JEE Main 2021 Physics February 26 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
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 ONLY ONE
is correct.
Choose the correct answer :
1. In a Young’s double slit experiment two slits are
separated by 2 mm and the screen is placed
one meter away. When a light of wavelength
500 nm is used, the fringe separation will be :
(1) 0.25 mm
(2) 0.75 mm
(3) 0.50 mm
(4) 1 mm
Answer (1)
Sol.
?
??
D
d
?
?
??
?
?
9
3
500 10 1
210
= 250 × 10
–6
= 0.25 mm
2. A planet revolving in elliptical orbit has :
A. a constant velocity of revolution.
B. has the least velocity when it is nearest to
the sun.
C. its areal velocity is directly proportional to
its velocity.
D. areal velocity is inversely proportional to its
velocity.
E. to follow a trajectory such that the areal
velocity is constant.
Choose the correct answer from the options
given below :
(1) A only
(2) E only
(3) D only
(4) C only
Answer (2)
Sol.
dA
dt
 = constant according to Kepler’s law
3. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : Body ‘P’ having mass M moving
with speed ‘u’ has head-on collision elastically
with another body ‘Q’ having mass ‘m’ initially
at rest. If m << M, body ‘Q’ will have a maximum
speed equal to ‘2u’ after collision.
Reason R : During  elastic collision, the
momentum and kinetic energy are both
conserved.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct and R is the
correct explanation of A
(2) A is not correct but R is correct
(3) A is correct but R is not correct
(4) Both A and R are correct but R is NOT the
correct explanation of A
Answer (1)
Sol. V
2
 = 
?
?
2M
u
Mm
?
1
m<<M
?
?? ?
2
2M
Vu2u
M
4. LED is constructed from Ga-As-P
semiconducting material. The energy gap of
this LED is 1.9 eV. Calculate the wavelength of
light emitted and its colour.
[h = 6.63 × 10
–34
 Js and c = 3 × 10
8
 ms
–1
]
(1) 654 nm and red colour
(2) 1046 nm and blue colour
(3) 1046 nm and red colour
(4) 654 nm and orange colour
Answer (1)
Sol. ?E
g
 = 1.9 eV
?
??
1242
nm
1.9
= 654 nm of red colour
Page 2


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
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 ONLY ONE
is correct.
Choose the correct answer :
1. In a Young’s double slit experiment two slits are
separated by 2 mm and the screen is placed
one meter away. When a light of wavelength
500 nm is used, the fringe separation will be :
(1) 0.25 mm
(2) 0.75 mm
(3) 0.50 mm
(4) 1 mm
Answer (1)
Sol.
?
??
D
d
?
?
??
?
?
9
3
500 10 1
210
= 250 × 10
–6
= 0.25 mm
2. A planet revolving in elliptical orbit has :
A. a constant velocity of revolution.
B. has the least velocity when it is nearest to
the sun.
C. its areal velocity is directly proportional to
its velocity.
D. areal velocity is inversely proportional to its
velocity.
E. to follow a trajectory such that the areal
velocity is constant.
Choose the correct answer from the options
given below :
(1) A only
(2) E only
(3) D only
(4) C only
Answer (2)
Sol.
dA
dt
 = constant according to Kepler’s law
3. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : Body ‘P’ having mass M moving
with speed ‘u’ has head-on collision elastically
with another body ‘Q’ having mass ‘m’ initially
at rest. If m << M, body ‘Q’ will have a maximum
speed equal to ‘2u’ after collision.
Reason R : During  elastic collision, the
momentum and kinetic energy are both
conserved.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct and R is the
correct explanation of A
(2) A is not correct but R is correct
(3) A is correct but R is not correct
(4) Both A and R are correct but R is NOT the
correct explanation of A
Answer (1)
Sol. V
2
 = 
?
?
2M
u
Mm
?
1
m<<M
?
?? ?
2
2M
Vu2u
M
4. LED is constructed from Ga-As-P
semiconducting material. The energy gap of
this LED is 1.9 eV. Calculate the wavelength of
light emitted and its colour.
[h = 6.63 × 10
–34
 Js and c = 3 × 10
8
 ms
–1
]
(1) 654 nm and red colour
(2) 1046 nm and blue colour
(3) 1046 nm and red colour
(4) 654 nm and orange colour
Answer (1)
Sol. ?E
g
 = 1.9 eV
?
??
1242
nm
1.9
= 654 nm of red colour
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
5. Assume that a tunnel is dug along a chord of
the earth, at a perpendicular distance (R/2)
from the earth’s centre, where ‘R’ is the radius
of the Earth. The wall of the tunnel is
frictionless. If a particle is released in this
tunnel, it will execute a simple harmonic motion
with a time period :
(1)
g
2R ?
