JEEAdvanced2021
304,Kaset
t
yHei
ght
s,AyyappaSoci
et
y,Madhapur
,Hyder
abad-500081
PHYSICS
1.
Max. Marks: 60
The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of
the Vernier scale correspond to nine divisions of the main scale. The figure below on the
left shows the reading of this calipers with no gap between its two jaws. The figure on
the right shows the reading with a solid sphere held between the jaws. The correct
diameter of the sphere is
(A) 3.07 cm
(B) 3.11 cm
(C) 3.15 cm
(D) 3.17 cm
Key: C
Sol
O
5
O
30
32
e
O
6
e  5  MSD  6  VSD
Required diameter  e  R
  6VSD  5MSD   32MSD  1VSD
 27MSD  5VSD  27mm  4.5mm
 3.15 cm
R
O
1
R  1 VSD  32  MSD 
2021-Jee-Advanced
2.
Question Paper-1_Key & Solutions
An ideal gas undergoes a four step cycle as shown in the P – V diagram below. During
this cycle, heat is absorbed by the gas in
(A) steps 1 and 2 (B) steps 1 and 3 (C) steps 1 and 4 (D) steps 2 and 4
Key: C
Sol
 in the path (1) volume of an ideal gas is increasing at constant pressure.
 V  T  dQ is +ve (absorbed)
In the path (4) pressure of an ideal gas is increasing at constant volume  P  T  dQ is
+ve (absorbed)
3.
An extended object is placed at point O, 10 cm in front of a convex lens L1 and a
concave lens L 2 is placed 10 cm behind it, as shown in the figure. The radii of curvature
of all the curved surfaces in both the lenses are 20 cm. The refractive index of both the
lenses is 1.5. The total magnification of this lens system is
(A) 0.4
(B) 0.8
(C) 1.3
(D) 1.6
Key: B
For lens L1
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Sol
2021-Jee-Advanced
Question Paper-1_Key & Solutions
1 1 1
 
v u f1
1 1
1
 
v1 10
20
1
1
v

 m1  1  2
v1
20
u1
For lens L 2
1
1
1
1
 

v2
20 30
12
v
12
m2  2 
u 2 30
24 4
m  m1  m 2 
  0.8
30 5
4.
A heavy nucleus Q of the half-life 20 minutes undergoes alpha-decay with probability of
60% and beta-decay with probability of 40%. Initially, the number of Q nuclei is 1000.
The number of alpha-decays of Q in the first one hour is
(A) 50
(B) 75
(C) 350
(D) 525
Key: D
at t = 0 N 0  1000 ;  -decay and  -decay are in 3 : 1 ratio.
At t = 20 min 
  decay
300
At t = 40 min  300 + 150
At t = 60 min 
300  150  75
525
  decay
200
200 + 100
200  100  50
350
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Sol
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Question Stem for Question Nos. 5 and 6
Question Stem
A projectile is thrown from a point O on the ground at an angle 45 from the vertical and
with a speed 5 2 m / s . The projectile at the highest point of its trajectory splits into two
equal parts. One part falls vertically down the ground, 0.5 s after the splitting. The other
part, t seconds after the splitting, falls to the ground at a distance x meters from the point
O. The acceleration due to gravity g  10 m / s2 .
5.
The value of t is _______ .
Key: 0.5
6.
The value of x is _______ .
Key: 7.5
5,6. Sol
y
5 2
5 m/s 45
O 5 m/s
m/2
g  10 m / s2
A m/2
Q
P
 from O to H time of ascent 
Uy
g

x
5
 0.5 sec
10
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HB
2021-Jee-Advanced
Question Paper-1_Key & Solutions
 at highest point mv x 
m 1
vx
2
v1x  10 m / s
 one part A of mass m/2 is falling vertically down in 0.5 sec
 part B is not having vertical component of velocity, it is falling freely in gravity
 t  0.5 sec
OQ  OP  PQ  5  0.5  10  0.5  7.5m
Question Stem for Question Nos. 7 and 8
Question Stem
In the circuit shown below, the switch S is connected to position P for a long time so that
the charge on the capacitor becomes q1 C . Then S is switched to position Q. After a
long time, the charge on the capacitor is q 2 C .
7.
The magnitude of q1 is _______ .
Key: 1.33
8.
The magnitude of q 2 is _______ .
Key: 0.66 to 0.67
7,8. Sol
Before shifting the switch S from P to Q.
t 
2
B
1V
i1
1 F
2V
A
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1
2021-Jee-Advanced
i1 
1
Amp
3
Question Paper-1_Key & Solutions
 VA  2  2i1  VB  VAB  2 
2 4
 V
3 3
q1  VAB  C
q1 
4
C
3
 after shifting the switch S from P to Q.
t 
1
2
B
i2
2V
A
i2 
2
Amp
3
VA  2  2 
VAB 
2
 VB
3
2
V
3
2
q 2  VAB  C  C
3
Question Stem for Question Nos. 9 and 10
Question Stem
Two point charges  Q and  Q / 3 are placed in the xy-plane at the origin (0, 0) and a
point (2, 0), respectively, as shown in the figure. This results in an equipotential circle of
radius R and potential V = 0 in the xy-plane with its center at (b, 0). All lengths are
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measured in meters.
2021-Jee-Advanced
9.
Question Paper-1_Key & Solutions
The value of R is _______ meter.
Key: 1.73
10.
The value of b is ______ meter.
Key: 3
9,10. Sol
Let ‘P’ be the point (2, 0) and CP = x
y
Rx
 2, 0 
 0, 0  A
d
VA  VB  0
VB 
Q
3
R
x
B
C
1
3
VA 
kQ
kQ

0
d  R  3 R  x 
kQ
kQ

0
d  R  3 R  x 
 d  3R, x  d  2 
R
3
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 R 3, d  3
Page-7
Q
2021-Jee-Advanced
11.
Question Paper-1_Key & Solutions
A horizontal force F is applied at the centre of mass of a cylindrical object of mass m and
radius R, perpendicular to its axis as shown in the figure. The coefficient of friction
between the object and the ground is  . The centre of mass of the object has an
acceleration a. The acceleration due to gravity is g. Given that the object rolls without
slipping, which of the following statement(s) is(are) correct ?
(A) For the same F, the value of does not depends on whether the cylinder is solid or
hollow
(B) For a solid cylinder, the maximum possible value of is 2 g
(C) The magnitude of the frictional force on the object due to the ground is always mg
(D) For a thin-walled hollow cylinder,
Key: BD
Sol
 R
a
F
f
F  f  ma
f  R  I
a  R
FR  mR 2  I
 
FR
I  mR 2
FR 2
I  mR 2
IF
Ia
f

I  mR 2 R 2
a
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mR 2
(B) for solid cylinder I 
2
2021-Jee-Advanced
Question Paper-1_Key & Solutions
FR
2F
fR
mgR

 a

 2 g
2
3
mR
3m
I
2
mR
2
2
(C) f  N  mg is the limiting friction and not always the value.
a
2
2
(D) for hollow cylinder, a 
12.
2
FR 2
F
; I  mR 2 
2
2mR
2m
A wide slab consisting of two media of refractive indices n1 and n 2 is placed in air as
shown in the figure. A ray of light is incident from medium n1 to n 2 at an angle  , where
sin  is slightly larger than 1 / n1 . Take refractive index of air as 1. Which of the
following statement(s) is(are) correct ?
(A) The light ray enters air if n 2  n1
(B) The light ray is finally reflected back into the medium of refractive index n1 if
n 2  n1
(C) The light ray is finally reflected back into the medium of refractive index n1 if
n 2  n1
(D) The light ray is reflected back into the medium of refractive index n1 if n 2  1
Key: BCD
Sol n1 sin   n 2 sin r
sin  is given slightly greater than
1
. If 'n 2 ' medium was not present then ray would
n1
have suffered TIR at n1 and air interface.
air
r
n2

 sin  
n2
sin r
n1
n2
1
sin r 
n1
n1
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n1
2021-Jee-Advanced
sin r 
Question Paper-1_Key & Solutions
1
1
, the ray will come back into n 2 and
 As sin r is slightly greater than
n2
n2
finally n1 medium irrespective of value of n 2 . So both B, C options are correct.
Also if n1  n 2 the light ray does not enter into air
13.
A particle of mass M = 0.2 kg is initially at rest in the xy-plane at a point  x  , y  h 
, where   10 m and h  1 m . The particle is accelerated at time t = 0 with a constant
acceleration a  10 m / s2 along the positive x-direction. Its angular momentum and


torque with respect to the origin, in SI units, are represented by L and  , respectively.
ˆi, ˆj and k̂ are unit vectors along the positive x, y and z-directions, respectively. If
kˆ  ˆi  ˆj then which of the following statement(s) is(are) correct ?
(A) The particle arrives at the point  x  , y  h  at time t = 2 s

(B)   2 kˆ when the particle passes through the point  x  , y  h 

(C) L  4 kˆ when the particle passes through the point  x  , y  h 

(D)   k̂ when the particle passes through the point  x  0, y  h 
Key: ABC
Sol
y
O
P
x
VR  10 m / s a  10m / s
 10,  1
2
Q
10,  1
R
VQ  20 m / s
1 2 1
at   10  t 2  20
2
2
t  2 sec ,
  ˆj  ma ˆi  ma kˆ  2 kˆ
PQ 
L  ˆj  0.2  20 ˆj  4 kˆ
Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom
?
(A) The ratio of the longest wavelength to the shortest wavelength in Balmer series is 9/5
(B) There is an overlap between the wavelength ranges of Balmer and Paschen series
1
(C) The wavelengths of Lyman series are given by  1  2   0 , where  0 is the shortest

m 
wavelength of Lyman series and m is an integer
(D) The wavelength ranges of Lyman and Balmer series do not overlap
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14.
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Key: AD
Sol
(A)
1
1 
 1
 R 2  2 

2 n 
1
 1 1  5R
 R 2  2  
2
 2 3  36
(1)
1
1 R
 R 2  
3
2  4
(2)
 2  2
1  3
R
9
 4  (A) is correct
5R 5
36
(B) Balmer series belongs to visible
Pasehen series belongs to I.R region.
(C)
0
1
1
1
1 
1


 R 1  2  ;
 R when m     
1  
1 
R

0
 m 
1  2  1  2 
 m   m 
(D) Lyman series belongs to U.V
Balmer series belongs to visible.
A long straight wire carries a current, I = 2 ampere. A semi-circular conducting rod is
placed beside it on two consuding parallel rails of negligible resistance. Both the rails are
parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as
shown in the figure. Two ends of the semi-circular rod are at distances 1 cm and
4 cm from the wire. At time t = 0, the rod starts moving on the rails with a speed v = 3.0
m/s (see the figure).
A resistor R  1.4  and a capacitor C0  5.0 F are connected in series between the rails.
At time t = 0, C 0 is uncharged. Which of the following statement(s) is(are) correct ?
 0  4  107 SI units. Take ln 2  0.7 
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15.
2021-Jee-Advanced
Question Paper-1_Key & Solutions
6
(A) Maximum current through R is 1.2 10 ampere
(B) Maximum current through R is 3.8 106 ampere
(C) Maximum charge on capacitor C 0 is 8.4 1012 coulomb
(D) Maximum charge on capacitor C 0 is 2.4 1012 coulomb
Key: AC
Sol
PV
i
A
++
x
B
dx
1 cm
C0
R
4 cm

Here V is instantaneous velocity
Emf AB    V  B   dx

 V
 0i
dx
2 x
 0iV
n  4 
2

i1
R
  i1R 

C0
q
dq
 i1 
C
dt
(1)
 0i
 n4  V 
2
For maximum charge
From (1)  
dq
 i1  0
dt
q
C
 0iV
n  4   C 0
2
 8.4  10 12 C
I max 
For maximum current q = 0
i1 
 iV n  4 

 1.2  106 Amp
; 0
R
2
R
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
2021-Jee-Advanced
16.
Question Paper-1_Key & Solutions
A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving
down with a constant acceleration along a fixed inclined plane with angle   45 . P1
and P2 are pressure at points 1 and 2, respectively, located at the base of the tube Let
   P1  P2  /  gd  , where  is density of water, d is the inner diameter of the tube and g
is the acceleration due to gravity. Which of the following statement(s) is(are) correct ?
(A)   0 when a  g / 2
(C)  
(B)   0 when a  g / 2
2 1
when a  g / 2
2
(D)  
1
when a  g / 2
2
Key: AC
Sol
d
g/ 2
2
2d
P1  

