JEE Main 2021 Physics February 26 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
PART–A : PHYSICS
SECTION - I
Multiple Choice Questions: This section contains
20 multiple choice questions. Each question has
4 choices (1), (2), (3) and (4), out of which ONL Y ONE
is correct.
Choose the correct answer :
1. The incident ray, reflected ray and the outward
drawn normal are denoted by the unit vectors
a,b andc
 
 respectively. Then choose the
correct relation for these vectors.
(1)ba 2c ??
 
(2)b2a c ??
 
(3)ba 2(ac)c ?? ?
 
(4)ba c ??
 
Answer (3)
Sol.ba 2cosc ?? ?
 
?
?
b

a

c

ba 2cosc ?? ?
 
ba 2(ac)c ?? ?
 
2. A radioactive sample is undergoing ? decay. At
any time t
1
, its activity is A and another time t
2
,
the activity is 
A
5
. What is the average life time
for the sample?
(1)
21
tt
ln5
?
(2)
21
ln(tt )
2
?
(3)
21
ln5
tt ?
(4)
12
tt
ln5
?
Answer (1)
Sol.
21
(t t )
A
Ae
5
?? ?
?
21
(t t )
1
e
5
?? ?
?
? ln 5 = ?(t
2
 – t
1
)
?
21
avg
(t t ) 1
t
ln5
?
??
?
3. An inclined plane making an angle of 30° with
the horizontal is placed in a uniform horizontal
electric field 
N
200
C
 as shown in the figure.
A body of mass 1 kg and charge 5 mC is
allowed to slide down from rest at a height of
1 m. If the coefficient of friction is 0.2, find the
time taken by the body to reach the bottom.
2
13
g 9.8m/s ;sin30 ;cos30
22
??
??? ??
??
??
??
1 kg, 5 mC
1 m
30°
E = 200 N/C
(1) 2.3 s (2) 1.3 s
(3) 0.92 s (4) 0.46 s
Answer (2)
Sol.
3
2
?N
N
9.8
2
9.8 3 1
22
?
?
30°
9.8 3 1 3
a 4.9 (0.2)
22
??
? ??
?? ?
??
??
??
= 2.24 m/s
2
21
T1.3s
(sin30) 2.24
?
??
?
4. Given below are two statements :
Statement I : An electric dipole is placed at the
centre of a hollow sphere. The flux of electric
field through the sphere is zero but the electric
field is not zero anywhere in the sphere.
Statement II : If R is the radius of a solid
metallic sphere and Q be the total charge on it.
The electric field at any point on the spherical
surface of radius r (< R) is zero but the electric
flux passing through this closed spherical
surface of radius r is not zero.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
Answer (3)
Page 2


JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
PART–A : PHYSICS
SECTION - I
Multiple Choice Questions: This section contains
20 multiple choice questions. Each question has
4 choices (1), (2), (3) and (4), out of which ONL Y ONE
is correct.
Choose the correct answer :
1. The incident ray, reflected ray and the outward
drawn normal are denoted by the unit vectors
a,b andc
 
 respectively. Then choose the
correct relation for these vectors.
(1)ba 2c ??
 
(2)b2a c ??
 
(3)ba 2(ac)c ?? ?
 
(4)ba c ??
 
Answer (3)
Sol.ba 2cosc ?? ?
 
?
?
b

a

c

ba 2cosc ?? ?
 
ba 2(ac)c ?? ?
 
2. A radioactive sample is undergoing ? decay. At
any time t
1
, its activity is A and another time t
2
,
the activity is 
A
5
. What is the average life time
for the sample?
(1)
21
tt
ln5
?
(2)
21
ln(tt )
2
?
(3)
21
ln5
tt ?
(4)
12
tt
ln5
?
Answer (1)
Sol.
21
(t t )
A
Ae
5
?? ?
?
21
(t t )
1
e
5
?? ?
?
? ln 5 = ?(t
2
 – t
1
)
?
21
avg
(t t ) 1
t
ln5
?
??
?
3. An inclined plane making an angle of 30° with
the horizontal is placed in a uniform horizontal
electric field 
N
200
C
 as shown in the figure.
A body of mass 1 kg and charge 5 mC is
allowed to slide down from rest at a height of
1 m. If the coefficient of friction is 0.2, find the
time taken by the body to reach the bottom.
2
13
g 9.8m/s ;sin30 ;cos30
22
??
??? ??
??
??
??
1 kg, 5 mC
1 m
30°
E = 200 N/C
(1) 2.3 s (2) 1.3 s
(3) 0.92 s (4) 0.46 s
Answer (2)
Sol.
3
2
?N
N
9.8
2
9.8 3 1
22
?
?
30°
9.8 3 1 3
a 4.9 (0.2)
22
??
? ??
?? ?
??
??
??
= 2.24 m/s
2
21
T1.3s
(sin30) 2.24
?
??
?
4. Given below are two statements :
Statement I : An electric dipole is placed at the
centre of a hollow sphere. The flux of electric
field through the sphere is zero but the electric
field is not zero anywhere in the sphere.
Statement II : If R is the radius of a solid
metallic sphere and Q be the total charge on it.
The electric field at any point on the spherical
surface of radius r (< R) is zero but the electric
flux passing through this closed spherical
surface of radius r is not zero.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
Answer (3)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol. 
inc
Tot
0
q
??
?
q
inc
 = 0 ??
Tot
 = 0 +q –q
E0 ?
 
