇒2 2.r =
1
2lim (1 tan 2 1)+→∞
= + − x
xP x log =p
212(1 tan 1)2 2
00
(tan ) 1limlim2( ) 2
+ −++ →→= = =
xx
xx
x
xP e e e
1
21
log P loge2
= =
( ) (1 )f x f x= −
1/ 2x x= +
f1 1
2 2x f x
+ = −
( 1/ 2)f x + ( 1/ 2)sinf x x+
'( ) '(1 )f x f x= − − 1/ 2,x =
'(1/ 2) 0.f =1 0
sin sin
1/ 2 1/ 2
(1 ) ( )π π− = −∫ ∫t tf t e dt f y e dy
1 ).t y− =
(1/ 4) 0, (3/ 4) 0.′ ′= =f f (1/ 2) 0′ =f
⇒ ( ) 0f x′ =
217
4
(2cosec )x dx
π
π∫
2cosec ,4
u ue e x xπ−+ = =
⇒ ln(1 2),2
u xπ
= + =
⇒ 0u =
⇒ cosec cot ux x e+ = cosec cot ux x e−− =
⇒ cot2
u ue ex
−−( ) 2cosec cotu ue e dx x xdx−− = −
⇒ 17 ( )( )
2cosec cot
u uu u e e
e e dux x
−− −
− +∫0
16
ln(1 2 )
2 ( )u ue e du−
+
= − +∫ln(1 2 )
16
0
2( )u u
e e du
+−= +∫
3 5
2 4
1 1
2 12
dxx x
x x
−
− +∫
2 4
2 12 z
x x− + =
1
4
dz
z∫
⇒1
2z c× +
⇒2 4
1 2 12
2c
x x− + +
23
4b∆ =
sin120 sin30
a b
° °=
⇒ 3a b= 3s∆ =1
( 2 )2
s a b= +
⇒3
( 2 )2
a b∆ = +
(12 7 3)∆ = +
2 2(x 0) (y 2) x 0λ− + − + =
∴ 1 4 0 5λ λ+ − = ⇒ =
∴ 2 2x y 5x 4y 4 0+ + − + =
y 0= ⇒ x 1, 4= − −
∴
1/4 3/4
1/2
Mathematics182
16. (a) Any point on the line can be taken as
(1 3 ), ( 1), (5 2)µ µ µ≡ − − +Q
3 2, 3, 5 4µ µ µ= − − − −PQ
Now 1( 3 2) 4( 3) 3(5 4) 0µ µ µ− − − − + − =
⇒ 3 2 4 12 15 12 0µ µ µ− − − + + − =
8 2µ = ⇒ 1/ 4µ =
17. (d)
Equation is
Let
⇒
⇒
⇒
18. (a) Locus is parabola
Equation of AB Is
19. (b)
Area of region shaded area
20. (c) Distance of (1, –3) from
29 4y mx m= + +
24 3 9 4m m− = +
2 2 12 116 9 24 9 4
24 2m m m m+ − = + ⇒ = =
14 ( 3)
2y x− = −
2 8 3− = −y x
⇒ 2 5 0x y− + =
( , )B α β=
⇒ 1 09 4
α β+ − =
x y
/ 9 / 4 1
1 2 5
α β −= = ⇒
−9
,5
α = −8
5β =
9 8, .
5 5B
≡ −
3 41
9 4
x y+ =
⇒ 1 3 3 03
xy x y+ = ⇒ + − =
22 2 ( 3 3)
( 3) ( 4)10
x yx y
+ −− + − =
2 2 2 210 90 – 60 10 160 – 80 9 9 6 – 6 – 18x x y y x y xy x y+ + + = + + +2 2 2 210 90 – 60 10 160 – 80 9 9 6 – 6 – 18x x y y x y xy x y+ + + = + + +
⇒ 2 2 9 – 6 – 54 – 62 241 0.x y xy x y+ + =
1 2 3∩ ∩ =S S S
2 24 4
4 6
π π× ×= +
2 1 14
4 6π = +
20
3
π=
3 0+ =y x
3 3 1
2
− + ×>
3 3
2
−>
3 0+ =y x
2 2 16+ <x y60°
60°
183Mock Test-4
JEE-MAIN: MATHEMATICS MOCK TEST-4
1. The number of solutions of is:
a. 3 b. 1 c. 2 d. 0
2. The real roots of the equation are
a. – 1, 4 b. 1, 4
c. – 4, 4 d. None of these
3. If is a root of equation then
its real roots are
a. 1, 1 b. – 1, – 1 c. 1, – 1 d. 1, 2
4. Let , then the adjoint of A is
a. b.
c.
d. None of these
5. The value of is
a. b.
c. d.
6. The sum of the coefficients of even power of x in the
expansion of is
a. 256 b. 128 c. 512 d. 64
7. In the expansion of the coefficient of
will be
a. b.
c. d.
8. The total number of seven digit numbers the sum of
whose digits is even is
a. 9000000 b. 4500000
c. 8100000 d. None of these
9. If E and F are events with and
then
a. occurrence of E occurrence of F
b. occurrence of F occurrence of E
c. non-occurrence of E non-occurrence of F
d. None of the above implication holds
10.
a. b. c. d.
11. equals
a. b.
c. d.
12. The length of the shadow of a pole inclined at 10° to the
vertical towards the sun is 2.05 metres, when the elevation
of the sun is 38°. The length of the pole is
a. 2.05 sin 38
sin 42
°°
b. 2.05 sin 42
sin 38
°°
c. 2.05 cos 38
cos 42
°°
d. None of these
13. If then
a. b.
c. f is continuous at d. None of these
14. If ( )( )
,( )( )
x a x by
x c x d
− −=
− − then
dy
dx=
a. 1 1 1 1
2
y
x a x b x c x d
+ − − − − − −
b. 1 1 1 1
yx a x b x c x d
+ − − − − − −
c. 1 1 1 1 1
2 x a x b x c x d
+ − − − − − −
d. None of these
15. If
2
2
1( ) ,
1
xf x
x
−=
+ for every real number x, then the
minimum value of f
a. Does not exist because f is unbounded
b. Is not attained even though f is bounded
c. Is equal to 1
d. Is equal to –1
log 4( 1) log 2( 3)x x− = −
2 5 | | 4 0x x+ + =
1 3
2
i+4 3 1 0x x x− + − =
1 0 0
5 2 0
1 6 1
A
= −
2 5 32
0 1 6
0 0 2
− −
1 0 0
5 2 0
1 6 1
− − − −
1 0 0
5 2 0
1 6 1
− − − − −
0.234
232
990
232
9990
232
990
232
9909
2(1 x x+ + 3 5)x+
21 2 3,
x
x x
e
− + 5x
71
120
71
120−
31
40
31
40−
( ) ( )P E P F≤ ( ) 0,P E F∩ >
⇒
⇒
⇒
tan tan4 4
π πθ θ + − − =
2 tan 2θ 2cot 2θ tan 2θ cot 2θ
tanh( )x y+
tanh tanh
1 tanh tanh
x y
x y
+−
tanh tanh
1 tanh tanh
x y
x y
++
tanh tanh
1 tanh tanh
x y
x y
−−
tanh tanh
1 tanh tanh
x y
x y
−+
2 1sin , when 0
( ) ,
0, when 0
x xf x x
x
≠=
=
(0 0) 1f + = (0 0) 1f − =
0x =
Mathematics184
16.
a. b.
c. d.
17. The area bounded by the circle line
and x-axis lying in the first quadrant, is
a. b.
c. d.
18. The general solution of is
a. b.
c. d.
19. The length of perpendicular from the point
upon the straight line
is
a. b. c. d.
20. The circle passing through point of intersection of the
circle 0S = and the line 0P = is
a. 0S Pλ+ =
b. 0S Pλ− = and 0S Pλ + =
c. 0P Sλ− =
d. All of these
21. Axis of a parabola is and vertex and focus are at a
distance and respectively from the origin. Then
equation of the parabola is-
a. b.
c. d.
22. If and be unlike vectors, then
a. b.
c. 0 d. None of these
23. A line makes the same angle with each of the x and
z-axis. If the angle which it makes with y-axis is such
that then equals
a. b.
c. d. None of these
24. is equal to
a. b. c. d.
25. If and are the roots of the equation
and then
a. –8 b. –16 c. 16 d. 8
26. If are in H.P., then is equal to
a. b.
c. d. None of these
27. If then the value of x in terms of
y is
a. b.
c. d.
28. The number of arrangements of the letters of the word
BANANA in which the two N’s do not appear adjacently,
is
a. 40 b. 60
c. 80 d. 100
29. Differential coefficient of1sin x−
w.r.t 1 2cos 1 x− − is
a. 1 b. 2
1
1 x+
c. 2 d. None of these
30. The interval in which the function 2 xx e− is non decreasing, is
a. ( , 2]−∞ b. [0, 2]
c. [2, )∞ d. None of these
21
x
x
adx
a=
−∫
11sin
log
xa ca
− + 1sin xa c− +
11cos
log
xa ca
− + 1cos xa c− +
2 24,x y+ = 3x y=
2
π4
π3
π π
2 2dy
xdx
=
2y c
x= +
2y c
x= − 2y cx=
2
3y c
x= −
( cos , sin )α αa a
tan , 0α= + >y x c c
cosαc 2sin αc 2sec αc 2cos αc
y x=
2 2 2
2( ) 8( 2)x y x y− = + − 2( ) 2 ( 2)x y x y+ = + −2( ) 4 ( 2)x y x y− = + − 2( ) 2 ( 2)x y x y+ = − +
ar
br
a b⋅ =rr
| | | |a brr
| | | |a b−rr
,θ
,β2 2sin 3sin ,β θ= 2
cos θ
3
5
2
3
1
5
(~ (~ )) ∧p q
qp ∧~ ∧p q ~∧p q ~ ~∧p q
α β2 6 0x x λ+ + = 3 2 20,α β+ = − λ =
, , ,a b c d ab bc cd+ +
3ad ( )( )a b c d+ +
3ac
2 33 6 10 ....,y x x x= + + +
1/ 31 (1 )y −− − 1/ 31 (1 )y− +1/ 3
1 (1 )y−+ + 1/ 3
1 (1 )y−− +
Space for rough work
185Mock Test-4
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. For any integer k, let where
The value of the expression is
2. Let m be the smallest positive integer such that the
coefficient of 2x in the expansion of
2 3 49 50(1 ) (1 ) ... (1 ) (1 )x x x mx+ + + + + + + + is
51
3(3 1)n C+
for some positive integer n. Then the value of n is
3. The minimum number of times a fair coin needs to be
tossed, so that the probability of getting at least two heads
is at least 0.96 is
4. For any real number x, let |x| denote the largest integer less
than or equal to x. Let f be a real valued function defined
on the interval [−10, 10] by f (x) = [ ] if [ ] is odd
1 [ ] if [ ] is even
x x x
x x x
-ÏÌ+ -Ó
Then the value of is
5. The value of is ________
6. Let f be a real-valued differentiable function on R (the set
of all real numbers) such that f(1) = 1. If the y-intercept of
the tangent at any point P(x, y) on the curve y = f(x) is
equal to the cube of the abscissa of P, then the value of
f (−3) is equal to
7. The line is tangent to the hyperbola
If this line passes through the point of
intersection of the nearest directrix and the x-axis, then the
eccentricity of the hyperbola is
8. If ,a brr
and cr
are unit vectors satisfying
2 2 2
9,a b b c c a− + − + − =r rr r r r
then 2 5 5a b c+ +rr r
is.
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. If then at is equal to
a. b. c. 1 d.
10. For function f(x)
a. for atleast one x in interval
b.
c. for all x in the interval
d. f’ (x) is strcily decreasing in the interval
11. Let 1/
( )(1 )
=+ n n
xf x
xfor 2≥n and
occurs times
( ) ( )( ).= o oKo1442443
f n
g x f f f x Then 2 ( )−∫ nx g x dx equals
a.
111
(1 )( 1)
−+ +
−n nnx K
n n b.
111
(1 )1
−+ +
−n nnx K
n
c.
111
(1 )( 1)
n nnx Kn n
++ +
+ d.
111
(1 )1
n nnx Kn
++ +
+
12. Let be a function which is continuous on
[0, 2] and is differentiable on (0, 2) with Let
for If for all
then equals
a. b. c. d.
13. The function is the solution of the differential
equation in (–1,1) satisfying
Then is
a. b. c. d.
