sri chaitanya · 2019. 6. 4. · 2019-jee-advanced 2 question paper-_key & solutions sri...
TRANSCRIPT
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-1
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-2
PHYSICS Max Marks: 62 SECTION-1 (Maximum Marks : 32)
This section contains EIGHT (08) questions Each question has FOUR options. ONE MORE THAN ONE of these four option(s) is(are) correct answer(s) For each question, choose the option(s) corresponding to (all) the correct answer(s). 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 and both of which are correct ; Partial Marks : + 1 If two or more options are correct but ONLY one option is chosen and it is a correct option; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks : -1 In all other cases. For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answers, then Choosing ONLY (A), (B), and (D) will get +4 marks ; Choosing ONLY (A) and (B) will get +2 marks ; Choosing ONLY (A) and (D) will get +2 marks ; Choosing ONLY (B) and (D) will get +2 marks ; Choosing ONLY (A) will get +1 marks; Choosing ONLY (B) will get +1 marks; Choosing ONLY (D) will get +1 marks; Choosing no option (i.e. the question is unanswered) will get 0 marks ; and Choosing any other combination of options will get -1 mark. Question ID: 337911170
1. A free hydrogen atom after absorbing a photon of wavelength a gets excited from the
state n = 1 to the state n = 4. Immediately after that the electron jumps to n=m state by
emitting a photon of wavelength e . Let the change in momentum of atom due to the
absorption and the emission are ap and ep , respectively. If a e1/5
, which of the
option (s) is /are correct ?
[Use hc = 1242 eV nm; 1 nm = 10-9 m, h and c are Planck’s constant and speed of light,
respectively]
1) a e1p / p2
2) The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is 14
3) m = 2
4) e = 418 nm
Key: 2,3
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-3
Sol: 1 1116a
R
2 2
2 2
2 2
2
1 1 14
1 114
15 516
1 1 3164
1 1 24
e
a
e
Rm
m
m
mm
Kinetic energy 21n
21
1 1142
1 1 124213.64 163 124213.6
16
m
e
e
KK
Question ID: 337911165
2. A small particle of mass m moving inside a heavy, hollow and straight tube along the
tube axis undergoes elastic collision at two ends. The tube has no friction and it is
closed at one end by a flat surface while the other end is fitted with a heavy movable
flat piston as shown in figure. When the distance of the piston from closed end is L =
L0 the particle speed is v = v0. The piston is moved inward at a very low speed V such
that V << dLL
v0, where dL is the infinitesimal displacement of the piston. Which of the
following statement(s) is/are correct ?
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-4
1) If the piston moves inward by dL, the particle speed increases by 2v dLL
2) After each collision with the piston, the particle speed increases by 2V
3) The rate at which the particle strikes the piston is v/L
4) The particle’s kinetic energy increases by a factor of 4 when the piston is moved
inward from Lo to 12
L0
Key: 2,4
Sol:
Before the collision speed is v.
After the collision speed is v + 2V.
Change in speed is 2V.
If V1 is the instantaneous speed and ‘x’ remaining distance then V1x = constant.
V1 = 2v.
Question ID: 337911167
3. A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid
diatomic gas is initially at pressure P0, volume V0, and temperature T0. If the gas
mixture is adiabatically compressed to a volume V0/4, then the correct statement(s)
is/are, (Given 21.2 = 2.3 ; 23.2 9.2 ; R is gas constant)
1) The work W done during the process is 13RT0
2) The final pressure of the gas mixture after compression is in between 9P0 and 10P0
3) Adiabatic constant of the gas mixture is 1.6
4) The average kinetic energy of the gas mixture after compression is in between 18RT0
and 19RT0
Key: 1,2,3
Sol: 15 o
o
RTPV
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-5
2o
o
RTPV
1 2
1 11
mixture 1 11
6
5 15 1
oo
o
P P
V V
RTP P PV
C CC C
5 75 12 2
3 55 12 2
32 1.620
R R
R R
1.61.6
1.6 3.2
constant
P / 4
4 2
9.2
o o o
o o
o
PV
V P V
P P P
P P
2 2 1 19.2
41 1 1.6
2.3 1 1.3 60.6 0.6
13
oo o o
o o o
o
VP PVPV PVW
PV RT
W RT
Question ID: 337911163
4. A thin and uniform rod of mass M and length L is held vertical on a floor with large
friction. The rod is released from rest so that it falls by rotating about its contact-point
with the floor without slipping. Which of the following statement(s) is/are correct,
when the rod makes an angle 600 with vertical ?
[g is the acceleration due to gravity]
1) The normal reaction force from the floor on the rod will be Mg16
2) The angular acceleration of the rod will be 2gL
3) The angular speed of the rod will be 3g2L
4) The radial acceleration of the rod’s center of mass will be 3g4
Key: 1,3,4
Sol: 2
sin602 3l mlmg
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-6
34 3
g l 3 3
4g
l
2
2
2
2
11 cos602 2 3
4 632
l mlmg
g l
gl
radial acceleration 2
2la
34ga
3 1 3 3 34 2 4 2
3 98 16
1516 16
g gMg N M ll
Mg N Mg
Mg MgN Mg
Question ID: 337911166
5. An electric dipole with dipole moment 0p i j2
is held fixed at the origin O in the
presence of an uniform electric field of magnitude E0. If the potential is constant on a
circle of radius R centered at the origin as shown in figure, then the correct
statements(s) is/are :
( 0 is permittivity of free space. R >>dipole size)
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-7
1) Total electric field at point B is BE 0
2) Total electric field at point A is A 0E 2E i j
3) The magnitude of total electric field on any two points of the circle will be same
4) 1/3
0
0 0
pR4 E
Key: 1,4
Sol:
B A
At ‘B’ electric field due to dipole is
3
3
13
42 3
oB
oo
o
o o
A o o o
KPER
KP ER
PRE
E E E E
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-8
Question ID: 337911168
6. Three glass cylinders of equal height H = 30 cm and same refractive index n = 1.5 are
placed on a horizontal surface as shown in figure. Cylinder I has a flat top, cylinder II
has a convex top and cylinder III has a concave top. The radii of curvature of the two
curved tops are same (R = 3 m). If H1, H2, and H3 are the apparent depths of a point X
on the bottom of the three cylinders, respectively, the correct statement(s) is/are :
1) 0.8 cm < (H2 – H1) < 0.9 cm
2) H3 > H1
3) H2 > H1
4) H2 > H3
Key: 3,4
Sol: (I)
12 2 30 20
3 3HH cm
(II)
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-9
1 1.5 1 1.5
1 1 1.52 3 0.3
V H R
V
1 1 56
1 2966 0.193529
V
V
V
(III)
1 1 56631
V
V
Question ID: 337911164
7. A block of mass 2M is attached to a massless spring with spring-constant k. This block
is connected to two other blocks of masses M and 2M using two massless pulleys and
strings. The acceleration of the blocks are a1, a2 and a3 as shown in the figure. The
system is released from rest with the spring in its unstretched state. The maximum
extension of the spring is x0. Which of the following option(s) is/are correct ?
