2018-jee-advanced - sri chaitanya · 2018-jee-advanced 2 question paper-_key & solutions sri...

2018-Jee-Advanced Question Paper-2_Key & Solutions

Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081

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JEE (ADVANCED) 2018 PAPER 2 PART I-PHYSICS

SECTION 1 (Maximum Marks: 24)

This section contains SIX (06) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four

option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of which are correct options.

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is acorrect option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : -2 In all other cases.

For Example: If first, third and fourth are the ONLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the three correct options (either first or third or fourth option) ,without selecting any incorrect option (second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in -2 marks.

Q.1 A particle of mass m is initially at rest at the origin. It is subjected to a force and starts

moving along the x-axis. Its kinetic energy K changes with time as dK tdt

, where is a

positive constant of appropriate dimensions. Which of the following statements is (are) true? A) The force applied on the particle is constant B) The speed of the particle is proportional to time C) The distance of the particle from the origin increases linearly with time D) The force is conservative Key: (A, B, D)

Sol: dk tdt

as 212

k mv dk dvmv tdt dt

0 0

v t

m vdv tdt

2 2

2 2mv t

v tm

dvadt m

Constant Since F=ma F m m

m Constant

Q.2 Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity 0u . Which of the following statements is (are) true?

A) The resistive force of liquid on the plate is inversely proportional to h B) The resistive force of liquid on the plate is independent of the area of the plate C) The tangential (shear) stress on the floor of the tank increases with 0u D) The tangential (shear) stress on the plate varies linearly with the viscosity of the liquid Key: (A, C, D)

0uh




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Sol: viscous force is given by dvF Ady

since h is very small therefore; magnitude of viscous

force is given by dvF Ady

00&AuF F F u

h

1 ,F F Ah

Q.3 An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density . It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 120° at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is 0 . Which of the following statements is (are) true?

A) The electric flux through the shell is 03 /R B) The z component of the electric field is zero at all the points on the surface of the shell C) The electric flux through the shell is 02 /R D) The electric field is normal to the surface of the shell at all points Key: (A, B)

P

Q

O060

0120

R

z-axis

Sol: filed due to straight wire is perpendicular to the wire & radially outward. Hence 0zE length,

2 sin 60 3PQ R R according to gauss’s law

Total flux=0 0

3. inq RE ds

Q.4 A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale.)




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A)

B)

C)

D) Key: (D)

A

B

F

f

045

/ 2f

Sol: distance of a point A is 2f

Let A’ is the image of A from mirror, for this image

1 1 1

2fv f

1 2 1 1v f f f

Image of line AB should be perpendicular to the principle axis & image of F will form at infinity, therefore correct image diagram is

'A

'B

or

f-xx f-x

2

1

f hf u h

2

f f xh

f x

2h f

Q.5 Ina radioactive decay chain, 23290Th nucleus decay to 212

82 Pb nucleus. Let N and N be the

number of and particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

A) 5N

B) 6N

C) 2N

D) 4N

Key: (A, C)




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Sol: 23290 Th is converting into

21282 Pb

Change in mass number (A) =20

Number of particle =20 54

Due to 5 particle, z will be change by 10 unit. Since given change is 8, therefore number of particle is 2 Q.6 In an experiment to measure the speed of sound by a resonating air column, a tuning fork of

frequency 500Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7cm and 83.9 cm. Which of the following statements is (are) true?

A) The speed of sound determined from this experiment is 332m/s B) The end correction in this experiment is 0.9cm C) The wavelength of the sound wave is 66.4cm D) The resonance at 50.7cm corresponds to the fundamental harmonic Key: (A, C) Sol: let 1n harmonic is corresponding to 50.7 cm & 2n harmonic is corresponding 83.9 cm.

(83.9 50.7)2 cm 33.2

2 cm

66.4 cm 16.64 cm

Length corresponding to fundamental mode must be close to 4

& 50.7 cm must be an odd

multiple of this length 16.6 3 49.8 cm. therefore 50.7 is 3rd harmonic If end correction is e, then

350.74

e cm Speed of sound, v f

500 66.4v cm/sec 332.000 m/s


This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded -off to the second decimal

place; e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) using the mouse and the on- screen virtual numeric keypad in the place designated to enter the answer.

Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct numerical value is entered as answer.

Zero Marks : 0 In all other cases.

Q.7 A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass

0.4m kg is at rest on this surface. An impulse of 1.0Ns is applied to the block at time 0t so that it starts moving along the x-axis with a velocity /

0tv t v e , where

0v is a constant and 4s . The displacement of the block, in metres, at t is . Take 1 0.37e .




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Key: 6.30

J=1 m=0.4

Sol: /0

tV V e 0 2.5JVm

m/s

/0

tV V e /0

tdx V edt

/0

0 0 1

x xt x edx V e dt e dx

0

0

/

1

tx V

e

1 02.5( 4)( )X e e 2.5 4 0.37 1X 6.30X Q.8 A ball is projected from the ground at an angle of 45° with the horizontal surface. It

reaches a maximum height of 120m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in meters, is .

Key: 30.00

045

u

1 120H m030 2H

2V u

Sol: 2 2

1sin 45 1202

uHg

2

1204ug

……..(i)

When half of kinetic energy is lost V=2

u

22

2

2

sin 302

2 16

uuH

g g

……..(ii) From (i) & (ii)

2H = 30.00 Q.9 A particle, of mass 310 kg and charge 1.0C , is initially at rest. At time 0t , the

particle comes under the influence of an electric field 0 sinE t E t i

, where

0 1.0 /E N C and 310 /rad s . Consider the effect of only the electrical force on the particle. Then the maximum speed, in m/s , attained by the particle at subsequent times is .