14. Let (0, 0), (3, 4), (6, 0)O P Q be the vertices of the triangle
OPQ. The point R inside the triangle OPQ is such that the
triangles OPR, PQR, OQR are of equal area. The
coordinates of R are
cos sin ,7 7
k
k ki
π πα
= +
1.i = −
12
11
3
4 1 4 21
| |
| |
k kk
k kk
α α
α α
+=
− −=
Σ −
Σ −
102
10
( )cos10
f x xdxπ
π−∫
1 23 2 5
2
0
4 (1 )
−
∫d
x x dxdx
2 1x y+ =
2 2
2 21.
x y
a b− =
1sec(tan ),y x−= dy
dx1x =
1
2
1
22
1x cos ,x 1,
x= ≥
[1, ),f (x 2) f (x) 2∞ + − <
xlim f '(x) 1
→∞=
[1, ), f (x 2) f (x) 2∞ + − >
[1, )∞
: [0,2]f R→
(0) 1.=f
2
0
( ) ( )
x
F x f t dt= ∫ [0,2].∈x ( ) ( )′ ′=F x f x
(0,2),∈x (2)F
21−e
41−e 1−e
4e
( )=y f x
2 1+ =
−dy xy x x
dx x
4
2
2
1 1
++ =
−
dy xy x x
x
(0) 0.=f
3
2
3
2
( )∫ f x dx
3
3 2
π−
3
3 4
π−
3
6 4
π−
3
6 2
π−
Mathematics186
a. 4
, 33
b. 2
3,3
c. 4
3,3
d. 4 2
,3 3
15. Tangents are drawn to the hyperbola parallel
to the straight lien The points of contact of the
tangents on the hyperbola are.
a. b.
c. d.
16. If the vectors and are the
sides of a triangles then the length of the median
through is
a. b.
c. d.
17. In consider the planes and Let
be a plane, different from and which passes
through the intersection of and If the distance of the
point (0, 1, 0) from is 1 and the distance of a point
from is 2, then which of the following
relations is (are) true?
a.
b.
c.
d.
18. The negation of is equivalent to
a. b.
c. d.
SECTION 3 Contains 2 Match The Following Type Questions
You will have to match entries in Column I with the entries in
Column II.
19. Match the statements given in Column I with the
interval/union of intervals given in Column II
Column I Column II
(A) The set 2
2izRe : z
1 z
−
is
a complex number,
| z | 1, z 1= ≠ ± is
1. ( , 1) (1, )−∞ − ∪ ∞
(B) The domain of the
function 1f (x) sin −=
x 2
2(x 1)
8(3)
1 3
−
−
− is
2. ( , 0) (0, )−∞ ∪ ∞
(C) If
1 tan 1
f ( ) tan 1 tan ,
1 tan 1
θθ θ θ
θ= −
− −
then the set
f ( ) : 02
πθ θ ≤ <
is
3. [2, )∞
(D) If 3 / 2f (x) x (3x 10),= −
x 0,≥ then f(x) is
increasing in
4. ( , 1] [1, )−∞ − ∪ ∞
5. ( ,0] [2, )−∞ ∪ ∞
20. Let Match the conditions/expressions
in Column I with statements in Column II
Column I Column II
(A) If then
satisfies
1.
(B) If then
satisfies
2.
(C) If then
satisfies
3.
(D) If then
satisfies
4.
2 2
1,9 4
x y− =
2 1.x y− =
9 1,
2 2 2
9 1,
2 2 2
−
3 3, 2 2− 3 3,2 2−
ˆˆ3 4AB i k= + ˆˆ ˆ5 2 4AC i j k= − +
,ABC
A
18 72
33 45
3,R 1 : 0=P y 2 : 1.+ =P x z
3P 1P 2 ,P
1P 2 .P
3P
(α, β, γ) 3P
2α + β + 2γ 2 0+ =
2α – β + 2γ 4 0+ =
2α + β – 2γ – 10 0=
2α – β + 2γ – 8 0=
~ (~ )s r s∨ ∧
~s r∧ ( ~ )s r s∧ ∧
( ~ )s r s∨ ∨ s r∧
2
2
6 5( )
5 6
x xf x
x x
− +=
− +
1 1,x− < <
( )f x
0 ( ) 1f x< <
1 2,x< <
( )f x
( ) 0f x <
3 5,x< <
( )f x
( ) 0f x >
5,x > ( )f x ( ) 1f x <
Space for rough work
187Mock Test-4
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. The least period of the function
is then the value
of must be (where denotes the greatest integer
function)
2. If 1tan
ax by
bx a
− − = + then the value of
1
(2008)x
dy
dx =−
must be
3. The indicated horse power I of an engines is calculated
from the formula where
Assuming that error of 10% may have been made in
measuring P, L, N and d. If the greatest possible in I is
then must be
4. If
then the value of must be
5. A particle moves in a straight line with a velocity given
by (x is the distance travelled). If the time
taken by a particle to traverse a distance of 99 m is
then the value of must be
6. If a triangle has its orthocenter at (1, 1) and circum
centre at and if centroid and nine point centre are
and respectively, then the value of
must be
7. If the radius of a circle which passes through the point
and whose centre is the limit of the point of
intersection of the lines and
as is then the value of
must be
8. A, B, C, D are any four points in the space. If
(area of )
then the value of must be
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. Let and be two sets containing 2 elements and 4
elements respectively. The number of subsets of
having 3 or more elements is
a. 256 b. 220
c. 219 d. 211
10. The real number for which the equation,
has two distinct real roots in [0, 1]
a. lies between 1 and 2
b. lies between 2 and 3
c. lies between –1 and 0
d. does not exist
11. A complex number z is said to be unimodular if
Suppose and are complex numbers such that
is unimodular and is not unimodular. Then
the point lies on a
a. straight line parallel to x-axis
b. straight line parallel to y-axis
c. circle of radius 2
d. circle of radius
12. If the adjoint of a 3 3× matrix P is
1 4 4
2 1 7 ,
1 1 3
then the
possible value(s) of the determinant of P is (are)
a. –2 b. –1
c. 1 d. 2
13. If the sum of the first ten term of the series
2 2 23 2 1
1 2 35 5 5
+ + +
24 +2
44 ...,
5
+
is then
m is equal to:
a. 102 b. 101
c. 100 d. 99
14. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6
and cards are to be placed in envelopes so that each
envelope contains exactly one card and no card is placed
in the envelope bearing the same number and moreover
the card numbered 1 is always placed in envelope
numbered 2. Then the number of ways it can be done is
a. 264 b. 265 c. 53 d. 67
[ ] [ ]sin cos tan
12 4 3
x x xπ π π + +
,λ
201λ [ ]⋅
33000
PLANI = 2.
4A d
π=
%λ λ
1 1
2
2 2 2 2sin ( 1) tan
3(4 8 13)
x xdx x
x x
− − + + = + + +
∫2
ln(4 8 13) ,x x cλ+ + + + 4λ−
( 1)dx
xdt
= +
.λ
1020 log eλ
3 3,
2 4
( , )α β ( , )γ δ
6 12 4 8α β γ δ+ + +
(2, 0)
3 5 1x y+ =
2(2 ) 5 1c x c y+ + = 1c → ,25
λ
λ
| |AB CD BC AD CA BD λ× + × + × =uuur uuur uuur uuur uuur uuur
ABC∆
125λ
A B
A B×
k
32 + 3 0x x k+ =
| | 1.=z
1z 2z
1 2
1 2
2
2
−−
z z
z z2z
1z
2
16,
5m
Mathematics188
15. Let , 0,2θ ϕ π∈ be such that 2cos 1 sinθ ϕ−
2sin tan cot cos 12 2
θ θθ ϕ = + −
tan 2 0π θ− > and
31 sin
2θ− < < − .
Then ϕ cannot satisfy:
a. 02
πϕ< <
b. 4
2 3
π πϕ< <
c. 4 3
3 2
π πϕ< <
d. 3
22
πϕ π< <
16. If then 1/ 2
2 1 1 21 cos(cot ) sin(cot ) 1x x x x− − + + −
is equal to
a. b.
c. d.
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem.
Two multiple choice questions will be asked based on this paragraph.
One or more than one option can be correct.
Paragraph for Question No. 17 to 18
Let denote the number of all n-digit positive integers formed
by the digits 0, 1 or
both such that no consecutive digits in them are 0. Let = the
number of such n-digit
integers ending with digit 1 and = the number of such
n–digit integers ending with digit 0.
17. Which of the following is correct?
a. b.
c. d.
18. The value of is
a. 7 b. 8
c. 9 d. 11
Paragraph for Question No. 19 to 20
Box 1 contains three cards bearing numbers 1, 2, 3; box 2
contains five cards bearing numbers 1, 2, 3, 4, 5: and box 3
contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card
is drawn from each of the boxes/ Let be the number on the
card drawn from the box,
19. The probability that is odd, is
a. b.
c. d.
20. The probability that are in an arithmetic
progression, is
a. b.
c. d.
0 1,x< <
21
x
x+x
21x x+ 21 x+
na
nb
nc
17 16 15a a a= + 17 16 15c c c≠ +
17 16 16b b c≠ + 17 17 16a c b= +
6b
ix
thi 1, 2,3.=i
1 2 3,+x x x
29
105
53
105
57
105
1
2
1 2 3, ,x x x
9
105
10
105
11
105
7
105
Space for rough work
189Mock Test-4
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
b d c d a c b b d a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
b a c a d a c b a d
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
a b a b b a d a a b
1. (b) Given
But at given log is not defined.
2. (d)
which is not possible. Hence, the given
equation has no real root.
3. (c)
or
So its real roots are 1 and
4. (d)
.
5. (a)
=
6. (c)
Therefore the required sum of coefficients
Note: Sum of all the binomial coefficients in
the 2nd
bracket in which all the powers of x are even.
7. (b)
∴ The coefficient of
8. (b) Suppose represents a seven digit number.
Then takes the value and all
take values 0, 1, 2, 3, , 9. If we keep fixed,
then the sum is either even or odd. Since
takes 10 values 0, 1, 2, ,9, five of the numbers so
formed will be even and 5 odd.
Hence the required number of numbers
9. (d) It is given that
and . . .(ii)
(a) Occurrence of E occurrence of F [from Eq.(i)]
(b) Occurrence of F occurrence of E [from Eq.(ii)]
(c) Non-occurrence of E non-occurrence of F [from
Eq.(i)]
10. (a)
.
11. (b) It is understandable.
12. (a)
sin 38 sin( )
2.05
o SPO
l=
= sin(180 38 90 10 )
2.05
o o o o− − −⇒
2.05sin 38
sin 42
o
ol =
log( 1) log( 3)
2 log 2 log 2
x x− −=
⇒ 2( 1) ( 3)x x− = −
⇒ 2,5x =
2,x =
25 | | 4 0x x+ + =
⇒ 2| | 5 | | 4 0x x+ + =
⇒ | | 1, 4,x = − −
4 3 1 0x x x− + − =
⇒3 ( 1) 1( 1) 0x x x− + − =
1 0x− = 31 0x + =
⇒1 3 1 3
1, 1, ,2 2
i ix
+ −= −
1.−
1 0 0
5 2 0
1 6 1
A
= −
⇒
2 5 32 2 0 0
( ) 0 1 6 5 1 0
0 0 2 32 6 2
T
adj A
− = − = − −
0.234 0.2343434.....=
0.2 0.034 0.00034 0.0000034 ...+ + + +
34 34 340.2 .....
1000 100000 10000000+ + + + ∞
3 5 7
2 1 1 134 ........
10 10 10 10
= + + + + ∞
32 1/10 2 1 10034 34
10 1 1/1000 10 1000 99
= + = + × × −
2 34 232.
10 990 990= + =
2 3 5 5 2 5(1 ) (1 ) (1 )x x x x x+ + + = + +
2 3 4 5(1 5 10 10 5 )x x x x x= + + + + +
2 4 6 8 10(1 5 10 10 5 )x x x x x× + + + + +
5(1 10 5).2 16 32 512= + + = × =
52 2n = =
2(1 2 3 ) xx x e−− +2 3
2(1 2 3 ) 11! 2! 3!
x x xx x
= − + − + − +
K
5x
1 1 11 ( 2) 3
5 ! 4 ! 3!
= − + − + −
1 1 1 71.
120 12 2 120= − − − = −
1 2 3 4 5 6 7x x x x x x x
1x 1, 2, 3, ,9K
2 3 7, ,x x xK
K
1 2 6, , ,x x xK
1 2 6x x x+ + +K
7x K
9 .10 .10.10 .10.10 . 5 4500000.= =
( ) ( )P E P F E F≤ ⇒ ⊆
( ) 0P E F E F∩ > ⇒ ⊂
⇒
⇒
⇒
1 tan 1 tantan tan
4 4 1 tan 1 tan
π π θ θθ θ
θ θ
+ − + − − = − − +
2 2
4 tan 2 tan2 2 tan 2
1 tan 1 tan
θ θθ
θ θ
= = = − −
38°
2.05 m S
10°
O
Q P
l
Mathematics190
13. (c)
but and
Therefore,
Hence is continuous at
14. (a) ( )( )
( )( )
x a x by
x c x d
− −= − −
⇒ 1
log [log( ) log( ) log( ) log( )]2
y x a x b x c x d= − + − − − − −
Differentiating w.r.t. x we get
1 1 1 1 1 1
2 ( ) ( ) ( ) ( )
dy
y dx x a x b x c x d
= + − − − − − −
Thus 1 1 1 1
.2 ( ) ( ) ( ) ( )
dy y
dx x a x b x c x d
= + − − − − − −
15. (d)
∴
and as
Hence has minimum value –1 and also there is no
maximum value.
Alternate:
There is only one critical point having minima.
Hence has least value at
16. (a) Put then
17. (c) Required area
Trick: Area of sector made
by an arc
18. (b) ⇒ Now integrate it.
