JEE Advanced 2021 Question Paper - 1 with Solutions | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


Answers & Solutions
f o r
JEE (Advanced)-2021
Time : 3 hrs. Max. Marks: 180
PAPER - 1
DATE : 03/10/2021
PART-I : PHYSICS
SECTION - 1
? 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. 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
Answer (C)
Page 2


Answers & Solutions
f o r
JEE (Advanced)-2021
Time : 3 hrs. Max. Marks: 180
PAPER - 1
DATE : 03/10/2021
PART-I : PHYSICS
SECTION - 1
? 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. 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
Answer (C)
JEE (ADVANCED)-2021 (Paper-1)
Sol. Least count of Vernier calipers = 0.01 cm
Error in scale = 4 LC
= 0.04 cm
Reading = 3.1 cm + 1 L.C
= 3.1 cm + 0.01
= 3.11 cm
So correct diameter of the sphere
= (3.11 + 0.04) cm
= 3.15 cm
So, option (C)
2. 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
P
V
3
2
4
1
(A) steps 1 and 2
(B) steps 1 and 3
(C) steps 1 and 4
(D) steps 2 and 4
Answer (C)
Sol. Given P – V diagram
For process (1)
?Q
1
 = nC
P
?T
P
V
3
2
4
1
As P = constant and V increases
so T will increase
So ?Q
1
 > 0
For process (2)
?Q
2
 = nC
V
?T
V = constant, P
?
, So T
?
For process (3), ?Q
3
 = nC
P
?T < 0
For process (4), ?Q
4
 = nC
P
?T
As ?T > 0
So ?Q
4
 > 0
Page 3


Answers & Solutions
f o r
JEE (Advanced)-2021
Time : 3 hrs. Max. Marks: 180
PAPER - 1
DATE : 03/10/2021
PART-I : PHYSICS
SECTION - 1
? 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. 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
Answer (C)
JEE (ADVANCED)-2021 (Paper-1)
Sol. Least count of Vernier calipers = 0.01 cm
Error in scale = 4 LC
= 0.04 cm
Reading = 3.1 cm + 1 L.C
= 3.1 cm + 0.01
= 3.11 cm
So correct diameter of the sphere
= (3.11 + 0.04) cm
= 3.15 cm
So, option (C)
2. 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
P
V
3
2
4
1
(A) steps 1 and 2
(B) steps 1 and 3
(C) steps 1 and 4
(D) steps 2 and 4
Answer (C)
Sol. Given P – V diagram
For process (1)
?Q
1
 = nC
P
?T
P
V
3
2
4
1
As P = constant and V increases
so T will increase
So ?Q
1
 > 0
For process (2)
?Q
2
 = nC
V
?T
V = constant, P
?
, So T
?
For process (3), ?Q
3
 = nC
P
?T < 0
For process (4), ?Q
4
 = nC
P
?T
As ?T > 0
So ?Q
4
 > 0
JEE (ADVANCED)-2021 (Paper-1)
3. An extended object is placed at point O, 10 cm in front of a convex lens L
1
 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
L
1
10 cm 10 cm
O
L
2
(D) 1.6
Answer (B)
Sol.
1
1 1 1
10 20 v
? ?
?
?
1
1 1 1 1
20 10 20 v
?
? ? ?
1
1 3 2
1
2 20 f
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? 1 20cm v ? ?   
1
20
?
1
1
20
2
10
v
m
u
?
? ? ?
?
again 
2
1 1 1
30 20 v
? ?
? ?
?
2
1 1 1 5 1
30 20 60 12 v
? ? ? ? ? ? ?
2
12 2
30 5
m ? ? ?
?
1 2
2
2
5
m m m ? ? ? ?
?= 0.8
4. A heavy nucleus Q of 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
Answer (D)
Page 4


Answers & Solutions
f o r
JEE (Advanced)-2021
Time : 3 hrs. Max. Marks: 180
PAPER - 1
DATE : 03/10/2021
PART-I : PHYSICS
SECTION - 1
? 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. 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
Answer (C)
JEE (ADVANCED)-2021 (Paper-1)
Sol. Least count of Vernier calipers = 0.01 cm
Error in scale = 4 LC
= 0.04 cm
Reading = 3.1 cm + 1 L.C
= 3.1 cm + 0.01
= 3.11 cm
So correct diameter of the sphere
= (3.11 + 0.04) cm
= 3.15 cm
So, option (C)
2. 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
P
V
3
2
4
1
(A) steps 1 and 2
(B) steps 1 and 3
(C) steps 1 and 4
(D) steps 2 and 4
Answer (C)
Sol. Given P – V diagram
For process (1)
?Q
1
 = nC
P
?T
P
V
3
2
4
1
As P = constant and V increases
so T will increase
So ?Q
1
 > 0
For process (2)
?Q
2
 = nC
V
?T
V = constant, P
?
, So T
?
For process (3), ?Q
3
 = nC
P
?T < 0
For process (4), ?Q
4
 = nC
P
?T
As ?T > 0
So ?Q
4
 > 0
JEE (ADVANCED)-2021 (Paper-1)
3. An extended object is placed at point O, 10 cm in front of a convex lens L
1
 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
L
1
10 cm 10 cm
O
L
2
(D) 1.6
Answer (B)
Sol.
1
1 1 1
10 20 v
? ?
?
?
1
1 1 1 1
20 10 20 v
?
? ? ?
1
1 3 2
1
2 20 f
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? 1 20cm v ? ?   
1
20
?
1
1
20
2
10
v
m
u
?
? ? ?
?
again 
2
1 1 1
30 20 v
? ?
? ?
?
2
1 1 1 5 1
30 20 60 12 v
? ? ? ? ? ? ?
2
12 2
30 5
m ? ? ?
?
1 2
2
2
5
m m m ? ? ? ?
?= 0.8
4. A heavy nucleus Q of 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
Answer (D)
JEE (ADVANCED)-2021 (Paper-1)
Sol: t
1/2
 
