JEE Mains 24 Feb 2021 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


 
 
Section - A 
1. Four identical particles of equal masses 1 kg made to move along the circumference of a circle 
of radius 1 m under the action of their own mutual gravitational attraction. The speed of each 
particle will be -  
 (1)
(1 2 2)G
2
?
 (2) G(1 2 2) ? (3)
? ?
G
2 2 1
2
? (4) 
? ?
G
1 2 2
2
? 
Sol. (1) 
 
 M
v
F2sin45°
45°
FC
F1
F2cos45°
F2cos45°
F2
M 
v
M
v 
F2sin45°
F2
R
v
M
2R
 
 ? ?By resolving force F
2
, we get 
 ? F
1
 + F
2
  cos 45° + F
2
 cos 45° 
 ? F
1
 + 2F
2
 cos 45° = F
c
 
 F
c
 = centripital force = 
2
MV
R
 
 ?
2
2
GM
(2R)
 + 
? ?
2
2
2GM
cos45
2R
? ?
? ?
?
? ?
? ?
? ?
 = 
2
MV
R
 
 ?
2
2
GM
4R
 + 
2
2
2GM
2 2R
 = 
2
MV
R
 
 ?
GM
4R
 + 
GM
2.R
 = V
2
 
 ? V = 
GM GM
4R
2.R
? 
 ? V = 
GM 1 2 2
R 4
? ?
?
? ?
? ?
? ?
 
 ? V = 
1
2
? ?
GM
1 2 2
R
? 
 (given : mass = 1 kg, radius = 1 m) 
 ? ?v ? ? ?
1
2
G(1 2 2) ? ?
 
24
th
 Feb. 2021 | Shift - 1
PHYSICS
Page 2


 
 
Section - A 
1. Four identical particles of equal masses 1 kg made to move along the circumference of a circle 
of radius 1 m under the action of their own mutual gravitational attraction. The speed of each 
particle will be -  
 (1)
(1 2 2)G
2
?
 (2) G(1 2 2) ? (3)
? ?
G
2 2 1
2
? (4) 
? ?
G
1 2 2
2
? 
Sol. (1) 
 
 M
v
F2sin45°
45°
FC
F1
F2cos45°
F2cos45°
F2
M 
v
M
v 
F2sin45°
F2
R
v
M
2R
 
 ? ?By resolving force F
2
, we get 
 ? F
1
 + F
2
  cos 45° + F
2
 cos 45° 
 ? F
1
 + 2F
2
 cos 45° = F
c
 
 F
c
 = centripital force = 
2
MV
R
 
 ?
2
2
GM
(2R)
 + 
? ?
2
2
2GM
cos45
2R
? ?
? ?
?
? ?
? ?
? ?
 = 
2
MV
R
 
 ?
2
2
GM
4R
 + 
2
2
2GM
2 2R
 = 
2
MV
R
 
 ?
GM
4R
 + 
GM
2.R
 = V
2
 
 ? V = 
GM GM
4R
2.R
? 
 ? V = 
GM 1 2 2
R 4
? ?
?
? ?
? ?
? ?
 
 ? V = 
1
2
? ?
GM
1 2 2
R
? 
 (given : mass = 1 kg, radius = 1 m) 
 ? ?v ? ? ?
1
2
G(1 2 2) ? ?
 
24
th
 Feb. 2021 | Shift - 1
PHYSICS
 
 
2. Consider two satellites S
1
 and S
2
 with periods of revolution 1 hr. and 8 hr. respectively revolving 
around a planet in circular orbits.The ratio of angular velocity of satellite S
1
 to the angular 
velocity of satellite S
2
 is -  
 (1)8 : 1  (2)1 : 8  (3)2 : 1  (4) 1 : 4  
Sol. (1) 
 We know that ? = 
2
T
?
 
