JEE Advance 2023 Question Paper - 2 with Solutions | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


 
- 1 - 
 
 
Answers & Solutions 
for 
JEE (Advanced)-2023 (Paper-2) 
  
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) 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. 
1. An electric dipole is formed by two charges +q and –q located in xy-plane at (0, 2) mm and (0, -2) mm, 
respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V
0
. The 
charges +q and –q are then moved to the points (-1, 2) mm and (1, -2) mm, respectively. What is the value of 
electric potential at P due to the new dipole? 
Date: 04/06/2023 
Time : 3 hrs. Max. Marks: 180 
PART-I : PHYSICS 
Page 2


 
- 1 - 
 
 
Answers & Solutions 
for 
JEE (Advanced)-2023 (Paper-2) 
  
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) 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. 
1. An electric dipole is formed by two charges +q and –q located in xy-plane at (0, 2) mm and (0, -2) mm, 
respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V
0
. The 
charges +q and –q are then moved to the points (-1, 2) mm and (1, -2) mm, respectively. What is the value of 
electric potential at P due to the new dipole? 
Date: 04/06/2023 
Time : 3 hrs. Max. Marks: 180 
PART-I : PHYSICS 
 
 
   
  
 
 (A) 
0
4
V
 (B) 
0
2
V
 
 (C) 
0
2
V
 (D) 
0
3
4
V
 
Answer (B) 
Sol. 
11
1
3
1
cos p
V
r
?
? 
  
 
22
2
3
2
cos p
V
r
?
? 
  
 
2 2 2
1 1 1
cos
cos
Vp
Vp
?
=
?
 
 
( ) ( )
2
1
ˆ ˆ ˆ ˆ
(–2 4 ) ·( ) 1
ˆ ˆ ˆ ˆ 2
0 4 ·
V q i j i j
V
q i j i j
++
==
++
 
 ? 
0
2
2
V
V = 
Page 3


 
- 1 - 
 
 
Answers & Solutions 
for 
JEE (Advanced)-2023 (Paper-2) 
  
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) 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. 
1. An electric dipole is formed by two charges +q and –q located in xy-plane at (0, 2) mm and (0, -2) mm, 
respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V
0
. The 
charges +q and –q are then moved to the points (-1, 2) mm and (1, -2) mm, respectively. What is the value of 
electric potential at P due to the new dipole? 
Date: 04/06/2023 
Time : 3 hrs. Max. Marks: 180 
PART-I : PHYSICS 
 
 
   
  
 
 (A) 
0
4
V
 (B) 
0
2
V
 
 (C) 
0
2
V
 (D) 
0
3
4
V
 
Answer (B) 
Sol. 
11
1
3
1
cos p
V
r
?
? 
  
 
22
2
3
2
cos p
V
r
?
? 
  
 
2 2 2
1 1 1
cos
cos
Vp
Vp
?
=
?
 
 
( ) ( )
2
1
ˆ ˆ ˆ ˆ
(–2 4 ) ·( ) 1
ˆ ˆ ˆ ˆ 2
0 4 ·
V q i j i j
V
q i j i j
++
==
++
 
 ? 
0
2
2
V
V = 
   
 
 
   
  
2. Young’s modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck’s constant h and the speed of light c, as Y = c
?
h
?
G
?
. Which of the following is the correct 
option? 
 (A) ? = 7, ? = –1, ? = –2 (B) ? = –7, ? = –1, ? = –2 
 (C) ? = 7, ? = –1, ? = 2 (D) ? = –7, ? = 1, ? = –2 
Answer (A) 
Sol. Y = c
?
h
?
G
?
 
 [M
1
L
–1
T
–2
] = [M
0
L
1
T
–1
]
?
 [M
1
L
2
T
–1
]
?
 [M
–1
L
3
T
–2
]
?
 