19. (a) Here, equation of line is tan , 0y x c cα= + >
Length of the perpendicular drawn on line from point
)sin,cos( αα aa
2
sin cos tan; cos
sec1 tan
a a c cp p c
α α αα
αα
− + += = =
+
20. (d) It is a fundamental concept.
21. (a)
Since, distance of vertex from origin is and focus is
V(1, 1) and F(2, 2) (ie, lying on y = x)
where, length of latusrectum
By definition of parabola
Where, PN is length of perpendicular upon
(ie, tangent at vertex).
22. (b) ,
23. (a) Here,
Now,
2
0
1lim ( ) sin ,x
f x xx+→
=
11 sin 1
x− ≤ ≤ 0x →
0 0lim ( ) 0 lim ( ) (0)x x
f x f x f+ −→ →
= = =
( )f x 0.x =
2 2
2 2 2
1 1 2 2( ) 1
1 1 1
x xf x
x x x
− + −= = = −
+ + +
( ) 1f x x< ∀
1≥ −2
22
1x≤
+
∴ 1 ( ) 1f x− ≤ <
( )f x
2 2
2 2 2 2
( 1)2 ( 1)2 4( )
( 1) ( 1)
x x x x xf x
x x
+ − −′ = =
+ +
( ) 0 0f x x′ = ⇒ =
2 2 2
2 4
( 1) 4 4 .2( 1)2( )
( 1)
x x x xf x
x
+ − +′′ =
+
2 2
2 3 2 3
( 1)4 16 ( ) 12 4
( 1) ( 1)
x x x x
x x
+ − − += =
+ +
∴ (0) 0f ′′ >
∴
( )f x 0.x =
min
1(0) 1.
1f f
−= = = −
log ,x x
ea t a a dx dt= ⇒ =
2 2
1
log1 1
x
xe
a dtdx
aa t=
− −∫ ∫
111 sin ( )
sin ( ) .log log
x
e e
at c c
a a
−−= + = +
3 22
0 34
3
xdx x dx= + −∫ ∫
3 222 1
30
1 44 sin
2 2 2 23
x x xx
− = + − +
3 3 2.
2 2 3 3
π ππ
= + − − =
2
2
c Rθ=
4. .
6 2 3
π π= =
2
2dy
dx x=
2
2,dy dx
x=
2
2 2.
∴
4a= 4 2= ( 2 )a =Q
∴ 2 (4 )( )PM a PN=
2 0x y+ − =
⇒2( ) 2
4 22 2
x y x y− + − =
⇒ 2( ) 8( 2)x y x y− = + −
| | | |a b a b⋅ = −r rr r
( cos 1)θ = −Q
cos , cos , cos ,l m nθ β θ= = = ( )l n=Q
2 2 21l m n+ + =
x 'x
'y
y
N
P
M
F (2,2)
y = x
(1,1) V
O
x + y – 2 = 0
yx 3=
(2,0) X
Y
191Mock Test-4
Given,
⇒
24. (b) .
25. (b) . . .(i)
. . .(ii)
and given . . .(iii)
Solving (i) and (iii), we get
Substituting these values in (ii), we get .
26. (a) Since , , ,a b c d are in H.P., therefore b is the H.M. of
a and c 2ac
ba c
=+
and c is the H.M. of b and d
i.e. 2
,bd
cb d
=+
∴2 2
( )( ) .ac bd
a c b db c
+ + =
⇒ 4ab ad bc cd ad+ + + = ⇒ 3 .ab bc cd ad+ + =
Trick : Check for .
27. (d) We have
28. (a) Total number of arrangements of word BANANA
The number of arrangements of words BANANA in
which two N’s appear adjacently
Required number of arrangements = 60 – 20 = 40
29. (a) Let1
1siny x−= and 1 2
2cos 1y x−= −
Differentiating w.r.t. x of 1y and
2 ,y we get
1
2
1
1
dy
dx x=
−
2 2
2 21
1 1( 2 ) 11.
2 11 (1 ) 1
dy dyx
dx dyxx x
−= − = ⇒ =
−− − −
30. (b) Let
⇒
Hence for every therefore it is non-
decreasing in [0, 2].
JEE Advance Paper-I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
4 5 3 4 2 9 2 3 a b,c,d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
a b b c a,b c b,d d b a
1. (4)
2. (5) Coeff. 2x
⇒ 2 3 4 49 50 2 51
2 2 2 2 2 3....... (3 1)C C C C C m n C+ + + + + = +
⇒ 3 3 4 49 50 2 51
3 2 2 2 2 3....... (3 1)C C C C C m n C+ + + + + = +
⇒ 1
1
n n n
r r rC C C+−+ = ⇒ 50 50 2 51
3 2 3(3 1)C C m n C+ ⋅ = +
⇒ 50 50 2 50 50 51
3 2 2 2 3
51( 1) 3
3C C m C n C C+ + − = ⋅ +
⇒ 51 2 50 50 51
3 2 2 3( 1) 51C m C n C C+ − = ⋅ +
⇒ 2 1 51m n− = 2 51 1m n= +
Min value of 2m for 51 1n+ is integer for 5.n =
3. (8) Let coin was tossed ‘n’ times
Probability of getting at least two heads
4. (4)
f (x) is periodic with period 2
put
⇒ 2 22cos cos 1θ β+ =
⇒ 2 22cos sinθ β= 2 2sin 3sinβ θ=
⇒ 2 22cos 3sinθ θ= 2
5cos 3,θ =
∴ 2 3cos .
5θ =
qpqp ∧=∧))(~(~
6α β+ = −
αβ λ=
3 2 20α β+ = −
2, 8β α= = −
16λ = −
..ei
4
1,
3
1,
2
1,1 ==== dcba
2 33 6 10 ....y x x x= + + +
∴ 2 31 1 3 6 10 ...y x x x+ = + + + +
⇒ 3 1/31 (1 ) 1 (1 )y x x y− −+ = − ⇒ − = + ⇒ 1/31 (1 )x y −= − +
6!60
3!2!= =
5!20
3!= =
2( ) xy f x x e−= =
2 22 (2 )x x xdyxe x e e x x
dx
− − −= − = −
'( ) 0f x ≥ [0, 2],x ∈
127 7
1
3(4 2) 7
1
112
43
1
π π
π
=
−
=
Σ −
= =
Σ −
ki i
k
ii k
k
e e
e e
11
2 2n n
n = − +
⇒1
1 0.962
n
n + − ≥
⇒2
251
n
n≥
+⇒ 8n ≥
1, 1 2( )
1 , 0 1
x xf x
x x
− ≤ <=
− ≤ <
∴10
10
( ) cosI f x x dxπ−
= ∫10 2
0 0
2 ( ) cos 2 5 ( ) cosf x x dx f x x dxπ π= = ×∫ ∫1 2
1 2
0 1
10 (1 )cos ( 1)cos 10( )x xdx x x dx I Iπ π
= − + − = + ∫ ∫
2
2
1
( 1)cosI x x dxπ= −∫ 1x t− =
1
1 – 2 – 1 2 3
Mathematics192
5. (2)
6. (9) 1 1( )y y m x x− = −
Put 0,x = to get y intercept 3
1 1 1y mx x= =
3
1 1 1
dyy x x
dx− =
3dy
x y xdx
− = −
2dy y
xdx x
− = −
1
ln 1f dxxxe e
x
−= =
21 1
y x dxx x
× = − ×∫
2
2
y y xxdx c
x x= − ⇒ = − +∫
⇒
3 3( ) ( 3) 9.
2 2
xf x x f= − + ∴ − =
7. (2) Substituting in
Also,
Also,
8. (3) 2 2 2
9a b b c c a− + − + − =r rr r r r
⇒ 2 2 2 2 2 2 9a b b c c a− ⋅ + − ⋅ + − ⋅ =r rr r r r
⇒ 3
2a b b c c a− = ⋅ + ⋅ + ⋅
r rr r r r . . .(i)
Now, 2| | 0a b c+ + ≥
rr r
⇒ 1 1 1 2 0a b b c c a+ + + ⋅ + ⋅ + ⋅ ≥r rr r r r
⇒ 3
2a b b c c a⋅ + ⋅ + ⋅ ≥ −
r rr r r r . . .(ii)
Equation (i) and (ii) are simultaneously true
If 1
2a b b c c a⋅ = ⋅ = ⋅ ≥ −
r rr r r r
Now, 2
2 5 5a b c+ +rr r
4 25 25 20 50 20( )a b b c a c= + + + ⋅ + ⋅ + ⋅r rr r r r
54 10 25 10 9= − − − = ⇒ 2 5 5 3a b c+ + =rr r
9 (a)
10. (b,c,d) For
also
is decreasing for
Also,
1
2
0
cosI t t dtπ= −∫1 1
1
0 0
(1 )cos cos( )I x x dx x x dxπ π= − − = −∫ ∫
∴1
0
10 2 cosI x x dxπ
= −
∫1
2
0
sin cos20
x xx
π ππ π
= − +
2
2 2 2
1 1 4020 4
10I
ππ π π
= − − − = ∴ =
1 23 2 5
2I II0
4 (1 )−∫d
x x dxdx
1 1
3 2 5 2 2 5
0 0
4 (1 ) 12 (1 ) = − − − ∫
d dx x x x dx
dx dx
11 1
3 2 4 2 2 5 2 5
0 00
4 5(1 ) ( 2 ) 12 (1 ) 2 (1 )
= × − − − − − −
∫x x x x x x x dx
1
2 5
0
0 0 12[0 0] 12 2 (1 )= − − − + −∫ x x dx
12 6
0
(1 )12
6
−= × −
x
112 0 2
6
= + =
,0a
e
2 1y x= − +
20 1
a
e= − +
21
a
e=
2
ea =
2 2 21 a m b= −
2 21 a m b2= −2 21 4a b= −
224
14
eb= −
2 2 1.b e= = 2 2 1( 1)b a e= −
∴ 1, 2a e= =
1sec(tan )y x−=
⇒ 1 1
2
1sec(tan ).tan (tan ).
1
dyx x
dx x
− −=+
⇒1
2 1
1 1 2x
dy
dx =
= = +
1f (x) x cos ,x 1
x
= ≥
1 1 1f '(x) cos sin 1forx
x x x
= + → → ∞
2 2 3
1 1 1 1 1 1f '(x) sin sin cos
x x x x x x
= − −
3
1 1cos 0 forx 1
x x
= < ≥
⇒ f '(x) [1, )∞
⇒ f '(x 2) f '(x).+ <
x x
1 1lim f (x 2) f (x) lim (x 2)cos x cos 2
x 2 x→∞ →∞
+ − = + − = +
∴ f (x 2) f (x) 2 x 1+ − > ∀ ≥
193Mock Test-4
11. (a) Here 1/ 1/
( )( )
[1 ( ) ] (1 2 )= =
+ +n n n x
f x xff x
f x x
⇒ 1/
( )(1 3 )
=+ n n
xfff x
x
⇒ 1/
times
( ) ( )( )(1 )n n
n
xg x f f f x
nx= =
+o oKo
Hence
12
1/( )
(1 )
−−= =
+∫ ∫n
n
n n
x dxI x g x dx
nx
2 1
2 1/ 2 1/
(1 )1 1
(1 ) (1 )
− += =
+ +∫ ∫n
n
n n n n
dnx
n x dx dx dxn nx n nx
∴
111
(1 )( 1)
−= + +
−n nI nx K
n n
12. (b)
13. (b)
This is a linear differential equation
I.F. =
Solution is
or
Now, (Using property)
=
(Taking )
14. (c) Since, ∆ is isosceles, hence centroid is the desired
point.
15. (a,b) Slope of tangent = m = 2 Equation of tangent in
slope form is
and point of contact is
16. (c)
Area of base (PQRS)
Height = proj. of PT on
Volume
17. (b, d) Let the required plane be
Distance of from is 2
18. (d)
(0) 0F =
( ) 2 ( ) ( )F x xf x f x′ ′= =2
( ) x cf x e +=2
( ) ( (0) 1)xf x e f= =Q
2
0
( )
x
xF x e dx= ∫
2
( ) 1xF x e= − ( (0) 0)F =Q ⇒ 4(2) 1F e= −
4
2 2
2
1 1
dy x x xy
dx x x
++ =
− −
22
1ln| 1|
21 2 1
xdx x
xe e x−
−∫ = = −
⇒3
2 2
2
( 2)1 1
1
x xy x x dx
x
+− = ⋅ −
−∫
52 4 2
1 ( 2 )5
xy x x x dx x c− = + = + +∫
(0) 0f = ⇒ 0c =
⇒5
2 2( ) 1
5
xf x x x− = +
3 / 2 3 / 2 2
23 / 2 3 / 2
( )1
xf x dx dx
x− −
=−
∫ ∫3 / 2 / 32 2
20 0
sin2 2 cos
cos1
xdx d
x
π θθ θ
θ=
−∫ ∫
sinx θ=/ 3/ 3
2
00
sin 22 sin
2 4d
ππ θ θθ θ = = − ∫
3 32 2 .