At P,
E = 0 and ? = 0
r
P
Q
5. A scooter accelerates from rest for time t
1
 at
constant rate a
1
 and then retards at constant
rate a
2
 for time t
2
 and comes to rest. The
correct value of 
1
2
t
t
 will be
(1)
12
1
aa
a
?
(2)
1
2
a
a
(3)
2
1
a
a
(4)
12
2
aa
a
?
Answer (3)
Sol. We have,
a
1
t
1
 = a
2
t
2
?
1 2
21
t a
ta
?
6. Find the peak current and resonant frequency
of the following circuit (as shown in figure).
120 ?
V = 30 sin 100t
100 mH
100 F ?
(1) 0.2 A and 50 Hz (2) 2 A and 100 Hz
(3) 2 A and 50 Hz (4) 0.2 A and 100 Hz
Answer (1)
Sol. ??
??
max
max
22
LC
V 30
I
Z
R(X X)
X
L
 = 100 × 0.1 = 10 ?
?
???
?
C
4
1
X 100
100 10
??
?
max
22
30
I0.2A
(120) (90)
?? ???
?
11
f 50Hz
LC 2 LC
7. The length of metallic wire is l
1
 when tension in
it is T
1
. It is l
2
 when the tension is T
2
. The original
length of the wire will be
(1)
?
?
11 22
21
Tl T l
TT
(2)
?
12
ll
2
(3)
?
?
21 12
12
Tl Tl
TT
(4)
?
?
21 12
21
Tl Tl
TT
Answer (4)
Sol. T
1
 = K(l
1
 – l
0
) ...(1)
T
2
 = K(l
2
 – l
0
) ...(2)
From (1) and (2),
?
?
?
21 12
0
21
Tl Tl
l
(TT )
8. The trajectory of a projectile in a vertical plane
is y = ?x – ?x
2
, where ? and ? are constants and
x & y are respectively the horizontal and vertical
distances of the projectile from the point of
projection. The angle of projection ? and the
maximum height attained H are respectively
given by
(1)
?
?
?
?
2
1
4
tan , (2)
?
?
?
?
2
1
tan ,
2
(3)
?
?? ??
??
??
??
2
1
tan , (4)
?
?
?
?
2
1
tan ,
4
Answer (4)
Sol. ?? ? ?
2
y xx
?
? ??? ? ?? ?
1
tan tan
also,  ?? ? ?
dy
2x
dx
?
? ? ?
?
dy
0x
dx 2
22 2
y
24 4
?? ?
???
?? ?
at x
2
?
?
?
9. An aeroplane, with its wings spread 10 m, is
flying at a speed of 180 km/h in a horizontal
direction. The total intensity of earth’s field at
that part is 2.5 × 10
–4
 Wb/m
2
 and the angle of dip
is 60°. The emf induced between the tips of the
plane wings will be _________ .
(1) 108.25 mV (2) 88.37 mV
(3) 62.50 mV (4) 54.125 mV
Answer (1)
Page 3


JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
PART–A : PHYSICS
SECTION - I
Multiple Choice Questions: This section contains
20 multiple choice questions. Each question has
4 choices (1), (2), (3) and (4), out of which ONL Y ONE
is correct.
Choose the correct answer :
1. The incident ray, reflected ray and the outward
drawn normal are denoted by the unit vectors
a,b andc
 
 respectively. Then choose the
correct relation for these vectors.
(1)ba 2c ??
 
(2)b2a c ??
 
(3)ba 2(ac)c ?? ?
 
(4)ba c ??
 
Answer (3)
Sol.ba 2cosc ?? ?
 
?
?
b

a

c

ba 2cosc ?? ?
 
ba 2(ac)c ?? ?
 
2. A radioactive sample is undergoing ? decay. At
any time t
1
, its activity is A and another time t
2
,
the activity is 
A
5
. What is the average life time
for the sample?
(1)
21
tt
ln5
?
(2)
21
ln(tt )
2
?
(3)
21
ln5
tt ?
(4)
12
tt
ln5
?
Answer (1)
Sol.
21
(t t )
A
Ae
5
?? ?
?
21
(t t )
1
e
5
?? ?
?
? ln 5 = ?(t
2
 – t
1
)
?
21
avg
(t t ) 1
t
ln5
?
??
?
3. An inclined plane making an angle of 30° with
the horizontal is placed in a uniform horizontal
electric field 
N
200
C
 as shown in the figure.
A body of mass 1 kg and charge 5 mC is
allowed to slide down from rest at a height of
1 m. If the coefficient of friction is 0.2, find the
time taken by the body to reach the bottom.
2
13
g 9.8m/s ;sin30 ;cos30
22
??
??? ??
??
??
??
1 kg, 5 mC
1 m
30°
E = 200 N/C
(1) 2.3 s (2) 1.3 s
(3) 0.92 s (4) 0.46 s
Answer (2)
Sol.
3
2
?N
N
9.8
2
9.8 3 1
22
?
?
30°
9.8 3 1 3
a 4.9 (0.2)
22
??
? ??
?? ?
??
??
??
= 2.24 m/s
2
21
T1.3s
(sin30) 2.24
?
??
?
4. Given below are two statements :
Statement I : An electric dipole is placed at the
centre of a hollow sphere. The flux of electric
field through the sphere is zero but the electric
field is not zero anywhere in the sphere.
Statement II : If R is the radius of a solid
metallic sphere and Q be the total charge on it.
The electric field at any point on the spherical
surface of radius r (< R) is zero but the electric
flux passing through this closed spherical
surface of radius r is not zero.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
Answer (3)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol. 
inc
Tot
0
q
??
?
q
inc
 = 0 ??
Tot
 = 0 +q –q
E0 ?
 