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Key: 2.00 Sol: 310 1 0n kg q C t

0 sinE E t Force on particle will be 0 sinF qE qE t At max' 0, 0 sin 0V a F qE t 0 sinF qE t

0 sindv Eq tdt m

/

0

0 0

sinV qEdv t dt

m

/0

00 cosqEV

m t

00 cos cos0qEV

m

3 3

1 1 2 210 10

V

m/s =2.00

Q.10 A moving coil galvanometer has 50 turns and each turn has an area 4 22 10 m . The magnetic field produced by the magnet inside the galvanometer is 0.02T . The torsional constant of the suspension wire is 4 110 Nmrad . When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2rad . The resistance of the coil of the galvanometer is 50 . This galvanometer is to be converted into an ammeter capable of measuring current in the range 0 1.0A . For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms,is…..

Key: 5.55 Sol: n= 50 turns A= 4 22 10 m B=0.02 T 410K mQ =0.2 rad 50gR AI =0-1.0A ,MB C M nIA BINA=C 4 40.02 1 50 2 10 10 0.210 gI =01.A For galvanometer, resistance is to be connected to ammeter in shunt.

0.1gI 50gR

gI I S g g gI R I I S 0.1 50 1 0.1 S

50 5.559

S

Q.11 A steel wire of diameter 0.5mm and Young’s modulus 11 22 10 /N m carries a load of mass M. The length of the wire with the load is 1.0m . A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0mm , is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2kg , the vernier scale division which coincides with a main scale division is . Take

210 /g m s and 3.2 .




-8

Key: 3.00 Sol: 0.5d mm 112 10Y 1l m

2 24 11

1.2 10 1

5 10 2 1044

Fl mglldAy y

8 11

1.2 103.2 25 10 2 104

l

=

3 3

12 12 0.30.8 25 2 10 40 10

mm

So 3rd division of a vernier scale will coincicle with main scale. Q.12 One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its

volume becomes eight times its initial value. If the initial temperature of the gas is 100K and the universal gas constant 1 18.0R J mol K , the decrease in its internal energy, in Joule , is .

Key: 900

Sol: iV V 8FV V For adiabatic process 5{3

for monotonic process

1 11 1 2 2.TV T V

23100 V =

23

2 8T V

2 25T k 1 100 25 12 75 9002v

FRU nc T

Joule

Q.13 In a photoelectric experiment a parallel beam of monochromatic light with power of 200W is incident on a perfectly absorbing cathode of work function6.2eV . The frequency of light is just above the thres hold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force 410F n N due to the impact of the electrons. The value of n is . Mass of the electron 319 10em kg and

191.0 1.6 10eV J . Key: 24 Sol: power = nhv n= number of protons per second Since KE= 0, hv

19200 6.25 1.6 10n joule 19

2001.6 10 6.25

n

As photon just above threshold frequency maxKE is zero and they are accelerated by

potential difference of 500V. fKE q V 2

22P q V P m Vm

Since efficiency is 100%, number of electrons = number of photons per second As photon is completely absorbed force exerted = nmv

31 19

19

200 2 9 10 1.6 10 5006.25 1.6 10

254

19

3 200 10 1600 2 40 10 3 246.25 1.6 10 6.25 1.6




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Q.14 Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the 2 1n to n transition has energy 74.8eV higher than the photon emitted in the 3 2n to n transition. The ionization energy of the hydrogen atom is 13.6eV . The value of Z is .

Key: 3

Sol: 2 22 1

1 313.6 1 13.64 4

E z z

2 23 2

1 1 513.6 13.64 9 36

E z z

2 1 3 2 74.8E 2 23 513.6 13.6 74.8

4 36z z

2 3 513.6 74.84 36

z 2 9z 3z


This section contains FOUR (04) questions. Each question has TWO (02) matching lists: LIST-I and LIST-II.

FOUR options are given representing matching of elements from LIST-I and LIST-II. ONLY ONE of these four options corresponds to a correct matching.

For each question, choose the option corresponding to the correct matching. For each question, marks will be awarded according to the following marking scheme:

Full Marks : +3 If ONLY the option corresponding to the correct matching is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : -1 In all other cases.

Q.15 The electric field E is measured at a point 0,0,P d generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II.

LIST-1 LIST-II P. E is independent of d 1. A point charge Q at the origin

Q. 1Ed

2. A small dipole with point charges Q at 0,0,l and -

At 0,0, l take 2l d

R. 21E

d 3. An infinite line charge coincident with the x-axis,

with uniform linear charge density

S. 31E

d 4. Two infinite wires carrying uniform linear charge

density parallel to the x- axis.The one along 0,y z l has a charge density and the one along 0,y z l has a charge density . Take 2l d

5. Infinite plane charge coincident with the xy-plane with uniform surface charge density

A) P → 5; Q → 3, 4; R → 1; S → 2 B) P → 5; Q → 3; R → 1, 4; S → 2 C) P → 5; Q → 3; R → 1, 2; S → 4 D) P → 4; Q → 2, 3; R → 1; S → 5



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Page

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Key: (B)

Sol: (i) 2 2

1KQE Ed d

(ii) dipole

23

2 1 3coskpEd

3

1Ed

for dipole

(iii) for line charge 2kE

d

1Ed

(iv) 2 2K KEd l d l

2 22 d l d lKl

d l

ˆ ˆdr t

V i jdt

2

1Ed

(V) Electric filed due to sheet 02

V is independent of r

Q.16 A planet of mass M, has two natural satellites with masses 1m and 2m . The radii of their

circular orbits are 1R and 2R respectively. Ignore the gravitational force between the

satellites. Define 1 1 1 1, ,v L K and T to be, respectively, the orbital speed, angular momentum,

kinetic energy and time period of revolution of satellite1; and 2 2 2 2, ,v L K and T to be the

corresponding quantities of satellite 2.Given 1

22m

m and 1

2

14

RR

, match the ratios in List-I to

the numbers in List-II.

