JEE Main 2022 Question Paper 28 June Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


 1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 28
th
 June, 2022)   TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Given below are two statements : One is labelled
as Assertion A and the other is labelled as Reason
R.
Assertion A : Product of Pressure (P) and time (t)
has the same dimension as that of coefficient of
viscosity.
Reason R:
Coefficient of viscosity = 
Force
Velocity gradient
Question: Choose the correct answer from the 
options given below :
(A) Both A and R true, and R is correct
explanation of A. 
(B) Both A and R are true but R is NOT the correct
explanation of A. 
(C) A is true but R is false.
(D) A is false but R is true.
Official Ans. by NTA (C) 
 
Sol. Pressure and time 
2
N
P : ,Time : Sec
m
2
N sec
Pt
m
?
2
F N Nsec
::
6 rv m.m / sec m
??
?
2. A particle of mass m is moving in a circular path
of constant radius r such that its centripetal
acceleration (a) is varying with time t as a = k
2
rt
2
.
where k is a constant. The power delivered to the
particle by the force acting on it is given as
(A) zero
(B) mk
2
r
2
t
2
(C) mk
2
r
2
t
(D) mk
2
rt
Official Ans. by NTA (C) 
 
Sol. 
2
22
V
a k rt
r
?? 
V krt ?
t
dv
a kr
dt
??
tt
F ma mkr ??
P F.V ?
F cos V ??
t
FV ? ? ? mkr krt ?
22
P mk r t ?
3. Motion of a particle in x-y plane is described by a
set of following equations x 4sin t
2
???
? ? ?
??
??
m and 
? ? y 4sin t m. ?? The path of particle will be – 
(A) circular
(B) helical
(C) parabolic
(D) elliptical
Official Ans. by NTA (A) 
 
Sol. x 4sin t
2
???
? ? ?
??
??
? ? y 4cos t ??
? ? x 4cos t ?? ? ? y 4sin t ??
Eliminate ‘t’ to find relation between x and y 
2 2 2 2 2 2 2
x y y cos t y sin t 4 ? ? ? ? ? ?
2 2 2
x y 4 ??
Page 2


 1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 28
th
 June, 2022)   TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Given below are two statements : One is labelled
as Assertion A and the other is labelled as Reason
R.
Assertion A : Product of Pressure (P) and time (t)
has the same dimension as that of coefficient of
viscosity.
Reason R:
Coefficient of viscosity = 
Force
Velocity gradient
Question: Choose the correct answer from the 
options given below :
(A) Both A and R true, and R is correct
explanation of A. 
(B) Both A and R are true but R is NOT the correct
explanation of A. 
(C) A is true but R is false.
(D) A is false but R is true.
Official Ans. by NTA (C) 
 
Sol. Pressure and time 
2
N
P : ,Time : Sec
m
2
N sec
Pt
m
?
2
F N Nsec
::
6 rv m.m / sec m
??
?
2. A particle of mass m is moving in a circular path
of constant radius r such that its centripetal
acceleration (a) is varying with time t as a = k
2
rt
2
.
where k is a constant. The power delivered to the
particle by the force acting on it is given as
(A) zero
(B) mk
2
r
2
t
2
(C) mk
2
r
2
t
(D) mk
2
rt
Official Ans. by NTA (C) 
 
Sol. 
2
22
V
a k rt
r
?? 
V krt ?
t
dv
a kr
dt
??
tt
F ma mkr ??
P F.V ?
F cos V ??
t
FV ? ? ? mkr krt ?
22
P mk r t ?
3. Motion of a particle in x-y plane is described by a
set of following equations x 4sin t
2
???
? ? ?
??
??
m and 
? ? y 4sin t m. ?? The path of particle will be – 
(A) circular
(B) helical
(C) parabolic
(D) elliptical
Official Ans. by NTA (A) 
 