P1  P2

 d 
2d a 
g
2
45

2ag
45
a

2 d  P2

P1  P2 
2a 
 
 1
g
gd


g
if   0  a 
2

2 1
g
then  
2
2
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if a 
2021-Jee-Advanced
17.
Question Paper-1_Key & Solutions
An  -particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially
at rest. They are accelerated through a potential V and then allowed to pass into a region
of uniform magnetic field which is normal to the velocities of the particles. Within this
region, the  -particle and the sulphur ion move in circular orbits of radii r and rS ,
respectively. The ratio  rS / r  is _______ .
Key: 4
Sol
q   2e ; q s   e
p2
 qV  P  qm
2m
r
m
q
r

rs
m qs

ms q 
rs
4
r
A thin rod of mass M and length a is free to rotate in horizontal plane about a fixed
vertical axis passing through point O. A thin circular disc of mass M and of radius a /4 is
pivoted on this rod with its center at a distance a /4 from the free end so that it can rotate
freely about its vertical axis, as shown in the figure. Assume that both the rod and the
disc have uniform density and they remain horizontal during the motion. An outside
stationary observer finds the rod rotating with an angular velocity  and the disc
rotating about its vertical axis with angular velocity 4 . The total angular momentum of
 Ma 2  
the system about the point O is 
n .
 48 


The value of n is _______ .
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18.
P

q
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Key: 49
2
Sol
2
ma 2
ma 
 3a 
    4  m   
I0 
3
2 4
 4
1 1 9 
16  6  3  9  49

ma 2 
 ma 2       ma 2  

48
 3 8 16 

 48
19.
 n  49
A small object is placed at the centre of a large evacuated hollow spherical container.
Assume that the container is maintained at 0 K. At time t = 0, the temperature of the
object is 200 K. The temperature of the object becomes 100 K at t  t1 and 50 K at t  t 2 .
Assume the object and the container to be ideal black bodies. The heat capacity of the
object does not depend on temperature. The ratio  t 2 / t1  is _____ .
Key: 9
Sol
Ok
A
200k
Ae  T 4  O 4   m s

dT
T
4
dT
dt
  dt
T1  100 k
1
1
 3  k t1
3
T0 T1
T2  50 k
1
1
 3  k t2
3
T0 T2
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t2
9
t1
2021-Jee-Advanced
CHEMISTRY
1.
Question Paper-1_Key & Solutions
Max. Marks: 60
The major product formed in the following reaction is
(A)
(B)
(C)
(D)
Key. B
Sol:
CH  C   CH2 4  C  C   CH2 2  CH3
NaNH 2
Na C  C   CH2 4  C  C   CH2 2 CH3
Na in liq. NH3
 Birch Reduction 
Na C  C-  CH2 4
H
C
C
H
(CH2)2CH3
Birch reduces alkynes to trans alkenes trans
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except terminal alkyne  Answer is B.
2021-Jee-Advanced
2.
Question Paper-1_Key & Solutions
Among the following, the conformation that corresponds to the most stable conformation
of meso-butane-2,3-diol is
(A)
(B)
(C)
(D)
Key B
CH3
Sol: Meso – Butane-2, 3-diol is
H
OH
H
OH
CH3
H
H
H
OH
OH
H
O
H
O
Me
CH 3
Me
CH3
H
gauche
For the given close packed structure of a salt made of cation X and anion Y shown below
(ions of only one face are shown for clarity), the packing fraction is approximately
(packing fraction =
packing efficiency
)
100
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3.
2021-Jee-Advanced
(A) 0.74
Question Paper-1_Key & Solutions
(B) 0.63
(C) 0.52
(D) 0.48.
Key B
Sol: Number of Y atoms per unit cell = 1
Number of X atoms per unit cell = 3
2ry  a
ry
2rx  2ry  2a
rx 
a

1
2
2a  a  2  1 
 
 a
2
 2 
4 3
4
r y  3  rx3
3
P.F.  3
3
a
P.F 
4.
3
4  1  2  1  
P.F.    3 

3  8  2  


4  1  0.21   1.21


3  8 
6
P.F.  0.63
3
3
The calculated spin only magnetic moments of Cr  NH3 6  and  CuF6  in BM,
respectively, are
(Atomic numbers of Cr and Cu are 24 and 29, respectively)
(A) 3.87 and 2.84
(B) 4.90 and 1.73
(C) 3.87 and 1.73
(D) 4.90 and 2.84
Sol:
3
Cr  NH3 6  , Cr 3  t 32g eg o
  3  3  2   3.87
3
CuF6  ;
  2  2  2   2.84
Cu 3  t 62g eg 2
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Key A
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Question Stem for Question Nos. 5 and 6
Question Stem
For the following reaction scheme, percentage yields are given along the arrow:
Mg 2C3
H 2O
NaNH 2
MeI
P
 4.0g 
Hg 2 / H 
Q
red hot
iron tube
873K
R
 x g
40%
75%
100%
333K
S
Ba  OH 2
NaOCl
T
80%
heat
80%
U
 y g
 decolourises



 Baeyer 's reagent 
X g and y g are mass of R and U, respectively.
(Use: Molar mass  in g mol 1  of H, C and O as 1, 12 and 16, respectively)
5.
The value of x is ______.
Key 1.62 gm
3CH  C  CH 3 
 3CH 3  C  C  CH 3 

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Sol:
2021-Jee-Advanced
Question Paper-1_Key & Solutions
4
 0.1
40
1
3
0.1
0.75  0.1
1


 0.4  0.75   162 gm 
3


= 1.62 gm
6.
The value of y is ______.
Key 3.2
Sol: Molar mass of u = 100 gm
For 100 % yield
From 80 gm of P 
 100 gm of u is obtained
From 4 gm of P 

100
 4gm of u will be obtained
80
= 5 gm of u will be obtained
But as the yield is
100 80 80


100 100 100
u obtained is = 5 
100 80 80


100 100 100
=
58 2
5 5
=3.2 gm
Question stem for Question Nos. 7 and 8
Question Stem
pz
p
versus
104
is given below (in
T
solid line), where p z is the pressure (in bar) of the gas Z at temperature T and
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p  1bar
.
Page-20
 Y  s   Z  g  , the plot of
For the reaction, X  s  
ln
2021-Jee-Advanced
Question Paper-1_Key & Solutions
d lnK
H

p
R
 1
d 
K z
T
p
(Given,  
, where the equilibrium constant,
and the gas constant,
R  8.314 J K 1 mol1 )
7.
The value of standard enthalpy, H ( in kJ mol1 ) for the given reaction is_____.
Key 166.28
Sol:
G  H  TS
P 
RTn  z   H  TS
 P0 
H S
P 
n  z   

RT
R
 P 
Slope = 
H 
4
 104   
R
2
H  20 103  R J
H  20  8.314 KJ
H  166.28KJ / mol
8.
The value of S ( in kJ mol1 ) for the given reaction, at 1000 K is_____.
Key : 141.34
 P  H S
n  z  

RT
R
 P 
At T  1000 K 
104
 10
T
P 
  n  z   3
 P 
3 
H S

RT
R
3 
20  8.314  103 S

8.314  1000
R
S  17  8.314  141.338
S  141.33J / K / mol Truncated value
S  141.34 J / K / mol round off value
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Sol:
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Question Stem for Question Nos. 9 and 10
Question Stem
The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x C . To
this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added
to make a new solution B. The difference in the boiling points of water in the two
solutions A and B is y 102 C .
(Assume: densities of the solutions A and B are the same as that of water and the soluble
salts dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), K b  0.5 K kg mol1 ; Boiling point
of pure water as 100C .)
9.
The value of x is_____.
Key 100.1
m  0.1 molal AgNO3
A
Sol:
1 t solution
 Tb A  iK b m  2  0.5  0.1  0.1
X  100.1C
10.
The value of y is_____.
Key 2.5
m
B
Sol:
0.1
molal AgNO3
2
m
0.1
molal BaCl2
2
 Tb B  K b i1m1  i 2 m 2 
 Tb B 

1  0.1
0.1 
2


3

2 
2
2 

1
 0.1  0.15
2
0.25
 0.125
2
 Tb B   Tb A  0.125  0.1
 0.025  2.5 102
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y  2.5
2021-Jee-Advanced
Given:
The compound(s), which on reaction with HNO3 will give the product having degree of
rotation,   D  52.7 is(are)
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11.
Question Paper-1_Key & Solutions
2021-Jee-Advanced
Question Paper-1_Key & Solutions
(A)
(B)
(C)
(D)
Key CD
Sol: D-Glucose on oxidation with HNO3 forms. D-Glucaric or saccharic acid. As the question
wants to produce L-Glucaric acid we have start with L-Glucose which is non
superimposable mirror image of D-Glucose. i.e., C & D also form the required product.
The reaction of Q with PhSNa yields an organic compound (major product) that gives
positive Carius test on treatment with Na 2O 2 followed by addition of BaCl2 . The correct
option(s) for Q is(are)
(A)
(B)
(C)
(D)
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12.
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Key AD
F
S
NO 2
Ph
NO 2

PhSNa
SN AR
Sol:
NO 2
NO 2
Cl
H2C
SPh
PhSNa


SN 2
SMe
NO 2
13.
SMe
NO 2
The correct statement(s) related to colloids is(are)
(A) The process of precipitating colloidal sol by an electrolyte is called peptization.
(B) colloidal solution freezes at higher temperature than the true solution at the same
concentration.
(C) Surfactants form micelle above critical micelle concentration (CMC).CMC depends
on temperature.
(D) Micelles are macromolecular colloids.
Key BC
Sol: (A) Incorrect
(B) correct
(C) correct
(D) Incorrect
An ideal gas undergoes a reversible isothermal expansion from state I to state II followed
by a reversible adiabatic expansion from state II to state III. The correct plot(s)
representing the changes from state I to state III is(are)
(p: pressure, V: volume, T: temperature, H : enthalpy, S: entropy)
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14.
2021-Jee-Advanced
Question Paper-1_Key & Solutions
(A)
(B)
(C)
(D)
Key ABD
Sol: (A)
Correct
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(B)
2021-Jee-Advanced
Question Paper-1_Key & Solutions
For II to III
P1r Tr  constant
r
P T 1 r  constant
1
dP
 r  r 1
 constant 
 T   ve
dr
 r 1 
correct
(C)
Since enthalpy change of ideal gas only depends upon temperature I 
 II (correct)
But In adiabatic expansion temperature will decrease so enthalpy should decrease
II 
 III (incorrect)
So the graph is incorrect
(D)
I 
 II correct
II 
 III correct
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So the graph is correct
2021-Jee-Advanced
15.
Question Paper-1_Key & Solutions
The correct statement(s) related to the metal extraction processes is(are)
(A) A mixture of PbS and PbO undergoes self-reduction to produce Pb and SO 2 .
(B) In the extraction process of copper from copper pyrites, silica is added to produce
copper silicate.
(C) Partial oxidation of sulphide ore of copper by roasting, followed by self-reduction
produces blister copper.
(D) In cyanide process, zinc powder is utilized to precipitate gold from Na  Au  CN 2  .
Key ACD
Sol:
 PbS  PbO 
 Pb  SO 2
 Zn  Na  Au  CN 2  
 Au  Na 2  Zn  CN 4 
16.
A mixture of two salts is used to prepare a solution S, which gives the following results:
White
precipitate(s)
only
Dilute NaOH(aq)
S
Room temperature
(aq solution
of the salts)
Dilute HCl (aq)
Room temperature
White
precipitate(s)
only
The correct option(s) for the salt mixture is(are)
(A) Pb  NO 3 2 and Zn  NO 3 2
(B) Pb  NO 3 2 and Bi  NO 3 3
(C) AgNO3 and Bi  NO 3 3
(D) Pb  NO 3 2 and Hg  NO3 2
Key AB
Ag   OH  
 Ag 2 O 
white
Brown
Bi3  OH  
 Bi  OH 3 
Hg 2  OH  
 HgO 
Brownish Red.
white
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Sol:
Pb 2  Cl 
 PbCl2 
2021-Jee-Advanced
17.
Question Paper-1_Key & Solutions
The maximum number of possible isomers (including stereosiomers) which may be
formed on mono-bromination of 1-methylcyclohex-1-ene using Br2 and UV light is
_____.
Key 13
Br
CH3
Br
Br2

uv
*

:2
:1
Br

Br
:2
:2
Br
:2 
:2
Br
:2
Br
Total isomers obtained = 13
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Sol:
2021-Jee-Advanced
18.
Question Paper-1_Key & Solutions
In the reaction given below, the total number of atoms having sp 2 hybridization in the
major product P is____.
1.O3  excess 
then Zn / H 2 O
P
2.NH2OH  excess 
Key 8
Sol:
(1)O3 excess
then Zn/H 2O
O
O
O
O