LIST–I LIST–II

P. 1

2

vv

1. 18

Q. 1

2

LL

2.1

R. 1

2

KK

3.2

S. 1

2

TT

4.8

A) P → 4; Q → 2; R → 1; S → 3

B) P → 3; Q → 2; R → 4; S → 1

C) P → 2; Q → 3; R → 1; S → 4

D) P → 2; Q → 3; R → 4; S → 1




Page

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Key: (B)

2m1R

M

1

2

2mm

1

2

14

RR

1R

1m

Sol: 2

1 1 121 1

GMm mVR R

2 21 2

1 2

,GM GMV VR R

2

1 22

2 1

4V RV R

(P) 1

2

2VV

(Q) L mvR 1 1 1 1

2 2 2 2

12 2 14

L m v RL m v R

(R) 212

K mv 2

21 1 12

2 2 2

2 2 8K m vK m v

(S) 2 /T R V 1 1 2 1 2

2 1 2 2 1

1 1 14 2 8

T R v R vT v R R v

Q.17 One mole of a monatomic ideal gas undergoes four thermodynamic processes as

shown schematically in the PV-diagram below. Among these four processes, one is

isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the

processes mentioned in List-1 with the corresponding statements in List-II.

LIST–I LIST–II P. In process I 1. Work done by the gas is zero Q. In process II 2. Temperature of the gas remains unchanged R. In process III 3. No heat is exchanged between the gas and

its surroundings S. In process IV 4. Work done by the gas is 0 06PV

A) P → 4; Q → 3; R → 1; S → 2 B) P → 1; Q → 3; R → 2; S → 4 C) P → 3; Q → 4; R → 1; S → 2 D) P → 3; Q → 4; R → 2; S → 1 Key: (C)




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Sol: process-I is an adiabatic process

Q U W 0Q W U

Volume of gas is decreasing 0W 0U

Temperature of gas increases

No heat is exchanged between the gas and surrounding.

Process –II is an isobaric process (pressure remain constant)

0 0 0 0 03 3 6W P V P V V PV

Process-III is an isochoric process ( volume remain constant)

Q U W Q U

Process-IV is an isothermal process (temperature remains constant)

Q U W 0U

Q.18 In the List-I below, four different paths of a particle are given as functions of time. In these

functions, and are positive constants of appropriate dimensions and . In each case,

the force acting on the particle is either zero or conservative. In List-II, five physical

quantities of the particle are mentioned: p is the linear momentum, L is the angular

momentum about the origin, K is the kinetic energy,U is the potential energy and E is the

total energy. Match each path in List-I with those quantities in List-II, which are conserved

for that path.

LIST–I LIST-II

P. r t t i t j 1. p

Q. cos sinr t t i t j 2. L

R. cos sinr t t i t j 3. K

S. 2

2r t t i t j 4. U

5. E

A) P → 1, 2, 3, 4, 5; Q → 2, 5; R → 2, 3, 4, 5; S → 5 B) P → 1, 2, 3, 4, 5; Q → 3, 5; R → 2, 3, 4, 5; S → 2, 5 C) P → 2, 3, 4; Q → 5; R → 1, 2, 4; S → 2, 5 D) P → 1, 2, 3, 5; Q → 2, 5; R →2, 3, 4, 5; S → 2, 5




Page

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Key: (A)

Sol: (P) ˆ ˆr t ti tj

ˆ ˆdr t

V i jdt

{Constant}

0dvadt

P mv

(Remain constant)

212

k mv {Remain constant} ˆ ˆ 0U UF i ix y

U Constant E= K+ U

0dL r Fdt

L

=constant

(Q) ˆ ˆcos sinr t i t j

2 ˆ ˆcos sindva t i t jdt

2a r 2U r

U depends on r hence it will change with time

Total energy remain constant because force is central

(R) ˆ ˆ( ) cos sinr t ti t j

( ) ˆ ˆ( ) sin cosdr tv t t i t jdt

2 ˆ ˆcos sina t t i t j

(S) 2ˆ ˆ2

r ti t j ˆdva j

dt

Constant F ma

Constant

0

ˆ ˆ ˆ. .t

U F dr m j i tj dt

212

E k U m [Remain constant]




Page

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PART-2:CHEMISTRY SECTION -1 (Maximum Marks: 24)

Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct option(s).

For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme:



Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is acorrect option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : -2 In all other cases.

For Example: If first, third and fourth are the ONLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct

options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the three correct options (either first or third or fourth option) ,without selecting any

incorrect option (second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in -2 marks

1. The correct option(s) regarding the complex [Co(en) (NH3)3(H2O)]3+ :- (en = H2NCH2CH2NH2) is (are) A) It has two geometrical isomers B) It will have three geometrical isomers if bidentate 'en' is replaced by two cyanide ligands C) It is paramagnetic D) It absorbs light at longer wavelength as compared to [Co(en) (NH3)4]3+

Key. ABD

Sol. A) [Co(en)(NH3)3(H2O)]+3 complex is type of [M(AA)b3c] have two G.I.