 1 = ? – ? 
 –1 = ? + 2? + 3? 
 –2 = – ? – ? – 2? 
 Solving 
 ? = 7, ? = –1, ? = –2 
3. A particle of mass m is moving in the xy-plane such that its velocity at a point (x, y) is given as 
ˆˆ ( 2 ), v yx xy = ? + 
where ? is a non-zero constant. What is the force F acting on the particle? 
 (A) ( )
2
ˆˆ 2 F m xx yy = ? + (B) ( )
2
ˆˆ 2 F m yx xy = ? + 
 (C) ( )
2
ˆˆ 2 F m yx xy = ? + (D) ( )
2
ˆˆ 2 F m xx yy = ? + 
Answer (A) 
Sol. 
dv
Fm
dt
= 
 ( ) ˆˆ 2 v yx xy = ? + 
 ˆˆ 2
dv dy dx
xy
dt dt dt
??
= ? +
??
??
 
  = 
( )
ˆˆ 2
yx
v x v y ?+ 
  = ? ?
ˆˆ 22 x x yy ? ? + ? 
  = ? ?
2
ˆˆ 2 xx yy ?+ 
 ? ?
2
ˆˆ 2 F m xx yy = ? + 
4. An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is n. 
The internal energy of one mole of the gas is U
n
 and the speed of sound in the gas is v
n
. At a fixed temperature 
and pressure, which of the following is the correct option? 
 (A) v
3
 < v
6
 and U
3
 > U
6
 (B) v
5
 > v
3
 and U
3
 > U
5
 
 (C) v
5
 > v
7
 and U
5
 < U
7
 (D) v
6
 < v
7
 and U
6
 < U
7
 
Answer (C) 
Page 4


 
- 1 - 
 
 
Answers & Solutions 
for 
JEE (Advanced)-2023 (Paper-2) 
  
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) 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. 
1. An electric dipole is formed by two charges +q and –q located in xy-plane at (0, 2) mm and (0, -2) mm, 
respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V
0
. The 
charges +q and –q are then moved to the points (-1, 2) mm and (1, -2) mm, respectively. What is the value of 
electric potential at P due to the new dipole? 
Date: 04/06/2023 
Time : 3 hrs. Max. Marks: 180 
PART-I : PHYSICS 
 
 
   
  
 
 (A) 
0
4
V
 (B) 
0
2
V
 
 (C) 
0
2
V
 (D) 
0
3
4
V
 
Answer (B) 
Sol. 
11
1
3
1
cos p
V
r
?
? 
  
 
22
2
3
2
cos p
V
r
?
? 
  
 
2 2 2
1 1 1
cos
cos
Vp
Vp
?
=
?
 
 
( ) ( )
2
1
ˆ ˆ ˆ ˆ
(–2 4 ) ·( ) 1
ˆ ˆ ˆ ˆ 2
0 4 ·
V q i j i j
V
q i j i j
++
==
++
 
 ? 
0
2
2
V
V = 
   
 
 
   
  
2. Young’s modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck’s constant h and the speed of light c, as Y = c
?
h
?
G
?
. Which of the following is the correct 
option? 
 (A) ? = 7, ? = –1, ? = –2 (B) ? = –7, ? = –1, ? = –2 
 (C) ? = 7, ? = –1, ? = 2 (D) ? = –7, ? = 1, ? = –2 
Answer (A) 
Sol. Y = c
?
h
?
G
?
 
 [M
1
L
–1
T
–2
] = [M
0
L
1
T
–1
]
?
 [M
1
L
2
T
–1
]
?
 [M
–1
L
3
T
–2
]
?
 
 1 = ? – ? 
 –1 = ? + 2? + 3? 
 –2 = – ? – ? – 2? 
 Solving 
 ? = 7, ? = –1, ? = –2 
3. A particle of mass m is moving in the xy-plane such that its velocity at a point (x, y) is given as 
ˆˆ ( 2 ), v yx xy = ? + 
where ? is a non-zero constant. What is the force F acting on the particle? 
 (A) ( )
2
ˆˆ 2 F m xx yy = ? + (B) ( )
2
ˆˆ 2 F m yx xy = ? + 
 (C) ( )
2
ˆˆ 2 F m yx xy = ? + (D) ( )
2
ˆˆ 2 F m xx yy = ? + 
Answer (A) 
Sol. 
dv
Fm
dt
= 
 ( ) ˆˆ 2 v yx xy = ? + 
 ˆˆ 2
dv dy dx
xy
dt dt dt
??
= ? +
??
??
 