[g is the acceleration due to gravity. Neglect friction]
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-10
1) At an extension of 0x4
of the spring, the magnitude of acceleration of the block
connected to the spring is 3g10
2) When spring achieves an extension of 0x2
for the first time, the speed of the block
connected to the spring is 3g M5k
3) 04Mgx
k
4) a2 – a1 = a1 – a3 Key: 4
Sol:
2 1 1 3a a a a
Effective mass of pulley-block system is 4 2 8
3 3M M M
M .
163oMgxK
22
oo
xx A A
Acceleration of block is 2 4 3 22 3 14 7A Mg k g
k M
Speed = 3 8 32 14 3 14
ox k Mg KAM K M
Question ID: 337911169 8. In a Young’s double slit experiment, the slit separation d is 0.3 mm and the screen
distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the
slits at angle as shown in figure. On the screen, the point O is equidistant from the
slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct ?
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-11
1) Fringe spacing depends on
2) For 0.36
degree, there will be destructive interference at point O.
3) For 0 , there will be constructive interference at point P.
4) For 0.36
degree, there will be destructive interference at point P.
Key: 4
Sol: sind x
2
3 70.36 36 100.3 10 6 105 180 180
d
Fringe width 7
46 10 1 2
3 10D mmd
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-12
SECTION-2 (Maximum Marks : 18) This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual numeric
keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the correct numerical value is entered. Zero Marks : +0 In all other cases Question ID: 337911172
1. A 10 cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair
of horizontal rails of zero resistance. One side of the rails is connected to an inductor
L = 1 mH and a resistance R = 1 as shown in figure. The horizontal rails, L and R lie
in the same plane with a uniform magnetic field B = 1 T perpendicular to the plane. If
the key S is closed at certain instant, the current in the circuit after 1 millisecond is
x 10-3A, where the value of x is _____
[Assume the velocity of wire PQ remains constant (1 cm’s) after key S is closed.
Given: e-1 = 0.37, where e is base of the natural logarithm]
Key: 0.63 (0.62 – 0.64)
Sol: EMF = BVl = 2 1 31 1 10 10 10 VV
Time constant = 31 10 1 .sec
1L mR
/5
31
3
1
10 11
10 0.63 0.63
toi i e
i e
mA
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-13
Question ID: 337911171
2. A ball is thrown from ground at angle with horizontal and with an initial speed u0.
For the resulting projectile motion, the magnitude of average velocity of the ball up to
the point when it hits the ground for the first time is V1. After hitting the ground, the
ball rebounds at the same angle but with a reduced speed of 0u / . Its motion
continues for a long time as shown in figure. If the magnitude of average velocity of
the ball for entire duration of motion is 0.8 V1, the value of is _____
Key: 4
Sol: 20
1sin 2uRg
20 1
2 2 21 sin 2u RR
g
01
12
11 2 1
2 sin
........
1..... 1 11
utg
tt
tT t t t
Average speed = 1
1
Rt
Average speed = RT
4 .
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-14
Question ID: 337911176 3. An optical bench has 1.5 m long scale having four equal divisions in each cm. While
measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is _____
Key: 1.39 (1.35 - 1.45)
Sol: 1 1 1v u f
2 2 2
1 1 1 2060 30
f cmf
df dv duf v u
Question ID: 337911174 4. A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass
system of angular frequency such that 24 24 M 10 mh
with h as Planck’s
constant. N photons of wavelength 68 10 m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 m . If the value of N is x 1012, then the value of x is ______
[Consider the spring as massless]
Key: 1
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-15
Sol: Km
KM
Momentum change for single photon = 2h
Momentum change = 2Nh
12
12
2
2
452 10 2
10
NhA MV
NhMA
M Nh
N
Question ID: 337911173 5. A monochromatic light is incident from air on a refracting surface of a prism of angle
750 and refractive index n0 = 3 . The other refracting surface of the prism is coated by a thin film of material of refractive index n as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of 060 . The value of n2 is
Key: 1.5 Sol: 11 sin60 3 sin r
1 1
2
2
1sin 302
451 3 sin90 1.52
o
o
o
r r
r
n n
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-16
Question ID: 337911175
6. Suppose a 22688 Ra nucleus at rest and in ground state undergoes -decay to a 222
86 Rn
nucleus in its excited state. The kinetic energy of the emitted particle is found to be
4.44 MeV. 22286 Rn nucleus then goes to its ground state by -decay. The energy of the
emitted photon is ____ keV.
[Given : atomic mass of 22688 Ra = 226.005 u, atomic mass of 222
86 Rn = 222.000 u,
atomic mass of particle = 4.000 u, 1 u = 931 MeV/c2, c is speed of the light]
Key: 135 (130 – 140)
Sol: 226 222 488 86 2Ra Rn He
226.005 222 4 0.005Rnm mRa m m
u
Q-Value = 35 10 931 4.655MeV MeV ,K.E. of 22286
4.44 4 0.08222
Rn
Energy of 4.655 4.52 135MeV KeV
SECTION-3 (Maximum Marks : 12) This section contains TWO (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II. List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Questions based on List-I and List-II and ONLY ONE of these four
options satisfies the condition asked in the Multiple Choice Question. Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the option corresponding to the correct combination 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 Question ID: 337911177
1. Answer the following by appropriately matching the lists based on the information given in the paragraph.
A musical instrument is made using four different metal strings 1, 2, 3 and 4 with mass per unit length ,2 ,3 and 4 respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string-1 ( ) at free length L0 and tension T0 the fundamental mode frequency is f0.