At P,
E = 0 and ? = 0
r
P
Q
5. A scooter accelerates from rest for time t
1
 at
constant rate a
1
 and then retards at constant
rate a
2
 for time t
2
 and comes to rest. The
correct value of 
1
2
t
t
 will be
(1)
12
1
aa
a
?
(2)
1
2
a
a
(3)
2
1
a
a
(4)
12
2
aa
a
?
Answer (3)
Sol. We have,
a
1
t
1
 = a
2
t
2
?
1 2
21
t a
ta
?
6. Find the peak current and resonant frequency
of the following circuit (as shown in figure).
120 ?
V = 30 sin 100t
100 mH
100 F ?
(1) 0.2 A and 50 Hz (2) 2 A and 100 Hz
(3) 2 A and 50 Hz (4) 0.2 A and 100 Hz
Answer (1)
Sol. ??
??
max
max
22
LC
V 30
I
Z
R(X X)
X
L
 = 100 × 0.1 = 10 ?
?
???
?
C
4
1
X 100
100 10
??
?
max
22
30
I0.2A
(120) (90)
?? ???
?
11
f 50Hz
LC 2 LC
7. The length of metallic wire is l
1
 when tension in
it is T
1
. It is l
2
 when the tension is T
2
. The original
length of the wire will be
(1)
?
?
11 22
21
Tl T l
TT
(2)
?
12
ll
2
(3)
?
?
21 12
12
Tl Tl
TT
(4)
?
?
21 12
21
Tl Tl
TT
Answer (4)
Sol. T
1
 = K(l
1
 – l
0
) ...(1)
T
2
 = K(l
2
 – l
0
) ...(2)
From (1) and (2),
?
?
?
21 12
0
21
Tl Tl
l
(TT )
8. The trajectory of a projectile in a vertical plane
is y = ?x – ?x
2
, where ? and ? are constants and
x & y are respectively the horizontal and vertical
distances of the projectile from the point of
projection. The angle of projection ? and the
maximum height attained H are respectively
given by
(1)
?
?
?
?
2
1
4
tan , (2)
?
?
?
?
2
1
tan ,
2
(3)
?
?? ??
??
??
??
2
1
tan , (4)
?
?
?
?
2
1
tan ,
4
Answer (4)
Sol. ?? ? ?
2
y xx
?
? ??? ? ?? ?
1
tan tan
also,  ?? ? ?
dy
2x
dx
?
? ? ?
?
dy
0x
dx 2
22 2
y
24 4
?? ?
???
?? ?
at x
2
?
?
?
9. An aeroplane, with its wings spread 10 m, is
flying at a speed of 180 km/h in a horizontal
direction. The total intensity of earth’s field at
that part is 2.5 × 10
–4
 Wb/m
2
 and the angle of dip
is 60°. The emf induced between the tips of the
plane wings will be _________ .
(1) 108.25 mV (2) 88.37 mV
(3) 62.50 mV (4) 54.125 mV
Answer (1)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol.?? ?
ind v v Total
(B )LV, B B sin60
?
?? ? ? ?
4
5
(2.5 10 )(sin60) 10 180
18
= 108.25 mV
10. If ‘C’ and ‘V’ represent capacity and voltage
respectively then what are the dimensions of ?
where C/V = ??
(1) [M
–1
 L
–3
 I
–2
 T
–7
](2)[M
–2
 L
–4
 I
3
 T
7
]
(3) [M
–2
 L
–3
 I
2
 T
6
](4)[M
–3
 L
–4
 I
3
 T
7
]
Answer (2)
Sol. ?
2
1
UCV
2
?
? ??
33
CU FL
V
VV
?
?
FL
V
IT
??
? ?
?
33
33
CFLIT
V
FL
    
??
?
2473
[M L T I ]
11. The recoil speed of a hydrogen atom after it
emits a photon in going from n = 5 state to n = 1
state will be
(1) 4.34 m/s (2) 2.19 m/s
(3) 4.17 m/s (4) 3.25 m/s
Answer (3)
Sol. ?
E
P
C
?
E
mv
C
? ?
E
v
mC
?
?
?? ?
?
?? ??
19
27 8
(13.6) 24 1.6 10
25 1.66 10 3 10
 v4.17m/s
12. Given below are two statements: one is labeled
as Assertion A and the other is labeled as
Reason R.
Assertion A : For a simple microscope, the
angular size of the object equals the angular
size of the image.
Reason R : Magnification is achieved as the
small object can be kept much closer to the eye
than 25 cm and hence it subtends a large angle.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) A is false but R is true
(2) A is true but R is false
(3) Both A and R are true but R is NOT the
correct explanation of A
(4) Both A and R are true and R is the correct
explanation of A
Answer (4)
Sol. Though image size is bigger than object size,
the angular size of the image is equal to the
angular size of object
13. Two masses A and B, each of mass M are fixed
together by a massless spring. A force acts on
the mass B as shown in figure. If the mass A
starts moving away from mass B with
acceleration ‘a’, then the acceleration of mass
B will be
FBA
(1)
? Ma F
M
(2)
?
MF
FMa
(3)
? FMa
M
(4)
? FMa
M
Answer (4)
Sol.
F
f
B M
f
a
f = Ma
F + f = Ma
1
? ??
? ?
??
??
1
Ff
a
M
 