B) If (en) is replaced by two cynide ligand, complex will be type of [Ma3b2c] and have 3 G.I.

C) [Co(en)(NH3)3(H2O)]3+ have d6 configuration (t 62g) on central metal with SFL therefore it

is dimagnetic in nature.

D)Complex [Co(en)(NH3)3(H2O)]3+ have lesser CFSE ( O) value than [Co(en)(NH3)4]3+

therefore complex [Co(en)(NH3)3(H2O)]+ absorbs longer wavelength for d–d transition.

2. The correct option(s) to distinguish nitrate salts of Mn2+ and Cu2+ taken separately is (are) :-

A) Mn2+ shows the characteristic green colour in the flame test

B) Only Cu2+ shows the formation of precipitate by passing H2S in acidic medium

C) Only Mn2+ shows the formation of precipitate by passing H2S in faintly basic medium

D) Cu2+/Cu has higher reduction potential than Mn2+/Mn (measured under similar conditions)

Key. BD

Sol. A) Cu+2 and Mn+2 both gives green colour in flame test and cannot distinguished. B) Cu+2 belongs to group-II of cationic radical will gives ppt. of CuS in acidic medium. C) Cu+2 and Mn+2 both form ppt. in basic medium. D) Cu+2/Cu = +0.34 V (SRP)

Mn+2/Mn = – 1.18 V (SRP)




Page

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3. Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4) at 288 K to give P (51%), Q

(47%) and R (2%). The major product(s) the following reaction sequence is (are) :-

22 22 3 2

233 22

1) /1) ,2) /2) ,

3) , /273 2783)4)4) , /273 2798

5) ,

Sn HClAc O pyridineBr H O excessBr CH CO H

NaNO HCl KH OH PONaNO HCl K

EtOH

R S major product s

A)

Br

Br

Br

Br B)

Br

Br

Br

Br C)

Br

Br Br D)

Br

BrBr

Br

Key. D

Sol.

3 2 4/HNO HSO

Br

2ACOPy

2 3 2/Br CHCOH

BrBrBr Br

2NaNO HCl

00 5 C 3 2HPO

4. The Fischer presentation of D-glucose is given below.

CHO

OHH

H

H

OH

OH

HHO

2CHOH

D-glucose The correct structure(s) of -glucopyranose is (are) :-

A)

O

H

H

HH

H

HO

OH

OH

HO

2CHOH

B)

O

HH

HH

H

HO

OH

OH2CHOH

OH C)

O

H

H

H

H

H

HOOH

OH

HO

2CHOH

D)

O

H

H

H

H

H

HO

OH

OHHO

2CHOH

Key. D




Page

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Sol.

CH=O

OHH

H

H

OH

OH

HHO

2CHOH

D-glucose

O

H

H

H

H

H

HO

OH

OHHO

2CHOH

L glucopyranose

OOH

OH

OH

OH

OH H

D glucopyranose

5. For a first order reaction A(g) 2B(g) + C(g) at constant volume and 300 K, the total pressure

at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A is present with

concentration [A]0, and t1/3 is the time required for the partial pressure of A to reach 1/3rd of its

initial value. The correct option(s) is (are) :-

(Assume that all these gases behave as ideal gases)

A) Time

0ln 3 tP P

B) 0A

13

t

C) Time

0ln tP P

D) 0A

Rat

e co

nsta

nt

Key. AD

Sol. 2A B C

0

0

02

t Pt t P P P P

0 2 tP P P 0 0

000

1 1ln ln

2t

P PKP Pt P P t P

00 0

0

21ln ln 2 ln 33 t

t

PK Kt P P Pt P P

and 01

03

1 1ln ln 3 tan/ 3

Pt cons tK P K

Rate constant does not depends on concentration

6. For a reaction, ,A P the plots of [A] and [P] with time at temperatures 1T and 2T are given

below.

If T2 > T1, the correct statement(s) is (are)




Page

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(Assume H and S are independent of temperature and ratio of lnK at T1 to lnK at T2 is

greater than 2

1

TT

. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium

constant, respectively.)

A) 0, 0H S B) 0, 0G H

C) 0, 0G S D) 0, 0G S

Key. AC

Sol. A P given 2 1T T

1 2

2 1

lnln

K TK T

1 1 2 2ln lnT k T k 0 0

1 2G G

1 2H T S H T S 1 2T S T S 0S






7. The total number of compounds having at least one bridging oxo group among the molecules

given below is__.

2 3 2 5 4 6 4 7 4 2 5 5 3 10 2 2 3 2 2 5, , , , , , ,N O N O P O P O H P O H PO H S O H S O

Key. 5 or 6

Sol:




Page

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8. Galena (an ore) is partially oxidized by passing air through it at high temperature. After some

time, the passage of air is stopped, but the heating is continued in a closed furnance such that the

contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is

______ .

(Atomic weights in g mol–1 : O = 16, S = 32, Pb = 207)

Key. 6.47

Sol: 2 2PbS O Pb SO 1000 1000 207

32 32mol gm

Mol of Pb=mol of 2O 100032

mol

1000 20732

mass of Pb g 207 6.4732

kg kg

9. To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted

to KMnO4 using the reaction,

MnCl2 + K2S2O8 + H2OKMnO4 + H2SO4 + HCl (equation not balanced).

Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic

acid(225 g) was added in portions till the colour of the permanganate ion disappeard. The

quantity of MnCl2 (in mg) present in the initial solution is _____.