Sol. x 4sin t
2
???
? ? ?
??
??
? ? y 4cos t ??
? ? x 4cos t ?? ? ? y 4sin t ??
Eliminate ‘t’ to find relation between x and y 
2 2 2 2 2 2 2
x y y cos t y sin t 4 ? ? ? ? ? ?
2 2 2
x y 4 ??
2 
4. Match List-I with List-II
List-I List-II 
A 
Moment of inertia of 
solid sphere of radius R 
about any tangent 
I 
2
5
MR
3
B 
Moment of inertia of 
hollow sphere of radius 
(R) about any tangent
II 
2
7
MR
5
C 
Moment of inertia of 
circular ring of radius 
(R) about its diameter.
III 
2
1
MR
4
D 
Moment of inertia of 
circular disc of radius 
(R) about any diameter.
IV 
2
1
MR
2
Question: Choose the correct answer from the 
options given below 
(A) A-II, B-II, C-IV, D-III
(B) A-I, B-II, C-IV, D-III
(C) A-II, B-I, C-III, D-IV
(D) A-I, B-II, C-III, D-IV
Official Ans. by NTA (A) 
 
Sol. Solid sphere 
R
I
0
I
com
2
0 com
I I MR ?? (Parallel Axis theorem) 
22
0
2
0
2
I MR MR
5
7
I MR
5
??
?
Hollow sphere 
R
I
0
I
com
2
0 com
I I MR ??
2 2 2
25
MR MR MR
33
? ? ?
I
2
I
3
I
1
×
1 2 3
I I I ?? (Perpendicular axis theorem) 
By symmetry MOI 
About 1” and 2” Axis are same i.e. 
12
II ?
?
2
13
2I I MR ??
? ?
2
com
I MR ?
2
1
MR
I
2
?
Similarly in disc 
22
1 com
MR MR
2I I
22
??
??
??
??
??
??
2
1
MR
I
4
?
Page 3


 1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 28
th
 June, 2022)   TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Given below are two statements : One is labelled
as Assertion A and the other is labelled as Reason
R.
Assertion A : Product of Pressure (P) and time (t)
has the same dimension as that of coefficient of
viscosity.
Reason R:
Coefficient of viscosity = 
Force
Velocity gradient
Question: Choose the correct answer from the 
options given below :
(A) Both A and R true, and R is correct
explanation of A. 
(B) Both A and R are true but R is NOT the correct
explanation of A. 
(C) A is true but R is false.
(D) A is false but R is true.
Official Ans. by NTA (C) 
 
Sol. Pressure and time 
2
N
P : ,Time : Sec
m
2
N sec
Pt
m
?
2
F N Nsec
::
6 rv m.m / sec m
??
?
2. A particle of mass m is moving in a circular path
of constant radius r such that its centripetal
acceleration (a) is varying with time t as a = k
2
rt
2
.
where k is a constant. The power delivered to the
particle by the force acting on it is given as
(A) zero
(B) mk
2
r
2
t
2
(C) mk
2
r
2
t
(D) mk
2
rt
Official Ans. by NTA (C) 
 
Sol. 
2
22
V
a k rt
r
?? 
V krt ?
t
dv
a kr
dt
??
tt
F ma mkr ??
P F.V ?
F cos V ??
t
FV ? ? ? mkr krt ?
22
P mk r t ?
3. Motion of a particle in x-y plane is described by a
set of following equations x 4sin t
2
???
? ? ?
??
??
m and 
? ? y 4sin t m. ?? The path of particle will be – 
(A) circular
(B) helical
(C) parabolic
(D) elliptical
Official Ans. by NTA (A) 
 
Sol. x 4sin t
2
???
? ? ?
??
??
? ? y 4cos t ??
? ? x 4cos t ?? ? ? y 4sin t ??
Eliminate ‘t’ to find relation between x and y 
2 2 2 2 2 2 2
x y y cos t y sin t 4 ? ? ? ? ? ?
2 2 2
x y 4 ??
2 
4. Match List-I with List-II
List-I List-II 
A 
Moment of inertia of 
solid sphere of radius R 
about any tangent 
I 
2
5
MR
3
B 
Moment of inertia of 
hollow sphere of radius 
(R) about any tangent
II 
2
7
MR
5
C 
Moment of inertia of 
circular ring of radius 
(R) about its diameter.
III 
2
1
MR
4
D 
Moment of inertia of 
circular disc of radius 
(R) about any diameter.
IV 
2
1
MR
2
Question: Choose the correct answer from the 
options given below 
(A) A-II, B-II, C-IV, D-III
(B) A-I, B-II, C-IV, D-III
(C) A-II, B-I, C-III, D-IV
(D) A-I, B-II, C-III, D-IV
Official Ans. by NTA (A) 
 