(2)
2R
g
?
(3)
R
2
g
?
(4)
1 g
2R ?
Answer (3)
Sol.
x
r
??
?? ? ?
??
??
3
GMr x
Fm
r
R
?
??
2
23
dx GM
x
dt R
? ?? ??
3
RR
T2 2
GM g
6. A particle is moving with uniform speed along
the circumference of a circle of radius R under
the action of a central fictitious force F which
is inversely proportional to R
3
. Its time period
of revolution will be given by :
(1) ?
4
3
TR (2) ?
3
2
TR
(3) ?
2
TR (4) ?
5
2
TR
Answer (3)
Sol.
?
3
K
F
R
??
?
2
3
mv K
R R
??
?
1
v
R
?
?
?
2R
T
v
?
?
2
TR
7. The normal density of a material is ? and its bulk
modulus of elasticity is K. The magnitude of
increase in density of material, when a pressure
P is applied uniformly on all sides, will be :
(1)
?P
K
(2)
?K
P
(3)
?
K
P
(4)
?
PK
Answer (1)
Sol.
?
?
? ??
??
??
P
K
V
–
V
?
???
?
?
V
–
V
?
?
?
?? ??
??
?
??
P
K
?
?? ?
?
?
P
K
?
?
?? ?
P
K
8. An alternating current is given by the equation
??? ?
12
iisin t icos t . The rms current will be :
(1)
?? ?
2
12
1
ii
2
(2)
?? ?
12
1
ii
2
(3) ??
?
1
22
2
12
1
ii
2
(4) ??
?
1
22
2
12
1
ii
2
Answer (4)
Sol.??? ?
12
iisin t icos t
? ?? ?
?
T
2
2
0
rms
idt
i
T
??
?? ?? ? ? ?
?
?
T
22 2 2
12 12
0
i sin t i cos t 2ii sin t cos t dt
T
?? ?
2 2
1 2
i i
0
22
?
??
??
1
22
2
rms 1 2
1
iii
2
Page 3


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
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 ONLY ONE
is correct.
Choose the correct answer :
1. In a Young’s double slit experiment two slits are
separated by 2 mm and the screen is placed
one meter away. When a light of wavelength
500 nm is used, the fringe separation will be :
(1) 0.25 mm
(2) 0.75 mm
(3) 0.50 mm
(4) 1 mm
Answer (1)
Sol.
?
??
D
d
?
?
??
?
?
9
3
500 10 1
210
= 250 × 10
–6
= 0.25 mm
2. A planet revolving in elliptical orbit has :
A. a constant velocity of revolution.
B. has the least velocity when it is nearest to
the sun.
C. its areal velocity is directly proportional to
its velocity.
D. areal velocity is inversely proportional to its
velocity.
E. to follow a trajectory such that the areal
velocity is constant.
Choose the correct answer from the options
given below :
(1) A only
(2) E only
(3) D only
(4) C only
Answer (2)
Sol.
dA
dt
 = constant according to Kepler’s law
3. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : Body ‘P’ having mass M moving
with speed ‘u’ has head-on collision elastically
with another body ‘Q’ having mass ‘m’ initially
at rest. If m << M, body ‘Q’ will have a maximum
speed equal to ‘2u’ after collision.
Reason R : During  elastic collision, the
momentum and kinetic energy are both
conserved.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct and R is the
correct explanation of A
(2) A is not correct but R is correct
(3) A is correct but R is not correct
(4) Both A and R are correct but R is NOT the
correct explanation of A
Answer (1)
Sol. V
2
 = 
?
?
2M
u
Mm
?
1
m<<M
?
?? ?
2
2M
Vu2u
M
4. LED is constructed from Ga-As-P
semiconducting material. The energy gap of
this LED is 1.9 eV. Calculate the wavelength of
light emitted and its colour.
[h = 6.63 × 10
–34
 Js and c = 3 × 10
8
 ms
–1
]
(1) 654 nm and red colour
(2) 1046 nm and blue colour
(3) 1046 nm and red colour
(4) 654 nm and orange colour
Answer (1)
Sol. ?E
g
 = 1.9 eV
?
??
1242
nm
1.9
= 654 nm of red colour
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
5. Assume that a tunnel is dug along a chord of
the earth, at a perpendicular distance (R/2)
from the earth’s centre, where ‘R’ is the radius
of the Earth. The wall of the tunnel is
frictionless. If a particle is released in this
tunnel, it will execute a simple harmonic motion
with a time period :
(1)
g
2R ?