6 8 3 4
π π = − = −
2 2 2, 2 4 2y mx a m b y x= ± − = ±
2 2
,ma b
c c
−−
2 9 4 9 1, ,
4 2 4 2 2 2 2
× ≡ − − ≡ ± ± ± ±
ˆˆ ˆ1 1
| | 3 1 22 2
1 3 4
i j k
PR SQ= × = −
− −
uuur uuur
1 ˆ ˆˆ ˆ ˆ ˆ| 10 10 10 | 5 | | 5 32
i j k i j k= − + − = − + =
1 2 3 2ˆˆ ˆ3 3
i j k− +
− + = =
2(5 3) 10 cu. units
3
= =
1 0x z yλ+ + − =
⇒2
| 1 | 11
22
λλ
λ
−= ⇒ = −
+
⇒ 3 2 2 2 0P x y z≡ − + − =
3P ( , , )α β γ| 2 2 2 |
24 1 4
α β γ− + −=
× +
⇒ 2 2 4 0 and 2 2 8 0α β λ α β λ− + + = − + − =
~ [~ (~ )]s r s∨ ∧
~ (~ ) ~ (~ )s r s= ∧ ∧
( ~ )s r s= ∧ ∨
( ) ( ~ )s r s s= ∧ ∨ ∧
s r F= ∧ ∨ s r= ∧
S P
QR
T( , , )α β γ
(6, 0)
(3, 4)
(0, 0)
Mathematics194
19. (A) 2 2 2
2 ( ) 2 ( )
1 ( ) 1 ( 2 )
+ += =
− + − − +i x iy i x iy
zx iy x y ixy
Using 2 21− =x y
2
2 2 1.
2 2
−= = −
−ix y
Zy ixy y
Q 1 1
1 1 1 1.− ≤ ≤ ⇒ − ≤ − − ≥y ory y
(B) For domain 2
2( 1)
8.31 1
1 3
−
−− ≤ ≤−
x
x
⇒ 2
2 2
3 31 1.
1 3
−
−
−− ≤ ≤
−
x x
x
Case (i): 2
2 2
3 31 0
1 3
−
−
−− ≤
−
x x
x
⇒ 2
2 2
(3 1)(3 1)0
(3 1)
−
−
− −≥
−
x x
x
⇒ ( ,0] (1, ).∈ −∞ ∪ ∞x
Case (ii): 2
2
3 31 0
1 3 2
−−+ ≥
− −
x x
x
⇒ 2
2
(3 1)(3 1)0
(3 .3 1)
−
−
− +≥
−
x x
x x
⇒ ( , 1) [2, ).∈ −∞ ∪ ∞x
So, ( , 0] [2, )∈ −∞ ∪ ∞x
(C) 1 1 3→ +R R R
0 0 2
( ) tan 1 tan
1 tan 1
θ θ θθ
= −
− −
f 2 22(tan 1) 2sec .θ θ= + =
(D) 1/ 2 3/ 2 1/23 15
( ) ( ) (3 10) ( ) 3 ( ) ( 2)2 2
′ = − + × = −f x x x x x x
Increasing, when 2.≥x
20. . The graph of is shown
(A) If
(B) If
(C) If
(D) If
JEE Advance Paper-II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
4824 1004 50 3 40 30 1601 500 c d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c a,d b c a,c,d c a b b c
1. (4824) The period of is as
Similarly the period of is 3 and the period of
is
Hence, the period of the given function
LCM of 24, 8, 3 = 24
2. (1004)
3. (50)
4. (3) Let
( 1)( 5)( )
( 2)( 3)
x xf x
x x
− −=
− −( )f x
1 1x− < < ⇒ 0 ( ) 1f x< <
1 2x< < ⇒ ( ) 0f x <
3 5x< < ⇒ ( ) 0f x <
5x > ⇒ ( ) 1f x <
[ ]sin
12
xπ
224
/12
ππ
=
[ 24] ([ ] 24)sin sin
12 12
x xπ π+ + =
[ ] [ ]sin 2 sin
12 12
x xπ ππ = + =
[ ]tan
3
xπ
cos4
xπ
28
/ 4
ππ
=
λ =
∴ 201 201 24 4824λ = × =
Q 1 1tan tan
1
bx
ax b aybbx a
xa
− −
− − = = + + ⋅
1 1tan tan
bx
a
− − = −
∴2
10
1
dy
dx x= −
+
∴2
1
1 1 1
1 ( 1) 1 1 2x
dy
dx =−
= = =+ − +
⇒1
1(2008) 2008 1004
2x
dy
dx =−
= × =
2
4
33000 33000
PL d NPLAN
I
π = =
∴2I P L d N
I P L d N
∆ ∆ ∆ ∆ ∆= + + +
⇒ 100 100 100 2 100 100I P L d N
I P L d N
∆ ∆ ∆ ∆ ∆ × = × + × + × + ×
10% 10% 2 10% 10% 50%= + + × + =
∴ 50λ =
1
2
2 2sin
(4 8 13)
xI dx
x x
− + = + +
∫
y
x 1 2 3 5
y =
195Mock Test-4
Put
Then
Hence
Then,
5. (40)
Putting we get
For
6. (30) Since, centroid divides the orthocenter and circum
center in the ratio (internally) and if centroid
then
and
Centroid is and nine point centre is the mid point
of orthocenter and circumcentre.
∴ Nine point centre is
i.e.,
and
7. (1601) Solving the equation and
then,
∴
or
Therefore the centre of the required circle is
but circle passes through (2,0)
Radius of the required circle
8. (500) Let PV of A, B, C and D be and
and
Adding all we get
(Area of )
Then,
9. (c)
Having elements
Total subsets of is
Total no. of subsets of having 3 or more elements
1
2 2
2 2sin
(2 2) 3
xI dx
x
− + = + +
∫
2 2 3tanx θ+ =
∴ 22 3secdx dθ θ=
23sec 3 tan lnsec
2 2
dI c
θ θθ θ θ θ= ⋅ = ⋅ − +∫
2
13 2 2 2 2 2 2tan ln 1
2 3 3 3
x x xc−
+ + + = ⋅ − + +
1 22 2 3( 1) tan ln(4 8 13)
3 4
xx x x c− + = + − + + +
3
4λ = −
4 3λ− =
1dx
xdt
= +
⇒1
dxdt
x=
+
⇒ ln( 1)x t c+ = +
0, 0t x= = 0c =
⇒ ln( 1)t x= +
99, ln100 2log 10e
x t= = =
∴10 10
20 log 20 2log 10 log 40e
e eλ = × × =
2 : 1
( , ),G x y2 1
13 2
(1, ) ( , ) ,2 4
O G x y C
32 1 1
42
2 1 3x
× + ×= =
+
32 1 1
54
2 1 6y
× + ×= =
+
∴4 5
,3 6
( ) ( )1 3/ 2 1 3/ 4, ,
2 2
+ +
5 7,
4 8
⇒4 5
,3 6
a β= =5 7
,4 8
γ δ= =
∴ 6 12 4 8α β γ δ+ + +
4 5 5 76 12 4 8 8 10 5 7 30
3 6 4 8= × + × + × + × = + + + =
2(2 ) 5 1c x c y+ + =
3 5 1,x y+ =
2 1 3(2 ) 5 1,
5
xc x c
− + + =
2(2 ) (1 3 ) 1c x c x+ + − =
2
2
1
2 3
cx
c c
−=
+ −(1 )(1 ) 1
(3 2)(1 ) 3 2
c c cx
c c c
+ − += =
+ − +
∴1
1lim
3 2c
cx
c→
+=
+⇒
2
5x =
∴ 1 3 1 (6 / 5) 1
5 5 25
xy
− −= = = −
2 1,
5 25
−
∴2 2
2 12 0
5 25
= − + − −
64 1 1601 1601
25 625 625 25 25
λ= + = = =
∴ 1601λ =
, ,a b crr r
0r
∴ ( )AB CD b a c b c a c× = − × − = − × + ×uuur uuur r rr r r r r
( )BC AD c b a c a b a× = − × − = − × + ×uuur uuur r rr r r r r
( )CA BD a c b a b c b× = − × − = − × + ×uuur uuur r r rr r r r
2( )AB CD BC AD CA BD a b b c c a× + × + × = − × + × + ×uuur uuur uuur uuur uuur uuur r rr r r r
∴ | | 2 | |AB CD BC AD CA BD a b b c c a× + × + × = × + × + ×uuur uuur uuur uuur uuur uuur r rr r r r
2 | ( ) ( ) |c a b a= − × −rr r r
2 | |AC AB= ×uuur uuur
14 | | 4
2AC AB= ⋅ × =uuur uuur
ABC∆
∴ 4λ =
125 500λ =
[ , ]A x y= , , , dB a b c=
A B× 2 4 8× =
A B× 82 256=
∴ A B×
8
2null set single ton set subset having
2 elements
256 1 8 C
= − + +
256 1 5 28 219= − − − =
Mathematics196
10. (d)
Not possible. As condition for two distinct real root is
(where are roots of
11. (c)
But
Hence, z lies on a circle of radius 2 centered at origin.
12. (a,d)
1 4 4
2 1 7 | | 4
1 1 3
adj P adj P
= ⇒ =
We know, 1| | | |nadj P P −= (where n is order of matrix)
⇒ 2| | | |adj P P= ⇒ 24 | |P= ⇒ | | 2P = ±
13. (b)
2 2 2 2 28 12 16 20 24
...5 5 5 5 5
+ + + + +
2 2 2 2 2
2 2 2 2 2
8 12 16 20 24....
5 5 5 5 5= + + + + +
14. (c) Number of required ways
=
15. (a,c,d,) Conditions:
tan( ) 0 tan 0θ θ− > ⇒ < and 3
1 sin2
θ− < < −
∴ 3 5 1
, 0 cos2 3 2
π πθ θ ∈ ⇒ < <
Also, 2 12cos (1 sin ) sin cos 1
sin cos2 2
θ φ θ φθ θ
− = −
⇒ 2cos 2cos sin 2sin cos 1θ θ φ θ φ− = −
⇒ 1 2cos 2sin( )θ θ φ+ = +
⇒ 1
sin( ) cos2
θ φ θ+ = +
⇒ 1 4
sin( ) 12 2 3
π πθ φ φ< + < ⇒ < <
16. (c)
17. (a)
is following Fibonacci series. Hence
18. (b)
Total no of ways exactly two consecutive two is = 5 ways
Total no of ways exactly three consecutive two is = 2
ways
Total no of ways exactly four consecutive two is = 1 ways
So,
19. (b) Case (i): One odd, 2 even
Total number of ways = 2×2×3+1×3×3+1×2×4 = 29.
Case (ii): AII 3 odd
Number of ways = 2×3×4 = 24
Favourable ways = 53
Required probability =
20. (c) Here
even
Hence number of favourable ways
3( ) 2 3f x x x k= + +2( ) 6 3f x x′ = + ( ) 0f x′ =
⇒ 2 1
2x = −
( ) ( ) 0f fα β = ,α β ( ) 0)f x′ =
1 2
1 2
21
2
−=
−z z
z z⇒ 2 2
1 2 1 2| 2 | | 2 |− = −z z z z
⇒ 1 2 1 2 1 2 1 2( 2 )( 2 ) (2 )(2 )− − = − −z z z z z z z z
⇒ 1 1 2 2 1 1 2 24 4+ = +z z z z z z z z
⇒ 2 2 2 2
1 2 2 14 | | | | 4 | | | | 0z z z z+ − − =
⇒ 2 2
1 2(| | 1) ( | | 4) 0z z− ⋅ − =
2| | 1,z ≠ ∴2| | 2z =
2
2
(4 4)
5
+=
n
nT
10 102 2
21 1
1 1616( 1) ( 2 1)
255n
n n
S n n n= =
= + = + +∑ ∑
16 10 11 21 2 10 11 16 1610 505
25 6 2 25 5
× × × × = + + = × = m
⇒ 101=m
4 4
2 35! 4 4! 3! 2! 1 53.C C− ⋅ − ⋅ + ⋅ − =
1/ 22 1 1 21 ( cos cot sin cot ) 1x x x x− − + + −
1/ 22
2 1 1
2 2
11 cos cos sin sin 1
1 1
xx x
x x
− − = + + − + +
1/ 22
22
2 2
11 1
1 1
xx
x x
= + + − + +
2 2 1/ 2 21 ( 1 1) 1 .x x x x= + + − = +
11a =
22a =
33a =
45a =
na
17 16 15.a a a= +
1 1
65 2 1b = + + 8=
53 53.
3 5 7 105=
× ×
2 1 32x x x= +
⇒1 3x x+ =
2 4 1 1 3
1 2 1 11.C C C C= ⋅ + ⋅ =
197Mock Test-5
JEE-MAIN: MATHEMATICS MOCK TEST-5
1. The domain of definition of the function given by
the equation is:
a. b.
c. d.
2. A real root of the equation is
a. 1 b. 2 c. 3 d. 4
3. is equal to
a. b.
c. d.