? ??
?
??
??
FMa
M
14. A tuning fork A of unknown frequency
produces 5 beats/s with a fork of known
frequency 340 Hz. When fork A is filed, the beat
frequency decreases to 2 beats/s. What is the
frequency of fork A?
(1) 335 Hz (2) 338 Hz
(3) 345 Hz (4) 342 Hz
Answer (1)
Sol. f
A
 = 340 ± 5
If f
A
 increases, then beat frequency decreases
? f
A
 = 335
15. A wire of 1 ? has a length of 1 m. It is
stretched till its length increases by 25%. The
percentage change in resistance to the nearest
integer is
(1) 76% (2) 56%
(3) 12.5% (4) 25%
Answer (2)
Page 4


JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
PART–A : PHYSICS
SECTION - I
Multiple Choice Questions: This section contains
20 multiple choice questions. Each question has
4 choices (1), (2), (3) and (4), out of which ONL Y ONE
is correct.
Choose the correct answer :
1. The incident ray, reflected ray and the outward
drawn normal are denoted by the unit vectors
a,b andc
 
 respectively. Then choose the
correct relation for these vectors.
(1)ba 2c ??
 
(2)b2a c ??
 
(3)ba 2(ac)c ?? ?
 
(4)ba c ??
 
Answer (3)
Sol.ba 2cosc ?? ?
 
?
?
b

a

c

ba 2cosc ?? ?
 
ba 2(ac)c ?? ?
 
2. A radioactive sample is undergoing ? decay. At
any time t
1
, its activity is A and another time t
2
,
the activity is 
A
5
. What is the average life time
for the sample?
(1)
21
tt
ln5
?
(2)
21
ln(tt )
2
?
(3)
21
ln5
tt ?
(4)
12
tt
ln5
?
Answer (1)
Sol.
21
(t t )
A
Ae
5
?? ?
?
21
(t t )
1
e
5
?? ?
?
? ln 5 = ?(t
2
 – t
1
)
?
21
avg
(t t ) 1
t
ln5
?
??
?
3. An inclined plane making an angle of 30° with
the horizontal is placed in a uniform horizontal
electric field 
N
200
C
 as shown in the figure.
A body of mass 1 kg and charge 5 mC is
allowed to slide down from rest at a height of
1 m. If the coefficient of friction is 0.2, find the
time taken by the body to reach the bottom.
2
13
g 9.8m/s ;sin30 ;cos30
22
??
??? ??
??
??
??
1 kg, 5 mC
1 m
30°
E = 200 N/C
(1) 2.3 s (2) 1.3 s
(3) 0.92 s (4) 0.46 s
Answer (2)
Sol.
3
2
?N
N
9.8
2
9.8 3 1
22
?
?
30°
9.8 3 1 3
a 4.9 (0.2)
22
??
? ??
?? ?
??
??
??
= 2.24 m/s
2
21
T1.3s
(sin30) 2.24
?
??
?
4. Given below are two statements :
Statement I : An electric dipole is placed at the
centre of a hollow sphere. The flux of electric
field through the sphere is zero but the electric
field is not zero anywhere in the sphere.
Statement II : If R is the radius of a solid
metallic sphere and Q be the total charge on it.
The electric field at any point on the spherical
surface of radius r (< R) is zero but the electric
flux passing through this closed spherical
surface of radius r is not zero.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
Answer (3)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol. 
inc
Tot
0
q
??
?
q
inc
 = 0 ??
Tot
 = 0 +q –q
E0 ?
 