(2)
2R
g
?
(3)
R
2
g
?
(4)
1 g
2R ?
Answer (3)
Sol.
x
r
??
?? ? ?
??
??
3
GMr x
Fm
r
R
?
??
2
23
dx GM
x
dt R
? ?? ??
3
RR
T2 2
GM g
6. A particle is moving with uniform speed along
the circumference of a circle of radius R under
the action of a central fictitious force F which
is inversely proportional to R
3
. Its time period
of revolution will be given by :
(1) ?
4
3
TR (2) ?
3
2
TR
(3) ?
2
TR (4) ?
5
2
TR
Answer (3)
Sol.
?
3
K
F
R
??
?
2
3
mv K
R R
??
?
1
v
R
?
?
?
2R
T
v
?
?
2
TR
7. The normal density of a material is ? and its bulk
modulus of elasticity is K. The magnitude of
increase in density of material, when a pressure
P is applied uniformly on all sides, will be :
(1)
?P
K
(2)
?K
P
(3)
?
K
P
(4)
?
PK
Answer (1)
Sol.
?
?
? ??
??
??
P
K
V
–
V
?
???
?
?
V
–
V
?
?
?
?? ??
??
?
??
P
K
?
?? ?
?
?
P
K
?
?
?? ?
P
K
8. An alternating current is given by the equation
??? ?
12
iisin t icos t . The rms current will be :
(1)
?? ?
2
12
1
ii
2
(2)
?? ?
12
1
ii
2
(3) ??
?
1
22
2
12
1
ii
2
(4) ??
?
1
22
2
12
1
ii
2
Answer (4)
Sol.??? ?
12
iisin t icos t
? ?? ?
?
T
2
2
0
rms
idt
i
T
??
?? ?? ? ? ?
?
?
T
22 2 2
12 12
0
i sin t i cos t 2ii sin t cos t dt
T
?? ?
2 2
1 2
i i
0
22
?
??
??
1
22
2
rms 1 2
1
iii
2
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
9. The temperature ? at the junction of two
insulating sheets, having thermal resistances R
1
and R
2
 as well as top and bottom temperatures
?
1 
and ?
2
 (as shown in figure) is given by :
?
2
?
?
1
R
2
R
1
(1)
??
12 2 1
21
R– R
R–R
(2)
??
22 1 1
21
R– R
R–R
(3)
???
?
11 2 2
12
RR
RR
(4)
???
?
12 2 1
12
RR
RR
Answer (4)
Sol.
R
2
?
2
?
?
1
R
1
?? ??
?
1 2
21
– –
RR
???? ?? ?
21 1 2 1 2
R– R R – R
?
???
??
?
12 2 1
12
RR
RR
10. In a typical combustion engine the work done by
a gas molecule is given by 
?
?? ?
2
x
–
2
kT
We
, where x
is the displacement, k is the Boltzmann constant
and T is the temperature, If ? and ? are
constants, dimensions of ? will be :
(1) [MLT
–1
](2)[M
0
LT
0
]
(3) [MLT
–2
](4)[M
2
LT
–2
]
Answer (2)
Sol.
?
?? ?
2
x
–
2
kT
We
??
??
?? ? ?
??
??
2–2
–2
22
kT ML T
MT
xL
?? ?? ?
??
22–2
MLT
?
?? ??
??
2–2
2
–2
MLT
MT
? ? ? ??
00
MLT
11. Find the gravitational force of attraction
between the ring and sphere as shown in the
diagram, where the plane of the ring is
perpendicular to the line joining the centres. If
8R is the distance between the centres of a
ring (of mass ‘m’) and a sphere (mass ‘M’)
where both have equal radius ‘R’
M
R
Y
m
X
R
8R
(1)
?
2
1GMm
R 38
(2) ?
8GmM
9R
(3)
?
2
8 GmM
27
R
(4)
?
2
22 GMm
3
R
Answer (3)
Sol. Sphere can be supposed to be concentrated at
centre.
m
R
8R
M
3/2
22
GMm 8R
F
R8R
?
??
?
??
2
8GMm
F
27R
?
12. If ?
1
 and ?
2
 are the wavelengths of the third
member of Lyman and first member of the
Paschen series respectively, then the value of
?
1
 : ?
2 
 is
(1) 1 : 3 (2) 7 : 108
(3) 7 : 135 (4) 1 : 9
Answer (3)
Sol.
??
?? ???
??
? ??
1
1
111 16
R
116 15R
??
?? ???
??
? ??
2
2
1 1 1 144
R
916 7R
?