List-I gives the above four strings while list-II lists the magnitude of some quantity.
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-17
List-I List-II
I) String-1 ( ) P) 1
II) String-2 (2 ) Q) 1/2
III) String-3 (3 ) R) 1 / 2
IV) String-4 (4 ) S) 1 / 3
T) 3/16
U) 1/16
If the tension in each string is T0, the correct match for the highest fundamental
frequency in f0 units will be,
1) IP, IIQ, IIIT, IVS 2) IQ, IIP, IIIR, IVT
3) IP, IIR, IIIS, IVQ 4) IQ, IIS, IIIR, IVP
Key: 3
Sol: 12
oo
o
TfL
1
2
3
4
2
3
2
o
o
o
o
f fff
ff
ff
Question ID: 337911178
2. Answer the following by appropriately matching the lists based on the information
given in the paragraph.
A musical instrument is made using four different metal strings 1, 2, 3 and 4 with mass
per unit length ,2 ,3 and 4 respectively. The instrument is played by vibrating the
strings by varying the free length in between the range L0 and 2L0. It is found that in
string-1 ( ) at free length L0 and tension T0 the fundamental mode frequency is f0.
List-I gives the above four strings while list-II lists the magnitude of some quantity.
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-18
List-I List-II
I) String-1 ( ) P) 1
II) String-2 (2 ) Q) 1/2
III) String-3 (3 ) R) 1 / 2
IV) String-4 (4 ) S) 1 / 3
T) 3/16
U) 1/16
The length of the strings 1, 2, 3 and 4 are kept fixed at L0, 0 03L 5L,2 4
, and 07L4
,
respectively. Strings 1, 2, 3, and 4 are vibrated at their 1st, 3rd, 5th, and 14th harmonics,
respectively such that all the strings have same frequency. The correct match for the
tension in the four strings in the units of T0 will be
1) IP, IIQ, IIIT, IVU
2) IT, IIQ, IIIR, IVU
3) IP, IIQ, IIIR, IVT
4) IP, IIR, IIIT, IVU
Key: 1
Sol: 12
oo
o
TfL
1
22
33
12
232 2 3
453 2 5
o
o
o
o
TfL
TfL
TfL
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-19
Question ID: 337911179
3. Answer the following by appropriately matching the lists based on the information
given in the paragraph.
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed
by the gas is given by T X , where T is temperature of the system and X is the
infinitesimal change in a thermodynamic quantity X of the system. For a mole of
monatomic ideal gas A A
3 T VX R ln R ln2 T V
. Here R is gas constant, V is
volume of gas TA and VA are constants.
The List-I below gives some quantities involved in a process and List-II gives some
possible values of these quantities.
List-I List-II
I) Work done by the system in process1 2 3 P) 01 RT ln 23
II) Change in internal energy in process 1 2 3 Q) 01 RT3
III) Heat absorbed by the system in process 1 2 3 R) 0RT
IV) Heat absorbed by the system in process 1 2 S) 04 RT3
T) 01 RT 3 ln 23
U) 05 RT6
If the process carried out on one mole of monatomic ideal gas is as shown in figure in
the PV-diagram with P0V0 = 01 RT3
, the correct match is,
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-20
1) IQ, IIR, IIIP, IVU 2) IQ, IIR, IIIS, IVU
3) IQ, IIS, IIIR, IVU 4) IS, IIR, IIIQ, IVT
Key: 2
Sol: Work done in 1 2 = PoVo 2 3 = 0 Total work in 1 2 3 = P0V0
Q in 1 2 3 = 43 oRT
U in 1 2 3 = RTo Question ID: 337911180
4. Answer the following by appropriately matching the lists based on the information
given in the paragraph.
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed
by the gas is given by T X , where T is temperature of the system and X is the
infinitesimal change in a thermodynamic quantity X of the system. For a mole of
monatomic ideal gas A A
3 T VX R ln R ln2 T V
. Here R is gas constant, V is
volume of gas TA and VA are constants.
The List-I below gives some quantities involved in a process and List-II gives some
possible values of these quantities.
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-21
List-I List-II
I) Work done by the system in process1 2 3 P) 01 RT ln 23
II) Change in internal energy in process 1 2 3 Q) 01 RT3
III) Heat absorbed by the system in process 1 2 3 R) 0RT
IV) Heat absorbed by the system in process 1 2 S) 04 RT3
T) 01 RT 3 ln 23
U) 05 RT6
If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with
P0V0 = 01 RT3
, the correct match is,
1) IP, IIR, IIIT, IVS 2) IP, IIT, IIIQ, IVT 3) IS, IIT, IIIQ, IVU 4) IP, IIR, IIIT, IVP
Key: 4
Sol: Work done in 12 = ln 23
oRT
2 3 0 U in 1 2 = 0
O in 1 2 = ln 23
oRTW
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-22
CHEMISTRY Max Marks: 62 SECTION-1 (Maximum Marks : 32)
This section contains EIGHT (08) questions Each question has FOUR options. ONE MORE THAN ONE of these four option(s) is(are) correct answer(s) For each question, choose the option(s) corresponding to (all) the correct answer(s). 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 and both of which are correct ; Partial Marks : + 1 If two or more options are correct but ONLY one option is chosen and it is a correct option; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks : -1 In all other cases. For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answers, then Choosing ONLY (A), (B), and (D) will get +4 marks ; Choosing ONLY (A) and (B) will get +2 marks ; Choosing ONLY (A) and (D) will get +2 marks ; Choosing ONLY (B) and (D) will get +2 marks ; Choosing ONLY (A) will get +1 marks; Choosing ONLY (B) will get +1 marks; Choosing ONLY (D) will get +1 marks; Choosing no option (i.e. the question is unanswered) will get 0 marks ; and Choosing any other combination of options will get -1 mark.