At P,
E = 0 and ? = 0
r
P
Q
5. A scooter accelerates from rest for time t
1
 at
constant rate a
1
 and then retards at constant
rate a
2
 for time t
2
 and comes to rest. The
correct value of 
1
2
t
t
 will be
(1)
12
1
aa
a
?
(2)
1
2
a
a
(3)
2
1
a
a
(4)
12
2
aa
a
?
Answer (3)
Sol. We have,
a
1
t
1
 = a
2
t
2
?
1 2
21
t a
ta
?
6. Find the peak current and resonant frequency
of the following circuit (as shown in figure).
120 ?
V = 30 sin 100t
100 mH
100 F ?
(1) 0.2 A and 50 Hz (2) 2 A and 100 Hz
(3) 2 A and 50 Hz (4) 0.2 A and 100 Hz
Answer (1)
Sol. ??
??
max
max
22
LC
V 30
I
Z
R(X X)
X
L
 = 100 × 0.1 = 10 ?
?
???
?
C
4
1
X 100
100 10
??
?
max
22
30
I0.2A
(120) (90)
?? ???
?
11
f 50Hz
LC 2 LC
7. The length of metallic wire is l
1
 when tension in
it is T
1
. It is l
2
 when the tension is T
2
. The original
length of the wire will be
(1)
?
?
11 22
21
Tl T l
TT
(2)
?
12
ll
2
(3)
?
?
21 12
12
Tl Tl
TT
(4)
?
?
21 12
21
Tl Tl
TT
Answer (4)
Sol. T
1
 = K(l
1
 – l
0
) ...(1)
T
2
 = K(l
2
 – l
0
) ...(2)
From (1) and (2),
?
?
?
21 12
0
21
Tl Tl
l
(TT )
8. The trajectory of a projectile in a vertical plane
is y = ?x – ?x
2
, where ? and ? are constants and
x & y are respectively the horizontal and vertical
distances of the projectile from the point of
projection. The angle of projection ? and the
maximum height attained H are respectively
given by
(1)
?
?
?
?
2
1
4
tan , (2)
?
?
?
?
2
1
tan ,
2
(3)
?
?? ??
??
??
??
2
1
tan , (4)
?
?
?
?
2
1
tan ,
4
Answer (4)
Sol. ?? ? ?
2
y xx
?
? ??? ? ?? ?
1
tan tan
also,  ?? ? ?
dy
2x
dx
?
? ? ?
?
dy
0x
dx 2
22 2
y
24 4
?? ?
???
?? ?
at x
2
?
?
?
9. An aeroplane, with its wings spread 10 m, is
flying at a speed of 180 km/h in a horizontal
direction. The total intensity of earth’s field at
that part is 2.5 × 10
–4
 Wb/m
2
 and the angle of dip
is 60°. The emf induced between the tips of the
plane wings will be _________ .
(1) 108.25 mV (2) 88.37 mV
(3) 62.50 mV (4) 54.125 mV
Answer (1)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol.?? ?
ind v v Total
(B )LV, B B sin60
?
?? ? ? ?
4
5
(2.5 10 )(sin60) 10 180
18
= 108.25 mV
10. If ‘C’ and ‘V’ represent capacity and voltage
respectively then what are the dimensions of ?
where C/V = ??
(1) [M
–1
 L
–3
 I
–2
 T
–7
](2)[M
–2
 L
–4
 I
3
 T
7
]
(3) [M
–2
 L
–3
 I
2
 T
6
](4)[M
–3
 L
–4
 I
3
 T
7
]
Answer (2)
Sol. ?
2
1
UCV
2
?
? ??
33
CU FL
V
VV
?
?
FL
V
IT
??
? ?
?
33
33
CFLIT
V
FL
    
??
?
2473
[M L T I ]
11. The recoil speed of a hydrogen atom after it
emits a photon in going from n = 5 state to n = 1
state will be
(1) 4.34 m/s (2) 2.19 m/s
(3) 4.17 m/s (4) 3.25 m/s
Answer (3)
Sol. ?
E
P
C
?
E
mv
C
? ?
E
v
mC
?
?
?? ?
?
?? ??
19
27 8
(13.6) 24 1.6 10
25 1.66 10 3 10
 v4.17m/s
12. Given below are two statements: one is labeled
as Assertion A and the other is labeled as
Reason R.
Assertion A : For a simple microscope, the
angular size of the object equals the angular
size of the image.
Reason R : Magnification is achieved as the
small object can be kept much closer to the eye
than 25 cm and hence it subtends a large angle.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) A is false but R is true
(2) A is true but R is false
(3) Both A and R are true but R is NOT the
correct explanation of A
(4) Both A and R are true and R is the correct
explanation of A
Answer (4)
Sol. Though image size is bigger than object size,
the angular size of the image is equal to the
angular size of object
13. Two masses A and B, each of mass M are fixed
together by a massless spring. A force acts on
the mass B as shown in figure. If the mass A
starts moving away from mass B with
acceleration ‘a’, then the acceleration of mass
B will be
FBA
(1)
? Ma F
M
(2)
?
MF
FMa
(3)
? FMa
M
(4)
? FMa
M
Answer (4)
Sol.
F
f
B M
f
a
f = Ma
F + f = Ma
1
? ??
? ?
??
??
1
Ff
a
M
 
? ??
?
??
??
FMa
M
14. A tuning fork A of unknown frequency
produces 5 beats/s with a fork of known
frequency 340 Hz. When fork A is filed, the beat
frequency decreases to 2 beats/s. What is the
frequency of fork A?
(1) 335 Hz (2) 338 Hz
(3) 345 Hz (4) 342 Hz
Answer (1)
Sol. f
A
 = 340 ± 5
If f
A
 increases, then beat frequency decreases
? f
A
 = 335
15. A wire of 1 ? has a length of 1 m. It is
stretched till its length increases by 25%. The
percentage change in resistance to the nearest
integer is
(1) 76% (2) 56%
(3) 12.5% (4) 25%
Answer (2)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol.
2
R
A(Vol.)
??
??

2
1
R ()
V
? ??
?
??
??

2
2
5
R
V4
? ??? ?
?
??? ?
??? ?