4. Matrix is
a. Orthogonal b. Idempotent
c. Skew-symmetric d. Symmetric
5. For a sequence and Then is
a. b.
c. d. None of these
6. Coefficients of in the expansion of
a. b.
c. d. None of these
7.
a. e b. c. e/2 d. e/3
8. The number of ways in which the following prizes be
given to a class of 20 boys, first and second Mathematics,
first and second Physics, first Chemistry and first English
is.
a. b.
c. d. None of these
9. If the integers m and n are chosen at random between 1
and 100, then the probability that a number of the form
is divisible by 5, equals
a. b. c. d.
10.
a. 1 b. c. d.
11. equals
a. b.
c. d.
12. A tower subtends an angle α at a point A in the plane of
its base and the angle of depression of the foot of the
tower at a point l meters just above A is .β The height of
the tower is
a. tan cotl β α b. tan cotl α β
c. tan tanl α β d. cot cotl α β
13. The value of k so that the function
is continuous at is
a. 1 b. 2
c. 4 d. None of these
14. If (1 ) ,xy x= + then dy
dx=
a. (1 ) log1
x xx ex
x
+ + +
b. log(1 )1
xx
x+ +
+
c. (1 ) log(1 )1
x xx x
x
+ + + +
d. None of these
15. 1 tan
x
x x+ is maxima at
a. sinx x= b. cosx x=
c. 3
xπ
= d. tanx x=
16.
a. b.
c. d.
17. The area of the triangle formed by the tangent to the
hyperbola and co-ordinate axes is
a. b. c. d.
( )y x
2 2 2x y+ =
0 1x< ≤ 1x x≤ ≤
0x−∞ < ≤ 1.x−∞ < <
4 2log log ( 8 ) 0x x+ − =
cosh( ) cosh( )i iα β α β+ − −
2 sinh sinhα β 2 cosh coshα β
2 sinh sini α β 2 cosh cosα β
0 4 1
4 0 5
1 5 0
− − −
1, 2
na a< > = 1 1
.3
n
n
a
a
+ =20
1
r
r
a=∑
20[4 19 3]
2+ ×
20
13 1
3
−
202(1 3 )−
[0 ( 1)]rx r n≤ ≤ −1 2( 3) ( 3) ( 2)n nx x x− −+ + + +
3 2 1( 3) ( 2) ... ( 2)n nx x x− −+ + + + + +
(3 2 )n r n
rC − (3 2 )n n r n r
rC − −−
(3 2 )n r n r
rC −+
2 3 41
3! 5! 7!+ + + + ∞ =K
2e
4 220 19× 3 3
20 19×2 420 19×
7 7m n+
1
4
1
7
1
8
1
49
2sin( ) sin( ) cosecπ θ π θ θ+ − =
1– sinθ sinθ−
2sinh x
cosh 2 1x − 2cosh 1x +
1(cosh 2 1)
2x −
1(cosh 2 1)
2x +
2(2 ), when 0( )
cos , when 0
k x x xf x
x x
− <=
≥0,x =
1
1 logdx
x x=
+∫
3/ 22(1 log )
3x c+ + 3/ 2(1 log )x c+ +
2 1 log x c+ + 1 log x c+ +
2xy a=2a 22a 23a 24a
Mathematics198
18. The solution of is
a.
b.
c.
d. None of these
19. The distance of point (–2, 3) from the line is
a.
b.
c.
d.
20. If the coordinates of one end of the diameter of the circle
2 2 8 4 0x y x y c+ − − + = are (–3, 2), then the coordinates
of other end are
a. (5, 3) b. (6, 2)
c. (1, –8) d. (11, 2)
21. Let by any point on the parabola Let P be
the point that divides the line segment from to
in the ratio Then, the locus of P is
a.
b.
c.
d.
22. If are unit vectors such that then
a. 1
b. 3
c. – 3/2
d. 3/2
23. The direction cosines of three lines passing through the
origin are and The lines will
be coplanar, if
a. b.
c.
d. None of these
24. is equal to
a.
b.
c.
d.
25. If the sum of the roots of the equation
be equal to their product, then
a. 4
b. –4
c. 6
d. None of these
26. If and are the complex numbers representing
the vertices of two triangles such that and
where r is a complex number, than the
two triangles
a. have the same area
b. are similar
c. are congruent
d. None of these
27. The rank of the matrix, is
a. 2 b. 3
c. 1 d. Indeterminate
28. The coefficient of in the expansion of
is
a. 1
b. (–1)n
c. n
d. n + 1
29. Three boys and two girls stand in a queue. The probability
that the number of boys ahead of every girl is atleast one
more than the number of girls ahead of her, is
a. 1/2
b. 1/3
c. 2/3
d. 34
30. The imaginary part of is
a.
b.
c.
d. None of these
logdy
x xdx
=
22 log
2
xy x x c= − +
22log
2
xy x x c= − +
2 21 1log
2 2y x x x c= + +
5− =x y
5 2
2 5
3 5
5 3
( , )x y 2 4 .y x=
(0,0)
( , )x y 1: 3.
2x y=
2 2y x=
2y x=
2 2x y=
, ,a b crr r
0,a b c+ + =rr r
a b b c c a⋅ + ⋅ + ⋅ =r rr r r r
1 1 1 2 2 2, , ; , ,l m n l m n
3 3 3, , .l m n
1 1 1
2 2 2
3 3 3
0
l n m
l n m
l n m
=1 2 3
2 3 1
3 1 2
0
l m n
l m n
l m n
=
1 2 3 1 2 3 1 2 30l l l m m m n n n+ + =
~ ( (~ ))∨p q
~ ∨p q
(~ )∧p q
~ ~∨p p
~ ~∧p q
2 2 3 0x xλ λ+ + =
λ =
, ,a b c , ,u v w
(1 )c r a rb= − +
(1 ) ,w r u rv= − +
2 3 1 4
0 1 2 1
0 2 4 2
A
= − − −
nx
2(1 ....) nx x −+ + +
)sin(costan 1 θθ i+−
)(sintanh 1 θ−
)(tanh 1 ∞−
)(sintanh2
1 1 θ−
199Mock Test-5
JEE ADVANCE PAPER-I
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. A variable plane is at a constant distance p form the origin
and meets the axes in A, B and C. If the locus of the
centroid of the tetrahedron OABC is
then the value of must be
2. The normal to the parabola at the point (2, 4)
meets it again at (18, –12). If length of normal chord is
then the value of must be
3. An electric component manufactured by ‘RASU
electronics’ is tested for its defectiveness by a
sophisticated testing device. Let A denote the event “the
device is defective” and B the event “the testing device
reveals the component to be defective”. Suppose
and where < 1.
If the probability that the component is not defective,
given that the testing device reveals it to be defective, is
then the value of 2008 must be
4. The coefficient of in the polynomials after parenthesis
have been removed and like terms have been collected in
the expansion
is then the value of
must be
5. The value of
must be
6. Let and be three non-coplanar unit vectors such
that the angle between every pair of them is If
where and are
scalars, then the value of is _______
7. If then the value of
must be
8. If then the value
of must be
SECTION 2 Contains 10 Multiple Choice Questions
With one or more than one correct option
9. Let A and B be two sets containing four and two elements
respectively. Then the number of subsets of the set
each having at least three elements is:
a. 219 b. 256
c. 275 d. 510
10. The quadratic equation with real coefficients
has purely imaginary roots. Then the equation
has
a. only purely imaginary roots
b. all real roots
c. two real and two purely imaginary roots
d. neither real nor purely imaginary roots
11. A value of for which is purely imaginary, is:
a. b.
c. d.
12. Let be a complex cube root of unity with and
be a matrix with Then when
a. 57 b. 55
c. 58 d. 56
13. If m is the A.M. of two distinct real numbers and
and G1, G2 and G3 are three geometric means
between and n, then equals,
a. b.
c. d.
2 2 2 2x y z pλ− − − −+ + = 160λ
2 8y x=
,λ2λ
( )P A α= ( / ) ( '/ ') 1 ,P B A P B A α= = − 0 1.α< <
,λ λ
50x
1000(1 )x+ + 999(1 )x x+ 2x+ 998 1000(1 ) ...x x+ + +
998 1000(1 ) ...x x+ + +!
,! !v
λµ
2λ µ+ 3v+
( )v µ>
,a brr
cr
.3
π
,a b b c pa qb rc× + × = + +r r rr r r r
,p q r
2 2 2
2
2p q r
q
+ +
2 22 22 , 2 ,
8 8
y yf x x xy
+ − =
(60, 48) (80, 48) (13, 5)f f f+ +
(cos sin ); (sin cos ),x a t t t y a t t t= + = −
2
2
/3
120)
t
d y
dxπ=
,×A B
( ) 0=p x
( ( )) 0−p p x
θ 2 3 sin
1 2 sin
i
i
θθ
+−
3
π6
π
1 3sin
4
−
1 1sin
3
−
ω 1ω ≠
[ ]ijP P ×n n .ω += i j
ijp
2 0,≠P =n
l
( , 1)n n >l
l4 4 4
1 2 32G G G+ +24 mnl
24 m nl
24 mnl2 2 24 m nl
1 1 1 11 7 2 5216 sin sin 27cos cos 28 tan tan 200cot cos
6 3 4 4
π π π π− − − − − + + +
1 1 1 11 7 2 5216 sin sin 27cos cos 28 tan tan 200cot cos
6 3 4 4
π π π ππ
− − − − − + + +
Mathematics200
14. Let Then can take value(s)
a. 1056 b. 1088 c. 1120 d. 1332
15. If then
a. b.
c. d.
16. The number of integers greater than 6,000 that can be
formed, using the digits 3, 5, 6, 7 and 8, without
repetition, is:
a. 216 b. 192 c. 120 d. 72
17. A ship is fitted with three engines and The
engines function independently of each other with
respective probabilities and For the ship to be
operational at least two of its engines must function. Let X
denote the event that the ship is operational and let
and denote respectively the events that the
engines And are functioning. Which of the
following is (are) true?
a.
b. P [Exactly two engines of the ship are functioning
c.
d.
18. Let PQR be a triangle of area with and
where a, b and c are the lengths of the sides of the
triangle opposite to the angles at P, Q and R respectively.
Then equals
a. b. c. d.
SECTION 3 Contains 2 Match The Following Type Questions
You will have to match entries in Column I with the entries in
Column II.
19. Match the statements given in Column I with the
interval/union of intervals given in Column II
Column I Column II
(A) The set is a
complex number,
is
1.
(B) The domain of the function
is
2.
(C) If
then the set
is
3.
(D) If
then f(x) is
increasing in
4.
5.
20. Match the statements/expressions in Column I with the
values given in Column II.
Column I Column II
(A) The number of solutions of the
equation sin cos 0xxe x− = in the
interval 0,2
π
1. 1
(B) Value (s) of k for which the plane
4 0,4 2 0kx y z x ky z+ + = + + = and
2 2 0x y z+ + = intersect in a
straight line
2. 2
(C) Value (s) of k for which
| 1 | | 2 | | 1 | 2 | 4x x x x k− + − + + + =
has integer solution (s)
3. 3
(D) If 1y y′ + and (0) 1y = then value
(s) of y (ln2)
4. 4
5. 5
( 1)422
1
( 1) .k kn
n
k
S k+
=
= −∑ nS
13 4 ,−=x x =x
3
3
2 log 2
2 log 2 1− 2
2
2 log 3−
4
1
1 log 3−2
2
2 log 3
2 log 3 1−
1 2,E E 3.E
1 1,
2 4
1.
4
1 2,X X 3X
1 2,E E 3E
1
3[ | ]
16
cP X X =
7[ ]
8X =
2
5[ | ]
16P X X =
1
7[ | ]
16P X X =
∆7
2,2
a b= =
5,
2c =
2sin sin 2
2sin sin 2
P P
P P
−
+
3
4∆45
4∆
23
4
∆
245
4
∆
2
2izRe : z
1 z
−
| z | 1, z 1= ≠ ±
( , 1) (1, )−∞ − ∪ ∞
1f (x) sin−=x 2
2(x 1)
8(3)
1 3
−
−
−
( , 0) (0, )−∞ ∪ ∞
1 tan 1
f ( ) tan 1 tan ,
1 tan 1
θθ θ θ
θ= −
− −
f ( ) : 02
πθ θ ≤ <
[2, )∞
3 / 2f (x) x (3x 10),= −
x 0,≥
( , 1] [1, )−∞ − ∪ ∞
( ,0] [2, )−∞ ∪ ∞
Space for rough work
201Mock Test-5
JEE ADVANCE PAPER-II
SECTION 1 Contains 8 Questions.
The answer to each question is a single digit integer ranging from 0 to
9 (both inclusive).
1. If the approximate value of 0 abcdef, It is
given that and then the
value of abcd must be
2. If then the
value of must be
3. The solution of the differential equation
satisfying 1
(0) ,8
y = y1 (0) = 0and y2 (0) = 1is
then the numerical value of λ must be
4. If and are the roots of the equation
and if area of the triangle
formed by the lines and is
then the value of 2008 must be
5. Tangents are drawn form to the circle
then the radius of the circle such that the area of the ∆ formed
by tangents and chore of contact is maximum must be
6. ‘P’ is any arbitrary point on the circum circle of the
equilateral triangle of side length 26 unit, then the value
of must be
7. The lines and
are coplanar for k is equal to
8. If a circle cuts a rectangular hyperbola in A, B, C
and D are the parameters of these four points be
and respectively, then the value of must be
SECTION 2 Contains 8 Multiple Choice Questions
With one or more than one correct option
9. Let where
Then value of is
a. b.
c. d.