Question ID: 337911182
1. With reference to aqua regia, choose the correct options(s)
1) Reaction of gold with aqua regia produces NO2 in the absence of air
2) Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state
3) Aqua regia is prepared by mixing conc. HCl and conc. HNO3 in 3 : 1 (v/v) ratio
4) The yellow colour of aqua regia is due to the presence of NOCl and Cl2
Key: 2,3,4
Sol: Aquaregia is a mixture of conc.HCl and conc.HNO3 in 3:1 ratio.
When gold dissolves in aquaregia, the species formed is 4AuCl in which gold is in +3
oxidation state.
In the absence of air the reaction between gold and aquaregia
3 4 24 2Au HNO HCl H AuCl H O NO
NO2 will not be produced. Yellow colour is due to the presence of NOCl and Cl2.
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-23
Question ID: 337911181 2. The cyanide process of gold extraction involves leaching out gold from its ore with
CN- in the presence of Q in water to form R. Subsequently, R is treated with T to obtain Au and Z. Choose the correct option(s)
1) Q is O2 2) T is Zn 3) Z is [Zn(CN)4]2- 4) R is [Au(CN)4]- Key: 1,2,3 Sol: Gold extraction:
2 2 2
22 4
4 8 2 4 4
4 2 2 4
R
TR Z
QAu NaCN H O O Na Au CN NaOH
Na Au CN Zn Na Zn CN Au
Question ID: 337911185 3. Which of the following reactions produce(s) propane as a major product ?
1)
2)
3)
4) Key: 1,3 Sol:
CH3
O
ONa NaOHCaO,Δ
CH3 CH3
CH3
O
ONa H2Oelectrolysis CH3
CH3
Cl
CH3Zn
dil.HCl CH3CH3
Br
CH3
Br
ZnCH2CH3
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-24
Question ID: 337911188
4. Choose the correct option(s) from the following.
1) Nylon-6 has amide linkages
2) Cellulose has only -D-glucose units that are joined by glycosidic linkages
3) Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst
at high pressure
4) Natural rubber is polyisoprene containing trans alkene units
Key: 1,3
Sol: Nylon-6 has amide linkages.
Cellulose has only β-D-glucose units that are joined by glycosidic linkages.
Teflon is prepared by heating tetrafluoromethane in presence of a persulphate catalyst
at high pressure.
Natural rubber is polyisoprene containing cis alkene units
Question ID: 337911186
5. Choose the correct option(s) that give(s) an aromatic compound as the major product.
1)
2)
3)
4)
Key: 2,3
Sol:
BrNaOEt
Anti aromatic
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-25
CH3
Br
Br
1.alcKOH2.NaNH2
3.redhot iron tube,873K
CH3
CH3 CH3Aromatic
NaOMe- Na+
Aromatic
+ Cl2 (excess) UV,500K
Cl
Cl
Cl
Cl
Cl
Cl
Non aromatic Question ID: 337911184
6. The ground state energy of hydrogen atom is – 13.6 eV. Consider an electronic state
of He+ whose energy, azimuthal quantum number and magnetic quantum number are –
3.4 eV, 2 and 0, respectively. Which of the following statement(s) is (are) true from the
state ?
1) It is a 4d state
2) It has 3 radial nodes
3) It has 2 angular nodes
4) The nuclear charge experienced by the electron in this state is less than 2e, where e
is the magnitude of the electronic charge
Key: 1,3
Sol: 2
2213.6 3.4En
2
2
2 2
213.6 3.4
4 4n
n n
Wave function corresponds to 4,2,0 represents 24z
d – orbital which has only one radial
nodes and two angular nodes. It experiences nuclear change of 2e units.
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-26
Question ID: 337911187
7. Choose the correct option(s) for the following reaction sequence
Consider Q, R and S as major products.
1)
2)
3)
4) Key: 1,4 Sol:
MeO C C CH2
CHO i)Hg2+,dilH2SO4 MeO C CH2 CH2
CHOOii)AgNO3 NH4OH
MeO C CH2 CH2
COOHO
MeO CH2 CH2 CH2
COOH
MeO CH2 CH2 CH2
COCl MeO
OMeO
iii)Zn-Hg.conc.HCl
i)SOCl2 pyridine
ii)AlCl3 Zn-Hgconc.HCl
'Q'
'R' 'S'
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-27
Question ID: 337911183
8. Consider the following reactions (unbalanced)
Zn + hot conc. H2SO4 G + R + X
Zn + conc. NaOH T + Q
G + H2S + NH4OH Z (a precipitate) + X + Y
Choose the correct option(s)
1) The oxidation state of Zn in T is + 1
2) Bond order of Q is 1 in its ground state
3) Z is dirty white in colour
4) R is a V-shaped molecule
Key: 2,3,4
Sol: 2 4Hot & Conc.
Zn H SO G R X
2 4 4 2 2
2 2 2
4 2 4 4 4 22
2
Conc.G R X
T Q
ZG XY
Zn H SO ZnSO SO H O
Zn NaOH Na ZnO H
ZnSO H S NH OH ZnS NH SO H O
1) Oxidation state of Zn in Na2ZnO2 is +2 2) Bond order of Q is one for H2(Q) 3) ZnS is white in colour 4) SO2 is angular in shape
SECTION-2(Maximum Marks : 18) This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual numeric
keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the correct numerical value is entered. Zero Marks : +0 In all other cases Question ID: 337911190
1. Total number of cis N – Mn – Cl bond angles (that is Mn – N and Mn – Cl bonds in cis
positions) present in a molecule of cis –[Mn(en)2Cl2] complex is ___
(en = NH2CH2CH2NH2)
Key: 6
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-28
Sol:
Mn
2H N Cl
Cl2NH
2CH 2CH
2NH
2 2CH CH
2NH
3 1
24
Similarly from bottom N-there will be another (2).
Total N – Mn – Cl angle are 6.