  = 
( )
ˆˆ 2
yx
v x v y ?+ 
  = ? ?
ˆˆ 22 x x yy ? ? + ? 
  = ? ?
2
ˆˆ 2 xx yy ?+ 
 ? ?
2
ˆˆ 2 F m xx yy = ? + 
4. An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is n. 
The internal energy of one mole of the gas is U
n
 and the speed of sound in the gas is v
n
. At a fixed temperature 
and pressure, which of the following is the correct option? 
 (A) v
3
 < v
6
 and U
3
 > U
6
 (B) v
5
 > v
3
 and U
3
 > U
5
 
 (C) v
5
 > v
7
 and U
5
 < U
7
 (D) v
6
 < v
7
 and U
6
 < U
7
 
Answer (C) 
 
 
   
  
Sol. 
1
22
n
n RT nRT
U
??
== 
 
2
1
n
RT
RT
n
v
MM
??
+
??
?
??
== 
 ? U
7
 > U
5
 and U
7
 > U
6
 
  and v
5
 > v
7
 
SECTION 2 (Maximum Marks : 12) 
• This section contains THREE (03) questions. 
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) 
correct answer(s). 
• For each question, choose the option(s) corresponding to (all) the correct answer(s). 
• Answer to each question will be evaluated according to the following marking scheme: 
 Full Marks  : +4  ONLY if (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. 
 5. A monochromatic light wave is incident normally on a glass slab of thickness d, as shown in the figure. The 
refractive index of the slab increases linearly from n1 to n2 over the height h. Which of the following statement(s) 
is(are) true about the light wave emerging out of the slab? 
 
Page 5


 
- 1 - 
 
 
Answers & Solutions 
for 
JEE (Advanced)-2023 (Paper-2) 
  
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) 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. 
1. An electric dipole is formed by two charges +q and –q located in xy-plane at (0, 2) mm and (0, -2) mm, 
respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is V
0
. The 
charges +q and –q are then moved to the points (-1, 2) mm and (1, -2) mm, respectively. What is the value of 
electric potential at P due to the new dipole? 
Date: 04/06/2023 
Time : 3 hrs. Max. Marks: 180 
PART-I : PHYSICS 
 
 
   
  
 
 (A) 
0
4
V
 (B) 
0
2
V
 
 (C) 
0
2
V
 (D) 
0
3
4
V
 
Answer (B) 
Sol. 
11
1
3
1
cos p
V
r
?
? 
  
 
22
2
3
2
cos p
V
r
?
? 
  
 
2 2 2
1 1 1
cos
cos
Vp
Vp
?
=
?
 
 
( ) ( )
2
1
ˆ ˆ ˆ ˆ
(–2 4 ) ·( ) 1
ˆ ˆ ˆ ˆ 2
0 4 ·
V q i j i j
V
q i j i j
++
==
++
 
 ? 
0
2
2
V
V = 
   
 
 
   
  
2. Young’s modulus of elasticity Y is expressed in terms of three derived quantities, namely, the gravitational 
constant G, Planck’s constant h and the speed of light c, as Y = c
?
h
?
G
?
. Which of the following is the correct 
option? 
 (A) ? = 7, ? = –1, ? = –2 (B) ? = –7, ? = –1, ? = –2 
 (C) ? = 7, ? = –1, ? = 2 (D) ? = –7, ? = 1, ? = –2 
Answer (A) 
Sol. Y = c
?
h
?
G
?
 
 [M
1
L
–1
T
–2
] = [M
0
L
1
T
–1
]
?
 [M
1
L
2
T
–1
]
?
 [M
–1
L
3
T
–2
]
?
 