NH 2OH / excess
*
OH
N
*
*
HO
N*
OH
*N
*
*
*N
OH
Total 8 atoms 4N  4C have sp 2 hybridization
As NCERT gives oxygen of phenol to be sp 3 so similarly here too it should be sp 3 .
19.
The total number of possible isomers for  Pt  NH 3 4 Cl2  Br2 is____.
Key 6
Sol: Considering structural and Sterio
(1)  Pt  NH 3 4 Br2  Cl2 
 cis & trans
(2)  Pt  NH 3 4 ClBr  ClBr 
 cis & trans
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(3)  Pt  NH 3 4 Cl2  Br2 
 cis & trans
2021-Jee-Advanced
Question Paper-1_Key & Solutions
MATHEMATICS
1.
Max. Marks: 60
Consider a triangle  whose two sides lie on the x-axis and the line x  y  1  0 . If the
orthocenter of  is (1, 1), then the equation of the circle passing through the vertices of
the triangle  is
(A) x 2  y 2  3x  y  0
(B) x 2  y 2  x  3y  0
(C) x 2  y 2  2y  1  0
(D) x 2  y 2  x  y  0
Key: B
SOL:
H 1,1  O.C
Property : Image of O.C w.r.t any side lie on circum-circle of the triangle
Given sides X-axis i.e. y = 0 & x  y  1  0 .
 2, 2 
1,1
x  y 1  0
H
y0
A  1,0 
1, 1
Image of H 1,1 w.r.t y = 0 is P 1, 1
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Image of H 1,1 w.r.t x  y  1  0 be Q  ,  
2021-Jee-Advanced
  1   1 2 1  1  1
1


1
12  12
Question Paper-1_Key & Solutions
 3 ,
  2,   2  Q  2, 2 
Point of Intersection of sides y = 0 & x  y  1  0 is A  1,0  .
 1
Now P 1, 1 , Q  2, 2  & A  1,0  lie on circumcircle slope of PA 
, slope of
2
 2
 
QA  .
Product of slopes = 1 .
 PA  QA
1

 PQ is diameter of circumcircle .
circumcircle is
 x  1 x  2    y  1 y  2   0
x 2  x  2  y 2  3y  2  0 ,
2.
x 2  y 2  x  3y  0
9


The area of the region  x, y  : 0  x  , 0  y  1, x  3y, x  y  2  is
4


(A)
11
32
(B)
35
96
(C)
37
96
(D)
13
.
32
Key: A
Sol:
y
2
1
3 1
 , 
2 2
A1      A 2
x
9
4
3
2
x
3y  x
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xy2
2021-Jee-Advanced
Question Paper-1_Key & Solutions
2
x

R.A  A1  A 2      2  x  dx 
3

3/2 
2
x
dx
3

2
2
1
 4x

 
 2  dx 
3
3

3/2 
 4x 2

1  x2 


2x


 
 6
3/2 3  2  2
2
  2 9  1  81 
   4   4      3    2 
3
  3 4  3  32

 8  12   9  18  1  81  64 



 3   6  3  32 


3.
9/4
9/4
9/4
 xdx
2
4 3 1 17 8  9
17
   

3 2 3 32
6
3  32
16  17
33
11

 .
3  32 3  32 32
Consider three sets E1  1, 2,3 , F1  1,3,4 and G1  2,3,4,5 . Two elements are
chosen at random, without replacement, from the set E1 , and let S1 denote the set of
these chosen elements. Let E 2  E1  S1 and F2  F1  S1 . Now two elements are chosen
at random, without replacement, from the set F2 and let S2 denote the set of these chosen
elements
Let G 2  G1  S2 . Finally, two elements are chosen at random, without replacement,
from the set G 2 and let S3 denote the set of these chosen elements.
Let E 3  E 2  S3 . Given that E1  E 3 , let p be the conditional probability of the event
S1  1, 2 . Then the value of p is
(A)
1
5
(B)
3
5
(C)
1
2
(D)
2
5
Key: A
1 1 1
 
3
2 10
Reqiured probability=
1 1 1 1 1
1 3 C2 1  1  2 1 
     1  4       
3 2 10 3  2
10
C2 6  3  3 10 
1
1 1 1
  
20 20 12 15
=
1
60
1


20  6  5  4  5
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
1
20
2021-Jee-Advanced
Question Paper-1_Key & Solutions
4.
Let 1 , 2 ,....,10 be positive valued angles (in radian) such that 1   2  ....  10  2 .
Define the complex numbers z1  ei1 , z k  z k 1eik for k = 2, 3, …, 10, where i  1 .
Consider the statements P and Q given below:
P : z 2  z1  z3  z 2  ...  z10  z9  z1  z10  2
2
2
Q : z 22  z12  z32  z 22  .....  z10
 z92  z12  z10
 4
Then,
(A) P is TRUE and Q is FALSE
(B) Q is TRUE and P is FALSE
(C) both P and Q are TRUE
(D) both P and Q are FALSE
Key: C
Given Z1  ei
1
Zk  Zk 1eik

For k = 2,3,4…10

2
 ei2 , 3  ei3 …….
Z1
Z2
10
 ei10
Z9
Clearly
Length of Arc  AB   chord AB
C
z3
B z2
3
2
A z1
O
Z 
 Z 2  Z1  Arg  2 
 Z1 
Z 
 Z3  Z2  Arg  3  …………
 Z2 
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Z 
Z1  Z10  Arg  10 
 Z1 
2021-Jee-Advanced
Question Paper-1_Key & Solutions
P = Z2  Z1  Z3  Z 2  ......  Z1  Z10
Z 
Z 
Z 
P  Arg  2   Arg  3   .....  Arg  10 
 Z1 
 Z2 
 Z1 
P  2  3  .......  10  2   2  3  .....  10 
 Z 2  Z1  Z3  Z 2  ......  Z1  Z10  2
z2
z1
3
2
O
X
z1
Z 
Z 
 Z2  Z1  Arg  2  and Z2    Z1   Z2  Z1  Arg  2 
 Z1 
 Z1 
 Z2 
Z 
 Z22  Z12  Arg  22   2 Arg  2 
 Z1 
 Z1 
Z 
z 22  z12  2 Arg  2 
 Z1 
 Z3 

 Z2 
Similarly Z32  Z22  2 Arg 
2
Q  Z 22  Z12  Z32  Z22  ...... Z12  Z10
 4
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Question stem for Question Nos. 5 and 6
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Three numbers are chosen at random, one after another with replacement, from the set S
= {1, 2, 3, …., 100}. Let p1 be the probability that the maximum of chosen numbers is
at least 81 and p 2 be the probability that the minimum of chosen numbers is at most 40.
5.
The value of
625
p1 is ____
4
Key: 76.25
P1  Probability that maximum of chosen number is at least 81
 P1  1  Probability of chosen numbers  80
 1
80 80 80


100 100 100
 P1  1 
6.
( with replacement)
64
61
625
305

 125P1  61 
 76.25
P1 
125 125
4
4
The value of
125
p 2 is ____.
4
Key: 24.5
P2  Probability of minimum of chosen number is atmost 40.
 P2  1  Probability of chosen numbers  41
 1
60 60 60


100 100 100
P2  1 
27
98

125 125
( with replacement)
 125P2  98 
125
98
 24.50
P2 
4
4
Question stem for Question Nos. 7 and 8
Question Stem-2
Let  ,  and  be real numbers such that the system of linear equations
x  2y  3z  
4x  5y  6z  
7x  8y  9z    1
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is consistent. Let M represent the determinant of the matrix
2021-Jee-Advanced
 2  
M    1 0


Question Paper-1_Key & Solutions
 1 0 1 
Let P be the plane containing all those  ,  ,   for which the above system of linear
equations is consistent, and D be the square of the distance of the point  0,1,0  from the
plane P.
7.
The value of M is ____.
Key: 1
8.
The value of D is ____.
Key: 1.5
7 & 8 Sol:
x  2y  3z  
4x  5y  6z  
7x  8y  9z    1
1 2 3
 4 5 6 0
7 8 9
 2 3
 2 1
1  
5 6 3 
5 2
 1 8 9
 1 8 3
 3   15  16   2  3  2   2   1 8  5  5  
 3    6  4   4  8  5   5
 3    2    1  3    2    1
1  3
1  1
2  4 
6 34 
2
7  1 9
7  1 3
 3  3  2   2    12  14   1 4   4  7  
 3 3  2   2  2  4   4  7
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 3  2  4  2   2   6  a  2    1
2021-Jee-Advanced
1 2 
3  4 5 
7 8  1
Question Paper-1_Key & Solutions
  5  5  8   2  4   4  7     32  35 
 5  5  8  8  8  14  3
 3  6  3  3
For consistent, 1   2   3  0
   2    1  0 ……………….(1)
 2 
M   1 0
1 0 1
(7)
  1  0   2    0     0  1
   2  
=1
(from (1) )
M 1
(8)
P is plane containing all point  , ,  
For the above system.
x  2y  z  1  0 ……….(2)
 Plane ‘P’ is
Perpendicular distance of (0,1,0) from (2) =
0  2 1  0  1
12   2   12
2

3
6
D = Square of the distance
2
 3  9 3

    1.5
 6 6 2
Question stem for Question Nos. 9 and 10
Question Stem
Consider the lines L1 and L 2 defined by
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L1 : x 2  y  1  0 and L 2 : x 2  y  1  0
2021-Jee-Advanced
Question Paper-1_Key & Solutions
For a fixed constant  , let C be the locus of a point P such that the product of the
distance of P from L1 and the distance of P from L 2 is  2 . The line y  2x  1 meets C
at two points R and S, where the distance between R and S is
270 .
Let the perpendicular bisector of RS meet C at two distinct points R ' and S' . Let D be
the square of the distance between R ' and S' .
9.
The value of  2 is ______.
Key: 9
Sol: Let P  h, k 
L1  x 2  y  1  0 , L 2  x 2  y  1  0
h
  h
2  k 1
3
h 2 
2
 
2  k 1
3
2
  k  1  3 2
2
Locus of P  h, k  is
C : 2x 2   y  1  3 2 …….(1)
2
y  2x  1 cuts curve (1) at R & S
 2x 2   2x   3 2
2
 2x 2  3 2 …… (2)
3 2
x1  x 2  0 , x 1x 2 
2
2
2
 y 1
2
& 2
   y  1  3 .
 2 
 y  12  6 2  y 2  2y  1  6 2  0 …..(3)
y1  y 2  2 , y1y 2  1  6 2
Given RS  270

 x 2  x1 2   y2  y1 2 

 x1  x 2 2  4x1x 2   y1  y2 2  4y1y2 
270 where R  x1, y1  & S  x 2 , y 2 
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270
2021-Jee-Advanced
 3 2 
Question Paper-1_Key & Solutions


2
0  4 
   2   4 1  6  270
 2 
2
6 2  4  4  24 2  270
30 2  270   2  9
2  9
Similarly
10.
The value of D is _____
Key: 77.14
 1
Sol: Slope of  r bisector of RS 
2
 x  x 2 y1  y 2 
,
Midpoint of RS =  1

2 
 2
 2
=  0,    0,1
 2
Equation  r bisector is
y 1  
1
1
 x  0   y  x  1 …..(4)
2
2
(4)Cuts curve (1) at R '& S'
 2x   y  1  3  27
2
2
2
2
 1 
 2x    x   27
 2 
2
x2
 2x 
 27  7x 2  108 ……..(5)
4
2
x1  x 2  0 , x1x 2 
108
7
& 2  4  y  1   y  1  27
2
2
7  y  1  27  7y 2  14y  20  0
2
20
7
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y1  y 2  2, y1y 2 
2021-Jee-Advanced
R 'S' 
Question Paper-1_Key & Solutions
 x1  x 2 2  4x1x 2   y1  y2 2  4y1y2
 108 
 20 
 0  4 
  4  4

 7 
 7 
D  R 'S' 

4  108  28  80
7
432  108
540

 77.142  8.783
7
7
D= 8.783
For any 3  3 matrix M, let M denote the determinant of M. Let
1 2 3 
1 0 0 
1 3 2 
E   2 3 4  , P  0 0 1  and F  8 18 13 .