Sol. Solid sphere 
R
I
0
I
com
2
0 com
I I MR ?? (Parallel Axis theorem) 
22
0
2
0
2
I MR MR
5
7
I MR
5
??
?
Hollow sphere 
R
I
0
I
com
2
0 com
I I MR ??
2 2 2
25
MR MR MR
33
? ? ?
I
2
I
3
I
1
×
1 2 3
I I I ?? (Perpendicular axis theorem) 
By symmetry MOI 
About 1” and 2” Axis are same i.e. 
12
II ?
?
2
13
2I I MR ??
? ?
2
com
I MR ?
2
1
MR
I
2
?
Similarly in disc 
22
1 com
MR MR
2I I
22
??
??
??
??
??
??
2
1
MR
I
4
?
3 
5. Two planets A and B of equal mass are having
their period of revolutions T
A
 and T
B
 such that
T
A
 = 2T
B
. These planets are revolving in the
circular orbits of radii r
A
 and r
B
 respectively.
Which out of the following would be the correct
relationship of their orbits?
(A) 
22
AB
2r r ?
(B) 
33
AB
r 2r ?
(C) 
33
AB
r 3r ?
(D) 
? ?
2
2 2 3 3
A B B A
T T r 4r
GM
?
? ? ?
Official Ans. by NTA (C) 
 
Sol. 
3
2
a
2
Tr
Gm
?
?
23
Tr ?
23
AA
BB
Tr
Tr
? ? ? ?
?
? ? ? ?
? ? ? ?
3
2
A
B
r 2
1r
??
??
??
??
??
??
??
 ? 
33
AB
r 4r ?
6. A water drop of diameter  cm is broken into
64 equal droplets. The surface tension of water is 
0.075 N/m. In this process the gain in surface 
energy will be : 
(A) 
4
2.8 10 J
?
? (B) 
3
1.5 10 J
?
?
(C) 
4
1.9 10 J
?
? (D) 
5
9.4 10 J
?
?
Official Ans. by NTA (A) 
 
Sol. d = 2cm; r = 1 cm; T = 0.075 
SE T ? ? ? ?
? ?
f1
0.075 A A ??
2
i
A 4 r ??
2
f0
A 4 r 64 ? ? ?
By volume conservation 
33
0
44
r 64 r
33
? ? ? ?
0
r
r
4
?
2
2
f
r
A 4 64 16 r
4
??
? ? ? ? ?
??
??
? ?
22
SE 0.075 16 r 4 r ? ? ? ? ?
? ?
? ?
2
0.075 12 0.01 ??
4
2.8 10 J
?
??
7. Given below are two statement :
Statement – I : What µ amount of an ideal 
gas undergoes adiabatic change from state 
? ?
1 1 1
P ,V ,T to state ? ?
2 2 2
P ,V ,T , the work done  
is 
? ?
21
1R T T
W
1
?
?
??
, where 
P
V
C
C
?? and 
R = universal gas constant,  
Statement — II: In the above case. when work is 
done on the gas. the temperature of the gas would 
rise. 
Choose the correct answer from the options given 
below: 
(A) Both statement —I and statement-II are true.
(B) Both statement —I and statement-II are false.
(C) Statement-I is true but statement-II is false.
(D) Statement-I is false but statement-II is true.
Official Ans. by NTA (A) 
 
Page 4


 1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 28
th
 June, 2022)   TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Given below are two statements : One is labelled
as Assertion A and the other is labelled as Reason
R.
Assertion A : Product of Pressure (P) and time (t)
has the same dimension as that of coefficient of
viscosity.
Reason R:
Coefficient of viscosity = 
Force
Velocity gradient
Question: Choose the correct answer from the 
options given below :
(A) Both A and R true, and R is correct
explanation of A. 
(B) Both A and R are true but R is NOT the correct
explanation of A. 
(C) A is true but R is false.
(D) A is false but R is true.
Official Ans. by NTA (C) 
 
Sol. Pressure and time 
2
N
P : ,Time : Sec
m
2
N sec
Pt
m
?
2
F N Nsec
::
6 rv m.m / sec m
??
?
2. A particle of mass m is moving in a circular path
of constant radius r such that its centripetal
acceleration (a) is varying with time t as a = k
2
rt
2
.
where k is a constant. The power delivered to the
particle by the force acting on it is given as
(A) zero
(B) mk
2
r
2
t
2
(C) mk
2
r
2
t
(D) mk
2
rt
Official Ans. by NTA (C) 
 