?? ?
?
1
2
16 7R 7
15R 144 135
Page 4


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
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 ONLY ONE
is correct.
Choose the correct answer :
1. In a Young’s double slit experiment two slits are
separated by 2 mm and the screen is placed
one meter away. When a light of wavelength
500 nm is used, the fringe separation will be :
(1) 0.25 mm
(2) 0.75 mm
(3) 0.50 mm
(4) 1 mm
Answer (1)
Sol.
?
??
D
d
?
?
??
?
?
9
3
500 10 1
210
= 250 × 10
–6
= 0.25 mm
2. A planet revolving in elliptical orbit has :
A. a constant velocity of revolution.
B. has the least velocity when it is nearest to
the sun.
C. its areal velocity is directly proportional to
its velocity.
D. areal velocity is inversely proportional to its
velocity.
E. to follow a trajectory such that the areal
velocity is constant.
Choose the correct answer from the options
given below :
(1) A only
(2) E only
(3) D only
(4) C only
Answer (2)
Sol.
dA
dt
 = constant according to Kepler’s law
3. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : Body ‘P’ having mass M moving
with speed ‘u’ has head-on collision elastically
with another body ‘Q’ having mass ‘m’ initially
at rest. If m << M, body ‘Q’ will have a maximum
speed equal to ‘2u’ after collision.
Reason R : During  elastic collision, the
momentum and kinetic energy are both
conserved.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct and R is the
correct explanation of A
(2) A is not correct but R is correct
(3) A is correct but R is not correct
(4) Both A and R are correct but R is NOT the
correct explanation of A
Answer (1)
Sol. V
2
 = 
?
?
2M
u
Mm
?
1
m<<M
?
?? ?
2
2M
Vu2u
M
4. LED is constructed from Ga-As-P
semiconducting material. The energy gap of
this LED is 1.9 eV. Calculate the wavelength of
light emitted and its colour.
[h = 6.63 × 10
–34
 Js and c = 3 × 10
8
 ms
–1
]
(1) 654 nm and red colour
(2) 1046 nm and blue colour
(3) 1046 nm and red colour
(4) 654 nm and orange colour
Answer (1)
Sol. ?E
g
 = 1.9 eV
?
??
1242
nm
1.9
= 654 nm of red colour
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
5. Assume that a tunnel is dug along a chord of
the earth, at a perpendicular distance (R/2)
from the earth’s centre, where ‘R’ is the radius
of the Earth. The wall of the tunnel is
frictionless. If a particle is released in this
tunnel, it will execute a simple harmonic motion
with a time period :
(1)
g
2R ?
(2)
2R
g
?
(3)
R
2
g
?
(4)
1 g
2R ?
Answer (3)
Sol.
x
r
??
?? ? ?
??
??
3
GMr x
Fm
r
R
?
??
2
23
dx GM
x
dt R
? ?? ??
3
RR
T2 2
GM g
6. A particle is moving with uniform speed along
the circumference of a circle of radius R under
the action of a central fictitious force F which
is inversely proportional to R
3
. Its time period
of revolution will be given by :
(1) ?
4
3
TR (2) ?
3
2
TR
(3) ?
2
TR (4) ?
5
2
TR
Answer (3)
Sol.
?
3
K
F
R
??
?
2
3
mv K
R R
??
?
1
v
R
?
?
?
2R
T
v
?
?
2
TR
7. The normal density of a material is ? and its bulk
modulus of elasticity is K. The magnitude of
increase in density of material, when a pressure
P is applied uniformly on all sides, will be :
(1)
?P
K
(2)
?K
P
(3)
?
K
P
(4)
?
PK
Answer (1)
Sol.
?
?
? ??
??
??
P
K
V
–
V
?
???
?
?
V
–
V
?
?
?
?? ??
??
?
??
P
K
?
?? ?
?
?
P
K
?
?
?? ?
P
K
8. An alternating current is given by the equation
??? ?
12
iisin t icos t . The rms current will be :
(1)
?? ?
2
12
1
ii
2
(2)
?? ?
12
1
ii
2
(3) ??
?
1
22
2
12
1
ii
2
(4) ??
?
1
22
2
12
1
ii
2
Answer (4)
Sol.??? ?
12
iisin t icos t
? ?? ?
?
T
2
2
0
rms
idt
i
T
??
?? ?? ? ? ?
?
?
T
22 2 2
12 12
0
i sin t i cos t 2ii sin t cos t dt
T
?? ?
2 2
1 2
i i
0
22
?
??
??