(Atomic weights in g mol–1 : Mn = 55, Cl = 35.5)

Key. 126

Sol: 2 2 2 8 2 4 2 4amole amole

MnCl K S O H O KMnO H SO HCl

2 4 4 2

2 4 4

2 0.225 / 90 51 55 71

126

H

eq eq

C O MnO COm of C O m of MnO

aa

mg

10. For the given compound X, the total number of optically active stereoisomers is____.

Key. 7




Page

-19

11. In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is____.

(A tomic weight in g mol–1: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis)

Key. 495 Sol:

12. The surface of copper gets tarnished by the formation of copper oxide. 2N gas was passed to

prevent the oxide formation during heating of copper at 1250 K. However, the 2N gas contains

1 mole % of water vapour as impurity. The water vapour oxidizes copper as per the reaction

given below: 2 2 22Cu s H O g Cu O s H g 2Hp is the minimum partial pressure of 2H

(in bar) needed to prevent the oxidation at 1250K. The value of 2HIn P is___

(Given : total pressure=1 bar, R(universal gas constant)= 1 18JK mol , In (10)=2.3. Cu s and

2Cu O s are mutually immiscible. At 1250K:

12 22 1/ 2 ; 78,000Cu s O g Cu O s G J mol

12 2 21/ 2 ; 1,78,000H g O g H O g G Jmol ; G is the Gibbs energy)

Key. -14.6

Sol:




Page

-20

13. Consider the following reversible reaction, A(g) + B(g) AB(g).

The activation energy of the backward reaction exceeds that of the forward reaction by 2RT (in J

1mol ). If the pre-exponential factor of the forward reaction is 4 times that of the reverse

reaction, the absolute value of G (in 1mol ) for the reaction at 300K is____

(Given ; ln (2) = 0.7, RT = 2500 J mol–1 at 300 K and G is the Gibbs energy)

Key. -8500

Sol:

14. Consider an electrochemical cell: A(s) 2, 2 ,1n nA aq M B aq M B s . The value of H for

the cell reaction is twice that of G at 300K. If the emf of the cell is zero, the S

(in 1 1JK mol ) of the cell reaction per mole of B formed at 300K is___

(Given : ln (2) = 0.7, R (universal gas constant) = 8.3 J K–1 mol–1. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)

Key. -11.62

Sol:




Page

-21






15. Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II LIST-I LIST-II P. 2dsp 1. 4

6FeF Q. 3sp 2. 2 33

Ti H O Cl

R. 3 2sp d 3. 3

3 6Cr NH

S. 2 3d sp 4. 24FeCl

5. 4Ni CO

6. 2

4Ni CN

The correct option is A) 5; 4,6; 2,3; 1P Q R S B) 5,6; 4; 3; 1,2P Q R S C) 6; 4,5; 1; 2,3P Q R S D) 4,6; 5,6; 1, 2; 3P Q R S Key. C Sol:




Page

-22

16. The desired product X can be prepared by reacting the major product of the reactions in LIST-I

with one or more appropriate reagents in LIST-II. (given, order of migratory aptitude: ary1>alky1>hydrogen)

Ph

Ph

O

OH

Me X LIST-I LIST-II

p. 1.

q. 2.

r. 3. Fehling solution

s. 4. HCHO, NaOH 5. NaOBr The correct option is

A) P 1; Q2,3; R1,4; S2,4 B) P1,5; Q3,4; R 4,5; S3 C) P1,5; Q3,4; R5; S2,4 D) P1,5; Q 2,3; R1,5; S2,3 Key. D Sol: Preparations of aromatic acids.




Page

-23

17. LIST-I contains reactions and LIST-II contains major products.

LIST-I LIST-II

P. 1.

Q. 2.

R. 3.

S. 4.

5.

Match each reaction in LIST-I with one or more product in LIST-II and choose the correct option. A) P1,5; Q2; R3; S4

B) P1,4; Q2; R4; S3

C) P1,4; Q1,2; R3,4; S4

D) P4,5; Q4; R4; S3,4

Key. B Sol:

P.

Q.

R.

S. 18. Dilution processes of different aqueous solutions; with water, are given in LIST-I. The effects of

dilution of the solutions on H are given in LIST-II

(Note: Degree of dissociation of weak acid and weak base is <<1; degree of hydrolysis of

salt <<1; H represents the concentration of H ions)




Page

-24

LIST-I LIST-II P. (10 mL of 0.1 M NaOH+20mL of 0.1 M

acetic acid) diluted to 60ML 1. the value of H does not change on

dilution

Q. (20mL of 0.1 M NaOH+20mL of 0.1 M

acetic acid) diluted to 80mL 2. the value of H changes to half of its

initial value on dilution

Q. (20mL of 0.1 M HCl+20mL of 0.1 M

ammonia solution) diluted to 80ML 3. the value of H changes to two times

of its initial value on dilution

S. 10mL saturated solution of 2Ni OH in

equilibrium with excess solid 2Ni OH is

diluted to 20mL (solid 2Ni OH is still

present after dilution)

4. the value of H changes to 12

times


5. the value of H changes to 2 times


Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is

A) P4; Q2; R3; S1 B) P4; Q3; R2; S3

C) P1; Q4; R5; S3 D) P1; Q5; R4; S1

Key. D Sol:

S. Because of dilution solubility does not change so [H+] = constant




Page

-25

PART-2:MATHEMATICS SECTION -1 (Maximum Marks: 24)

Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct option(s).