 given : Ratio of time period  
 
1
2
T
T
 = 
1
8
 
 ? ? ? ? ? ?
1
T
 
 ?
1
2
?
?
 = 
2
1
T
T
 
 ?
1
2
?
?
 = 
8
1
 
 ? ?
1
 : ?
2
 = 8 : 1  
 
3. n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following 
processes – 
 A ?B : Isothermal expansion at temperature T so that the volume is doubled from V
1
 to  
V
2
 = 2V
1
 and pressure changes from P
1
 to P
2
. 
 B ? C : Isobaric compression at pressure P
2
 to initial volume V
1
. 
 C ? A : Isochoric change leading to change of pressure from P
2
 to P
1
. 
 Total workdone in the complete cycle ABCA is –  
 
A
B
C 
V1 V2=2V1
V
P2 
P1 
P 
 
 (1)0 (2)nRT
1
ln2
2
? ?
?
? ?
? ?
 (3)nRTln2 (4) nRT 
1
ln2
2
? ?
?
? ?
? ?
 
Page 3


 
 
Section - A 
1. Four identical particles of equal masses 1 kg made to move along the circumference of a circle 
of radius 1 m under the action of their own mutual gravitational attraction. The speed of each 
particle will be -  
 (1)
(1 2 2)G
2
?
 (2) G(1 2 2) ? (3)
? ?
G
2 2 1
2
? (4) 
? ?
G
1 2 2
2
? 
Sol. (1) 
 
 M
v
F2sin45°
45°
FC
F1
F2cos45°
F2cos45°
F2
M 
v
M
v 
F2sin45°
F2
R
v
M
2R
 
 ? ?By resolving force F
2
, we get 
 ? F
1
 + F
2
  cos 45° + F
2
 cos 45° 
 ? F
1
 + 2F
2
 cos 45° = F
c
 
 F
c
 = centripital force = 
2
MV
R
 
 ?
2
2
GM
(2R)
 + 
? ?
2
2
2GM
cos45
2R
? ?
? ?
?
? ?
? ?
? ?
 = 
2
MV
R
 
 ?
2
2
GM
4R
 + 
2
2
2GM
2 2R
 = 
2
MV
R
 
 ?
GM
4R
 + 
GM
2.R
 = V
2
 
 ? V = 
GM GM
4R
2.R
? 
 ? V = 
GM 1 2 2
R 4
? ?
?
? ?
? ?
? ?
 
 ? V = 
1
2
? ?
GM
1 2 2
R
? 
 (given : mass = 1 kg, radius = 1 m) 
 ? ?v ? ? ?
1
2
G(1 2 2) ? ?
 
24
th
 Feb. 2021 | Shift - 1
PHYSICS
 
 
2. Consider two satellites S
1
 and S
2
 with periods of revolution 1 hr. and 8 hr. respectively revolving 
around a planet in circular orbits.The ratio of angular velocity of satellite S
1
 to the angular 
velocity of satellite S
2
 is -  
 (1)8 : 1  (2)1 : 8  (3)2 : 1  (4) 1 : 4  
Sol. (1) 
 We know that ? = 
2
T
?
 
 given : Ratio of time period  
 
1
2
T
T
 = 
1
8
 
 ? ? ? ? ? ?
1
T
 
 ?
1
2
?
?
 = 
2
1
T
T
 
 ?
1
2
?
?
 = 
8
1
 
 ? ?
1
 : ?
2
 = 8 : 1  
 
3. n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following 
processes – 
 A ?B : Isothermal expansion at temperature T so that the volume is doubled from V
1
 to  
V
2
 = 2V
1
 and pressure changes from P
1
 to P
2
. 
 B ? C : Isobaric compression at pressure P
2
 to initial volume V
1
. 
 C ? A : Isochoric change leading to change of pressure from P
2
 to P
1
. 
 Total workdone in the complete cycle ABCA is –  
 
A
B
C 
V1 V2=2V1
V
P2 
P1 
P 
 
 (1)0 (2)nRT
1
ln2
2
? ?
?
? ?
? ?
 (3)nRTln2 (4) nRT 
1
ln2
2
? ?
?
? ?
? ?
 