8 13 18
0 1 0 
 2 4 3 
If Q is a nonsingular matrix of order 3  3 , then which of the following statements is(are)
TRUE?
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11.
2021-Jee-Advanced
Question Paper-1_Key & Solutions
1 0 0 
(A) F  PEP and P 2  0 1 0 


0 0 1 
(B) EQ  PFQ 1  EQ  PFQ 1
(C)  EF   EF .
3
2
(D) Sum of the diagonal entries of P 1EP  F is equal to the sum of diagonal entries of
E  P 1FP .
Key: ABD
Sol:
1 0 0
A. P   0 0 1 
0 1 0


1 2 3 


E  2 3 4 
 8 13 18 


1 3 2 


F   8 18 13 
2 4 3 


P2  I
 1 0 0  1 2 3   1 0 0 




PEP   0 0 1  2 3 4   0 0 1 
 0 1 0  8 13 18   0 1 0 




1 2 3 


 2 3 4   F
 8 13 18 


PEP  F
P 2 EP  PF  EP  PF
B. EQ  PFQ 1  E Q  EPQ 1
 E Q  PQ 1
0
( E 0
RHS = EQ  PFQ 1
 0  EPQ 1  0
(EP = PF)
EQ  PFQ 1  EQ  PFQ 1
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C. since E  0
2021-Jee-Advanced
3
 EF   0
Question Paper-1_Key & Solutions
EF  0
2
 EF 
3
  EF 
2
is false
1 0 0
D. P   0 0 1 
0 1 0


1
P 1EP  F
F  P 1EP  F  F  2F
 1 0 0  1 3 2  1 0 0 




E  P FP  E   0 0 1  8 18 13  0 0 1 
 0 1 0  2 4 3  0 1 0 




1
2 4 6


 4 6 8 
 16 26 36 


Tr  2F   Tr  E  P 1FP   44
12.
Let f :    be defined by
x 2  3x  6
f x  2
x  2x  4
Then which of the following statements is (are) TRUE ?
(A) f is decreasing in the interval  2, 1
(B) f is increasing in the interval 1,2 
(C)f is onto
 3 
(D) Range of f is   ,2 
 2 
Key: AB
f ' x  
5x  x  4 
x
2
 2x  4

2
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Sol:
x 2  3x  6
f x  y  2
x  2x  4
2021-Jee-Advanced


4
Question Paper-1_Key & Solutions

0
f is increasing
 , 4    0,   and
decreasing  4,0 
x 2  3x  6
y 2
x  2x  4
 y  1 x 2   2y  3 x  4y  6  0
y  1,   0
 12y 2  4y  33  0
 3 11 
y   ,   1 .
 2 6
13.
Let E, F and G be three events having probabilities
1
1
1
1
P  E   ,P  F   and P  G   , and let P  E  F  G   .
8
6
10
4
For any event H, if Hc denotes its complement, then which of the following statements
is(are) TRUE?


(A) P E  F  G c 
(C) P  E  F  G  
1
40
13
24


(B) P E C  F  G 


1
15
(D) P E c  Fc  G c 
5
12
Key : ABC
1
1
1
P  E   , P  G   P  F1  
8
4
6
P  E  F1  G  
1
10
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Sol:
2021-Jee-Advanced
Question Paper-1_Key & Solutions
F
E
a
c
b
h
d
e
f
G
P E  a  b  d  h 
1
8
P  F  b  c  h  e 
1
6
1 1
1
  …………..(2)
6 10 15
 b+c+e =
P G 
1 1
1
abd   
…………..(1)
8 10 40
1
dhef
4
def 
1 1
3
 
…………….(3)
4 10 20
From (1) b 
1
40
P  E  F  Gc  
From (2) e 
1
40
1
15
P  F  G  Ec  
1
15
1 1 1 13
P E  F  G    
8 4 6 24
PE  F  G 
13
24
P  E c  G e  Fc   1  P  E  F  G 
13
24

11
24
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 1
2021-Jee-Advanced
14.
Question Paper-1_Key & Solutions
For any 3  3 matrix M, let M denote the determinant of M. Let I be the 3  3 identity
matrix. Let E and F be two 3  3 matrices such that  I  EF  is invertible. If
G   I  EF  , then which of the following statements is (are) TRUE ?
1
(A) FE  I  FE FGE
(B)  I  FE 1  FGE   I
(D)  I  FE  I  FGE   I
(C) EFG  GEF
Key: ABC
Sol: (A) 1  EF  G  I  G  I  EFG 
 G  EFG  I  G  GEF
(B) G   I  EF 
 EFG  GEF
1
 FGE  F  I  EF  E
1

 E 1  I  EF  F1


1
 FGE  E 1F1  I

1
 FGE  E 1F1  I  1
 FGE  E 1F1  I FE  FE
 FGE I  FE  FE
 FGE I  FE  FE
(C) 1  FE  . I  FGE 
 I  FGE  FE  FE.FGE  0
 I  FGE  FE  F  G  I  E
 I  FGE  FE  FGE  FE  I
(B) FGE  I  FE  FE .
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(D)  I  FE  . I  FGE   I .
2021-Jee-Advanced
15.
For any positive integer n, let Sn :  0,    be defined by
Question Paper-1_Key & Solutions
 1  k  k  1 x 2 
Sn  x    k 1 cot 
 ,

x


1
n
  
where for any x   , cot 1  x    0,   and tan 1  x     ,  . Then which of the
 2 2
following statement is (are) TRUE?
(A) S10  x  
 1  11x 2 
 tan 1 
 , for all x > 0
2
10x



(B) lim cot  Sn  x    x , for all x  0
n 
(C) The equation S3  x  

4
has a root in  0, 
1
(D) tan  Sn  x    , for all n  1 and x  0
2
Key: AB
Sn :  0,    R
n


x
Sn  x    tan 1 
 1  k  k  1 x 2 
k 1


n
  k  1 x  kx  n
 tan 1  KH  x
Sn  x    tan 1 
 1  k  kH  x 2  
k 1

 k 1
Sn  tan 1  n  1 x  tan 1 x .
S10  x   tan 111x  tan 1 x
 10x  

1  10x
 tan 1 


cot


2
2 
 1  11x  2
 1  11x 
(A) 
1  1  11x

 tan 
2
 10x

2



(B) lim cot tan 1  n  1 x  tan 1 x
n 

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Sol:
2021-Jee-Advanced
Question Paper-1_Key & Solutions



nx
 x 
 lim cot  tan 1 
 cot tan 1  2 


2



n 
x 
 1   n  1 x  

 cot cot x  x
(C) S3  x  

4
 tan 1 4x tan 1 x 

4
3x
 3x  
 
 1  4x 2  3x  1  0
tan 1 
2 
2
1  4x
 1  4x  4
9  4 4  0
(D)

16.
nx
1

2
1
x
1   n  1 x
x
nx
n
n
1
1  n 1
For any complex number w  c  id , let arg  w   ( ,  ] , where i  1 . Let  and 
 z 
be real numbers such that for all complex numbers z  x  iy satisfying arg 
z
 
 4 ,

the ordered pair  x, y  lies on the circle
x 2  y 2  5x  3y  4  0
Then which of the following statements is (are) TRUE?
(A)   1
(B)   4
(C)   4
(D)   4
Key: BD
 z 
Sol: arg 
z
 
 4

y=0
x 2  5x  4  0
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x  4, 1
2021-Jee-Advanced
Question Paper-1_Key & Solutions
Z
45
 5 3
C   2 , 2 
90
B
  ,0
A
  ,0
17.
For x   , the number of real roots of the equation 3x 2  4 x 2  1  x  1  0 is ____.
Key: 4
3x 2  4 x 2  1  x  1  0
1
1
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Sol:
2021-Jee-Advanced
Question Paper-1_Key & Solutions
3x 2  x  1  4 x 2  2
18.
In a triangle ABC, let AB  23 , BC = 3 and CA = 4. Then the value of
cot A  cot C
cot B
is _____
Key: 2
Sol:
AB  23 , BC  3 , CA  4
b 2  c2  a 2
cos A 
2bc
cot A  cot C
2b 2
 2 2
2
cot B
a  c  a2
19.
 



Let u , v and w be vectors in three-dimensional space, where u and v are unit vectors
which are not perpendicular to each other and
 
 
 
u  w  1, v  w  1, w  w  4
If the volume of the parallelepiped, whose adjacent sides are represented by the vectors
 
 

u , v and w , is 2 , then the value of 3u  5v is ______
Key: 7
 
Sol: u  v
 
 
u.w  1; v.w  1
 
w.w  4
 
uv  0
  
u  u u.v
 
 
 u v w 2  v.u v.v
   
v.w v.w

1 u.v 1

u.v 1 1  2
1
1 4
 
u.w
 
v.w  2
 
w.w
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 
 2
4  u.v  u.v  1  1  4  u.v   2
2021-Jee-Advanced
Question Paper-1_Key & Solutions



 1
4  u.v   2u.v  0; u.v  0  u.v 
2


1
3u  5v  9  25  30  =
2
 4,0
  ,0
 1,0 
y  0; x 2  5x  4  0
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  ,0
34  15  7
JEEAdvanced2021
304,Kaset
t
yHei
ght
s,AyyappaSoci
et
y,Madhapur
,Hyder
abad-500081
PHYSICS




1.
Max. Marks: 60
SECTION-1(Maximum Marks: 24)
One or More Type
This section contains SIX (06) questions.
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN
ONE of these four option(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, 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 unanswered;
Negative Marks : −2 In all other cases.
One end of a horizontal uniform beam of weight W and length L is hinged on a
vertical wall at point O and its other end is supported by a light inextensible rope. The
other end of the rope is fixed at point Q, at a height L above the hinge at point O. A
block of weight aW is attached at the point P of the beam, as shown in the figure (not
( )
to scale). The rope can sustain a maximum tension of 2 2 W . Which of the
following statement(s) is(are) correct?
A) The vertical component of reaction force at O does not depend on a
B) The horizontal component of reaction force at O is equal to W for a = 0.5
C) The tension in the rope is 2W for a = 0.5
D) The rope breaks if a > 1.5
Ans. ABD
2021-Jee-Advanced
Question Paper-2_Key & Solutions
T
L
O
L
q
w´
Sol.
w
L
2

T
L
L    L
2
2

1
1
T  2     . For   T   2
2
2

T

 R Y      R Y 
2
2

T
1
Rx 
    
2

2
2.
1
For    R x  
2
For   1.5  T  2 2
A source, approaching with speed u towards the open end of a stationary pipe of
length L, is emitting a sound of frequency fS . The farther end of the pipe is closed.
The speed of sound in air is u and f 0 is the fundamental frequency of the pipe. For
which of the following combination(s) of u and fS , will the sound reaching the pipe
lead to a resonance ?
A) u = 0.8 u and fS = f 0
B) u = 0.8 u and f S = 2f 0
C) u = 0.8 u and f S = 0.5f 0
D) u = 0.5 u and f S = 1.5 f 0
Ans. AD
V
V
and   4L  f 0 
4L

lapp = v - u . lres = V
fs
n
f0
f0 
for resonance ,
l app = l Þ l res =
( odd )
V-u
= 4L
fs
0.2V V
=
fs
5fs
V
= lres
For fs = f 0 Þ l app =
5f 0
But u = 0.8v Þ l app =
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Sol.
2021-Jee-Advanced
Question Paper-2_Key & Solutions
f
V
¹ lres
For fs = 0 Þ l app =
2
10f 0
f
2V
¹ l res
For fs = 0 Þ l app =
2
5fs
3f
V
For u  and fs  0
2
2
3.
Þ
l app =
V
V
=
= l res
2fs 3f 0
For a prism of prism angle q = 600 , the refractive indices of the left half and the right
half are, respectively, n1 and n 2 (n 2 ³ n1 ) as shown in the figure. The angle of
incidence i is chosen such that the incident light rays will have minimum deviation if
n1 = n 2 = n = 1.5 . For the case of unequal refractive indices, n1 = n and n 2 = n + Dn
(where Dn << n ), the angle of emergence e = i + De . Which of the following
statement(s) is(are) correct ?
A) The value of De (in radians) is greater than that of Dn
B) De is proportional to Dn
C) De is lies between 2.0 and 3.0 mill radians, if Dn = 2.8´10-3
D) De is lies between 1.0 and 1.6 mill radians, if Dn = 2.8´10-3
Ans. BC
AD
Sol. i 
2

i
AD
sin
3
2

A
2
sin
2
n2
 A  D

 2 
sin 

3
4
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n1
2021-Jee-Advanced
Question Paper-2_Key & Solutions
A
2
sin i
n  n
 sin e 
as n 
sin r
2
i1  i 2  A  D and r1  r2 
sin i  e 

 sin i  cos e  cosisin  e 
n  n
2
n  n
2
n
3n
n n
1
e  

2
2
2 2
2
e   n 
7
3
For n  2.8  103 and n 
2
e  2.15  103
  



A physical quantity S is defined as S = E ´ B / m 0 , where E is electric field, B is

magnetic field and m 0 is the permeability of free space. The dimensions of S are the