Question ID: 337911189
2. The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by
conc.HNO3 to a compound with the highest oxidation state of sulphur is ____ (Given
data : Molar mass of water = 18g mol-1) s
Key: 288 (288.00 – 288.30)
Sol: S8 + HNO3 H2SO4 + NO2 + H2O
Rhombic sulphur
Balancing the equation
S8 + 48HNO3 8H2SO4 + 48NO2 + 16H2O
Rhombic sulphur
1 mole S8 produces 16 x 18 gm of H2O i.e. 288 gms of H2O.
Question ID: 337911191
3. The decomposition reaction 2 5 2 4 22 2N O g N O g O g is started in a closed
cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After 310Y s the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of
the reaction is 4 15 10 s , assuming ideal gas behavior, the value of Y is __
Key: 2.30
Sol: 2 2 4 252 2gN O N O g O g
Given rate constant of the reaction = 4 15 10 sec
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-29
2 5
2 5
2 5
2 5
overall
overall
12
2
N ON O
N ON O
dPK P
dtdp
K Pdt
2 5 2 4 21 /2
2 2xx x
N O g N O g O g
3
4
1 1.4520.90 atm
2.303 110 log0.12 5 10
23.03 2.3010
x
x
y
y
Question ID: 337911193
4. Total number of hydroxyl groups present in a molecule of major product P is ___
Key: 6
Sol:
i)H2,Pd-BaSO4,quinoline
ii)dil.KMnO4(excess),273K
OH
OH
OH
OH
OHOH
Question ID: 337911194
5. Total number of isomers, considering both structural and stereoisomers of cyclic ethers
with the molecular formula 4 8C H O is____
Key: 10
Sol:
OO
CH3
O
CH3
O
CH3CH3
O
CH3 CH3
O
CH3
* * * *
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-30
Question ID: 337911192
6. The mole fraction of urea in an aqueous urea solution containing 900g of water is 0.05.
If the density of the solution is 1.2g cm-3, the molarity of urea solution is ___
(Given data: Molar masses of urea and water are 60g mol-1 and 18g mol-1, respectively)
Key: 2.98 (2.80 – 3.05)
Sol: Xurea = 0.05 Mass of water = 900 gm
Number of moles of water = 2
900 5018H On Let moles of urea = Un
0.05
505019
U
U
U
nn
n
Mass of urea = 50 6019
Given density (d) of solution = 1.2 g/ml.
50 60900massof solution 19density / 1.2solutionV
g cc
50191000 2.98
50 60900 1000191.2
U
ml
nMV
SECTION-3(Maximum Marks : 12) This section contains TWO (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II. List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Questions based on List-I and List-II and ONLY ONE of these four
options satisfies the condition asked in the Multiple Choice Question. Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the option corresponding to the correct combination 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
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-31
Question ID: 337911195
1. Answer the following by appropriately matching the lists based on the information
given in the paragraph.
Consider the Bohr’s model of a one – electron atom where the electron moves around
the nucleus. In the following List – I contains some quantities for the nth orbit of the
atom and List – II contains options showing how they depend on n
List – I List – II
(I) Radius of the nth orbit (P) 2n
(II) Angular momentum of the electron in the nth orbit (Q) 1n
(III) Kinetic energy of the electron in the nth orbit (R) 0n
(IV)Potential energy of the electron in the nth orbit (S) 1n
(T) 2n
(U) 1/2n
Which of the following options has the correct combination considering List – I and List – II?
1. (II), (R) 2. (II), (Q) 3. (I), (P) 4. (I), (T) Key: 4
Sol: 2nr
Z
2
2
2
2
L n
ZKEn
ZPEn
Question ID: 337911196
2. Answer the following by appropriately matching the lists based on the information
given in the paragraph.
Consider the Bohr’s model of a one – electron atom where the electron moves around
the nucleus. In the following List – I contains some quantities for the nth orbit of the
atom and List – II contains options showing how they depend on n
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-32
List – I List – II
(I) Radius of the nth orbit (P) 2n
(II) Angular momentum of the electron in the nth orbit (Q) 1n
(III) Kinetic energy of the electron in the nth orbit (R) 0n
(IV) Potential energy of the electron in the nth orbit (S) 1n
(T) 2n
(U) 1/2n
Which of the following options has the correct combination considering List – I and
List – II?
1. (III), (S) 2. (IV), (Q) 3. (III), (P) 4. (IV), (U)
Key: 3
Sol: 2nr
Z
2
2
2
2
L n
ZKEn
ZPEn
Question ID: 337911197
3. Answer the following by appropriately matching the lists based on the information
given in the paragraph List – I includes starting materials and reagents of selected
chemical reactions. List – II gives structures of compounds that may be formed as
intermediate products and/ or final products from the reactions of List – I
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-33
List – I List – II
(I) (P)
(II) (Q)
(III) (R)
(IV) (S)
(T)
(U)
Which of the following options has correct combination considering List – I and List –
II?
1. (I), (S), (Q), (R) 2. (II), (P), (S), (U)
3. (I), (Q), (T), (U) 4. (II), (P), (S), (T)
Key: 2
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-34
Sol:
CN
O
O
i)DIBAL-Hii)dilHCl
CHO
CHO
NaBH4
iii)conc.H2SO4
OH
OH
O
i)O3
ii)Zn,H2O
iv)conc.H2SO4
iii)NaBH4CH2
COOH
O
COOH
OH
COOH
O
O
Cl
COOCH3
i)KCN ii)H3O+,Δ iii)LiAlH4CN
COOCH3
COOH
COOH OH
OH
O
iv)conc.H2SO4
COOMe
COOMe
i)LiAlH4 iI)conc.H2SO4OH
OH
O=
O
Question ID: 337911198
4. Answer the following by appropriately matching the lists based on the information
given in the paragraph List – I includes starting materials and reagents of selected
chemical reactions. List – II gives structures of compounds that may be formed as
intermediate products and/ or final products from the reactions of List – I
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-35
List – I List – II
(I) (P)
(II) (Q)
(III) (R)
(IV) (S)
(T)
(U)
Which of the following options has correct combination considering List – I and List –
II?