 
 
Sol. (4) 
  
 
 P
A
B 
C
V1 V2=2V1
V
P2
P1
 
 A ? B = isotheraml process 
 B ? ?C = isobaric process 
 C ? ?A = isochoric process 
 also, V
2
 = 2V
1
 
 work done by gas in the complete cycle ABCA is –  
 ? w = w
AB
 + w
BC
 + w
CA
 ....(1) 
 ? ?w
CA
 = 0, as isochoric process 
 ?w
AB
 = 2P
1
V
1
 ln 
2
1
v
v
? ?
? ?
? ?
 = 2 nRT ln (2) 
 ?w
BC
 = P
2
 (V
1
 – V
2
) = P
2
 (V
1
 – 2V
1
) = -P
2
V
1
 = -nRT 
 ?Now put the value of w
AB
, w
BC
 and w
CA
 in equation, we get 
 ? w = 2nRT ln (2) – nRT + 0 
 ? w = nRT [2ln (2) – 1] 
? w = nRT [ln (2) - 
1
2
] ?
 
4. Two equal capacitors are first connected in series and then in parallel. The ratio of the 
equivalent capacities capacities in the two cases will be -  
 (1)2 : 1  (2)1 : 4  (3)4 : 1  (4) 1 : 2 
Sol. (2) 
 Given that first connection 
 
 C C
2 1
 
 ? ?
12
1
C
 = 
1
C
 + 
1
C
? C
12
 = 
C
2
 
 Second connection  
 C
34
 = C + C = 2 C  
Page 4


 
 
Section - A 
1. Four identical particles of equal masses 1 kg made to move along the circumference of a circle 
of radius 1 m under the action of their own mutual gravitational attraction. The speed of each 
particle will be -  
 (1)
(1 2 2)G
2
?
 (2) G(1 2 2) ? (3)
? ?
G
2 2 1
2
? (4) 
? ?
G
1 2 2
2
? 
Sol. (1) 
 
 M
v
F2sin45°
45°
FC
F1
F2cos45°
F2cos45°
F2
M 
v
M
v 
F2sin45°
F2
R
v
M
2R
 
 ? ?By resolving force F
2
, we get 
 ? F
1
 + F
2
  cos 45° + F
2
 cos 45° 
 ? F
1
 + 2F
2
 cos 45° = F
c
 
 F
c
 = centripital force = 
2
MV
R
 
 ?
2
2
GM
(2R)
 + 
? ?
2
2
2GM
cos45
2R
? ?
? ?
?
? ?
? ?
? ?
 = 
2
MV
R
 
 ?
2
2
GM
4R
 + 
2
2
2GM
2 2R
 = 
2
MV
R
 
 ?
GM
4R
 + 
GM
2.R
 = V
2
 
 ? V = 
GM GM
4R
2.R
? 
 ? V = 
GM 1 2 2
R 4
? ?
?
? ?
? ?
? ?
 
 ? V = 
1
2
? ?
GM
1 2 2
R
? 
 (given : mass = 1 kg, radius = 1 m) 
 ? ?v ? ? ?
1
2
G(1 2 2) ? ?
 
24
th
 Feb. 2021 | Shift - 1
PHYSICS
 
 
2. Consider two satellites S
1
 and S
2
 with periods of revolution 1 hr. and 8 hr. respectively revolving 
around a planet in circular orbits.The ratio of angular velocity of satellite S
1
 to the angular 
velocity of satellite S
2
 is -  
 (1)8 : 1  (2)1 : 8  (3)2 : 1  (4) 1 : 4  
Sol. (1) 
 We know that ? = 
2
T
?
 