4.
(
)
same as the dimensions of which of the following quantity (ies)
Energy
Force
A)
B)
Ch arg e ´ Current
Length ´ Time
Energy
Power
C)
D)
Volume
Area
Ans. BD
 
 E  B

Sol. S 
0
 E   B
S     
 
 
 0

CB  B

 
 
 0
?
 B2 
   C
  0 
B2
has dimensions of energy density
0


 


ML2T 2 
 MT 3
3
1 
L LT 
 

A heavy nucleus N, at rest, undergoes fission N ® P + Q , where P and Q are two
lighter nuclei. Let d = M N - M P - M Q , where M P , M Q and M N are the masses of P,
Q and N, respectively. E P and E Q are the kinetic energies of P and Q, respectively.
The speeds of P and Q are uP and uQ , respectively. If c is the speed of light, which
of the following statement(s) is(are) correct?
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5.
S 
 
energy speed  

 

 volume 


2021-Jee-Advanced
A) E P + E Q = c 2 d
Question Paper-2_Key & Solutions
æ M
ö
P
ç
÷ c2d
B) E P =
çM + M ÷
Qø
è P
MQ
u
C) P =
uQ M P
D) The magnitude of momentum for P as well as Q is C 2md , where
m=
MP MQ
(M
P
+ MQ
)
Ans. ACD
Sol. N  P  Q
P
Q
VP
VQ
  M N  M P  M Q Q  c2

E P  E Q  C 2
Since there are no external forces acting on the system. Momentum has to be
conserved.
PP = PQ = P
M P VP = M Q VQ
VP M Q
=
VQ M P
P2
P2
\
+
= dc2
2M P 2M Q
P2
K.E =
2M
2
P =
2M P M Q .d c 2
MP + MQ
P = 2mdc 2
P = c 2md
Where m =
MP + MQ
Two concentric circular loops, one of radius R and the other of radius 2R, lie in the
xy-plane with the origin as their common centre, as shown in the figure. The smaller
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6.
MP MQ
2021-Jee-Advanced
Question Paper-2_Key & Solutions
loop carries current I1 in the anti-clockwise direction and the larger loop carries

current I2 is the clockwise direction, with I2 > 2I1 . B (x, y) denotes the magnetic
field at a point (x, y) in the xy-plane. Which of the following statement(s) is(are)
correct?
A)
B)
C)
D)

B (x, y) is perpendicular to the xy-plane at any point in the plane

B (x, y) depends on x and y only through the radial distance r = x 2 + y 2

B (x, y) is non-zero at all points for r < R

B (x, y) points normally outward from the xy-plane for all the points between the
two loops
Ans. AB

 ......... 
A
B
Sol.
0i1 0i2

,i2  2i1soBcenteris 
2R
4R

direction A B changes as we cross any wire
Bcenter 
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Due to radial symmetry B depends only on r.
consider point A just inside inner loop and B just outside it field at A due to inner
loop is outward so there must be a null point somewhere between O&A (for r <R)
Also field between the loops is in ward field just outside the outer loop is outward and
at far a was points it is inwards. So there is another null point outside the outer loop
two (for r > 2R)
2021-Jee-Advanced




Question Paper-2_Key & Solutions
SECTION-2(Maximum Marks: 12)
Paragraph with Numerical
This section contains THREE (03) question stems.
There are TWO (02) questions corresponding to each question stem.
The answer to each question is a NUMERICAL VALUE.
For each question, enter the correct numerical value corresponding to the answer in the
designated place using the mouse and the on-screen virtual numeric keypad.

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 : +2 If ONLY the correct numerical value is entered at the designated place;
Zero Marks : 0 In all other cases.
Question Stem for Question Nos. 7 and 8
Question Stem
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass
test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a
mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle
is sealed at atmospheric pressure p0 = 105 Pa so that the volume of the trapped air is
n 0 = 3.3cc . When the bottle is squeezed from outside at constant temperature, the
pressure inside rises and the volume of the trapped air reduces. It is found that the test
tube begins to sink at pressure p0 + Dp without changing its orientation. At this
pressure, the volume of the trapped air is n0 - Dn .
Let Dn = X cc and Dp = Y ´103 Pa .
7.
The value of X is _____.
Ans. 0.30
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8.
The value of Y is _____.
Ans. 10.00
2021-Jee-Advanced
Sol. 7&8
Question Paper-2_Key & Solutions
mass test tube = 5g
5g
volume 
 2cc
2.5g
cc
Buoyant force = 2g
ignoring mass of air trapped, minimum
value of volume of air trapped for
glass tube to float
WB

 3cc
Swater
Vin  3.3cc Vfin  3cc V  0.3cc  X
Pin  P0 Pfin  P0  P
isothermal condition, hence
Pin Vin  Pfin Vfin
3.3 P0  3 P0  P

P0  P  1.1P0
 Y  10
P  0.1P0  104 Pa
Question Stem for Question Nos. 9 and 10
Question Stem
A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string
of length L= 1.0m. It is suspended from a fixed point at height H = 0.9 m above a
frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at
rest vertically below the point of suspension. A horizontal impulse P = 0.2 kg-m/s is
imparted to the bob at some instant. After the bob slides for some distance, the string
becomes taut and the bob lifts off the floor. The magnitude of the angular momentum
of the pendulum about the point of suspension just before the bob lifts off is
J kg- m2 / s . The kinetic energy of the pendulum just after the lift-off is K Joules.
9.
The value of J is _____.
Ans. 0.18
10. The value of K is _____.
Ans. 0.16
P  mv  v 
0.2
 2m1
0.1
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Sol. 9&10
2021-Jee-Advanced
Question Paper-2_Key & Solutions
L  J  mvH   0.9 0.2  0.18

1
0.9
vcos
 v
v cos   1.8ms1
2
1
\
K.E = m (1.8) = 0.162
2
Question Stem for Question Nos. 11 and 12
Question Stem
In a circuit, a metal filament lamp is connected in series with a capacitor of
capacitance C mF across a 200 V, 50 Hz supply. The power consumed by the lamp is
500 W while the voltage drop across it is 100 V. Assume that there is no inductive
load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle
(in degrees) between the current and the supply voltage is j . Assume, p 3 » 5 .
11. The value of C is _____.
Ans. 100
100

v02  v2R  vC2
 200
2
2
 100  vC2
vC  100 3
P   vR iR
i 100  500
i5
i´ x c = vc
1
 100 3 ,
2 50 C
5
C 4
10  3
1
C  100F
C 4 ,
10
5
C
5
100100 3
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Sol.
200
2021-Jee-Advanced
12. The value of j is _____.
Ans. 60
v
100 3
Sol. tan   c 
vR
100

Question Paper-2_Key & Solutions
  600
SECTION-3(Maximum Marks: 12)
Paragraph with Single Answer Type
This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02)
questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the
correct answer.

For each question, choose the option corresponding to the correct answer

Answer to each question will be evaluated according to the following marking scheme:
Full Marks
: +3 If ONLY the correct option 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.
Paragraph
A special metal S conducts electricity without any resistance. A closed wire loop,
made of S, does not allow any change in flux through itself by inducing a suitable
current to generate a compensating flux. The induced current in the loop cannot
decay due to its zero resistance. This current gives rise to a magnetic moment which
in turn repels the source of magnetic field or flux. Consider such a loop, of radius a,
with its centre at the origin. A magnetic dipole of moment m is brought along the
axis of this loop from infinity to a point at distance r(>>a) from the centre of the
loop with its north pole always facing the loop, as shown in the figure below.
The magnitude of magnetic field of a dipole m, at a point on its axis at distance r, is
m0 m
, where m 0 is the permeability of free space. The magnitude of the force
2p r 3
between two magnetic dipoles with moments,
m1 and m2 , separated by a distance
r on the common axis, with their north poles facing each other, is
k m1m 2
, where k
4
r
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is a constant of appropriate dimensions. The direction of this force is along the line
joining the two dipoles.
2021-Jee-Advanced
13.
Question Paper-2_Key & Solutions
When the dipole m is placed at a distance r from the centre of the loop (as shown in the
figure), the current induced in the loop will be proportional to
A) m / r 3
B) m 2 / r 2
Ans. A
Sol. Suppose self-inductance the loop = L
æ m0m
2ö
÷÷ - Li = 0
.
p
a
3
è 2 pr
ø
æm ö
Þ i µçç 3 ÷÷
èr ø
D)
m2 / r
The work done in bringing the dipole from infinity to a distance r from the canter of the
loop by the given process is proportional to
A)
Ans. C
m / r5
B)
m2 / r 5
r
Sol.
m / r2
.......... (1)
Then ç
ç
14.
C)
W=
ò Fdr
Also
F=
Mind
æ m 0 m ö pa 2
= pa i = pa . çç 3 ÷÷
è 2pr ø L
K
\ W =ò nm .
r
2
pa 2
2pL
0
D)
m2 / r 7
rn
(from equal)
2
( )m
m2 / r 6
Km.mind
r =¥
2
C)
m
. 3 .dr
r
m2
µ ò 7 dr
r =¥ r
r
m2
µ 6
r
Paragraph
A thermally insulating cylinder has a thermally insulating and frictionless movable
partition in the middle, as shown in the figure below. On each side of the partition,
there is one mole of an ideal gas, with specific heat at constant volume, CV = 2R .
Here, R is the gas constant. Initially, each side has a volume V0 and temperature T0 .
The left side has an electric heater, which is turned on at very low power to transfer
heat Q to the gas on the left side. As a result the partition moves slowly towards the
right reducing the right side volume to V0 / 2 . Consequently, the gas temperatures
on the left and the right sides become TL and TR , respectively. Ignore the changes
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in the temperatures of the cylinder, heater and the partition. N’/
2021-Jee-Advanced
T
15. The value of R is
T0
A)
Ans. A
16.
2
Question Paper-2_Key & Solutions
3
B)
The value of
(
C) 2
D) 3
Q
is
RT0
(
)
A) 4 2 2 + 1
(
)
B) 4 2 2 - 1
(
)
C) 5 2 + 1
)
D) 5 2 - 1
Ans. B
Sol. 15&16
Right chamber is undergoing adiabatic process.
3
g=
2
T0 V0
TR = 2 T0
\
3
-1
2
3
æ V0 ö 2 -1
= TR çç ÷÷
è 2 ø
TR
= 2
T0
æ 3V ö
Pf çç 0 ÷÷ = 1RTL
è 2 ø
æV ö
Pf çç 0 ÷÷ = 1RTL \ TL = 3TR
è 2 ø
TL = 3 2T6
Q = (dU L ) + (dU)R
Q = nC v dTL + nC v dTR
( )(3
=1 2R
)
2T0 - T0 + 1(2R ).
(
)
(
) (
)
2T6 - T0 = 2R 4 2 - 2 T6
Q = 4RT6 2 2 - 1
Q
= 4 2 2 -1
RT6
)
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(
2021-Jee-Advanced
Question Paper-2_Key & Solutions
SECTION-4(Maximum Marks: 12)
Non-Negative Integer Answer Type



This section contains THREE (03) questions.
The answer to each question is a NON-NEGATIVE INTEGER.
For each question, enter the correct integer corresponding to the answer 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
: +4 If ONLY the correct integer is entered;
Zero Marks
: 0 In all other cases.
17.
In order to measure the internal resistance r1 of a cell of emf E, a meter bridge of wire
resistance R 0 = 50 W , a resistance R 0 / 2 , another cell of emf E/2 (internal resistance r)
and a galvanometer G are used in a circuit, as shown in the figure. If the null point is
found at l  72 cm , then the value of r1 = __ W .
Ans. 3
E
R
R 0  0  r1
2
P.D between the points where the secondary cell is connected is
R
 78R 0 
28R 0 
E
E
 i 0 