1. (IV), (Q), (R) 2. (IV), (Q), (U) 3. (III), (S), (R) 4. (III), (T), (U)
Key: 1
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-36
Sol:
CN
O
O
i)DIBAL-Hii)dilHCl
CHO
CHO
NaBH4
iii)conc.H2SO4
OH
OH
O
i)O3
ii)Zn,H2O
iv)conc.H2SO4
iii)NaBH4CH2
COOH
O
COOH
OH
COOH
O
O Cl
COOCH3
i)KCN ii)H3O+,Δ iii)LiAlH4CN
COOCH3
COOH
COOH OH
OH
O
iv)conc.H2SO4
COOMe
COOMe
i)LiAlH4 iI)conc.H2SO4OH
OH
O=
O
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-37
MATHEMATICS Max Marks: 62 SECTION-1 (Maximum Marks : 32)
This section contains EIGHT (08) questions Each question has FOUR options. ONE MORE THAN ONE of these four option(s) is(are) correct answer(s) For each question, choose the option(s) corresponding to (all) the correct answer(s). 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 and both of which are correct ; Partial Marks : + 1 If two or more options are correct but ONLY one option is chosen and it is a correct option; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks : -1 In all other cases. For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct answers, then Choosing ONLY (A), (B), and (D) will get +4 marks ; Choosing ONLY (A) and (B) will get +2 marks ; Choosing ONLY (A) and (D) will get +2 marks ; Choosing ONLY (B) and (D) will get +2 marks ; Choosing ONLY (A) will get +1 marks; Choosing ONLY (B) will get +1 marks; Choosing ONLY (D) will get +1 marks; Choosing no option (i.e. the question is unanswered) will get 0 marks ; and Choosing any other combination of options will get -1 mark. Question ID: 337911204
1. For , 1a a , let
3 3
7/32 2 2
1 2 ......lim 541 1 1....
1 2
n
n
nan an an n
Then the possible value(s) of a is/are
1. – 9 2. – 6 3. 7 4. 8
Key: 1,4
Sol:
1/33
11
7/32 2
1 1
1
lim lim1 1 1
nn
rrn nn n
r r
rr n n
nn ran r a
n
=
11/3 4/3 1
00
1102
0
3 34 4 541 11
1
x dx x l
ldx a x a aa x
1 8 9 9 8a a or 8 9a or
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-38
Question ID: 337911202
2. Let :f be a function. We say that f has
PROPERTY 1 if
0
0limh
f h fh
exists and is finite, and
PROPERTY 2 if 20
0limh
f h fh
exists and is finite
Then which of the following options is/are correct ?
1. f x x has PROPERTY 1
2. sinf x x has PROPERTY 2
3. f x x x has PROPERTY 2
4. 2/3f x x has PROPERTY 1
Key: 1,4
Sol: If 0 0
0lim lim 0h h
hf x x h
h
If 20
sinh 0sin lim .h
f x x D Eh
If 20
0lim .h
h hf x x x D E
h
If
2/31/32/3
1/30 0
0lim lim 0h h
hf x x hh
Question ID: 337911201
3. For non – negative integers n, let
0
2
0
1 2sin sin2 2
1sin2
n
kn
k
k kn nf n
kn
Assuming cos-1x takes values in 0, which of the following options is/are correct?
1. If 1tan cos 6f ,then 2 2 1 0
2. 1sin 7cos 5 0f
3. 342
f
4. 1lim2n
f n
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-39
Key: 1,2,3
Sol: 0
0
2 3cos cos2 2
2 21 cos2
n
k
n
k
kn nf n
kn
1sin
2 31 cos cos2 2 2sin
21
sin2 21 cos
2 2sin2
nn nn
n n nn
nn nn
n nn
2 cos
2 cos2 2
nn
n n
.
26 cos 2 1 2 1 08 8
f Tan
1sin 7cos 5 sin 7. 07
f
34 cos
6 2f
lim 1n
f n
Question ID: 337911205
4. Let :f be given by 1 2 5f x x x x . Define
0
, 0x
F x f t dt x
Then when of the following options is/are correct?
1. F has a local minimum at x = 1
2. 0F x for all 0,5x
3. F has a local maximum at x = 2
4. F has two local maxima and one local minimum in 0,
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-40
Key: 1,2,3
Sol: 1 1 2 5F x f x x x x
At 1x and 5x F x has local minima
At 2x , F x has local maxima and 0 0,5F x x
1 2 5
Question ID: 337911199
5. Let
1 2 3
1 0 0 1 0 0 0 1 00 1 0 , 0 0 1 , 1 0 00 0 1 0 1 0 0 0 1
P I P P
4 5 6
0 1 0 0 0 1 0 0 10 0 1 , 1 0 0 , 0 1 01 0 0 0 1 0 1 0 0
P P P
and 6
1
2 1 31 0 23 2 1
Tk k
k
X P P
Where TkP denotes the transpose of the matrix kP . Then which of the following options
is/ are correct?
1. X is a symmetric matrix
2. The sum of diagonal entries of X is 18
3. X – 30I is an invertible matrix
4. If 1 11 1 ,1 1
X
then 30
Key: 1,2,4
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-41
Sol: Let 6
1
2 1 31 0 2 ,3 2 1
Tk k
k
A X P AP
Here T Tk k k kP P P P I for all k =1,2,3,4,5,6.
1 &T Tk k kP P P A A
6 6
1 1
TT T Tk k k k
k k
X P AP P AP X
Let 6 6 6
1 1 1
1 2 2 2 30 11 2 2 2 30 30 1 301 2 2 2 30 1
TK K K K
K K KB XB P AP B P AB P AB AB B
30 and 30X I is singular
6
1
18Tk k
k
Tr P AP
Question ID: 337911203
6. Let 2sin , 0xf x x
x
Let x1 < x2 < x3 < …….. < xn <……. be all the points of local maximum of f
and y1 < y2 < y3 < ……. < yn < ….. be all the points of local minimum of f .
Then which of the following options is/are correct ?
1. 1 1x y
2. 1 2n nx x
3. 12 ,22nx n n
for every n
4. 1n nx y for every n
Key: 2,3,4
Sol: 2
14
cos 2 sin , 0x x x xf x xx
4
2 cos tan2 0
xx x xcos x
x
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-42
y
0 1/23/2 3
1y 1x2y
Question ID: 337911200
7. Let x and let
1 1 1 20 2 2 , 0 4 00 0 3 6
x xP Q
x x
and 1R PQP
Then which of the following options is/are correct?