?
?? ? 
1
R9
100 100 56%
R16
16. Draw the output signal Y in the given
combination of gates.
A
0
1 2 3 4 5 t(s)
B
0 12 3 4 5 t(s)
A
B
Y
(1)
0 1234 5 t(s)
(2)
0 1234 5 t(s)
(3)
0 1234 5 t(s)
(4)
0 1234 5 t(s)
Answer (4)
Sol. A = 0 ? Y = 0 for B = 0 or B = 1
A = 1 ? Y = 0 for B = 1
A = 1 ? Y = 1 for B = 0
Hence output is 1 only when A = 1 and B = 0
?
0 1234 5 t(s)
17. A particle executes S.H.M., the graph of
velocity as a function of displacement is
(1) a parabola (2) a helix
(3) a circle (4) an ellipse
Answer (4)
Sol. ?? ?
22
vA x
? ??
?
2
22
2
v
xA
22
22
xv
1
A(A)
? ??
?
 .... Ellipse equation
18. The internal energy (U), pressure (P) and
volume (V) of an ideal gas are related as
U = 3PV + 4. The gas is
(1) diatomic only
(2) either monoatomic or diatomic
(3) polyatomic only
(4) monoatomic only
Answer (3)
Sol. U = (3)nRT + 4
dU = 3nRdT ? f = 6
19. Given below are two statements :
Statement I : A second’s pendulum has a time
period of 1 second.
Statement II : It takes precisely one second to
move between the two extreme positions.
In the light of the above Statement, choose the
correct answer from the options given
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Answer (4)
Sol. Time period of second pendulum is 2 sec
?? ?
T
t1
2
20. A cord is wound round the circumference of
wheel of radius r. The axis of the wheel is
horizontal and the moment of inertia about it is
I. A weight mg is attached to the cord at the
end. The weight falls from rest. After falling
through a distance ‘h’, the square of angular
velocity of wheel will be
(1) 2gh (2)
?
2
2mgh
I2mr
(3)
?
2
2mgh
Imr
(4)
?
2
2gh
Imr
Answer (3)
Sol. ?PE = ?KE
?? ? ?
22
11
mgh I m(r )
22
? ??
?
2
2
2mgh
Imr
Page 5


JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
PART–A : PHYSICS
SECTION - I
Multiple Choice Questions: This section contains
20 multiple choice questions. Each question has
4 choices (1), (2), (3) and (4), out of which ONL Y ONE
is correct.
Choose the correct answer :
1. The incident ray, reflected ray and the outward
drawn normal are denoted by the unit vectors
a,b andc
 
 respectively. Then choose the
correct relation for these vectors.
(1)ba 2c ??
 
(2)b2a c ??
 
(3)ba 2(ac)c ?? ?
 
(4)ba c ??
 
Answer (3)
Sol.ba 2cosc ?? ?
 
?
?
b

a

c

ba 2cosc ?? ?
 
ba 2(ac)c ?? ?
 
2. A radioactive sample is undergoing ? decay. At
any time t
1
, its activity is A and another time t
2
,
the activity is 
A
5
. What is the average life time
for the sample?
(1)
21
tt
ln5
?
(2)
21
ln(tt )
2
?
(3)
21
ln5
tt ?
(4)
12
tt
ln5
?
Answer (1)
Sol.
21
(t t )
A
Ae
5
?? ?
?
21
(t t )
1
e
5
?? ?
?
? ln 5 = ?(t
2
 – t
1
)
?
21
avg
(t t ) 1
t
ln5
?
??
?
3. An inclined plane making an angle of 30° with
the horizontal is placed in a uniform horizontal
electric field 
N
200
C
 as shown in the figure.
A body of mass 1 kg and charge 5 mC is
allowed to slide down from rest at a height of
1 m. If the coefficient of friction is 0.2, find the
time taken by the body to reach the bottom.
2
13
g 9.8m/s ;sin30 ;cos30
22
??
??? ??
??
??
??
1 kg, 5 mC
1 m
30°
E = 200 N/C
(1) 2.3 s (2) 1.3 s
(3) 0.92 s (4) 0.46 s
Answer (2)
Sol.
3
2
?N
N
9.8
2
9.8 3 1
22
?
?
30°
9.8 3 1 3
a 4.9 (0.2)
22
??
? ??
?? ?
??
??
??
= 2.24 m/s
2
21
T1.3s
(sin30) 2.24
?
??
?
4. Given below are two statements :
Statement I : An electric dipole is placed at the
centre of a hollow sphere. The flux of electric
field through the sphere is zero but the electric
field is not zero anywhere in the sphere.
Statement II : If R is the radius of a solid
metallic sphere and Q be the total charge on it.
The electric field at any point on the spherical
surface of radius r (< R) is zero but the electric
flux passing through this closed spherical
surface of radius r is not zero.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true
Answer (3)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol. 
inc
Tot
0
q
??
?
q
inc
 = 0 ??
Tot
 = 0 +q –q
E0 ?
 