For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme:



Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is acorrect option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : -2 In all other cases. For Example: If first, third and fourth are the ONLY three correct options for a question with second option being an

incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the three correct options (either first or third or fourth option) ,without selecting any incorrect option (second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in -2 marks

Q1. For any positive integer n, define : 0,nf as 1

1

1tan1 1

n

nj

f xx j x j

for

all 0,x

(Here, the inverse trigonometric function 1tan x

assume values in ,2 2

.) Then, which of the following statement(s) is (are) TRUE?

A) 5

21

1tan 0 55

jf

B) 10

21

11 ' 0 sec 0 10j

jf f

C) For any fixed positive integer n, 1lim tan nxf x

n

D) For any fixed positive integer n, 2limsec 1nxf x

Key: D

Sol:

1

1

1tan

1 1

n

nj

x j x jf x

x j x j

1 1

1

tan tan 1n

nj

f x x j x j

1 1tan tannf x x n x

1 1tan tan tan tannf x x n x

tan1n

x n xf x

x x n

2tan1n

nf xx nx




Page

-26

2 2sec 1 tann nf x f x

2

22limsec lim1 1

1nx x

nf xx nx

2. Let T be the line passing through the points P(-2, 7) and Q(2, -5). Let 1F be the set of all pairs of circles 1 2,S S such that T is tangents to 1S at P and tangent to 2S at Q, and also such that 1S and 2S touch each other at a point, say, M. Let 1E be the set representing the locus of M as the pair 1 2,S S varies in 1F . Let the set of all straight line segments joining a pair of distinct points of 1E and passing through the point R(1, 1) be 2F . Let 2E be the set of the mid-points of the line segments in the set 2F . Then , which of the following statement(s) is (are) TRUE?

A) The point (-2, 7) lies in 1E B) The point 4 7,5 5

does NOT lie in 2E

C) The point 1 ,12

lies in 2E D) The point 30,2

does NOT lie in 1E Key. D Sol:

AP AQ AM

Locus of M is a circle having PQ as its diameter

Hence, 1 : 2 2 7 5 0E x x y y and 2x

Locus of B (midpoint) is a circle having RC as its diameter 22 : 1 1 0E x x y

Now, after checking the options, we get (D)

3. Let S be the of all column matrices1

2

3

bbb

such that 1b , 2b , 3b and the system of equations (in

real variables) Has at least one solution. Then, which of the following system(s) (in real

variables) has (have) at least one solution of each1

2

3

bb Sb

?

A) 1 22 3 ,4 5x y z b y z b and 32 6x y z b B) 1 23 ,5 2 6x y z b x y z b and 32 3x y z b C) 1 22 5 ,2 4 10x y z b x y z b and 32 5x y z b D) 1 22 5 ,2 3x y z b x z b and 34 5x y z b




Page

-27

Key. AD

Sol. We find D=0 & since no pair of planes are parallel, so there are infinite number of solution s.

Let 1 2 3P P P 1 2 37 13P P P 1 2 37 13b b b A) 0D unique solution for any 1 2 3, ,b b b B) 0D but 1 2 37 13P P P C) 0D Also 2 1 3 12 ,b b b b but 1 2 37 13b b b hence two simultaneous equations in 1 2 3, ,b b b can be considered as two different planes hence line of intersection has only common solutions other solutions are not common hence C is Not Correct D) 0D

4. Consider two straight lines, each of which is tangent to both the circle 2 2 12

x y and the

parabola 2 4y x . Let these lines intersect at the point Q. Consider the ellipse whose center is at the origin O(0, 0) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is 2 , then the which of the following statement(s) is (are) TRUE?

A) For the ellipse, the eccentricity is 12

and the length of the latus rectum is 1

B) For the ellipse, the eccentricity is 12

and the length of the latus rectum is 12

C) The area of the region bounded by the ellipse between the lines 12

x and 1x is

1 24 2

D) The area of the region bounded by the ellipse between the lines 12

x and 1x is

1 216

Key. AD Sol.

Let equation of common tangent is 1y mxm

4 2

2

10 0 1 2 0 121

m m m mm

Equation of common tangents are 1y x and 1y x




Page

-28

Point Q is 1,0

Equation of ellipse is 2 2

11 1/ 2x y

A) 1 112 2

e and 22 1bLR

a

C)

Area 12

2 2 1

1/ 21/ 2

1 12. . 1 2 1 9 sin2 22xx dx x x

1 1 22 24 4 8 8 4 4 2

Correct answer are (A) and (D) 5. Let s, t, r be the non-zero complex numbers and L be the set of solutions

, , 1z x iy x y i of the equation 0sz t z r , where z x iy .Then, which of the

following statements(s) is (are)TRUE?

A) If L has exactly one element, then s t

B) If s t , then L has infinitely many elements

C) The number of elements in : 1 5L z z i is at most 2

D) If L has more than one element, then L has infinitely many elements

Key. ACD Sol. Given

0sz tz r ….(1)

On taking conjugate 0s z tz r ……..(2)

From (1) and (2) eliminating z

2 2z s t rt rs

A) If s t then z has unique value

B) If s t then rt rs may or may not be zero so L may be empty set

C) locus of z is null set or singleton set or a line in all cases it will intersect given circle at most

two points

D) In this case locus of z is a line so L has infinite elements




Page

-29

6. Let : 0,f be a twice differentiable function such that 2sin sinlim sint

f x t f t xx

t x

for all 0,x . If 6 12

f

, then which of the following statement (s) is (are) TRUE?