1. det 2
det 0 4 0 85
x xR
x x
, for all x
2. For x = 1, there exists a unit vector i j k for which 000
R
3. There exists a real number x such that PQ = QP
4. For x = 0, if 1 1
6R a ab b
, then a + b = 5
Key: 1,4
Sol: 1 2 2
24 12 4 10 8 0 4 0 8
5
x xR PQP Q x x
x x
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-43
If 1 44 0 0x R
For 1 1 1 2 0 0 6 3 0
10 0 2 2 0 4 0 0 3 26
0 0 3 0 0 6 0 0 2x R
12 6 4
10 24 8
60 0 36
24 134
6 0 23
0 0 0
R I
2 44 0, 2 03 3b ba a
3, 2 5b a a b
If PQ = QP no value of x .
Question ID: 337911206 8. Three lines
1 : ,L r i
2 : ,L r k j
3 : ,L r i j vk v
are given . For which point(s) Q on L2 can we find a point P on L1 and a point R on L3 so that P, Q and R are collinear?
1. k 2. 12
k j 3. k j 4. 12
k j
Key: 2,4 Sol: , 0,0 , 0, ,1 , 1,1,P Q R
If P,Q,R are collinear 11 1
1 11 1,01
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-44
SECTION-2 (Maximum Marks : 18) This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual numeric
keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the correct numerical value is entered. Zero Marks : +0 In all other cases
Question ID: 337911207
1. Suppose
2
0 0
0 0
det 03
n nn
kk k
n nn n k
k kk k
k C k
C k C
Holds for some positive integer n. The 0 1
nnk
k
Ck equals____
Key: 6.20
Sol: 2 1
1
11 2 2 02
2 4
n n
n n
n nn n n
n
2 1 2 3 21 2 1 2 2 0n n nn n n n
4 1 1 2 0n n n n
2 3 4 0 4 0,n n n n n N
4 54
0
2 1 31 6.201 5 5k
k
Ck
Question ID: 337911210
2. The value of
10
1
0
11 7 7sec sec sec4 12 2 12 2k
kk
In the interval 3,4 4
equals ___
Key: 0
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-45
Sol: 10
0
7 7sec sec12 2 12 2 2k
k kS
10
0
72cos6k
ec k
10
1
0
2 1 cos6
k
k
ec
4 1 4
1 15sec sec 1 04
Question ID: 337911211
3. The value of the integral
/2
50
3 cos
cos sind
equals ____
Key: 0.5
Sol:
/2 /2
5 50 0
3 cos 3 sin
cos sin sin cosI d d
/2 /2 2
4 40 0
3 3sec2
sin cos 1I d d
Tan
Put, 2 2sec 2Tan t d tdt
4 3 4
0 0
6 1 16
1 1 1t dt dtt t t
3 2
0
1 1 1 1 13 3 0 1 0 1 0.503 2 3 2 2t t
I l
Question ID: 337911209
4. Let X denote the number of elements in a set X. Let S= {1, 2, 3, 4, 5, 6} be a sample
space, where each element is equally likely to occur. If A and B are independent events
associated with S, then the number of ordered pairs (A, B) such that 1 B A ,
equals ____
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-46
Key: 422
Sol: If n(B) = 1, n(A) = 6 no. of ordered pairs (A, B) = 6 If n(B) = 2, n(A) = 3 no. of ordered pairs (A, B) = 180 If n(B) = 2, n(A) = 6 no. of ordered pairs (A, B) = 15 If n(B) = 3, n(A) = 4 no. of ordered pairs (A, B) = 180 If n(B) = 3, n(A) = 6 no. of ordered pairs (A, B) = 20 If n(B) = 4, n(A) = 6 no. of ordered pairs (A, B) = 15 If n(B) = 5, n(A) = 6 no. of ordered pairs (A, B) = 6 Total number of ordered pairs (A, B) = 422.
Question ID: 337911208 5. Five persons A, B, C, D and E are seated in a circular arrangement. If each of them is
given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ___
Key: 30 Sol: Total number of ways = 2 3 5 30 Question ID: 337911212
6. Let 2a i j k and 2b i j k
be two vectors. Consider a vector
, ,c a b
. If the projection of c
on the vector a b
is 3 2 , then the
minimum value of .c a b c
equals_____
Key: 18 Sol: 3 3a b i j
.3 2 2
c a b
a b
2
18 . 18c c c a b
SECTION-3 (Maximum Marks: 12) This section contains TWO (02) List-Match sets. Each List-Match set has TWO (02) Multiple Choice Questions. Each List-Match set has two lists : List-I and List-II. List-I has Four entries (I), (II), (III) and (IV) and List-II has Six entries (P), (Q), (R), (S), (T) and (U). FOUR options are given in each Multiple Choice Questions based on List-I and List-II and ONLY ONE of these four
options satisfies the condition asked in the Multiple Choice Question. Answer to each question will be evaluated according to the following marking scheme : Full Marks : + 3 If ONLY the option corresponding to the correct combination 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
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-47
Question ID: 337911214
1. Answer the following by appropriately matching the lists based on the information given
in the paragraph. Let sin cosf x x and cos 2 sing x x be two functions
defined for x > 0. Define the following sets whose elements are written in the increasing
order :
: 0 ,X x f x 1: 0Y x f x
: 0 ,Z x g x 1: 0W x g x
List – I contains the sets X, Y, Z and W. List – II contains some information regarding
these sets.
List – I List – II
(I) X (P) 3, ,4 ,72 2
(II) Y (Q) an arithmetic progression
(III) Z (R) NOT an arithmetic progression
(IV) W (S) 7 13, ,6 6 6
(T) 2, ,3 3
(U) 3,6 4
Which of the following is the only CORRECT combination?