 given : Ratio of time period  
 
1
2
T
T
 = 
1
8
 
 ? ? ? ? ? ?
1
T
 
 ?
1
2
?
?
 = 
2
1
T
T
 
 ?
1
2
?
?
 = 
8
1
 
 ? ?
1
 : ?
2
 = 8 : 1  
 
3. n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following 
processes – 
 A ?B : Isothermal expansion at temperature T so that the volume is doubled from V
1
 to  
V
2
 = 2V
1
 and pressure changes from P
1
 to P
2
. 
 B ? C : Isobaric compression at pressure P
2
 to initial volume V
1
. 
 C ? A : Isochoric change leading to change of pressure from P
2
 to P
1
. 
 Total workdone in the complete cycle ABCA is –  
 
A
B
C 
V1 V2=2V1
V
P2 
P1 
P 
 
 (1)0 (2)nRT
1
ln2
2
? ?
?
? ?
? ?
 (3)nRTln2 (4) nRT 
1
ln2
2
? ?
?
? ?
? ?
 
 
 
Sol. (4) 
  
 
 P
A
B 
C
V1 V2=2V1
V
P2
P1
 
 A ? B = isotheraml process 
 B ? ?C = isobaric process 
 C ? ?A = isochoric process 
 also, V
2
 = 2V
1
 
 work done by gas in the complete cycle ABCA is –  
 ? w = w
AB
 + w
BC
 + w
CA
 ....(1) 
 ? ?w
CA
 = 0, as isochoric process 
 ?w
AB
 = 2P
1
V
1
 ln 
2
1
v
v
? ?
? ?
? ?
 = 2 nRT ln (2) 
 ?w
BC
 = P
2
 (V
1
 – V
2
) = P
2
 (V
1
 – 2V
1
) = -P
2
V
1
 = -nRT 
 ?Now put the value of w
AB
, w
BC
 and w
CA
 in equation, we get 
 ? w = 2nRT ln (2) – nRT + 0 
 ? w = nRT [2ln (2) – 1] 
? w = nRT [ln (2) - 
1
2
] ?
 
4. Two equal capacitors are first connected in series and then in parallel. The ratio of the 
equivalent capacities capacities in the two cases will be -  
 (1)2 : 1  (2)1 : 4  (3)4 : 1  (4) 1 : 2 
Sol. (2) 
 Given that first connection 
 
 C C
2 1
 
 ? ?
12
1
C
 = 
1
C
 + 
1
C
? C
12
 = 
C
2
 
 Second connection  
 C
34
 = C + C = 2 C  
 
 
 Now, the ratio of equivalent capacities in the two cases will be –  
 ?
12
34
C
C
 = 
C / 2
2C
?
12
34
C
C
 = 
1
4
 
 
 
C
3
4
C
?
  
 
5. A cell E
1
 of emf 6V and internal resistance 2 ? is connected with another cell E
2
 of emf 4V and 
internal resistance 8 ? (as shown in the figure). The potential difference across points X and Y is – 
 
P
E1
6V,2 ?
X
E2
4v,8 ?
Y
 
 (1)3.6V (2)10.0V (3)5.6V (4) 2.0V 
Sol. (3) 
 
 
P 
E1
6V,2 ?
X 
E2
4v,8 ?
Y
I
I
I
 
 emf of E
1
 = 6v 
 r
1
 = 2 ? 
 emf of  E
2
= 4 ? 
 r
2
 = 8 ? 
 |v
x
 - v
y
| = potential difference across points x and y 
 E
eff
 = 6 – 4 = 2 V 
 R
eq
 = 2 + 8 = 10 ? 
 So, current in the circuit will be 
 ? ? ?? ? ?
eff
eq
E
R
? I = 
2
10
 = 0.2 A  
 Now, potential difference across points X and Y is 
 |v
x
 - v
y
| = E + iR 
 ? |v
x
 - v
y
| = 4 + 0.2 × 8 = 5.6 V  
 ? |v
x
 - v
y
| = 5.6 v ?
Page 5


 
 