Sol. 
2
22
V
a k rt
r
?? 
V krt ?
t
dv
a kr
dt
??
tt
F ma mkr ??
P F.V ?
F cos V ??
t
FV ? ? ? mkr krt ?
22
P mk r t ?
3. Motion of a particle in x-y plane is described by a
set of following equations x 4sin t
2
???
? ? ?
??
??
m and 
? ? y 4sin t m. ?? The path of particle will be – 
(A) circular
(B) helical
(C) parabolic
(D) elliptical
Official Ans. by NTA (A) 
 
Sol. x 4sin t
2
???
? ? ?
??
??
? ? y 4cos t ??
? ? x 4cos t ?? ? ? y 4sin t ??
Eliminate ‘t’ to find relation between x and y 
2 2 2 2 2 2 2
x y y cos t y sin t 4 ? ? ? ? ? ?
2 2 2
x y 4 ??
2 
4. Match List-I with List-II
List-I List-II 
A 
Moment of inertia of 
solid sphere of radius R 
about any tangent 
I 
2
5
MR
3
B 
Moment of inertia of 
hollow sphere of radius 
(R) about any tangent
II 
2
7
MR
5
C 
Moment of inertia of 
circular ring of radius 
(R) about its diameter.
III 
2
1
MR
4
D 
Moment of inertia of 
circular disc of radius 
(R) about any diameter.
IV 
2
1
MR
2
Question: Choose the correct answer from the 
options given below 
(A) A-II, B-II, C-IV, D-III
(B) A-I, B-II, C-IV, D-III
(C) A-II, B-I, C-III, D-IV
(D) A-I, B-II, C-III, D-IV
Official Ans. by NTA (A) 
 
Sol. Solid sphere 
R
I
0
I
com
2
0 com
I I MR ?? (Parallel Axis theorem) 
22
0
2
0
2
I MR MR
5
7
I MR
5
??
?
Hollow sphere 
R
I
0
I
com
2
0 com
I I MR ??
2 2 2
25
MR MR MR
33
? ? ?
I
2
I
3
I
1
×
1 2 3
I I I ?? (Perpendicular axis theorem) 
By symmetry MOI 
About 1” and 2” Axis are same i.e. 
12
II ?
?
2
13
2I I MR ??
? ?
2
com
I MR ?
2
1
MR
I
2
?
Similarly in disc 
22
1 com
MR MR
2I I
22
??
??
??
??
??
??
2
1
MR
I
4
?
3 
5. Two planets A and B of equal mass are having
their period of revolutions T
A
 and T
B
 such that
T
A
 = 2T
B
. These planets are revolving in the
circular orbits of radii r
A
 and r
B
 respectively.
Which out of the following would be the correct
relationship of their orbits?
(A) 
22
AB
2r r ?
(B) 
33
AB
r 2r ?
(C) 
33
AB
r 3r ?
(D) 
? ?
2
2 2 3 3
A B B A
T T r 4r
GM
?
? ? ?
Official Ans. by NTA (C) 
 
Sol. 
3
2
a
2
Tr
Gm
?
?
23
Tr ?
23
AA
BB
Tr
Tr
? ? ? ?
?
? ? ? ?
? ? ? ?
3
2
A
B
r 2
1r
??
??
??
??
??
??
??
 ? 
33
AB
r 4r ?
6. A water drop of diameter  cm is broken into
64 equal droplets. The surface tension of water is 
0.075 N/m. In this process the gain in surface 
energy will be : 
(A) 
4
2.8 10 J
?
? (B) 
3
1.5 10 J
?
?
(C) 
4
1.9 10 J
?
? (D) 
5
9.4 10 J
?
?
Official Ans. by NTA (A) 
 