100 
2 
3R 0   100 
 2
 r1  2 
Sol. i in primary circuit =
r1 
r1 
3R 0 156R 0

2
100
156  50 3
  50 25
100 2
2
r1  78  75
r1  3
The distance between two stars of masses 3M S and 6MS is 9R. Here R is the mean
distance between the centres of the Earth and the Sun, and M S is the mass of the Sun.
The two stars orbit around their common centre of mass in circular orbits with period nT,
where T is the period of Earth’s revolution around the Sun.
The value of n is ____.
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18.
2021-Jee-Advanced
Question Paper-2_Key & Solutions
Ans. 9
G 3Ms  6Ms
Sol.
  6Ms  3R 2
2
9R 

GMs
2
T' 
3

81R
9R

19.
- - - - 6R
-
- 3R

6ms
com
81R 3
T '  2
GM
T '  9T .
In a photoemission experiment, the maximum kinetic energies of photoelectrons from
metals P, Q and R are E P , E Q and E R , respectively, and they are related by
E P = 2E Q = 2E R . In this experiment, the same source of monochromatic light is used
for metals P and Q while a different source of monochromatic light is used for the metal
R. The work functions for metals P, Q and R are 4.0 eV, 4.5 eV and 5.5 eV, respectively.
The energy of the incident photon used for metal R, in eV, is ____.
Ans. 6
Sol. E P = hn1 - 4
E Q = hn1 - 4.5
E R = hn 2 - 5.5
E P = 2E Q = 2E R
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hn1 - 4 = 2 (hn1 - 4.5)
9 - 4 = hn1
EQ = ER
5 - 4.5 = hn 2 = 5.5
Þ hn 2 = 6eV
Eincident photon = 6eV
2021-Jee-Advanced
CHEMISTRY




1.
Question Paper-2_Key & Solutions
Max. Marks: 60
SECTION-1(Maximum Marks: 24)
One or More Type
This section contains SIX (06) questions.
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN
ONE of these four option(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, 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 unanswered;
Negative Marks : −2 In all other cases.
The reaction sequence(s) that would lead to o-xylene as the major product is(are)
Me
NH2
Me
1. NaNO 2 /HCl 273K
2.CuCN
3.DIBAL-H then H3O +
4. N 2 H 4 , KOH heat
1.Mg, CO 2 ,H3O +
2.SOCl2
3.H 2 ,Pd-BaSO 4
4.Zn-Hg,HCl


A)

B)
Me
Br
1.i.BH 3
ii.H 2 O 2 ,NaOH
2.PBr3
3.Zn, dil.HCl

1.O3,Zn/H2O
2.N



2H4 ,KOH,heat
D)
C)
Ans : AB
Sol :
A)
Me
Me
NH2
Me
N2Cl
Me
CN
CuCN


NaNO2 /HCl 273K


CHO
DIBAL


N2H 4
Me
Me
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B)
2021-Jee-Advanced
Question Paper-2_Key & Solutions
Me
Me
Mg/CO2

H O+
Me
SOCl2


3
Br
COOH
COCl
H2 ,Pd-BaSO4
Me
CHO
Me
Zn-Hg


HCl
Me
C)
Me
Me
Me
Me




PBr3
i.BH3
ii.H 2O2 ,NaOH

Zn/HCl
OH
Br
D)
O
O3 ,Zn/H 2O


N 2H4


OH 
O
2.
Correct option(s) for the following sequence of reactions is(are)
V
Br2
CHCl3
1.Q

 P 

 R 
PhCH3 
light
KOH
2.H 2 ,Pd/C
i. KMnO 4
S
Foul smelling
KOH, heat
ii. H 3O 
W
3
U
T 
2.heat
1. NH
A) Q  KNO 2 , W  LiAlH 4
B) R = benzenamine, V = KCN
C) Q  AgNO2 , R  phenylmethanamine
D) W  LiAlH 4 , V  AgCN
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Ans : AD (As per standard ref) or CD (as per NCERT)
Sol :
2021-Jee-Advanced
Question Paper-2_Key & Solutions
V = AgCN
CHCl3
Br2
1.Q = KNO2
PhCH3 

 PhCH 2 Br 
 PhCH 2 NH 2 
 PhNC
light
2.H 2 ,Pd/C
KOH
S
W = LiAlH 4
PhCOOH  PhCONH 2
T
u
Note : As per the basic NCERT treatment, Q may be taken as AgNO 2 . But as per
mechanism, PhCH 2 Br shown preferably react by S N 1 mechanism. As Ag  facilitate
S N 1 mechanism more, KNO2 should give more nitro products
3.
For the following reaction
k
2 X  Y 
P
d  P
the rate of reaction is
 k  X . Two moles of X are mixed with one mole of Y to
dt
make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The
correct statement(s) about the reaction is(are)
(Use: ln 2 = 0.693)
A) The rate constant, k, of the reaction is 13.86  104 s –1.
B) Half-life of X is 50 s
d  X
C) At 50 s, 
 13.86  103 mol L–1s –1.
dt
d  Y
D) At 100 s, 
 3.46  103 mol L–1s –1.
dt
Ans : BCD
k
2 X  Y 
P
Sol :
t = 0 2 mol 1 ml 0
t = 50 s(2 – 1) mol 0.5 mol 0.5 mol
1 a
1 2
k  ln 0  ln
t at 50 1
1
 0.693
50
 0.01386 s 1

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13.86  103
2021-Jee-Advanced
After100s, k =
Question Paper-2_Key & Solutions
1
1
ln
100 a t
1
1
1
ln 2 
ln
50
100 at
 at  0.25
d  p
d  y

 13.86  103  0.25
dt
dt
 3.46  103
Some standard electrode potentials at 298 K are given below:
Pb 2 / Pb - 0.13 V

4.

Ni 2 / Ni - 0.24 V
Cd 2 / Cd
- 0.40 V
Fe 2 / Fe
- 0.44 V
To a solution containing 0.001 M of X2 and 0.1 M of Y 2 , the metal rods X and Y
are inserted (at 298 K) and connected by a conducting wire. This resulted in
dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given: Gas constant, R  8.314 J K –1mol –1 , Faraday constant, F  96500 C mol –1 )
A) Cd and Ni
B) Cd and Fe
C) Ni and Pb
D) Ni and Fe
Ans : ABC
Sol :
X
Y
0.001MX 2
0.01MY 2
0
EPb2 / Pb  EPb
 0.03log103
2
/ Pb
= - 0.13 – 0.03 × 3
= 0.13 – 0.09
= - 0.22 V
1
EPb
 0.13  0.03log10
2
/ Pb
= - 0.13 – 0.03
= - 0.16 V
E Ni / Ni  0.24  0.09  0.33V
2
E1Ni 2 / Ni  0.24  0.03  0.27
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ECd 2  /Cd  0.40  0.09  0.49V
2021-Jee-Advanced
Question Paper-2_Key & Solutions
1
ECd
 0.40  0.03  0.43V
2
/ Cd
EFe2  / Fe  0.40  0.09  0.53V
1
EFe
 0.44  0.03  0.47V
2
/ Fe
G   nFEcell
G of A   ve
G of B   ve
G of C   ve
G of D   ve
5.
The pair(s) of complexes where in both exhibit tetrahedral geometry is(are)
(Note: py = pyridine
Given: Atomic numbers of Fe, Co, Ni and Cu are 26, 27, 28 and 29, respectively)
B) [Co  CO 4 ]– and [CoCl4 ]2–
A) [FeCl4 ]– and [Fe  CO 4 ]2–
C) [Ni  CO 4  and  Ni  CN 4 ]2–
D) [Cu  py 4 ] and [Cu  CN 4 ]3–
Ans : ABD
Sol : A) FeCl4
Fe 2  3d 6 4 s 0
3d
FeCl4 
4s
4p
××
×× ××
××
sp 3
3d
Fe  CO 4 
2
4s
4p
××
×× ××
××
sp 3
3d
[Co  CO 4 ]– 
4s
4p
××
×× ××
××
sp 3
B)
3d
[CoCl4 ]2– 
4s
4p
××
×× ××
××
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sp 3
2021-Jee-Advanced
Question Paper-2_Key & Solutions
3d
[Cu  py 4 ] 
4s
4p
××
×× ××
××
sp 3
D)
3d
[Cu  CN 4 ] 
3–
4s
4p
××
×× ××
××
sp 3
6.
The correct statement(s) related to oxoacids of phosphorous is(are)
A) Upon heating, H3PO3 undergoes disproportionation reaction to produce H3PO 4
and PH3 .
B) While H3PO3 can act as reducing agent, H3PO 4 cannot.
C) H3PO3 is a monobasic acid.
D) The H atom of P–H bond in H3PO3 is not ionizable in water.
Ans : ABD
Sol : A) 4 H 3 PO3  3H 3 PO4  PH 3
B) H3PO3 is a reducing acid as it has P – H bond
C) H3PO3 is a dibasic acid
D) H 3 PO3  H 2O  H 3 PO3  H 3O 
H 3 PO3  H 2O  H 3 PO32  H 3O 




SECTION-2(Maximum Marks: 12)
Paragraph with Numerical
This section contains THREE (03) question stems.
There are TWO (02) questions corresponding to each question stem.
The answer to each question is a NUMERICAL VALUE.
For each question, enter the correct numerical value corresponding to the answer in the
designated place using the mouse and the on-screen virtual numeric keypad.

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 : +2 If ONLY the correct numerical value is entered at the designated place;
Zero Marks : 0 In all other cases.
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Question Stem for Question Nos. 7 and 8
At 298 K, the limiting molar conductivity of a weak monobasic acid is
4×102S cm 2 mol-1. At 298 K, for an aqueous solution of the acid the degree of
dissociation is  and the molar conductivity is y  102 S cm 2 mol-1. At 298 K, upon
2021-Jee-Advanced
Question Paper-2_Key & Solutions
20 times dilution with water, the molar conductivity of the solution becomes
3y  102 S cm 2 mol-1.
7.
The value of  is ___.
Ans : 0.22
Sol :   4  102 S cm 2 mol 1
For concentration C,
y  102 y


4  102 4
2
 y
C 
 4
C. 2
k

y
1 
1
4
C
,
For concentration
20
2
C  3y 
. 
20  4 
k
3y
1
4
44
y
 0.86
51
11
   0.21to 0.22
51
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8.
The value of y is ___.
Ans : 0.86
Sol :   4  102 S cm 2 mol 1
For concentration C,
y  102 y


4  102 4
2
 y
C 
 4
C. 2
k

y
1 
1
4
C
,
For concentration
20
2
C  3y 
. 
44
20  4 
y
 0.86
k
3y
51
1
4
11
   0.21to 0.22
51
2021-Jee-Advanced
Question Paper-2_Key & Solutions
Question Stem for Question Nos. 9 and 10
Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the
salt reacted with y g of nitrobenzene in the presence of required amount of HCl to
produce 1.29 g of an organic salt (quantitatively).
(Use Molar masses  in g mol –1  of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and
119, respectively).
9.
The value of x is ___.
Ans : 3.57
Sol : Sn  2 HCl  SnCl2  H 2

NH 3Cl 
NO2
3SnCl2  7 HCl 
3SnCl4  2 H 2O
1.29 g
1.29
 0.01
3
129
 nSnCl2  0.03
 nSn  0.03
x = 0.03 × 114 = 3.57
y = 0.01 × 123 = 1.23
nPhNH Cl  
10. The value of y is ___.
Ans : 1.23
Sol : S n  2 HCl  SnCl2  H 2

NH 3Cl 
NO2
3SnCl2  7 HCl 
3SnCl4  2 H 2O
1.29 g
1.29
 0.01
3
129
 nSnCl2  0.03
 nSn  0.03
x = 0.03 × 114 = 3.57
y = 0.01 × 123 = 1.23
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nPhNH Cl  
2021-Jee-Advanced
Question Paper-2_Key & Solutions
Question Stem for Question Nos. 11 and 12
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare
a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of
0.03 M KMnO 4 solution to reach the end point. Number of moles of Fe 2 present in
250 mL solution is x  102 (consider complete dissolution of FeCl2 ). The amount of
iron present in the sample is y% by weight.
(Assume: KMnO4 reacts only with Fe 2 in the solution
Use: Molar mass of iron as 56 g mol1 )
11. The value of x is ___.
Ans : 1.875
Sol : ngl of Fe 2  nge KMnO4
 12.5  103  0.03  5
 1.875  103
 nFe in 250 of SAn
2
 1.875  103  10
 1.875  102
 x  1.875
WFe  1.875  102  56
= 1.05 g
1.05
Fe 
 100  18.75
5.6
12. The value of y is ___.
Ans : 18.75
Sol : ngl of Fe 2  nge KMnO4
 12.5  103  0.03  5
 1.875  103
 nFe in 250 of SAn
2
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 1.875  103  10
 1.875  102
 x  1.875
WFe  1.875  102  56
= 1.05 g
1.05
Fe 
 100  18.75
5.6
2021-Jee-Advanced
Question Paper-2_Key & Solutions
SECTION-3(Maximum Marks: 12)
Paragraph with Single Answer Type

This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02)
questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the
correct answer.