= 20 min
In 60 min, no. of half-life = 3
?
3
1000
1000 0.6
2
A N
? ?
? ? ?
? ?
? ?
=
7
1000 0.6
8
? ?
= 525
SECTION - 2
? 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. 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 to
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/s
2
.
5. The value of t is ________.
Answer (00.50)
Sol.
2 2
sin
2
u
H
g
?
?
= 
50 1 5
2 10 2 4
? ?
?
2 2 5
4 10
H
t
g
?
? ?
?
? ?
1
s 0.5 sec
2
t
Ans. 00.50
Page 5


Answers & Solutions
f o r
JEE (Advanced)-2021
Time : 3 hrs. Max. Marks: 180
PAPER - 1
DATE : 03/10/2021
PART-I : PHYSICS
SECTION - 1
? 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. 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
Answer (C)
JEE (ADVANCED)-2021 (Paper-1)
Sol. Least count of Vernier calipers = 0.01 cm
Error in scale = 4 LC
= 0.04 cm
Reading = 3.1 cm + 1 L.C
= 3.1 cm + 0.01
= 3.11 cm
So correct diameter of the sphere
= (3.11 + 0.04) cm
= 3.15 cm
So, option (C)
2. 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
P
V
3
2
4
1
(A) steps 1 and 2
(B) steps 1 and 3
(C) steps 1 and 4
(D) steps 2 and 4
Answer (C)
Sol. Given P – V diagram
For process (1)
?Q
1
 = nC
P
?T
P
V
3
2
4
1
As P = constant and V increases
so T will increase
So ?Q
1
 > 0
For process (2)
?Q
2
 = nC
V
?T
V = constant, P
?
, So T
?
For process (3), ?Q
3
 = nC
P
?T < 0
For process (4), ?Q
4
 = nC
P
?T
As ?T > 0
So ?Q
4
 > 0
JEE (ADVANCED)-2021 (Paper-1)
3. An extended object is placed at point O, 10 cm in front of a convex lens L
1
 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
L
1
10 cm 10 cm
O
L
2
(D) 1.6
Answer (B)
Sol.
1
1 1 1
10 20 v
? ?
?
?
1
1 1 1 1
20 10 20 v
?
? ? ?
1
1 3 2
1
2 20 f
? ? ? ?
? ?
? ? ? ?
? ? ? ?
? 1 20cm v ? ?   
1
20
?
1
1
20
2
10
v
m
u
?
? ? ?
?
again 
2
1 1 1
30 20 v
? ?
? ?
?
2
1 1 1 5 1
30 20 60 12 v
? ? ? ? ? ? ?
2
12 2
30 5
m ? ? ?
?
1 2
2
2
5
m m m ? ? ? ?
?= 0.8
4. A heavy nucleus Q of 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
Answer (D)
JEE (ADVANCED)-2021 (Paper-1)
Sol: t
1/2
 
= 20 min
In 60 min, no. of half-life = 3
?
3
1000
1000 0.6
2
A N
? ?
? ? ?
? ?
? ?
=
7
1000 0.6
8
? ?
= 525
SECTION - 2
? 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. 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 to
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/s
2
.
5. The value of t is ________.
Answer (00.50)
Sol.
2 2
sin
2
u
H
g
?
?
= 
50 1 5
2 10 2 4
? ?
?
2 2 5
4 10
H
t
g
?
? ?
?
? ?
1
s 0.5 sec
2
t
Ans. 00.50
JEE (ADVANCED)-2021 (Paper-1)
6. The value of x is ________.
Answer (07.50)
Sol.
3
2
R
X ?
 as X
cm
 = R
2 2
sin 50
5
10
u
R
g
?
? ? ?
?
3 15
7.5 m
2 2
R
X ? ? ?
Ans.: 07.50
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 q
1
 µC. Then S is switched to position Q. After a long time, the charge on the capacitor is
q
2
 µC.
1 ? 2 ?
2V 1 ?F
Q
P
1V
S
7. The magnitude of q
1
 is_____.
Answer (01.33)
Sol. With switch S at position P after long time potential difference across capacitor branch
= 
2 1
2 1
1 1
2 1
2 2 4
3 3
v
? ?
? ?
?
? Charge on capacitor 
? ? ?
1
4
3
q C C
?
1
4
1.33
3
q ? ?
8. The magnitude of q
2
 is_____.
Answer (00.67)
Sol. With switch S at position Q after long time potential difference across capacitor
= potential difference across resistance of 1 ohm.
= 
2
3
v