Sol. d = 2cm; r = 1 cm; T = 0.075 
SE T ? ? ? ?
? ?
f1
0.075 A A ??
2
i
A 4 r ??
2
f0
A 4 r 64 ? ? ?
By volume conservation 
33
0
44
r 64 r
33
? ? ? ?
0
r
r
4
?
2
2
f
r
A 4 64 16 r
4
??
? ? ? ? ?
??
??
? ?
22
SE 0.075 16 r 4 r ? ? ? ? ?
? ?
? ?
2
0.075 12 0.01 ??
4
2.8 10 J
?
??
7. Given below are two statement :
Statement – I : What µ amount of an ideal 
gas undergoes adiabatic change from state 
? ?
1 1 1
P ,V ,T to state ? ?
2 2 2
P ,V ,T , the work done  
is 
? ?
21
1R T T
W
1
?
?
??
, where 
P
V
C
C
?? and 
R = universal gas constant,  
Statement — II: In the above case. when work is 
done on the gas. the temperature of the gas would 
rise. 
Choose the correct answer from the options given 
below: 
(A) Both statement —I and statement-II are true.
(B) Both statement —I and statement-II are false.
(C) Statement-I is true but statement-II is false.
(D) Statement-I is false but statement-II is true.
Official Ans. by NTA (A) 
 
4 
Sol. 
? ?
fi
adiabatic
NR T T
W statment 1
1
?
??
??
Q W U ? ? ?
0 W U ? ? ?
UW ? ? ?
If work is done on the gas, i.e. work is negative 
U ?? is positive.
? Temperature will increase.
8. Given below are two statements :
Statement-I : A point charge is brought in an
electric field. The value of electric field at a point
near to the charge may increase if the charge is
positive.
Statement-II : An electric dipole is placed in a
non-uniform electric field. The net electric force on
the dipole will not be zero.
Choose the correct answer from the options given
below :
(A) Both statement-I and statement-II are true.
(B) Both statement-I and statement-I are false.
(C) Statement-I is true but statement-II is false.
(D) Statement-I is false but statement-II is true.
Official Ans. by NTA (A) 
 
Sol. If the electric field is in the positive direction and 
the positive charge is to the left of that point then 
the electric field will increase. But to the left of 
the positive charge the electric field would 
decrease.  
If the dipole is kept at the point where the electric 
field is maximum then the force on it will be 
zero. 
9. The three charges q/2, q and q/2 are placed at the
corners A, B and C of a square of side 'a' as shown
in figure. The magnitude of electric field (E) at the
comer D of the square, is :
(A) 
2
0
q 1 1
2 4a 2
??
?
??
??
??
(B) 
2
0
q1
1
4a 2
??
?
??
??
??
(C) 
2
0
q1
1
4a 2
??
?
??
??
??
(D) 
2
0
q 1 1
2 4a 2
??
?
??
??
??
Official Ans. by NTA (A) 
 
Sol. 
q/2
q/2
D
K(q/2)/a
2
Kq/2/a
2
Kq/( 2a)
2
45°
D
Kq/2a
2
Kq/2a
2
Kq/2a
2
45°
? ?
net
22 D
kq 2kq
E
2a 2a
??
? ?
net
2 D
kq 1 1
E
2 a 2
??
??
??
??
? ?
net
2 D
0
q 1 1
E
2 4a 2
??
??
??
??
??
Page 5


 1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 28
th
 June, 2022)   TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. Given below are two statements : One is labelled
as Assertion A and the other is labelled as Reason
R.
Assertion A : Product of Pressure (P) and time (t)
has the same dimension as that of coefficient of
viscosity.
Reason R:
Coefficient of viscosity = 
Force
Velocity gradient
Question: Choose the correct answer from the 
options given below :
(A) Both A and R true, and R is correct
explanation of A. 
(B) Both A and R are true but R is NOT the correct
explanation of A. 
(C) A is true but R is false.
(D) A is false but R is true.
Official Ans. by NTA (C) 
 
Sol. Pressure and time 
2
N
P : ,Time : Sec
m
2
N sec
Pt
m
?
2
F N Nsec
::
6 rv m.m / sec m
??
?
2. A particle of mass m is moving in a circular path
of constant radius r such that its centripetal
acceleration (a) is varying with time t as a = k
2
rt
2
.
where k is a constant. The power delivered to the
particle by the force acting on it is given as
(A) zero
(B) mk
2
r
2
t
2
(C) mk
2
r
2
t
(D) mk
2
rt
Official Ans. by NTA (C) 
 