For each question, choose the option corresponding to the correct answer

Answer to each question will be evaluated according to the following marking scheme:
Full Marks
: +3 If ONLY the correct option 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.
Paragraph-1:
The amount of energy required to break a bond is same as the amount of energy
released when the same bond is formed. In gaseous state, the energy required for
homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond
Strength. BDE is affected by s-character of the bond and the stability of the
radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds
are given below:
H3C  H  g   H3C•  g  + H•  g  ΔH°=105Kcal mol-1
Cl  Cl  g   Cl•  g  +Cl•  g  ΔH°=58Kcal mol-1
H 3C  Cl  g   H 3C•  g  + Cl•  g  ΔH°=85Kcal mol-1
H  Cl  g   H •  g  +Cl•  g  ΔH°=103Kcal mol-1
Correct match of the C–H bonds (shown in bold) in Column J with their BDE in
Column K is
Column J
Molecule
Column K
BDE  kcal mol-1 
P) H – CH(CH 3 ) 2
i) 132
Q) H – CH 2 Ph
ii) 110
R) H – CH  CH 2
iii) 95
S) H – C  CH
iv) 88
A) P – iii, Q – iv, R – ii, S – i
C) P – iii, Q – ii, R – i, S – iv
Ans : A
Sol : Q < P < R < S
B) P – i, Q – ii, R – iii, S – iv
D) P – ii, Q – i, R – iv, S – iii
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13.
2021-Jee-Advanced
14. For the following reaction
Question Paper-2_Key & Solutions
light
CH 4  g   Cl 2  g  
 CH 3Cl  g   HCl  g 
the correct statement is
A) Initiation step is exothermic with H  –58 kcal mol –1
B) Propagation step involving  CH 3 formation is exothermic with
H  –2 kcal mol –1
C) Propagation step involving CH3Cl formation is endothermic with
H  27 kcal mol –1.
D) The reaction is exothermic with H  –25 kcal mol –1.
Ans : D
CH 4  Cl2  CH 3Cl  HCl
Sol :
105 58
85
103
H  105  58   85  103
 25k cal mol 1
Paragraph-2:
The reaction of K 3  Fe  CN  6  with freshly prepared FeSO 4 solution produces a dark
blue precipitate called Turnbull’s blue. Reaction of K 4  Fe  CN  6  with the FeSO4
solution in complete absence of air produces a white precipitate X, which turns
blue in air. Mixing the FeSO4 solution with NaNO3 , followed by a slow addition
of concentrated H 2SO 4 through the side of the test tube produces a brown ring.
15.
Precipitate X is
A) Fe 4 [Fe  CN  6 ]3 B) Fe[Fe  CN  6 ]
C) K 2 Fe[Fe  CN  6 ]
D) KFe[Fe  CN 6 ]
Ans : C
Sol : K 2 Fe[Fe  CN  6 ] + FeSO 4
 K 2 Fe[Fe  CN  6 ]  + K 2SO 4
white
16.
Among the following, the brown ring is due to the formation of
B) [Fe  NO 2 (H 2O) 4 ]3
A) [Fe  NO 2 (SO 4 ) 2 ]2–
C) [Fe  NO 4 (SO 4 ) 2 ]
D) [Fe  NO (H 2O)5 ]2
Ans : D
Sol : Fe 2+ +5H 2O + NO   Fe  H 2O 5  NO 
2+
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brown ring
2021-Jee-Advanced
Question Paper-2_Key & Solutions
SECTION-4(Maximum Marks: 12)
Non-Negative Integer Answer Type



This section contains THREE (03) questions.
The answer to each question is a NON-NEGATIVE INTEGER.
For each question, enter the correct integer corresponding to the answer 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
: +4 If ONLY the correct integer is entered;
Zero Marks
: 0 In all other cases.
17.
One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed
by II, as shown below. If the work done by the gas in the two processes are same,
the value of ln
V3
is ____.
V2
2250
( p1 ,V1 )
U
K
R
450
I
II
( p2 ,V2 )
( p3 ,V3 )

S J K 1 mol 1

(U: internal energy, S: entropy, p: pressure, V: volume, R: gas constant)
5
(Given: molar heat capacity at constant volume, CV ,m of the gas is R )
2
Ans : 10
u
 1800  u  1800  R  W
Sol :
R
V
1800 R  nRT ln 3
V2
For process I
Q=0
u  W
For process II
u  0
u  W  nRT ln
= 1800 R
V3
V2
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T1  900 K
2021-Jee-Advanced
Question Paper-2_Key & Solutions
1800 R  1 
u  nCv T
5R
T  900
2
T = 180 K
V
1  180ln 3  1800
V2
18.
ln
V3
 10
V2
Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The
change in the velocity  in cm s –1  of He atom after the photon absorption is ___.
(Assume: Momentum is conserved when photon is absorbed.
Use: Planck constant  6.6  10 –34 J s, Avogadro number  6  1023 mol –1 , Molar
mass of He  4 g mol –1 )
Ans : 30
h
6.6  1034  6  1023
Sol : V 

cms 1
3

m 4  10  330  10
19. Ozonolysis of ClO 2 produces an oxide of chlorine. The average oxidation state of
chlorine in this oxide is ___.
Ans : 6
Sol : ClO2  O3  ClO3  O2
Oxidation state of Cl in ClO3 is
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
2021-Jee-Advanced
Question Paper-2_Key & Solutions
MATHEMATICS




1.
Max. Marks: 60
SECTION-1(Maximum Marks: 24)
One or More Type
This section contains SIX (06) questions.
Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN
ONE of these four option(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, 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 unanswered;
Negative Marks : −2 In all other cases.
Let
S1   i, j, k  : i, j,k  1,2,....,10,
S2   i, j :1  i  j  2  10,i, j  1,2,.....,10 ,
S3   i, j,k, l  :1  i  j  k  l ,i, j,k, l  1,2,....,10
and
S4  {  i, j, k, l  : i, j, k and l are distinct elements in {1,2,....,10}}.
If the total number elements in the set Sr is n r ,r  1,2,3, 4, then which of the following
statements is(are) TRUE?
A) n1  1000
B) n 2  44
C) n 3  220
D)
n4
 420
12
Ans: ABD
Sol: n1  10  10  10
89
 44
2
i  1, j  1, 2,3,......8;i  2, j  1, 2,3,......8; i  3, j  2,3,....8 ,... and so on
n2  8  8  7  6  5  4  3  2  1  8 
10  9  8  7
 210 (Select 4 numbers and arrange in increasing order)
24
n
n 4  10 P4  10  9  8  7  4  420
12
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n 3  10C4 
2021-Jee-Advanced
Question Paper-2_Key & Solutions
2.
Consider a triangle PQR having sides of lengths p,q and r opposite to the angles P,Q
and R, respectively. Then which of the following statements is(are) TRUE?
p2
A) cos P  1 
2qr
 qr
 pr
cos P  
cosQ
B) cos R  

 p  q
 p  q 
C)
qr
sin Qsin R
2
p
sin P
D) If p  q and p  r, then cosQ 
p
p
and cos R 
q
r
Ans: AB
q2  r 2 p2
p2
Sol: A) cos P 

1
2qr
2qr
2qr
B) p  q  r   r cos Q  q cos R    p cos R  r cos P    p cos Q  q cos P 
q  r sin Q  sin R
sin Q.sin R

2
P
sin P
sin P
q
D) cosQ   Q is acute  R is acute (similarly)
r
q
p2  q2  r 2 q
cos Q  
  p 2  r 2  q 2  2p 2  p 2  q 2  r 2
r
2pr
r
C)
f  0  1 and
 /3
 f  t  dt  0
0
Then which of the following statements is(are) TRUE?
 
A) The equation f  x   3cos3x  0 has at least one solution in  0, 
 3
B) The equation f  x   3sin 3x  
6
has at least one solution in

 
 0, 
3
x
C) lim
x 0
x  f  t  dt
0
1  ex
2
 1
x
D) lim
x 0
sin x  f  t  dt
0
x2
 1
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3.
R is obtuse(contradiction)
  
Let f :   ,    be a continuous function such that
 2 2
2021-Jee-Advanced
Ans: ABC
Question Paper-2_Key & Solutions
 /3
Sol: f  0  1,  f  t  dt  0
0
x
A) Let g  x    f  x  dx  sin 3x
0
 
g  0  0  g   / 3  g '  x   0 has atleast one solution in  0, 
 3
x
B) Let g  x    f  t  dt  cos3x 
0
6x

g  0  1  g   / 3
 g '  x   0 has atleast one solution in  0,  / 3
x
C) lim
x  f  t  dt
0
x2
x
 f  t  dt  xf  x 
 lim 0
x 0
1 e
2xe x
f  x   xf ' x   f  x 
 lim
 1
x2
2
2
x 0

2 e  2x ex 
x 0
2
x

f
t
dt




 sin x   0
  1  1  1
D) lim 

x 0 
x  x 


4.
For any real numbers  and  , let y ,  x  , x   , be the solution of the differential
dy
 y  xex , y 1  1 .
dx
Let S  y  x  :  ,    . Then which of the following functions belong(s) to the set
equation


S?
x2 x 
1
A) f  x   e   e   e x

2
2
x2 x 
1
B) f  x    e   e   e  x

2
2
ex 
1 
e2   x
C) f  x    x     e   e
2
2 
4
ex  1
e2   x
 
D) f  x     x    e   e
 
2 2
4
Ans: AC
yex   x e  x dx
If     0 , then ye
x
x2

c
2
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Sol:
2021-Jee-Advanced
Question Paper-2_Key & Solutions
1
1
 c  c  e 
2
2
2
x
1
 y    e   e x
2
 2
1,1  e 
 x2
1
Put   1  y    e   e2 (A is correct)
2
 2
If     0 , then ye
x
xe  x
e  x


c
       2
Put     1
x x ex
y e  c
2
4
ex 
1
e2 
x 
1,1  y   x    e  e   (C is corret)
2
2
4

 1 




5.
Let O be the origin and OA  2i  2j  k,OB
 i  2j  2k and OC  OB   OA for
2
  9
some   0 . If OB  OC  , then which of the following statements is(are) True?
2


3
A) Projection of OC on OA is 
2
9
B) Area of the triangle OAB is
2
9
C) Area of the triangle ABC is
2
D) The acute angle between the diagonals of the parallelogram with adjacent sides



OA and OC is
3
Ans: ABC
 1
Sol: OC  1  2  i   21    j   2    k 
2
i
j
i
j
k
k
9 1
1
OB  OC  
1
2
2 
1
2
2
2 2
2
1  2 2  2 2  
2  1 2  2   2


1
6i  3 j   6  k
2
 9  36  9  36     1
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
2021-Jee-Advanced
Question Paper-2_Key & Solutions
 1
 OC  i  4j  k
2
OC.OA 2  8  1 3
A)


OA
2.3
2
i j k
1
1
9
B) ar  OAB  2 2 1  6i  3j  6k 
2
2
2
1 2 2


i
i
j
k
1
4 1  1 4 1
4
8 1
5 8 1
2 2
1
1
9
 12i  6j  12k  6i  3j  6k 
4
2
2
C) ar  ABC 
1
1
2
5
2
k
j
9
2
2
cos  


5
3 10
d2 . d2 3
2.
2
2
2
Let E denote the parabola y  8x. Let P   2, 4 , and let Q and Q' be two distinct
 
d1 d 2


D) d1  OA  OC,d 2  OA  OC
6.
9
points on E such that the lines PQ and PQ' are tangents to E. Let F be the focus of E.
Then which of the following statements is(are) TRUE?
A) The triangle PFQ is a right- angled triangle
B) The triangle QPQ' is a right – angled triangle
C) The distance between P and F is 5 2
D) F lies on the line joining Q and Q'
Ans: ABD
Sol: y 2  8x;p  2,4 lies on directrix
Q

P  2,4

Q
F  2,0
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A) Portion of tangent between POC and directrix substneds 90ºat focus
B) Tangents at ends of focal chords are perpendicular
C) 4  4  4 2
D) QQ' is a focal chord.
2021-Jee-Advanced
Question Paper-2_Key & Solutions
SECTION-2(Maximum Marks: 12)
Paragraph with Numerical




This section contains THREE (03) question stems.
There are TWO (02) questions corresponding to each question stem.
The answer to each question is a NUMERICAL VALUE.
For each question, enter the correct numerical value corresponding to the answer in the
designated place using the mouse and the on-screen virtual numeric keypad.