10. If the angles of elevation of the top of a tower from three
collinear points A, B and C, on a line leading to the foot
of the lower, are and respectively, then the
ratio, AB : BC is
a. b.
c. d. 2 : 3
11. Let and are
integers, and let be the left hand
derivative of at If then
a. b.
c. d.
12. Let where is a twice differentiable
positive function on such that
Then, for
a.
b.
c.
d.
13. Let Then for
an arbitrary constant the value of equals
a.
10log (4.04)
4log 4 0.6021=
10log 0.4343,e =
1 tan 1( tan cot ) tan ,
tan
xx x dx a c
b x
− − + = +
∫
4 5a b+
3 2
3 28 0
d y d y
dx dx− =
8
1
8 7,
xe xy
λ− +
=
1m
2m
2 ( 3 2) ( 3 1) 0x x+ + + − =
1 2 2, ,y m x y m x= = y c=
2( ) ,a b c+ 2 2
( )a b+
(6, 8)P2 2 2 ,x y r+ =
2 2 2| | | | | |PA PB PC+ +uuur uuur uuur
4 6 1
3 5 2
x y z+ + −= =
−3 2 5 0x y z− + + =
2 3 4x y z k= + + −
2xy c=
1 2 3, ,t t t
4t 1 2 3 416t t t t
1 1 1
2
2tan tan tan ,
1
xy x
x
− − − = + −
1| | .
3x <
a y
3
2
3
1 3
x x
x
−−
3
2
3
1 3
x x
x
+−
3
2
3
1 3
x x
x
−+
3
2
3
1 3
x x
x
++
30 , 45° ° 60°
3 :1 3 : 2
1: 3
m
( 1)( ) ;
log cos ( 1)
nxg x
x
−=
−0 2,x< < m n
0, 0,≠ >m n p
| 1|x− 1.x =1
lim ( ) ,x
g x p+→
=
1, 1n m= = 1, 1n m= = −
2, 2n m= = 2,> =n m n
( ) log( ( ))=g x f x ( )f x
(0, )∞ ( 1) ( ).+ =f x x f x
1, 2, 3, ...,=N
2
1 1 14 1 ...
9 25 (2 1)
− + + + +
− N
2
1 1 14 1 ...
9 25 (2 1)
+ + + +
− N
2
1 1 14 1 ...
9 25 (2 1)
− + + + +
+ N
2
1 1 14 1 ...
9 25 (2 1)
+ + + +
+ N
4 2 4 2, .
1 1
−
− −= =+ + + +∫ ∫
x x
x x x x
e eI dx J dx
e e e e
,C −J I
4 2
4 2
1 1log
2 1
− ++ + +
x x
x x
e eC
e e
Mathematics202
b.
c.
d.
14. The integral equals:
a.
b.
c.
d.
15. Let the population of rabbits surviving at a time t be
government by the differential equation
If then (p)t equals:
a.
b.
c.
d.
16. A straight line L through the point (3, –2) is inclined at an
angle to the line If L also intersects the
x-axis, then the equation of L is
a.
b.
c.
d.
SECTION 3 Contains 2 Paragraph Type Questions
Each paragraph describes an experiment, a situation or a problem.
Two multiple choice questions will be asked based on this paragraph.
One or more than one option can be correct.
Paragraph for Question No. 17 to 18
Consider the line
17. The unit vector perpendicular to both and is
a.
b.
c.
d.
18. The distance of the point (1, 1, 1) from the plane passing
through the point (−1, −2, −1) and whose normal is
perpendicular to both the lines and is
a.
b.
c.
d.
Paragraph for Question No. 19 to 20
Let ABCD be a square of side length 2 units, is the circle
through vertices, A, B, C, D and is the circle touching all the
sides of the square ABCD. L is a line through A.
19. If P is a point on and Q in another point on then
is equal to
a. 0.75
b. 1.25
c. 1
d. 0.5
20. A line M through A is drawn parallel to BD. Point S
moves such that its distances from the line BD and the
vertex A are equal. If locus of S cuts M at and and
AC at then area of is
a. sq. unit
b. sq. unit
c. 1 sq. unit
d. 2 sq. unit
2
2
1 1log
2 1
+ ++ − +
x x
x x
e eC
e e
2
2
1 1log
2 1
− ++ + +
x x
x x
e eC
e e
4 2
4 2
1 1log
2 1
+ ++ − +
x x
x x
e eC
e e
2
0
1 4sin 4sin2 2
x xdx
π
+ −∫4π −
24 4 3
3
π− −
4 3 4−
4 3 43
π− −
( ) 1( ) 200.
2
dp tp t
dt= − ( )0 100,p =
/ 2400 300
te−
/ 2300 200
te−−/ 2
600 500t
e−/ 2400 300 te−−
60° 3 1.x y+ =
3 2 3 3 0y x+ + − =
3 2 3 3 0y x− + + =
3 3 2 3 0y x− + + =
3 3 2 3 0y x+ − + =
1
1 2 1: ,
3 1 2
x y zL
+ + += =
2
2 2 3:
1 2 3
x y zL
− + −= =
1L 2L
ˆˆ ˆ7 7
99
i j k− + +
ˆˆ ˆ7 5
5 3
i j k− − +
ˆˆ ˆ7 5
5 3
i j k− + +
ˆˆ ˆ7 7
99
i j k− −
1L 2L
2
75
7
75
13
75
23
75
2C
1C
1C 2 ,C
2 2 2 2
2 2 2 2
PA PB PC PD
QA QB QC QD
+ + ++ + +
2T 3T
1,T 1 2 3T T T∆
1
2
2
3
203Mock Test-5
ANSWER & SOLUTIONS
JEE-Main
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
d a c c b b c a a b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c b d c b c b d a d
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
c c a b d b a b a c
1. (d)
⇒ for domain
2. (a)
⇒
⇒
⇒
3. (c)
4. (c) It is skew-symmetric.
5. (b) The sequence is a G.P. with common ratio
Now from
6. (b) We have
Therefore coefficient of in the given expression
Coefficient of in
7. (c)
Here
.
Trick: The sum of this series upto 4 terms is 1.359…and
this is value of e/2 approximately.
8. (a) Four first prizes can be given in ways since first
prize of Mathematics can be given in 20 ways, first prize
of Physics also in 20 ways, similarly first prizes of
Chemistry and English can be given in 20 ways each.
(Note that a boy can stand first in all the four subjects).
Then two second prizes can be given in ways since a
boy cannot get both the first and second prizes. Hence the
required number of ways
9. (a)
Therefore, for the number ends at unit place 7,
9, 3, 1, 7, …..
will be divisible by 5 if it end at 5 or 0.
But it cannot end at 5. Also for end at 0. For this m and n
should be as follows
m n
1
2
3
4
For any given value of m, there will be 25 values of n.
Hence, the probability of the required event is
Note: Power of prime numbers have cyclic numbers in
their unit place.
10. (b)
.
11. (c)
2 2 2y x= −
2log 2 2xy = − 2 2 1x x− ⇒ <
4 2log log ( 8 ) 0x x+ − =
( )0
24 log 8x x= + −
⇒ 12 8x x= + −
⇒ 24 8 2 8x x x x= + + − +
22 8 2 4x x x+ = +
⇒ 2 28 4 4x x x x+ = + +4 4x =
⇒ 1.x =
cosh( ) cosh( )i iα β α β+ − −
cosh cosh ( ) sinh sinh( )i iα β α β= +
cosh cosh ( ) sinh sinh ( )i iα β α β− +
2sinh sinh iα β= 2 sinh sin .i α β=
1.
3
20(1 ) 2[1 (1/3) ]
,1 1 (1/3)
na r
r
− −− −
20
13 1 .
3
= −
1 2( 3) ( 3) ( 2)n nx x x
− −+ + + + +3 2 1( 3) ( 2) .... ( 2)n n
x x x− −+ + + + +
( 3) ( 2)( 3) ( 2)
( 3) ( 2)
n nn nx x
x xx x
+ − += = + − +
+ − +
1 2 1 3 2 1( .... )n n
n n n nx ax x a x a a
x a
− − − −−= + + + +
−Q
rx
= rx [( 3) ( 2) ]n n
x x+ − +
3 2 (3 2 )n n r n n r n n r n r
r r rC C C− − − −= − = −
1 2 3 4
1! 3! 5 ! 7 ! (2 1) !
nS
n= + + + + + +
−K K
1 2 1 (2 1) 1.
2 (2 1) ! 2 (2 1) !n
n nT
n n
− += =
− −
1 1 1
2 (2 2) ! (2 1) !n n
= +
− −
⇒1 1
1
2 2 2 2n
e e e e eS T
− − + −= = + =
∑
420
219
4 220 19 .= ×
1 2 3 47 7,7 49,7 34,7 2401,.....= = = =
7 ,r r N∈
∴ 7 7m n+
4r 4 2r −
4 1r − 4 3r −
4 2r − 4r
4 3r − 4 1r −
100 25 1
100 100 4
×=
×
2sin( ) sin( )cosecπ θ π θ θ+ −
2
1sin sin 1
sinθ θ
θ= − = −
2 1sinh (cosh 2 1)
2x x= −
Mathematics204
12. (b)
From figure, we can deduce tan cotH l α β= .
13. (d) 0;
Hence no value of k can make
14. (d) (1 )xy x= +
Taking log on both sides, log log(1 )y x x= +
Differentiating w.r.t. x, we get
1 1
log(1 )(1 )
dyx x
y dx x= + +
+
Thus (1 ) log(1 )1
xdy xx x
dx x
= + + + +
15. (b) If is maxima, then its reciprocal
will be minima.
Let
∴
On putting
⇒
⇒
⇒
which is positive.
At is minimum.
So will be maximum.
16. (c) Put then
17. (b)
Given or . . .(i)
There are two points on the curve (a, a),(– a,– a)
The equation of the line at (a, a) is,
therefore, equation of the tangent at
is The interception of line
with x-axis is 2a and with y-axis is 2a.
Required area
18. (d)
⇒
⇒
⇒
19. (a) Distance of point (–2, 3) from the line 5x y− = is
2 3 5 105 2.
2 2
− − − −= =
20. (d) Obviously the centre of the circle is (4, 2) which
should be the middle point of the ends of diameter.
Hence the other end is (11, 2).
21. (c)
2
0(0 ) lim (2 ) 0
xf k x x
→ −− = − =
0(0 ) lim cos 1
xf x
→ ++ = =
∴ (0) cos 1f x= =
(0 ) 1.f − =
1 tan
x
x x+
1 tanx x
x
+
1 tan 1tan
x xy x
x x
+= = +
2
2
1sec ,
dyx
dx x= − +
2
2 3
22sec sec tan
d yx x x
dx x= +
0,dy
dx= 2
2
1sec 0x
x− + =
2
2
1sec x
x=
2 2cosx x=
cosx x=
∴2
2
2 3
22sec tan
cos
d yx x
dx x= +
22sec (sec tan ),x x x= +
cos ,x x=1 tanx x
x
+
1 tan
x
x x+
11 log ,t x dt dx
x= + ⇒ =
1/ 2 1/ 2
1/ 22 2(1 log ) .
1 log
dx dtt c x c
tx x= = + = + +
+∫ ∫
2xy a=2a
yx
=
( , )
( )a a
dyy a x a
dx
− = −
( , )
2
2( )
a a
ax a
x
− = −
( )y a x a− = − −
( , )a a 2 .x y a+ = 2x y a+ =
∴ 212 2 2 .
2a a a= × × =
logdy
x xdx
=
logdy x xdx=
logdy x xdx=∫ ∫2 2
log .2 4
x xy x c= − +
x O
y
1
3
(x,y)Q
P(h,k)
y2 = 4x
(0,0)
2a
2a
X
(–a,–a)
Y α β
A O
l
P
T
H
205Mock Test-5
By section formula,
Substituting in
Or is required locus.
22. (c) Squaring
we get
⇒
⇒
23. (a) Here, three given lines are coplanar if they have
common perpendicular
Let d.c.'s of common perpendicular be l, m, n
⇒ . . .(i)
. . .(ii)
and . . .(iii)
Solving (ii) and (iii), we get
⇒
Substituting in (i), we get
⇒
24. (b) .
25. (d) Under condition,
26. (b) Since a, b, c and u, v, w are the vertices of two
triangles.
Also,
and . . .(i)
Consider
Applying
[from Eq. (i)]= 0
Hence, two triangle are similar.
27. (a) Given ,
Since every minor of order 3 in A is 0 and there exists a
minor order 3 i.e. in A which is non-zero. Thus,
rank = 2.
28. (b) We have,
Coefficient of is
29. (a) Total number of ways to arrange 3 boys and 2 girls are
5!. According to given condition, following cases may
arise.