At P,
E = 0 and ? = 0
r
P
Q
5. A scooter accelerates from rest for time t
1
 at
constant rate a
1
 and then retards at constant
rate a
2
 for time t
2
 and comes to rest. The
correct value of 
1
2
t
t
 will be
(1)
12
1
aa
a
?
(2)
1
2
a
a
(3)
2
1
a
a
(4)
12
2
aa
a
?
Answer (3)
Sol. We have,
a
1
t
1
 = a
2
t
2
?
1 2
21
t a
ta
?
6. Find the peak current and resonant frequency
of the following circuit (as shown in figure).
120 ?
V = 30 sin 100t
100 mH
100 F ?
(1) 0.2 A and 50 Hz (2) 2 A and 100 Hz
(3) 2 A and 50 Hz (4) 0.2 A and 100 Hz
Answer (1)
Sol. ??
??
max
max
22
LC
V 30
I
Z
R(X X)
X
L
 = 100 × 0.1 = 10 ?
?
???
?
C
4
1
X 100
100 10
??
?
max
22
30
I0.2A
(120) (90)
?? ???
?
11
f 50Hz
LC 2 LC
7. The length of metallic wire is l
1
 when tension in
it is T
1
. It is l
2
 when the tension is T
2
. The original
length of the wire will be
(1)
?
?
11 22
21
Tl T l
TT
(2)
?
12
ll
2
(3)
?
?
21 12
12
Tl Tl
TT
(4)
?
?
21 12
21
Tl Tl
TT
Answer (4)
Sol. T
1
 = K(l
1
 – l
0
) ...(1)
T
2
 = K(l
2
 – l
0
) ...(2)
From (1) and (2),
?
?
?
21 12
0
21
Tl Tl
l
(TT )
8. The trajectory of a projectile in a vertical plane
is y = ?x – ?x
2
, where ? and ? are constants and
x & y are respectively the horizontal and vertical
distances of the projectile from the point of
projection. The angle of projection ? and the
maximum height attained H are respectively
given by
(1)
?
?
?
?
2
1
4
tan , (2)
?
?
?
?
2
1
tan ,
2
(3)
?
?? ??
??
??
??
2
1
tan , (4)
?
?
?
?
2
1
tan ,
4
Answer (4)
Sol. ?? ? ?
2
y xx
?
? ??? ? ?? ?
1
tan tan
also,  ?? ? ?
dy
2x
dx
?
? ? ?
?
dy
0x
dx 2
22 2
y
24 4
?? ?
???
?? ?
at x
2
?
?
?
9. An aeroplane, with its wings spread 10 m, is
flying at a speed of 180 km/h in a horizontal
direction. The total intensity of earth’s field at
that part is 2.5 × 10
–4
 Wb/m
2
 and the angle of dip
is 60°. The emf induced between the tips of the
plane wings will be _________ .
(1) 108.25 mV (2) 88.37 mV
(3) 62.50 mV (4) 54.125 mV
Answer (1)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol.?? ?
ind v v Total
(B )LV, B B sin60
?
?? ? ? ?
4
5
(2.5 10 )(sin60) 10 180
18
= 108.25 mV
10. If ‘C’ and ‘V’ represent capacity and voltage
respectively then what are the dimensions of ?
where C/V = ??
(1) [M
–1
 L
–3
 I
–2
 T
–7
](2)[M
–2
 L
–4
 I
3
 T
7
]
(3) [M
–2
 L
–3
 I
2
 T
6
](4)[M
–3
 L
–4
 I
3
 T
7
]
Answer (2)
Sol. ?
2
1
UCV
2
?
? ??
33
CU FL
V
VV
?
?
FL
V
IT
??
? ?
?
33
33
CFLIT
V
FL
    
??
?
2473
[M L T I ]
11. The recoil speed of a hydrogen atom after it
emits a photon in going from n = 5 state to n = 1
state will be
(1) 4.34 m/s (2) 2.19 m/s
(3) 4.17 m/s (4) 3.25 m/s
Answer (3)
Sol. ?
E
P
C
?
E
mv
C
? ?
E
v
mC
?
?
?? ?
?
?? ??
19
27 8
(13.6) 24 1.6 10
25 1.66 10 3 10
 v4.17m/s
12. Given below are two statements: one is labeled
as Assertion A and the other is labeled as
Reason R.
Assertion A : For a simple microscope, the
angular size of the object equals the angular
size of the image.
Reason R : Magnification is achieved as the
small object can be kept much closer to the eye
than 25 cm and hence it subtends a large angle.
In the light of the above statements, choose the
most appropriate answer from the options given
below :
(1) A is false but R is true
(2) A is true but R is false
(3) Both A and R are true but R is NOT the
correct explanation of A
(4) Both A and R are true and R is the correct
explanation of A
Answer (4)
Sol. Though image size is bigger than object size,
the angular size of the image is equal to the
angular size of object
13. Two masses A and B, each of mass M are fixed
together by a massless spring. A force acts on
the mass B as shown in figure. If the mass A
starts moving away from mass B with
acceleration ‘a’, then the acceleration of mass
B will be
FBA
(1)
? Ma F
M
(2)
?
MF
FMa
(3)
? FMa
M
(4)
? FMa
M
Answer (4)
Sol.
F
f
B M
f
a
f = Ma
F + f = Ma
1
? ??
? ?
??
??
1
Ff
a
M
 
? ??
?
??
??
FMa
M
14. A tuning fork A of unknown frequency
produces 5 beats/s with a fork of known
frequency 340 Hz. When fork A is filed, the beat
frequency decreases to 2 beats/s. What is the
frequency of fork A?
(1) 335 Hz (2) 338 Hz
(3) 345 Hz (4) 342 Hz
Answer (1)
Sol. f
A
 = 340 ± 5
If f
A
 increases, then beat frequency decreases
? f
A
 = 335
15. A wire of 1 ? has a length of 1 m. It is
stretched till its length increases by 25%. The
percentage change in resistance to the nearest
integer is
(1) 76% (2) 56%
(3) 12.5% (4) 25%
Answer (2)
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
Sol.
2
R
A(Vol.)
??
??