Sol. 
2
22
V
a k rt
r
?? 
V krt ?
t
dv
a kr
dt
??
tt
F ma mkr ??
P F.V ?
F cos V ??
t
FV ? ? ? mkr krt ?
22
P mk r t ?
3. Motion of a particle in x-y plane is described by a
set of following equations x 4sin t
2
???
? ? ?
??
??
m and 
? ? y 4sin t m. ?? The path of particle will be – 
(A) circular
(B) helical
(C) parabolic
(D) elliptical
Official Ans. by NTA (A) 
 
Sol. x 4sin t
2
???
? ? ?
??
??
? ? y 4cos t ??
? ? x 4cos t ?? ? ? y 4sin t ??
Eliminate ‘t’ to find relation between x and y 
2 2 2 2 2 2 2
x y y cos t y sin t 4 ? ? ? ? ? ?
2 2 2
x y 4 ??
2 
4. Match List-I with List-II
List-I List-II 
A 
Moment of inertia of 
solid sphere of radius R 
about any tangent 
I 
2
5
MR
3
B 
Moment of inertia of 
hollow sphere of radius 
(R) about any tangent
II 
2
7
MR
5
C 
Moment of inertia of 
circular ring of radius 
(R) about its diameter.
III 
2
1
MR
4
D 
Moment of inertia of 
circular disc of radius 
(R) about any diameter.
IV 
2
1
MR
2
Question: Choose the correct answer from the 
options given below 
(A) A-II, B-II, C-IV, D-III
(B) A-I, B-II, C-IV, D-III
(C) A-II, B-I, C-III, D-IV
(D) A-I, B-II, C-III, D-IV
Official Ans. by NTA (A) 
 
Sol. Solid sphere 
R
I
0
I
com
2
0 com
I I MR ?? (Parallel Axis theorem) 
22
0
2
0
2
I MR MR
5
7
I MR
5
??
?
Hollow sphere 
R
I
0
I
com
2
0 com
I I MR ??
2 2 2
25
MR MR MR
33
? ? ?
I
2
I
3
I
1
×
1 2 3
I I I ?? (Perpendicular axis theorem) 
By symmetry MOI 
About 1” and 2” Axis are same i.e. 
12
II ?
?
2
13
2I I MR ??
? ?
2
com
I MR ?
2
1
MR
I
2
?
Similarly in disc 
22
1 com
MR MR
2I I
22
??
??
??
??
??
??
2
1
MR
I
4
?
3 
5. Two planets A and B of equal mass are having
their period of revolutions T
A
 and T
B
 such that
T
A
 = 2T
B
. These planets are revolving in the
circular orbits of radii r
A
 and r
B
 respectively.
Which out of the following would be the correct
relationship of their orbits?
(A) 
22
AB
2r r ?
(B) 
33
AB
r 2r ?
(C) 
33
AB
r 3r ?
(D) 
? ?
2
2 2 3 3
A B B A
T T r 4r
GM
?
? ? ?
Official Ans. by NTA (C) 
 
Sol. 
3
2
a
2
Tr
Gm
?
?
23
Tr ?
23
AA
BB
Tr
Tr
? ? ? ?
?
? ? ? ?
? ? ? ?
3
2
A
B
r 2
1r
??
??
??
??
??
??
??
 ? 
33
AB
r 4r ?
6. A water drop of diameter  cm is broken into
64 equal droplets. The surface tension of water is 
0.075 N/m. In this process the gain in surface 
energy will be : 
(A) 
4
2.8 10 J
?
? (B) 
3
1.5 10 J
?
?
(C) 
4
1.9 10 J
?
? (D) 
5
9.4 10 J
?
?
Official Ans. by NTA (A) 
 
Sol. d = 2cm; r = 1 cm; T = 0.075 
SE T ? ? ? ?
? ?
f1
0.075 A A ??
2
i
A 4 r ??
2
f0
A 4 r 64 ? ? ?
By volume conservation 
33
0
44
r 64 r
33
? ? ? ?
0
r
r
4
?
2
2
f
r
A 4 64 16 r
4
??
? ? ? ? ?
??
??
? ?
22
SE 0.075 16 r 4 r ? ? ? ? ?
? ?
? ?
2
0.075 12 0.01 ??
4
2.8 10 J
?
??
7. Given below are two statement :
Statement – I : What µ amount of an ideal 
gas undergoes adiabatic change from state 
? ?
1 1 1
P ,V ,T to state ? ?
2 2 2
P ,V ,T , the work done  
is 
? ?
21
1R T T
W
1
?
?
??
, where 
P
V
C
C
?? and 
R = universal gas constant,  
Statement — II: In the above case. when work is 
done on the gas. the temperature of the gas would 
rise. 
Choose the correct answer from the options given 
below: 
(A) Both statement —I and statement-II are true.
(B) Both statement —I and statement-II are false.
(C) Statement-I is true but statement-II is false.
(D) Statement-I is false but statement-II is true.
Official Ans. by NTA (A) 
 