B G G B B
G G B B B
G B G B B
G B B G B
B G B G B
So, number of favourable ways 5 3! 2! 60= ¥ ¥ =
Required probability
30. (c) We know that
then the Imaginary part of be
,
0 0,
4 4
x yh k
+ += =
∴ 4 , 4x h y k= =2 2 24 , (4 ) 4(4 )y x k h k h= = ⇒ =
2y x=
( ) 0,a b c+ + =rr r
2 2 2 2 2 2 0a b c a b b c c a+ + + ⋅ + ⋅ + ⋅ =r r rr r r r r r
2 2 2| | | | | | 2( ) 0a b c a b b c c a+ + + ⋅ + ⋅ + ⋅ =
r r rr r r r r r
2( ) 3a b b c c a⋅ + ⋅ + ⋅ = −r rr r r r
⇒3
2a b b c c a⋅ + ⋅ + ⋅ = −
r rr r r r
1 1 1 0ll mm nn+ + =
2 2 2 0ll mm nn+ + =
3 3 3 0ll mm nn+ + =
2 3 2 3 2 3 3 2 2 3 3 2
l m nk
m n n m n l n l l m l m= = =
− − −
2 3 2 3 2 3 3 2 2 3 3 2( ), ( ), ( )l k m n n m m k n l n l n k l m l m= − = − = −
1 2 3 2 3 1 2 3 3 2 1 2 3 3 2( ) ( ) ( ) 0l m n n m m n l n l n l m l m− + − + − =
1 1 1
2 2 2
3 3 3
0
l m n
l m n
l m n
=
⇒
1 1 1
2 2 2
3 3 3
0
l n m
l n m
l n m
− =
qpqpqp ∧≡∧≡∨ )(~)(~~~))(~(~
23
λ− =
⇒2
3λ = −
( )1c r a rb= − +
( )1w r u rv= − +
111
a ub vc w
3 3 1 2(1 ) R R r R rR→ − − +
( ) ( ) ( )
11
1 1 1 1
a ub v
c r a rb w r u rv r r=
− − − − − − − − −
11
0 0 0
a ub v=
2 3 1 4
0 1 2 1
0 2 4 2
A
= − − −
2 2 3( 2 )R R R→ +
2 3 1 4
0 0 0 0
0 2 4 2
A
= − −
2 3
0 2
−
2 1(1 ...) [(1 ) ] (1 )n n nx x x x− − −+ + + = − = −2
0 1 2 ... ( 1) .n n n n n n
nC C x C x C x= − + + + −n
x ( 1) ( 1) .n n n
nC− = −
∴60 1
120 2= =
1 1 sintan (cos sin ) log ,
4 4 1 sin
ii
π θθ θ
θ− + + = + −
(cos ) 0θ >1tan (cos sin )iθ θ− +
11tanh (sin )
2θ− 1 1 1
tanh log2 1
xx
x
− + = − Q
Mathematics206
JEE Advance Paper-I
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
2560 3 1004 3954 139 4 112 2880 a d
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
d bcd b ad abc b bd c b a
1. (2560) Let the equation of the variable plane be
. . .(i)
Given that the plane is at a distance p form (0, 0, 0)
or . . .(ii)
Also, the plane (i) meets the axes in A, B and C. So the
coordinates of O, A, B and C are (0, 0, 0), (a, 0, 0), (0, b,
0) and (0, 0, c) respectively
Let be the centroid of the tetrahedron OABC,
then
Similarly and
or
Substituting these values in equation (ii), we get
Or
2. (3) Comparing the given parabola (i.e., 8x) with
Since, normal at to the parabola is
Here, and
Equation of normal is
. . . (i)
Solve equation (i) and then
and
Then and
Hence, point of intersection of normal and parabola are
and therefore normal meets the parabola
at and length of normal chord is distance
between their points
(given)
3. (1004) Given that
Thus,
and . . .(i)
. . .(ii)
But
from equation (i)
. . .(iii)
Putting the value of from equation (iii) in (ii), then
(given)
4. (3954) Using the formula for the sum of a geometric
progression, we find
Hence, the coefficient of
and
1x y z
a b c
+ + =
∴2 2 2
1
1 1 1
p
a b c
= + +
2 2 2 2
1 1 1 1
p a b c= + +
( , , )x y z
1 1(0 0)
4 4x a a= + + =
1
4y b=
1
4z c=
4 , 4 , 4a x b y c z= = =
2 2 2 2
1 1 1 1
16 16 16p x y z= + +
2 2 2 216x y z p− − − −+ + =
∴ 16λ =⇒ 160 160 16 2560λ = × =
2y x=2 4y ax=
∴ 4 8a =
∴ 2a =
1 1( , )x y 2 4y ax=
11 1( )
2
yy y x x
a− = −
12x =
14y =
∴4
4 ( 2)4
y x− = − −
⇒ 4 2y x− = − +
⇒ 6 0x y+ − =2 8y x= 2 8(6 )y y= −
⇒ 2 8 48 0y y+ − =
⇒ ( 12)( 4) 0y y+ − =
∴ 12y = − 4y =
18x = 2x =
(18, 12)− (2, 4)
(18, 12)−
2 2(18, 12) ( 12 4) 16 2PQ= = − + − − =
λ=
∴ 2 5 / 2λ = ( )2.5 3 .≈
( ) , ( / ) ( / ) 1P A P B A P B Aα α′ ′= = = −
( ) 1 ( )P A P A′ = − 1 α= −
( / ) 1 ( / ) 1 (1 )P B A P B A α α′ ′ ′= − = − − =
∴( ) ( ) ( )
( / )( ) ( )
P A B P B P A BP A B
P B P B
′∩ − ∩′ = =
( ) ( ) ( / )
( )
P B P A P B A
P B
−=
( )( / )
( )
P A BP B A
P A
∩=
Q
( ) (1 )
( )
P B
P B
α α− −=
( ) ( ) ( / ) ( ) ( / )P B P A P B A P A P B A′ ′= ⋅ + ⋅
(1 ) (1 )α α α α⋅ − + − ⋅
2 (1 )α α= −
( )P B
2 (1 ) (1 ) (1 ) 1( / )
2 (1 ) 2 (1 ) 2P A B
α α α α α αλ
α α α α− − − −′ = = = =
− −
∴1
2008 2008 10042
λ = × =
1000 999 2 998 1000(1 ) (1 ) (1 )x x x x x x+ + + + + + +K
10011001
1000 1000
1001 1001
(1 ) 1 (1 )1 (1 )
(1 )1
1(1 )1
x xx xx x
x xx xx
xx
+ − + − + + = = = + −+ − − ++
50 1001
50
1001!
50!951!x C= =
∴ 1001, 50λ µ= = v 951=
∴ 2 3 1001 100 2853 3954vλ µ+ + = + + =
207Mock Test-5
5. (139) ,
,
And
Hence, required value is
6. (4)
and
. . .(i)
. . .(ii)
. . .(iii)
⇒
7. (112)
8. (2880)
9. (a) Set A has 4 elements
Set B has 2 elements
Number of elements in set
Total number of subsets of
Number of subsets having 0 elements
Number of subsets having 1 element each
∴ Number of subsets having 2 elements each
Number of subsets having at least 3 elements
10. (d) with a, b of same sign.
If or
⇒
⇒
Hence real or purely imaginary number cannot satisfy
11. (d)
(For purely imaginary)
,
12. (b,c,d)
Null matrix if n is a multiple of 3
13. (b) Given m is A. M. between and n
Given in G.P.
Q 1 7 7sin sin
6 6 6
π π ππ− = − = −
1 2 2cos cos
3 3
π π− =
1 5 5tan tan
4 4 4
π π ππ− = − =
1 3cot cot
4 4 4
π π ππ− − = − =
1 2 3216 27 28 200
6 3 4 4
π π π ππ ×− + × + × + ×
36 18 7 150 139= − + + + =
| | | | | | 1= = =rr r
a b c
× + × = + +r r rr r r r
a b b c pa qb rc
( ) ( ) ( )⋅ × = + ⋅ + ⋅r rr r r r r
a b c p q a b r a c1
[ ]2
=rr r
a b c
[ ]2 2
+ + =rr rq r
p a b c
02 2+ + =
p rq
[ ]2 2+ + =
rr rp qr a b c
= = −p r q
⇒2 2 2
2
24
+ +=
p q r
q
∴2 2
2 22 , 28 8
y yf x x xy
+ − =
2 22 2
2 22 28 8
y yx x
= + − −
∴ (60, 48) (80, 48) (13, 5)f f f+ +
2 2 2 2 2 2(60) (48) (80) (48) (13) (5) 36 64 12= − + − + − = + +
112=
/ (cos ( sin ) cos )tan
/ ( sin cos sin )
dy dy dt a t t t tt
dx dx dt a t t t t
− − −= = =
− + +
⇒2
2(tan ) (tan )
d y d d dtt t
dx dx dt dx= = ⋅
32 1 sec
seccos
tt
at t at= =
2
2
/3
8 24
/ 3t
d y
dx a aπ
π π=
= =
∴2
2
/3
120 120 24 2880
t
d ya
dxπ
π=
= × =
∴ ( ) 4 2 8× = × =A B
∴ 8( ) 2 256× = =A B
08 1= =C
18 8= =C
2
8! 8 78 28
2!6! 2
×= = = =C
256 1 8 28 256 37 219= − − − = − =
2( )P x ax b= +2 2( ( )) ( )P P x a x b b= + +
x R∈ ix R∈2x R∈
( )P x R∈
⇒ ( ( )) 0P P x ≠
( ( )) 0.P P x =
2 3 sin 1 2 sin
1 2 sin 1 2 sin
θ θθ θ
+ +×
− +
i i
i i
22 6sin 0θ− =
2 1sin
3θ =
1sin
3θ = 1 1
sin3
θ −=
2 3 4 2
3 4 5 3
2 3 2 4
ω ω ω ωω ω ω ω
ω ω ω
+
+
+ + +
=
K
K
M M M M M
K K
n
n
n n n
P
4 6 5 7 9
5 7 9
2
4 6 2 4 2 6
ω ω ω ω ωω ω ω
ω ω ω ω+ + + +
+ + +
+ + =
+ +
K K K
K K K K
M M M K
K K K Kn n n n
P
2 =P
l
⇒ 2m n= +l
1 2 3, , , ,G G G nl
1/ 4
4n nr r
= ⇒ = l l
Mathematics208
So,
14. (a,d)
15. (a,b,c)
Rearranging, we get
Rearranging again,
16. (b) Case-1 Any 5 – digit number > 6000 is all 5-digits
number
Total number > 6000 using 5-digits = 5! = 120
Case-2: Using 4-digits
Can be 6.7 or 8 4 ways 3 ways 2 ways
i.e., 3 ways Total number
Total ways
17. (b,d)(a)
(b)
(c)
(d)
18. (c)
⇒
19. (A)
Using
(B) For domain
Case (i):
∴ 2 3
1 2 3, ,G G r G r= = =l l l
4 4 4 4 4 4 8
1 2 32 [1 2 ]G G G r r r+ + = + +l
2
4 1 2n n n = ⋅ + +
ll l l
2 23 3 2 2
2
( )1 (2 ) 4
n nn n n m m n
+ = + = = =
ll l l l
l l
( 1) ( 1)42 2 2 2 22
1 0
( 1) ((4 4) (4 3) (4 2) (4 1) )k k nn
n
k r
S k r r r r+ −
= =
= − + + + − + − +∑ ∑( 1)
0
(2(8 6) 2(8 4))n
r
r r−
=
= + + +∑( 1)
0
(32 20)n
r
r−
=
= +∑
16( 1) 20n n n= − +
4 (4 1)n n= +1056 for n 8
1332 for n 9
==
=
2 2log 3 ( 1) log 4 2( 1)= − = −x x x
⇒2log 3 2 2= −x x
⇒2
2
2 log 3=
−x
3
3
3
2 log 22
1 2 log 2 12
log 2
−= =
−−x
3 4
3 4
4
1
log 4 log 3 1.
1log 4 1 1 log 31
log 3
= = =− −−
x
3 4 3 2 72= × × × =
120 72 192= + =
1
1 1 112 4 4
1 1 3 1 1 1 1 1 3 1 1 1 8
2 4 4 4 4 2 2 4 4 2 4 4
cXP
X
× × =
× × + × × + × × + × ×
8 1 1 1
Exactly twoengingesarefunctioning 732 4 4 28X 8
32
P
− × × = =
2
1 1 3 1 1 1 1
54 2 4 2 4 2 4
1 1 1 1 3 1 1 1 3 8
4 2 4 2 4 2 4 2 4
XP
X
× + × + × = = × + × + × + ×
1
1 1 1 1 3 1 3
72 4 4 4 4 4 4
1 3 3 1 1 1 3 1 3 16
2 4 4 4 4 4 4 4 4
XP
X
× + × + × = = × + × + × + ×
2sin 2sin cos
2sin 2sin cos
P P P
P P P
−
+
⇒1 cos 3
1 cos 32
P
P
−=
+
29cos
35P =
1 5 7 8 6
2 2 2 35∆ = × × ×
⇒ 6∆ = ⇒
23
4
∆
2 2 2
2i(x iy) 2i(x iy)z
1 (x iy) 1 (x y 2ixy)
+ += =
− + − − +2 21 x y− =
2
2ix 2y 1Z .
y2y 2ixy
−= = −
−
Q1 1
1 y 1 1or 1.y y
− ≤ ≤ ⇒ − ≤ − − ≥
x 2
2(x 1)
8.31 1
1 3
−
−− ≤ ≤
−
⇒x x 2
2x 2
3 31 1.