 1 = ? – ? 
 –1 = ? + 2? + 3? 
 –2 = – ? – ? – 2? 
 Solving 
 ? = 7, ? = –1, ? = –2 
3. A particle of mass m is moving in the xy-plane such that its velocity at a point (x, y) is given as 
ˆˆ ( 2 ), v yx xy = ? + 
where ? is a non-zero constant. What is the force F acting on the particle? 
 (A) ( )
2
ˆˆ 2 F m xx yy = ? + (B) ( )
2
ˆˆ 2 F m yx xy = ? + 
 (C) ( )
2
ˆˆ 2 F m yx xy = ? + (D) ( )
2
ˆˆ 2 F m xx yy = ? + 
Answer (A) 
Sol. 
dv
Fm
dt
= 
 ( ) ˆˆ 2 v yx xy = ? + 
 ˆˆ 2
dv dy dx
xy
dt dt dt
??
= ? +
??
??
 
  = 
( )
ˆˆ 2
yx
v x v y ?+ 
  = ? ?
ˆˆ 22 x x yy ? ? + ? 
  = ? ?
2
ˆˆ 2 xx yy ?+ 
 ? ?
2
ˆˆ 2 F m xx yy = ? + 
4. An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is n. 
The internal energy of one mole of the gas is U
n
 and the speed of sound in the gas is v
n
. At a fixed temperature 
and pressure, which of the following is the correct option? 
 (A) v
3
 < v
6
 and U
3
 > U
6
 (B) v
5
 > v
3
 and U
3
 > U
5
 
 (C) v
5
 > v
7
 and U
5
 < U
7
 (D) v
6
 < v
7
 and U
6
 < U
7
 
Answer (C) 
 
 
   
  
Sol. 
1
22
n
n RT nRT
U
??
== 
 
2
1
n
RT
RT
n
v
MM
??
+
??
?
??
== 
 ? U
7
 > U
5
 and U
7
 > U
6
 
  and v
5
 > v
7
 
SECTION 2 (Maximum Marks : 12) 
• This section contains THREE (03) questions. 
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) 
correct answer(s). 
• For each question, choose the option(s) corresponding to (all) the correct answer(s). 
• Answer to each question will be evaluated according to the following marking scheme: 
 Full Marks  : +4  ONLY if (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. 
 5. A monochromatic light wave is incident normally on a glass slab of thickness d, as shown in the figure. The 
refractive index of the slab increases linearly from n1 to n2 over the height h. Which of the following statement(s) 
is(are) true about the light wave emerging out of the slab? 
 
   
 
 
   
  
  (A) It will deflect up by an angle
( )
22
21
1
tan
2
n n d
h
-
??
-
??
??
??
??
 
 (B) It will deflect up by an angle 
( )
21 1
tan
n n d
h
-
?? -
??
??
  
 (C) It will not deflect 
 (D) The deflection angle depends only on (n2 - n1) and not on the individual values of n1 and n2 
Answer (D) 
Sol.  
 Dashed line MN and PQ are incident wave front. 
 Just before entering the slab PQ is maxima. 
 P1 and Q1 are the points on other face of slab. 
 So, P1 will be at lead of n2d w.r.t. P 
 And Q? will be at lead of n1d w.r.t. Q 
 Right of the medium is homogeneous 
 So to have same path lead as P1.Q? must travel to Q1, so n2d = n1d + h sin? 
6. An annular disk of mass M, inner radius a and outer radius b is placed on a horizontal surface with coefficient of 
friction ?, as shown in the figure. At some time, an impulse 
0
ˆ Jx is applied at a height h above the center of the 
disk. If h = h m then the disk rolls without slipping along the x-axis. Which of the following statement(s) is(are) 
correct? 
 
  (A) For ? ? 0 and a ? 0, h m = b/2 
 (B) For ? ? 0 and a ? b, h m = b 
 (C) For h = h m, the initial angular velocity does not depend on the inner radius a 
 (D) For ? = 0 and h = 0, the wheel always slides without rolling 
Answer (A, B, C, D)