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 : +2 If ONLY the correct numerical value is entered at the designated place;
Zero Marks : 0 In all other cases.
Question Stem for Question Nos. 7 and 8
Consider the region R   x, y     : x  0 and y 2  4  x . Let F be the family of
all circles that are contained in R and have centers on the x-axis. Let C be the circle
that has largest radius among the circle in F. Let  ,  be a point where the circle C
meets the curve y 2  4  x .
7.
The radius of the circle C is _______.
Ans: 1.5
8.
The value of  is _________.
Ans: 2
Sol(7&8Q):
x  0, y 2  4  x

y2  4  x

h,0
 

0

4
Let equation of circle be  x  h   y 2  h 2
2
Solving with y 2  4  x
x 2  2hx  4  x  0
 x 2  x  2h  1  4  0
.....(1)
For touching, D = 0
  2h  1  16
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2
2021-Jee-Advanced
Question Paper-2_Key & Solutions
h
 2h  1  4
3
2
3
in (1)
2
x 2  4x  4  0
x2
So   2
Putting h 
Question Stem for Question Nos. 9 and 10
Let f1 :  0,    and f 2 :  0,    be defined by
x 21
f1  x      t  j dt,
j
x0
0 j1
and
f 2  x   98  x  1  600  x  1  2450, x  0,
50
49
where, for any positive integer n and real numbers a1 ,a 2 ,......,a n ,  in1 a i denotes the
product of a1 ,a 2 ,.....,a n . Let m i and n i , respectively, denote the number of points of
local minima and the number of points of local maxima of function fi ,i  1,2, in the
interval  0, 
9.
The value of 2m1  3n1  m1n1 is_______.
Ans: 57
10. The value of 6m 2  4n 2  8m 2 n 2 is _________.
Ans: 6
Sol (9&10Q):
x
f1  x     t  1  t  2  t  3  t  4 ...... t  21 dt
1
2
3
4
21
0
f11  x    x  1 x  2  x  3  x  4 .... x  21
2
3
4
21
Plotting wavycurve of f11  x 
-
1
+
2
+
3-
+
4
15 - 20 -
+
21
So for x  4k  1,  k  W  f11  x  changes sign from –ve to +ve
for x  4k  3,  k  W  f11  x  changes sign from +ve to –ve
So m1  no.of local minima = 6
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n1  no.of local maxima = 5
2021-Jee-Advanced
(Q9) 2m1  3n1  m1n1  12  15  30  57
Question Paper-2_Key & Solutions
f 2  x   98  x  1  600  x  1  2450
50
49
f 21  x   98  50  x  1  600  49  x  1
49
 98  50  x  1
48
 x  1  6
48
 98  50  x  1  x  7
Wavy curve of f 2'  x  is
48
7
1
Clearly m 2  1,n 2  0
(10) 6m 2  4n 2  8m 2 n 2  6  0  0  6
Question Stem for Question Nos. 11 and 12
  3 
  3 
Let g i :  ,   ,i  1,2 , and f :  ,    be functions such that
8 8 
8 8 
  3 
g1  x   1,g 2  x   4x   and f  x   sin 2 x, for all x   , 
8 8 
Define
Si 
3
8
 f  x  .g  x  dx,i  1,2
i

8
11.
The value of
16s1
is _______.

Ans: 2
12.
The value of
48s 2
is _________.
2
Ans: 1.5
Sol(11&12Q):
S1 

8
 sin
3
2
x.1.dx 

8

x sin 2x

2
4

8
1  cos 2x
dx
2


8
3

8
8

 1  3

  sin
 sin 
8 4
4
4


8
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3
2021-Jee-Advanced
16S1
2

(Q11)
3
S2 
8


Question Paper-2_Key & Solutions
 x   .sin 2 x dx
8
Using King’s Property
3
S2 
8


  4x .cos 2 xdx
8
3
Adding 2S2 



8

8
  4x dx
8
3
4
8
1   2
 2S2  area under graph  2    
2 8 2 16
48S
24
 1.5
(Q12)  2 2 

16

SECTION-3(Maximum Marks: 12)
Paragraph with Single Answer Type
This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02)
questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the
correct answer.

For each question, choose the option corresponding to the correct answer

Answer to each question will be evaluated according to the following marking scheme:
Full Marks
: +3 If ONLY the correct option 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.
Paragraph-1:
Let M   x, y     : x 2  y 2  r 2  ,
1
,n  1,2,3,..... Let S0  0
2n 1
and, for n  1 , let Sn denote the sum of the first n terms of this progression. For n  1 ,
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where r  0 . Consider the geometric progression a n 
2021-Jee-Advanced
Question Paper-2_Key & Solutions
let Cn denote the circle with center Sn 1 ,0 and radius a n , and D n denote the circle
with center Sn 1 ,Sn 1  and radius a n .
1025
. Let k be the number of all those circles Cn that are inside
513
M. Let l be the maximum possible number of circles among these k circles such that
no two circles intersect. Then
A) k+ 2l =22
B) 2k + l = 26
C) 2k + 3l = 34 D) 3k+ 2l = 40
Ans: D
13.
Consider M with r 
14.
2
Consider M with r 
99
M is
A) 198
Ans: B
Sol(13&14Q):
n
Sn   a n 
n 1

1
2
2198
. The number of all those circles D n that are inside
B) 199
C) 200
D) 201
1 1 1
1
 1  2  .....  n 1
0
2
2 2
2
1
n
2n  2  1

n 1
1
2
1
2
1
 2n 1  1 
For Cn , centre of circle is  n  2 ,0
 2

Radius of circle 
1
2n 1
Plotting circles,
An
Cn
Finding bigger x intercept of Cn
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2n 1  1
1
2n  2  1 2n  1
 n  2  n 1 
 n 1
2
2
2n 1
2
2021-Jee-Advanced
Question Paper-2_Key & Solutions
For Cn to be inside M for r 
1025 210  1

513 29  1
2n  1 210  1
 9
2n 1
2 1
1
1
1
210  1
 n 1  9
 2  n 1  9
2
2 1
2
2 1
n 1
9
 2  2  1  n  1  9  n  10
Hence no. Of circles possible = 10
For non intersecting pair of circles we need to choose alternating circles. Hence
maximum 5 circles can be chosen.
So. k = 10, l =5
3k + 2l = 40
 2n 1  1 2n 1  1
For D n centre is  n  2 , n 2 
2
 2

1
2n 1
Max distance of a point on D n from origin.
Radius 
= (Distance of centre from origin) + (radius)


2n  2 2  1
2n 1  1
1
 n 2
2  n 1 
2n 1
2
2
 2199  1
So for D n to be inside M with r   198  2
 2

2
n
2
2


2 1
2n 1
n 1
2

199
 2 2
1
n

1
2
2198
2
n 2

2 1
n 1
2

2
2
Hence 199 circles are possible
199

1
2
198
2
 n  200
Paragraph-II:
Let 1 :[0, )  ,  2 :[0, )  ,f :[0, )   and g :[0, )   be functions such
that f  0  g  0  0 ,
1  x   e x  x, x  0,
 2  x   x 2  2x  2e  x  2, x  0,
f  x 
x
  t  t e
2
 t2
dt, x  0
x
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and
2021-Jee-Advanced
g  x 
x
Question Paper-2_Key & Solutions
2

t e  t dt, x  0.
0
15.
Which of the following statements is TRUE?
1
A) f ln 3  g ln 3 
3
B) For every x  1, there exists an  1, x  such that 1  x   1  x

 

C) For every x > 0, there exists a   0, x  such that  2  x   2x  1    1
 3
D) f is an increasing function on the interval 0, 
 2
Ans: C
Sol:
1  x   e  x  x, x  0,
 2  x   x 2  2x  2e  x  2, x  0
f  x 
x
  t  t e
2
 t2
dt, x  0
x
x


x
 f  x   2 t  t e dt, x  0
2
0
g  x 
x


 2 t  t 2 e t dt ......1
 t2
2
0
2

t.e  t dt, x  0
0
x
tz
  z.e z .2z.dz
2
2
0
x
 2 z .e
2
0
 z2
x

2
0

f 1  x   2 x  x 2 e x
 2x 1  x  e  x
..... 2
 2 t 2 .e t .dt
.dz
2
2
f  for x  0,1
f  for x 1,   (Option D is wrong)
x
 f  x   g  x   2 t.e t dt
2
0
2

2

 f  x   g  x   1  e x
2
2
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x
   e  t    e  x   1  1  e  x
0
2021-Jee-Advanced

Question Paper-2_Key & Solutions
 

1 2
ln 3  1  
3 3
(Option A wrong)
1  x   e  x  x, x  0
f
ln 3  g
11  x   1  e  x x  0  1  x  is increasing
  0  1
11  0  0
1''  x   e  x  0
  1  x  is concave up
Method-I:
 2  x   x 2  2x  2e  x  2, x  0
 2  0  0
12  x   2x  2  2.e x

 2 x  1  e x



 ''2  x   2 1  e  x  0 (concave up)
 2  x   2  1    1 x
 2  x   mx  m  0
Option C correct
y  2  x 
y  mx
0
x

Method-2:
LMVT
  x    2  0
  x  0
 '2    2
 2  1    1  2
x0
x0
Which of the following statements is TRUE?
A) 1  x   1, for all x > 0
  2  x   2  1    1
B)  2  x   0, for all x > 0
2
2
2
 1
C) f  x   1  e x  x 3  x 5 , for all x  0, 
 2
3
5
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16.

2021-Jee-Advanced
Question Paper-2_Key & Solutions
2
2
1
 1
D) g  x   x 3  x 5  x 7 , for all x  0, 
 2
3
3
7
Ans: D
x
t4 
2
2
Sol: g  x   2 t 1  t  ..... dt
2! 

0
x
 2 4 t6 
 2  t  t  ..... dt
2! 

0
x
 t3 t5

t7
 2  
.... dt
 3 5 7.2!  0
 x3 x5 x7
 2x 3 2 5 x 7
...... 
 2  
 x  ......
5 7.2! 
3
5
7
3
(0,1)
2 3 2 5 x7
 x  x 
3
5
7
(0,0)
1  x   e  x  x, x  0
1  x   1



Option A wrong
 2  x   x 2  2x  2e  x  2, x  0
2  x  0
 Option B wrong also Option C wrong
SECTION-4(Maximum Marks: 12)
Non-Negative Integer Answer Type
This section contains THREE (03) questions.
The answer to each question is a NON-NEGATIVE INTEGER.
For each question, enter the correct integer corresponding to the answer 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
: +4 If ONLY the correct integer is entered;
Zero Marks
: 0 In all other cases.
17.
A number is chosen at random from the set 1, 2,3,....., 2000 . Let p be the probability
that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is
_______.
Ans: 214
Sol: 1, 2,3,......, 2000
E1  Event that it is a multiple of 3
E 2  Event that it is a multiple of 3
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 P  E1  E 2   P  E1   P  E 2   P  E1  E 2 
2021-Jee-Advanced

18.
Question Paper-2_Key & Solutions
856 856
 GE  500 

 214
2000
4
666  285  95 856

2000
2000
x 2 y2
Let E be the ellipse

 1 . For any three distinct points P,Q and Q' on E, let
16 9
M(P,Q) be the mid-point of the line segment joining P and Q, and M  P,Q ' be the
mid-point of the line segment joining P and Q' . Then the maximum possible value of
the distance between M  P,Q and M  P,Q ' , as P,Q and Q' vary on E, is ______
Ans: 4
Sol: Maximum chord = 2a = 8
P


Q




Q
Required distance between
1
M  P,Q ,M  P,Q'   8  4
2
19.
For any real number x, let [x] denote the largest integer less than or equal to x. If
10
 10x 
I
 dx , then the value of 9I is ________.
x 1
0 
Ans: 182
10x
100
  ' x   0   
  0  0,  10 
 9.01
x 1
11
10x
10x
 0,3.01
 1  or  4  or  9
x 1
x 1
9x  1
10x  4x  4
10x  9x  9
1
2
x
x
x9
9
3
Sol:   x  
GI 
2/3
9
10
0
1/9
2/3
9
 0.dx   1.dx   2.dx   3.dx
2
 2 1

 0      2  9    310  9
 3 9

3
5 50
5  150  27 182

3

 9I  182
9 3
9
9
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