A) 4 4 2

f

B) 4

2

6xf x x for all 0,x

C) There exists 0, such that ' 0f

D) " 02 2

f f

Key. BCD

Sol. 2sin sin

lim sint x

f x t f t xx

t x

By using L’Hopital

1

2cos sinlim sin

1t x

f x t f t xx

2cos sin sinf x x f x x x 1

2

sin cos1

sinf x x f x x

x

1

sinf x

dx

sinf x

x cx

Put &6 6 12

x f

0 sinc f x x x

A) 1

4 4 2f

B) sinf x x x

as 3 4

2sin , sin6 6x xx x x x x

42 0,

6xf x x x

C) 1 sin cosf x x x x

1 0 tanf x x x there exist 0, for which 1 0f

D) 11 2cos sinf x x x x 11

2 2f

,

2 2f

11 02 2

f f




Page

-30






7. The value of the integral

12

12 60 4

1 3

1 1x x

Key. 2

Sol.

12

1/42 60

1 3

1 1

dx

x x

12

1/460 2

6

1 3

11

1

dx

xx

x

Put 2

1 21 1

x dxt dtx x

1/31/3

6/41

1 3 1 3 2 1 3 3 1 22 2

dtI

t t

8. Let P be a matrix of order 3 3 such that all the entries in P are from the set 1,0,1 . Then the

maximum possible value of the determinant of P is_______.

Key. 4

Sol: 1 2 3

1 2 3 1 2 3 2 3 1 3 1 2 3 2 1 2 1 3 1 3 2

1 2 3x y

a a ab b b a b c a b c a b c a b c a b c a b cc c c

Now if 3x and 3y thecan be maximum 6 But it is not possible as 3x each term of 1x and 3y each term of 1y

3

1

1i i ii

a b c

and3

1

1i i ii

a b c

Which is contradiction. So now next possibility is 4

9. Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If is the

number of one- one functions form X to Y and is the number of onto functions from Y to X,

then the value of 15!

is ______.

Key. 119

Sol. 5n X 7n Y

Number of one-one function 75 5!C




Page

-31

Number of onto function Y to X

1,1,1,1,3 1,1,1,2,2

7 7 73 3 33

7! 7!5! 5! 3. 5! 4 5!3!4! 2! 3!

C C C

7 73 54 4 35 21 119

5!C C

10. Let :f be a differentiable function with 0 0f . If y f x satisfies the differential

equation 2 5 5 2dy y ydx

, then the value of limx

f x

is____.

Key. 0.4

Sol. 225 4dy ydx

So, 225 4dy dxy

Integrating,

21 1 5ln2 225 2

5 5

yx c

y

5 2ln 20

5 2y x cy

Now, 0c as 0 0f Hence 205 25 2

xy ey

205 25 2

x

x x

f xlet let e

f x

Now, RHS = 0 5 2 0xlet x x

25x

let f x

11. Let :f be a differentiable function with 1f O and satisfying the equation

' 'f x y f x f y f x f y for all ,x y. Then, then value of log 4e f is

__________

Key. 2

Sol. , : ' ' .P x y f x y f x f y f x f y x y R

0,0 : 0 0 ' 0 ' 0 0P f f f f f 1 2 ' 0f

,0 : . ' 0 ' . 0P x f x f x f f x f 1 '2

f x f x f x

1'2

f x f x 12

xf x e ln 4 2f

12. Let P be a point in the first octant, whose image Q in the plane 3x y (that is, the line segment PQ is perpendicular to the plane 3x y and the mid-point of PQ lies in the plane 3x y ) Lies on the z-axis. Let the distance of P from the x-axis be 5. If R is the image of P in the xy-plane, then the length of PR is_____

Key. 8




Page

-32

Sol. Let , ,P 0,0,Q & , ,R

Now, ˆ ˆ ˆ ˆ ˆ ˆ|| ||PQ i j i j i j

Also, mid point of PQ lies on the plane 3 6 32 2

Now, distance of point P from X-axis 2 2 5 2 2 225 16 as 3 as 4 Hence PR = 2 8 13. Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-

axis and z-axis, respectively, where O(0, 0, 0) is the origin. Let 1 1 1, ,2 2 2

S

be the centre of the

cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If , ,p SP q SQ r SR

and t ST

, then the value of p q r t

is _____

Key. 0.5

Sol.

1 1 1 1 ˆˆ ˆ, ,2 2 2 2

p SP i j k

1 1 1 1 ˆˆ ˆ, ,2 2 2 2

q SQ i j k

1 1 1 1 ˆˆ ˆ, ,

2 2 2 2r SR i j k

1 1 1 1 ˆˆ ˆ, ,2 2 2 2

t ST i j k

ˆ ˆˆ ˆ ˆ ˆ

1 11 1 1 1 1 14 4

1 1 1 1 1 1

i j k i j kp q r t

ˆ1 1ˆ ˆ ˆ ˆ2 2 2 2

16 2 2ki j i j

14. Let 2 2 2 210 10 10 101 2 3 102 3 ...... 10X C C C C , where 10 , 1, 2,.....10rC r denote binomial

coefficients. Then, the value of 11430

X is_____

Key. 646

Sol: 2

0. ; 10

nn

rr

X r C n

11

0. .

nn n

r rr

X n C C

11

0. .

nn n

n r rr

X n C C

2 11. ; 10n

nX n C n

19910.X C

199

1 .1430 143

X C = 646




Page

-33






15. Let 1 : 1 01

xE X x andx

and 1

2 1 : sin log1e

xE X is a real numberx

.