2
1
R ()
V
? ??
?
??
??

2
2
5
R
V4
? ??? ?
?
??? ?
??? ?

?
?? ? 
1
R9
100 100 56%
R16
16. Draw the output signal Y in the given
combination of gates.
A
0
1 2 3 4 5 t(s)
B
0 12 3 4 5 t(s)
A
B
Y
(1)
0 1234 5 t(s)
(2)
0 1234 5 t(s)
(3)
0 1234 5 t(s)
(4)
0 1234 5 t(s)
Answer (4)
Sol. A = 0 ? Y = 0 for B = 0 or B = 1
A = 1 ? Y = 0 for B = 1
A = 1 ? Y = 1 for B = 0
Hence output is 1 only when A = 1 and B = 0
?
0 1234 5 t(s)
17. A particle executes S.H.M., the graph of
velocity as a function of displacement is
(1) a parabola (2) a helix
(3) a circle (4) an ellipse
Answer (4)
Sol. ?? ?
22
vA x
? ??
?
2
22
2
v
xA
22
22
xv
1
A(A)
? ??
?
 .... Ellipse equation
18. The internal energy (U), pressure (P) and
volume (V) of an ideal gas are related as
U = 3PV + 4. The gas is
(1) diatomic only
(2) either monoatomic or diatomic
(3) polyatomic only
(4) monoatomic only
Answer (3)
Sol. U = (3)nRT + 4
dU = 3nRdT ? f = 6
19. Given below are two statements :
Statement I : A second’s pendulum has a time
period of 1 second.
Statement II : It takes precisely one second to
move between the two extreme positions.
In the light of the above Statement, choose the
correct answer from the options given
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Answer (4)
Sol. Time period of second pendulum is 2 sec
?? ?
T
t1
2
20. A cord is wound round the circumference of
wheel of radius r. The axis of the wheel is
horizontal and the moment of inertia about it is
I. A weight mg is attached to the cord at the
end. The weight falls from rest. After falling
through a distance ‘h’, the square of angular
velocity of wheel will be
(1) 2gh (2)
?
2
2mgh
I2mr
(3)
?
2
2mgh
Imr
(4)
?
2
2gh
Imr
Answer (3)
Sol. ?PE = ?KE
?? ? ?
22
11
mgh I m(r )
22
? ??
?
2
2
2mgh
Imr
JEE (MAIN)-2021 Phase-1 (26-02-2021)-E
SECTION - II
Numerical Value Type Questions: This section
contains 10 questions. In Section II, attempt any five
questions out of 10. The answer to each question is a
NUMERICAL VALUE. For each question, enter the
correct numerical value (in decimal notation,
truncated/rounded-off to the second decimal place;
e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using
the mouse and the on-screen virtual numeric keypad
in the place designated to enter the answer.
1. Two stream of photons, possessing energies
equal to twice and ten times the work function
of metal are incident on the metal surface
successively. The value of ratio of maximum
velocities of the photoelectrons emitted in the
two respective cases is x : y. The value of x is
_________.
Answer (01.00)
Sol.
??
2
1
mv E–
2
? V
2
 = ?? ?
2
E–
m
?
?? ??
2
1
2
V
m
...(1)
?? ??
2
2
2
V9
m
...(2)
From (1) & (2),
?
1
2
V 1
V3
?
?
x1
y 3
? (x = 01.00)
2. In the reported figure of earth, the value of
acceleration due to gravity is same at point A
and C but it is smaller than that of its value at
point B (surface of the earth). The value of
OA : AB will be x : y. The value of x is ________.
R = 6400 km
O
A
Earth
B
C
3200 km
Answer (04.00)
Sol. g
C
 = 
??
??
??
2
GM
R
R+
2
 = 
0
4
g
9
g
A
 = g
C
 ?? 
??
??
??
00
4AB
g=g 1–
9R
? AB = 
5R
9
? OA = 
4R
9
?
OA
AB
  = 
4
5
? x = 04.00
3. The zener diode has a V
z
 = 30 V. The current
passing through the diode for the following
circuit is _______mA.
90 V
5 k
4 k
Answer (09.00)
Sol. I
1
 = ?
30
6mA
5,000
90 V
5 k
4 k
I
1
I
I
D
also – 90 + 4000 I + 30 = 0
? I = 15 mA
? I
D
 =  15 – 6 = 9 mA
4. Time period of a simple pendulum is T. The time
taken to complete 
5
8
 oscillations starting from
mean position is T
?
?
. The value of ? is______.
Answer (07.00)
Sol. 
5
oscillation
8
 = 
1
2
 oscillation +
1
8
 oscillation
?t = 
TT
+
212
 = 
7T
12
? ? = 7