1. (IV), (P), (R), (S) 2. (III), (P), (Q), (U)
3. (III), (R), (U) 4. (IV), (Q), (T)
Key: 1
Sol: For x , 0 sin cos 0 cosf x x x n
cos 0 ( )1( ) 1x or or
,2
nx n N
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-48
For Y, 1 0 cos sin 0f x cos x x
sin 0 ( ) cos cos 0x or x
, ( ) cos 2 1 ,2
x n n N or x n n N
1 1cos ,2 2
x
22 ,2 , , 03 3
x n n n I x
For Z, 0 cos 2 sin 0 2 sin 2 1 ,2
g x x n n N
2 1 1 1 3 3sin , , ,4 4 4 4 4
nx
For W, 1 0 2 sin 2 sin cos 0g x x
sin 2 sin 0 cos 0x or x
2 1 , 2 sin2
x n n N or x n
1sin 1, ,02 2nx
2 1 , ,2
x n n n N
, 1 , 16 6
n nn n n
Question ID: 337911213
2. Answer the following by appropriately matching the lists based on the information
given in the paragraph
Let sin cosf x x and cos 2 sing x x be two functions defined for x > 0.
Define the following sets whose elements are written in the increasing order :
: 0 ,X x f x 1: 0Y x f x
: 0 ,Z x g x 1: 0W x g x
List – I contains the sets X, Y, Z and W. List – II contains some information regarding
these sets.
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-49
List – I List – II
(I) X (P) 3, ,4 ,72 2
(II) Y (Q) an arithmetic progression
(III) Z (R) NOT an arithmetic progression
(IV) W (S) 7 13, ,6 6 6
(T) 2, ,3 3
(U) 3,6 4
Which of the following is the only CORRECT combination?
1. (I), (P), (R) 2. (I), (Q), (U) 3. (II), (R), (S) 4. (II), (Q), (T)
Key: 4
Sol: For x , 0 sin cos 0 cosf x x x n
cos 0 ( )1( ) 1x or or
,2
nx n N
For Y, 1 0 cos sin 0f x cos x x
sin 0 ( ) cos cos 0x or x
, ( ) cos 2 1 ,2
x n n N or x n n N
1 1cos ,2 2
x
22 ,2 , , 03 3
x n n n I x
For Z, 0 cos 2 sin 0 2 sin 2 1 ,2
g x x n n N
2 1 1 1 3 3sin , , ,4 4 4 4 4
nx
For W, 1 0 2 sin 2 sin cos 0g x x
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-50
sin 2 sin 0 cos 0x or x 2 1 , 2 sin2
x n n N or x n
1sin 1, ,02 2nx 2 1 , ,
2x n n n N
, 1 , 16 6
n nn n n
Question ID: 337911215 3. Answer the following by appropriately matching the lists based on the information
given in the paragraph. Let the circles 2 2
1 : 9C x y and 2 22 : 3 4 16C x y , intersect at the points
X and Y. Suppose that another circle 2 2 23 :C x h y k r satisfies the
following conditions :
(i) Centre of 3C is collinear with the centres of 1C and 2C
(ii) 1C and 2C both lie inside 3C , and
(iii) 3C touches 1C at M and 2C at N
Let the line through X and Y intersect C3 at Z and W, and let a common tangent of C1
and C3 be a tangent to the parabola 2 8x y .
There are some expressions given in the List – I whose values are given in List – II
below
List – I List – II
(I) 2h k (P) 6
(II) Length of ZWLength of XY
(Q) 6
(III) Area of triangle MZNArea of triangle ZMW
(R) 54
(IV) (S) 215
(T) 2 6
(U) 103
Which of the following is the only CORRECT combination?
1. (I), (U) 2. (II), (T) 3. (II), (Q) 4. (I), (S)
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-51
Key: 3
Sol:
WN
M ZO(0,0)
Y
X
(3,4)
2 21
2 2 2 22 2
2 2 23
9 0,0
3 4 16 6 8 9 0 3, 4
C x y O
C x y x y x y C
C x h y k r
Let the centre of C3 be C.
29 12: 3: 2 ,5 5
OC CC C
2 6, 2 3 5 4 12 6h k r r
Eq. of common chord of C1 and C2 is 3x+4y-9=0. Let P be the point of intersection of
OC1 and common chord.
6 12 6 24, ,5 5 5
81 242 925 5
6
54
CP ZP WP MP
XY
ZWXYMZNZMW
Equation of the common chord of C1 and C3 is 103 4 15 03
x y .
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-52
Question ID: 337911216 4. Answer the following by appropriately matching the lists based on the information
given in the paragraph
Let the circles 2 21 : 9C x y and 2 2
2 : 3 4 16C x y , intersect at the points
X and Y. Suppose that another circle 2 2 23 :C x h y k r satisfies the
following conditions :
(i) Centre of 3C is collinear with the centres of 1C and 2C
(ii) 1C and 2C both lie inside 3C , and (iii) 3C touches 1C at M and 2C at N Let the line through X and Y intersect C3 at Z and W, and let a common tangent of C1
and C3 be a tangent to the parabola 2 8x y .
There are some expressions given in the List – I whose values are given in List – II
below
List – I List – II
(I) 2h k (P) 6
(II) Length of ZWLength of XY
(Q) 6
(III) Area of triangle MZNArea of triangle ZMW
(R) 54
(IV) (S) 215
(T) 2 6
(U) 103
Which of the following is the only INCORRECT combination?
1. (III), (R) 2. (IV), (U) 3. (IV), (S) 4. (I), (P)
Key: 3
2019-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081 www. srichaitanya.net, [email protected]
Page
-53
Sol:
WN
M ZO(0,0)
Y
X
(3,4)
2 21
2 2 2 22 2
2 2 23
9 0,0
3 4 16 6 8 9 0 3, 4
C x y O
C x y x y x y C
C x h y k r
Let the centre of C3 be C.
29 12: 3: 2 ,5 5
OC CC C
2 6, 2 3 5 4 12 6h k r r
Eq. of common chord of C1 and C2 is 3x+4y-9=0. Let P be the point of intersection of
OC1 and common chord.
6 12 6 24, ,5 5 5
81 242 925 5
6
54
CP ZP WP MP
XY
ZWXYMZNZMW
Equation of the common chord of C1 and C3 is 103 4 15 03
x y .