1( si, .2

n ,2

Here the inverse trigonometric function assumes valu inx es

Let 1:f E be the function defined by log1e

xf xx

and 2:g E be the function

defined by 1sing x log .1

xx

LIST-I LIST-II

P. The range of f is 1. 1, ,1 1

ee e

Q. The range of g contains 2. (0, 1)

R. The domain of f contains 3. 1 1,2 2

S. The domain of g is 4. ,0 0,

5. ,1

ee

6. 1,0 ,2 1

ee

The correct options is:

A) P-4;Q-2;R-1;S-1 B) P-3;Q-3;R-6;S-5 C) P-4;Q-2;R-1;S-6 D) P-4;Q-3;R-6;S-5

Key. A

Sol. 1 : 01

xEx

1 : ,0 1,E x

2 : 1 11

xE nx

11

x ee x




Page

-34

Now 1 01

xx e

1 1

01

e xe x

1, 1,

1x

e

also 01

x ex

10

1e x e

x

,1 ,

1ex

e

So 21: , ,

1 1eE

e e

as Range of1

xx

is 1R

Range of f is 0R or ,0 0,

Range of g is , \ 0 ,0 0,2 2 2 2

or

Now 4, 2, 1, 1P Q R S Hence a is correct 16. In a high school, a committee has to be formed form a group of 6 boys

1 2 3 4 5 6, , , , ,M M M M M M and 5girls 1 2 3 4 5, , , ,G G G G G . i) Let 1 be the total number of ways in which the committee can be formed such that the

committee has 5 members, having exactly 3 boys and 2 girls ii) Let 2 be the total number of ways in which the committee can be formed such that the

committee has at least 2 members, and having an equal number of boys and girls iii) Let 3 be the number of ways in which the committee can be formed such that the committee

has 5 numbers, at least 2 of them being girls iv) Let 4 be the total number of ways in which the committee can be formed such that the

committee has 4 members, having at lest 2 girls and such that both 1M and 1G are NOT in the committee together.

LIST-I LIST-II P. The value of 1 is 1.136 Q. The value of 2 is 2. 189 R. The value of 3 is 3. 192

S. The value of 4 is 4. 200 5. 381 6. 461




Page

-35

The correct options is: A) P-4;Q-6;R-2;S-1 B) P-1;Q-4;R-2;S-3 C) P-4;Q-6;R-5;S-2 D) P-4;Q-2;R-3;S-1 Key. C

Sol. (1) 1

6 5200

3 2

so 4P

(2) 2

6 5 6 5 6 5 6 5 6 51 1 2 2 3 3 4 4 5 5

111

5

46! So 6Q

(3) 3

5 6 5 6 5 6 5 62 3 3 2 4 1 5 0

11 5 6 5 65 0 5 1 4

= 381 So 5R

(4) 2

5 6 4 5 5 6 4 1 5189

2 2 1 1 3 1 2 1 4

So 2S

17. Let 2 2

2 2: 1x yHa b

, where 0a b , be a hyperbola in the xy-plane whose conjugate axis LM

subtends an angle of 60 at one of its vertices N, Let the area of the triangle LMN be 4 3 LIST-I LIST-II P. The length of the conjugate axis of H is 1. 8

Q. The eccentricity of H is 2. 43

R. The distance between the foci of H is 3. 23

S. The length of the latus rectum of H is 4. 4 The correct options is: A) P-4;Q-2;R-1;S-3 B) P-4;Q-3;R-1;S-2 C) P-4;Q-1;R-3;S-2 D) P-3;Q-4;R-2;S-1 Key. B

Sol.

tan 30 ba

3a b




Page

-36

Now are of 1 .. . 32

LMN b b 24 3 3b 2& 2 3b a

P. Length of conjugate axis = 2b = 4 So 4P

Q. Eccentricity 23

e So 3Q

R. distance between foci = 2ae 22 2 3 83

So 1R

S. Length of latus rectum 22 2 22 42 3 3

ba

So 2S

18. Let 21 2 3: , : , , 1, 2

2 2f f f e

and 4 :f be functions defied by

(i) 2

1 sin 1 xf x e

(ii) 12

sin0

tan1 0

xif xf x xif x

, where the inverse trigonometric function 1tan x assumes

values in ,2 2

(iii) 3 sin log 2ef x x , where for ,t t denotes the greatest integer less than or equal to t,

(iv)

2

4

1sin 0

0 0

x if xx

f x if x

LIST-I LIST-II P. the function 1f is 1. NOT continuous at x = 0 Q. The function 2f is 2. Continuous at x = 0 and NOT differentiable at x = 0 R. The function 3f is 3. Differentiable at x = 0 and is derivative is NOT continuous at x = 0

S. The function 4f is 4. Differentiable at x = 0 and its derivative is continuous at x = 0 The correct options is: A) P-2;Q-3;R-1;S-4 B) P-4;Q-1;R-2;S-3 C) P-4;Q-2;R-1;S-3 D) P-2;Q-1;R-4;S-3

Key. D

Sol. 2

sin 1 xi f x e




Page

-37

2 2

2

'1

1cos 1 . 0 . 22 1

x x

xf x e e x

e

at 0x '1f x does not exist

So. 2P

12

sin, 0

tan0 0

xxii f x xx

10

sinlim 1tanx

x xx x

2f x does not continuous at x = 0 So 1Q 3 sin 2 0iii f x n x /21 2x e

0 22

n x

0 sin 2 1n x 3 0f x So 4R

2

4

1sin , 0

0, 0

x xiv f x x

x

So 3S

2018-jee-advanced - sri chaitanya · 2018-jee-advanced 2 question paper-_key & solutions sri...

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