Section - A 
1. Four identical particles of equal masses 1 kg made to move along the circumference of a circle 
of radius 1 m under the action of their own mutual gravitational attraction. The speed of each 
particle will be -  
 (1)
(1 2 2)G
2
?
 (2) G(1 2 2) ? (3)
? ?
G
2 2 1
2
? (4) 
? ?
G
1 2 2
2
? 
Sol. (1) 
 
 M
v
F2sin45°
45°
FC
F1
F2cos45°
F2cos45°
F2
M 
v
M
v 
F2sin45°
F2
R
v
M
2R
 
 ? ?By resolving force F
2
, we get 
 ? F
1
 + F
2
  cos 45° + F
2
 cos 45° 
 ? F
1
 + 2F
2
 cos 45° = F
c
 
 F
c
 = centripital force = 
2
MV
R
 
 ?
2
2
GM
(2R)
 + 
? ?
2
2
2GM
cos45
2R
? ?
? ?
?
? ?
? ?
? ?
 = 
2
MV
R
 
 ?
2
2
GM
4R
 + 
2
2
2GM
2 2R
 = 
2
MV
R
 
 ?
GM
4R
 + 
GM
2.R
 = V
2
 
 ? V = 
GM GM
4R
2.R
? 
 ? V = 
GM 1 2 2
R 4
? ?
?
? ?
? ?
? ?
 
 ? V = 
1
2
? ?
GM
1 2 2
R
? 
 (given : mass = 1 kg, radius = 1 m) 
 ? ?v ? ? ?
1
2
G(1 2 2) ? ?
 
24
th
 Feb. 2021 | Shift - 1
PHYSICS
 
 
2. Consider two satellites S
1
 and S
2
 with periods of revolution 1 hr. and 8 hr. respectively revolving 
around a planet in circular orbits.The ratio of angular velocity of satellite S
1
 to the angular 
velocity of satellite S
2
 is -  
 (1)8 : 1  (2)1 : 8  (3)2 : 1  (4) 1 : 4  
Sol. (1) 
 We know that ? = 
2
T
?
 
 given : Ratio of time period  
 
1
2
T
T
 = 
1
8
 
 ? ? ? ? ? ?
1
T
 
 ?
1
2
?
?
 = 
2
1
T
T
 
 ?
1
2
?
?
 = 
8
1
 
 ? ?
1
 : ?
2
 = 8 : 1  
 
3. n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following 
processes – 
 A ?B : Isothermal expansion at temperature T so that the volume is doubled from V
1
 to  
V
2
 = 2V
1
 and pressure changes from P
1
 to P
2
. 
 B ? C : Isobaric compression at pressure P
2
 to initial volume V
1
. 
 C ? A : Isochoric change leading to change of pressure from P
2
 to P
1
. 
 Total workdone in the complete cycle ABCA is –  
 
A
B
C 
V1 V2=2V1
V
P2 
P1 
P 
 
 (1)0 (2)nRT
1
ln2
2
? ?
?
? ?
? ?
 (3)nRTln2 (4) nRT 
1
ln2
2
? ?
?
? ?
? ?
 
 
 
Sol. (4) 
  
 
 P
A
B 
C
V1 V2=2V1
V
P2
P1
 
 A ? B = isotheraml process 
 B ? ?C = isobaric process 
 C ? ?A = isochoric process 
 also, V
2
 = 2V
1
 
 work done by gas in the complete cycle ABCA is –  
 ? w = w
AB
 + w
BC
 + w
CA
 ....(1) 
 ? ?w
CA
 = 0, as isochoric process 
 ?w
AB
 = 2P
1
V
1
 ln 
2
1
v
v
? ?
? ?
? ?
 = 2 nRT ln (2) 
 ?w
BC
 = P
2
 (V
1
 – V
2
) = P
2
 (V
1
 – 2V
1
) = -P
2
V
1
 = -nRT 
 ?Now put the value of w
AB
, w
BC
 and w
CA
 in equation, we get 
 ? w = 2nRT ln (2) – nRT + 0 
 ? w = nRT [2ln (2) – 1] 
? w = nRT [ln (2) - 
1
2
] ?
 