4 
Sol. 
? ?
fi
adiabatic
NR T T
W statment 1
1
?
??
??
Q W U ? ? ?
0 W U ? ? ?
UW ? ? ?
If work is done on the gas, i.e. work is negative 
U ?? is positive.
? Temperature will increase.
8. Given below are two statements :
Statement-I : A point charge is brought in an
electric field. The value of electric field at a point
near to the charge may increase if the charge is
positive.
Statement-II : An electric dipole is placed in a
non-uniform electric field. The net electric force on
the dipole will not be zero.
Choose the correct answer from the options given
below :
(A) Both statement-I and statement-II are true.
(B) Both statement-I and statement-I are false.
(C) Statement-I is true but statement-II is false.
(D) Statement-I is false but statement-II is true.
Official Ans. by NTA (A) 
 
Sol. If the electric field is in the positive direction and 
the positive charge is to the left of that point then 
the electric field will increase. But to the left of 
the positive charge the electric field would 
decrease.  
If the dipole is kept at the point where the electric 
field is maximum then the force on it will be 
zero. 
9. The three charges q/2, q and q/2 are placed at the
corners A, B and C of a square of side 'a' as shown
in figure. The magnitude of electric field (E) at the
comer D of the square, is :
(A) 
2
0
q 1 1
2 4a 2
??
?
??
??
??
(B) 
2
0
q1
1
4a 2
??
?
??
??
??
(C) 
2
0
q1
1
4a 2
??
?
??
??
??
(D) 
2
0
q 1 1
2 4a 2
??
?
??
??
??
Official Ans. by NTA (A) 
 
Sol. 
q/2
q/2
D
K(q/2)/a
2
Kq/2/a
2
Kq/( 2a)
2
45°
D
Kq/2a
2
Kq/2a
2
Kq/2a
2
45°
? ?
net
22 D
kq 2kq
E
2a 2a
??
? ?
net
2 D
kq 1 1
E
2 a 2
??
??
??
??
? ?
net
2 D
0
q 1 1
E
2 4a 2
??
??
??
??
??
5 
10. An infinitely long hollow conducting cylinder with 
radius R carries a uniform current along its surface. 
Choose the correct representation of magnetic field 
(B) as a function of radial distance (r) from the axis
of cylinder. 
(A) 
(B) 
(C) 
(D) 
Official Ans. by NTA (D) 
 
Sol. 
r
P Ampere
 loop
R
1) r < R, B
p
 = 0
2) r > R, B
p
 =
0
µl
2r ?
p
1
B
r
?
11. A radar sends an electromagnetic signal of electric
field (E
0
) = 2.25 V/m and magnetic field
(B
0
) = 1.5 x 10
-8
 T which strikes a target on line of
sight at a distance of 3 km in a medium. After that,
a pail of signal (echo) reflects back towards the
radar vit1i same velocity and by same path. If the
signal was transmitted at time t
0
 from radar. then
after how much time echo will reach to the radar?
(A) 
5
2.0 10 s
?
?
(B) 
5
4.0 10 s
?
?
(C) 
5
1.0 10 s
?
?
(D) 
5
8.0 10 s
?
?
Official Ans. by NTA (B) 
 
Sol.
81 0
8
0
E 2.25
C 1.5 10 ms
B 1.5 10
?
?
? ? ? ?
?
3
5
8
6 10
t 4 10 s
1.5 10
?
?
? ? ?
?
12. The refracting angle of a prism is A and refractive
index of the material of the prism is cot (A/2).
Then the angle of minimum deviation will be -
(A) 180 – 2A (B) 90 – A
(C) 180 + 2A (D) 180 – 3A
Official Ans. by NTA (A) 
 
Sol. 
m
A
sin
2
µ
A
sin
2
?? ??
??
??
?
A
µ cot
2
?
?
m
A A
sin cos
22
?? ??
?
??
??
m
180 2A ? ? ?