1 3
−
−
−− ≤ ≤
−x x 2
2x 2
3 31 0
1 3
−
−
−− ≤
−
P
Q R
5/2 7/2
1B 3A
2B 2A
3B 1A
1/ 2 1/ 2= − =x x
O
/ 6π
209Mock Test-5
Case (ii):
So,
(C)
(D)
Increasing, when
20. (A)
So, one solution.
(B) Let (a, b, c) is direction ratio of the intersected line, then
We must have
(C) Let
⇒ k can take value 2, 3, 4, 5.
(D)
⇒
JEE Advance Paper-II
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
1 36 64 5522 5 1352 4 16 a a
11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
c a c d a b b c a c
1. (1) Let
Let
2. (36) Let
Put
Then,
We, get Then,
3. (64) We have,
⇒
Or
Putting we have,
∴
⇒ i.e.,
⇒x x 2
2x 2
(3 1)(3 1)0
(3 1)
−
−
− −≥
−
⇒ x ( ,0] (1, ).∈ −∞ ∪ ∞
x x 2
2x
3 31 0
1 3 2
−−+ ≥
− −
⇒x 2 x
x x 2
(3 1)(3 1)0
(3 .3 1)
−
−
− +≥
−
⇒ x ( , 1) [2, ).∈ −∞ ∪ ∞
x ( , 0] [2, ).∈ −∞ ∪ ∞
1 1 3R R R→ +
0 0 2
f ( ) tan 1 tan
1 tan 1
θ θ θθ
= −
− −
2 22(tan 1) 2 sec .θ θ= + =
1/ 2 3 / 2 1/ 23 15f (x) (x) (3x 10) (x) 3 (x) (x 2)
2 2′ = − + × = −
x 2.≥
'( ) 0, (0, / 2)π> ∀ ∈f x x
(0) 0and ( / 2) 0π< >f f
4 0+ + =ak b c
4 2 0+ + =a kb c
28 4 2 16= =
− − −a b c
k k k
22(8 ) 2(4 2 ) ( 16) 0− + − + − =k k k
⇒ 2,4.=k
( ) | 2 | | 1| | 1| | 2 |= + + + + − + −f x x x x x
1=
+∫ ∫dy
dxy
( ) 2 1= −xf x e
⇒ (ln 2) 3=f
10( ) logf x x=
⇒10
1( ) logf x e
x′ = ⋅
4, 0.04x xδ= =
∴ ( ) ( ) ( )f x x f x x f xδ δ ′+ = + ⋅
⇒10 10 10
1log ( ) log (0.04) logx x x e
xδ+ = + ×
⇒10 10 10
1log 4.04 log 4 (0.04) log e
x= + × ×
0.6021 (0.01)(0.4343)= +
0.6021 0.004343 0.606443= + = 1≈
( tan cot )I x x dx= +∫(sin cos )
(sin cos )
x xdx
x x
+= ∫
sin cosx x t− =
⇒ 21 sin 2x t− =
∴ (cos sin )x x dx dt+ =
1 1
222 sin 2 tan
11
2
dt tI t c
tt
− − = = = +
− −
∫
1 sin cos2 tan
sin 2
x xc
x
− − = +
1 tan 1
2 tan2 tan
xc
x
− − = +
2, 2a b= =
⇒4 5 4 32 36a b+ = + =
3 3
2 2
/8
/
d y dx
d y dx=
2
2ln 8
d yx c
dx= +
2ln 8y x c= +
0x =
2log (0) ln1 0c y= = =
2ln 8y x=
8
2
xy e=
8
18
xe
y D= +
4x− 4x
2 4x −6
4 2x−
2− 1− 1 2
Mathematics210
Again putting then
⇒
⇒
Putting we have
Hence,
4. (5522) Since and are the roots of the equation
Then
and coordinates of the vertices of the given triangle are
(0, 0) and
Hence, the required area of triangle
On comparing,
or
⇒
5. (5) Equation of chord of contact is
and
Then
Area of
For maximum or minimum then we get
( as P is outside the circle)
And
Area of triangle is also maximum at
6. (1352) Let PV of P, A, B and C are and
respectively and be the circumcircle of the
equilateral triangle ABC.
Then, . . .(i)
and unit . . .(ii)
Now,
Similarly and
[From equation (i) and (ii)]
0,x =1
1(0)
8y D= +
10
8D= +
∴ 1
8D = −
8
1
1
8 8
xe
y = − ⇒8
64 8
xe x
y E= − +
0,x = 1 1(0) 0
64 8y E= − + =
∴ 1 1 7
8 64 64E = − =
8 87 ( 8 7)
64 8 64 64
x xe x e x
y− +
= − + = ⇒ 64λ =
1m
2m
2 ( 3 2) ( 3 1) 0x x+ + + − =
1 2 1 2( 3 2), ( 3 1)m m m m+ = − + = −
∴ 2
1 2 1 2 1 2( ) 4 (3 4 4 3 4 3 4) 11m m m m m m− = + − = + + − + =
1( / , )c m c
2( / , )c m c
1 2
1
2
c cc c
m m= × − ×
2 12 2
1 2 1 2
1 1 1 1
2 2
m mc c
m m m m
− = − =
21 11
2 ( 3 1)c=
−
2 21 11( 3 1) 33 11
2 4( 3 1)( 3 1)c c
+ += = − +
33 11,
4 4a b= =
11 33,
4 4a b= =
∴ 2 2 33 11 44 11
16 16 16 4a b+ = + = =
2 2 112008( ) 2008
4a b+ = × 502 11 5522= × =
( )QR 26 8 0x y r+ − =
2 2
2 2
6 6 8 8 100
10(6 8 )
r rPM
⋅ + ⋅ − −= =
+
2 2
2 2
0 0
10(6 8 )
r rOM
+ −= =
+
42 2 22 2 ( ) ( ) 2
100
rQR QM OQ OM r
= ⋅ = − = −
1
2QPR QR PM∆ = ⋅ ⋅
∴24
21001
(say) 22 100 10
rrr
− ∆ = ⋅ − ⋅
∴2 2 3
2 (100 )(say)
1000
r rz
−∆ = =
∴ 2 2 2 2 31 3(100 ) ( 2 ) (100 ) 2
1000
dzr r r r r
dr= ⋅ − ⋅ − + − ⋅
2 22 22 (100 )
100 3 1000
r rr r
−= − −
0,dz
dr=
5,r = 10r ≠2
2
5r
d zve
dr=
= −
∴ 5.r =
,, a bprrr
cr
(0)Or
03
a b c+ +=
rr r
26| | | | | | | |
3p a b c= = = =
rr r r
2 2 2 2| | | | 2PA a p a p a p= − = + − ⋅uuur r r r r
2 2 2| | 2PB b p b p= + − ⋅uuur r r
2 2 2| | 2PC c p c p= + − ⋅uuur r r
∴ 2 2 2 2 2| | 3 2 ( )PA a b c p p a b cΣ = + + + − + +uuur rr r r
26 0p= −
22(26)
6 2(26) 13523
= = =
(6, 8)
Q
R
O
r r
M
P
(0, 0)
211Mock Test-5
7. (4) Any point on the first line in symmetrical form is
If the lines are coplanar, this
point must lie on both the planes which determine the
second line.
. . .(i)
and . . .(ii)
From equation (i), we get
Now substituting equation (ii), then
8. (16) Let the equation of circle
. . .(i)
and the equation of the rectangular hyperbola is
. . .(ii)
Put and in equation (i)
Then
This equation being fourth degree in t. Let the roots be
then
9. (a)
10. (a)
Similarly,
11. (c) From graph,
which holds if
12. (a)
..........
Summing up all terms
Hence,
13. (c)
where
14. (d)
(3 4, 5 6, 2 1).r r r− − − +
⇒ 3(3 4) 2(5 6) 2 1 5 0r r r− − − − + + =
2(3 4) 3(5 6) 4( 2 1) 0r r r k− + − + − + − =
2r =
2r = 4k =
2 2 2 2 0x y gx fy k+ + + + =
2xy c=
x ct=c
yt
=
22 2
2
22 0
c fcc t gct k
t t+ + + + =
⇒ 2 4 3 2 22 2 0c t gct kt fct c+ + + + =
1 2 3 4, , ,t t t t 1 2 3 4 1t t t t =
∴ 1 2 3 416 16 1 16t t t t = × =
21 1
2
2
1tan ( ) tan.2
11
xx
xyx x
x
− −
+ −= −
−
⇒3
1 1
2
3tan ( ) tan
1 3
x xy
x
− − −=
− ⇒
3
2
3
1 3
x xy
x
−=
−
1
3
h
AQ=
⇒ 3AQ h=
BQ h=
⇒3
hCQ =
∴ ( 3 1) 3
1
3
AB AQ BQ h
BC BQ CQ hh
− −= = =
− −
1p = −
⇒1
lim ( ) 1x
g x+→
= − ⇒0
lim (1 ) 1h
g h→
+ = −
⇒0
lim 1log cos
n
mh
h
h→
= −
⇒1 1
0 0lim lim 1,
( tan ) tan
n n
h h
n h n h
m h m h
− −
→ →
⋅ = − = − ⋅ −
2.n m= =
( 1) log( ( 1) log log( ( )) log ( )+ = + = + = +g x f x x f x x g x
⇒ ( 1) ( ) log+ − =g x g x x
⇒2
1( 1) ( )′′ ′′+ − = −g x g x
x
1 11 4
2 2
′′ ′′+ − = −
g g
1 1 42 1
2 2 9
′′ ′′+ − + = −
g g
2
1 1 4
2 2 (2 1)
′′ ′′+ − − = − − g N g N
N
2
1 1 1 14 1 ... .
2 2 9 (2 1)
′′ ′′+ − = − + + + − g N g
N
2 2
4 2 4 2
( 1) ( 1)
1 1
− −− = =
+ + + +∫ ∫x x
x x
e e ZJ I dx dz
e e z z
= xz e
2
2
11
1 1ln
2 111
−
−
− + − = = + + + + −
∫x x
x x
dze ez
ce e
zz
∴2
2
1 1ln .
2 1
− +− = + + +
x x
x x
e eJ I c
e e
2
0
1 4sin 4sin2 2
x xI
π
= + −∫0
1 2sin2
xI dx
π
= −∫
3
0
3
1 2sin 1 2sin2 2
xI dx dx
ππ
π
π = − + − − ∫ ∫
3
03
1 4cos 4cos2 2
x xx x
ππ
π
= + + − −
0 1
x–1 –x+1
A QCB
h
P
30° 45° 60°
Mathematics212
15. (a) Rearranging the equation we get,
. . .(i)
Integrating (1) on both sides we get
where k is a constant of integration.
Using we get
the relation is
16. (b) Inclination of line is
Inclination of line
Slope of line
Equation of = Line L
⇒
17. (b)
Hence unit vector will be
18. (c) Plane is given by
distance
19. (a) Let A, B, C and D be the complex numbers,
and respectively.
20. (c)
[as A is the foucs,
is the vertex and BD is the directrix of parabola].
Also is latus return
∴ Area of
3 31 4 1 4 0
3 2 3 2
π ππ
= + − − − − −
22 3 4 2 3 4 3 4
3 3 3I
π π π= + − − + = − −
( ) 1
( ) 400 2
dp tdt
p t=
−
/ 2( ) 400 ,
tp t k e= +
(0) 100,p = 300k = −
∴ 1/ 2( ) 400 300P t e= −
3 1x y+ = 150°
150 60L = ° ± ° 210 , 90= ° °
1tan 210 tan 30
3L = ° = ° =
12 ( 3)
3y x+ = −
⇒ 3 3 2 3 0y x− + + =
ˆˆ ˆ3 1 2 7 5
1 2 3
i j k
i j k= − − +
ˆˆ ˆ7 5.
5 3
i j k− − +
( 1) 7( 2) 5( 1) 0x y z− + − + + + =
⇒ 7 5 10 0x y z+ − + =
⇒ 1 7 5 10 13.
75 75
+ − += =
2, 2, 2i−
2i−
⇒2 2 2 2
2 2 2 2
PA PB PC PD
QA QB QC QD
+ + ++ + +
2 2 2 2 2
1 1 1 1 1
22 2 2 212 2 2 2
| 2 | | 2 | | 2 | | 2 | | | 2 3
| | 2 4| 2 | | 2 | | 2 | | 2 |
z z z i z i z
zz z z i z i
− + − + − + − += = =
++ + − + − + −
2AG =
∴ 1 1
1
2AT TG= =
1T
2 3T T
∴ 2 3
14
2T T = ×
1 2 3
1 1 41
2 2 2T T T∆ = × × =
D
G
C
B A
M
T1 T2