4. Two equal capacitors are first connected in series and then in parallel. The ratio of the 
equivalent capacities capacities in the two cases will be -  
 (1)2 : 1  (2)1 : 4  (3)4 : 1  (4) 1 : 2 
Sol. (2) 
 Given that first connection 
 
 C C
2 1
 
 ? ?
12
1
C
 = 
1
C
 + 
1
C
? C
12
 = 
C
2
 
 Second connection  
 C
34
 = C + C = 2 C  
 
 
 Now, the ratio of equivalent capacities in the two cases will be –  
 ?
12
34
C
C
 = 
C / 2
2C
?
12
34
C
C
 = 
1
4
 
 
 
C
3
4
C
?
  
 
5. A cell E
1
 of emf 6V and internal resistance 2 ? is connected with another cell E
2
 of emf 4V and 
internal resistance 8 ? (as shown in the figure). The potential difference across points X and Y is – 
 
P
E1
6V,2 ?
X
E2
4v,8 ?
Y
 
 (1)3.6V (2)10.0V (3)5.6V (4) 2.0V 
Sol. (3) 
 
 
P 
E1
6V,2 ?
X 
E2
4v,8 ?
Y
I
I
I
 
 emf of E
1
 = 6v 
 r
1
 = 2 ? 
 emf of  E
2
= 4 ? 
 r
2
 = 8 ? 
 |v
x
 - v
y
| = potential difference across points x and y 
 E
eff
 = 6 – 4 = 2 V 
 R
eq
 = 2 + 8 = 10 ? 
 So, current in the circuit will be 
 ? ? ?? ? ?
eff
eq
E
R
? I = 
2
10
 = 0.2 A  
 Now, potential difference across points X and Y is 
 |v
x
 - v
y
| = E + iR 
 ? |v
x
 - v
y
| = 4 + 0.2 × 8 = 5.6 V  
 ? |v
x
 - v
y
| = 5.6 v ?
 
 
 
6. If Y,K and ? are the values of Young’s modulus, bulk modulus and modulus of rigidity of any 
material respectively. Choose the correct relation for these parameters. 
 (1)K = 
Y
9 3Y
?
? ?
 N/m
2
 (2) ? = 
3YK
9K Y ?
 N/m
2 
 (3)Y = 
9K
3K
?
? ?
 N/m
2 
 (4) Y =  
9K
2 3K
?
? ?
N/m
2
 
Sol. (1) 
 ?y = 3k (1 – 2 ?) 
 ? ? ? ? ? ?
1
2
y
1
3k
? ?
?
? ?
? ?
 ....(1) 
 ? y = 2 ? (1 + ?) 
 ? ? ? = 
y
2 ?
– 1  ....(2) 
 by comparing equation (1) and (2), we get 
 ? ?
y
2 ?
– 1 = 
1
2
y
1
3k
? ?
?
? ?
? ?
 
 ?
y
?
– 2 = 1 –
y
3k
 
 ?
y
?
 = 1 + 2 –
y
3k
?
y
?
 = 3 –
y
3k
 
 ?
y
3k
 = 3 –
y
?
?
y
3k
 = 
3 y ? ?
?
 
 ? k = 
y
9 3y
?
? ?
?
 
7. Two stars of masses m and 2m at a distance d rotate about their common centre of mass in 
free space. The period of revolution is -  
 (1) ??
3
d
3Gm
 (2)
1
2 ?
3
3Gm
d
 (3)
1
2 ?
3
d
3Gm
 (4) 2 ?
3
3Gm
d
 
Sol. (1) 
 
 
2m m
d
 
 ? ?
2
G(m)(2m)
d
 = m ?
2
 × 
2d
3
 
 ?
2
2Gm
d
 = ?
2
 × 
2d
3