1
22
2
rms 1 2
1
iii
2
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
9. The temperature ? at the junction of two
insulating sheets, having thermal resistances R
1
and R
2
 as well as top and bottom temperatures
?
1 
and ?
2
 (as shown in figure) is given by :
?
2
?
?
1
R
2
R
1
(1)
??
12 2 1
21
R– R
R–R
(2)
??
22 1 1
21
R– R
R–R
(3)
???
?
11 2 2
12
RR
RR
(4)
???
?
12 2 1
12
RR
RR
Answer (4)
Sol.
R
2
?
2
?
?
1
R
1
?? ??
?
1 2
21
– –
RR
???? ?? ?
21 1 2 1 2
R– R R – R
?
???
??
?
12 2 1
12
RR
RR
10. In a typical combustion engine the work done by
a gas molecule is given by 
?
?? ?
2
x
–
2
kT
We
, where x
is the displacement, k is the Boltzmann constant
and T is the temperature, If ? and ? are
constants, dimensions of ? will be :
(1) [MLT
–1
](2)[M
0
LT
0
]
(3) [MLT
–2
](4)[M
2
LT
–2
]
Answer (2)
Sol.
?
?? ?
2
x
–
2
kT
We
??
??
?? ? ?
??
??
2–2
–2
22
kT ML T
MT
xL
?? ?? ?
??
22–2
MLT
?
?? ??
??
2–2
2
–2
MLT
MT
? ? ? ??
00
MLT
11. Find the gravitational force of attraction
between the ring and sphere as shown in the
diagram, where the plane of the ring is
perpendicular to the line joining the centres. If
8R is the distance between the centres of a
ring (of mass ‘m’) and a sphere (mass ‘M’)
where both have equal radius ‘R’
M
R
Y
m
X
R
8R
(1)
?
2
1GMm
R 38
(2) ?
8GmM
9R
(3)
?
2
8 GmM
27
R
(4)
?
2
22 GMm
3
R
Answer (3)
Sol. Sphere can be supposed to be concentrated at
centre.
m
R
8R
M
3/2
22
GMm 8R
F
R8R
?
??
?
??
2
8GMm
F
27R
?
12. If ?
1
 and ?
2
 are the wavelengths of the third
member of Lyman and first member of the
Paschen series respectively, then the value of
?
1
 : ?
2 
 is
(1) 1 : 3 (2) 7 : 108
(3) 7 : 135 (4) 1 : 9
Answer (3)
Sol.
??
?? ???
??
? ??
1
1
111 16
R
116 15R
??
?? ???
??
? ??
2
2
1 1 1 144
R
916 7R
?
?? ?
?
1
2
16 7R 7
15R 144 135
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
13. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : An electron microscope can
achieve better resolving power than an optical
microscope.
Reason R : The de Broglie’s wavelength of the
electrons emitted from an electron gun is much
less than wavelength of visible light.
In the light of the above statements, choose the
correct answer from the options given below :
(1) A is false but R is true.
(2) Both A and R are true and R is the correct
explanation of A.
(3) Both A and R are true but R is NOT the
correct explanation of A.
(4) A is true but R is false.
Answer (2)
Sol. Resolving power increases on decreasing the
wavelength.
14. Five equal resistances are connected in a
network as shown in figure. The net resistance
between the points A and B is :
D
C
R
R
E
R
R
R
A
B
(1)
R
2
(2)
3R
2
(3) R (4) 2R
Answer (3)
Sol. E and D are at same potential
So, 
??
eq
11 1
R2R2R
R
eq
 = R
15. If two similar springs each of spring constant
K
1
 are joined in series, the new spring constant
and time period would be changed by a
factor :
(1)
1
,2 2
4
(2)
1
,2 2
2
(3)
1
,2
2
(4)
1
,2
4
Answer (3)
Sol.
??
1
11 1
KKK
?
1
K
K
2
?
1
T
K
16. Four identical solid spheres each of mass ‘m’
and radius ‘a’ are placed with their centres on
the four corners of a square of side ‘b’. The
moment of inertia of the system about one side
of square where the axis of rotation is parallel
to the plane of the square is :
(1)
2
4
ma
5
(2)
22
8
ma +mb
5
(3)
22
8
ma +2mb
5
(4)
22
4
ma +2mb
5
Answer (3)
Sol.
b
????
22
2
Ima 42mb
5
??
22
8
ma 2mb
5
17. Consider the combination of 2 capacitors C
1
and C
2
, with C
2
 > C
1
, when connected in
parallel, the equivalent capacitance is 
15
4
times
the equivalent capacitance of the same
connected in series. Calculate the ratio of
capacitors, 
2
1
C
C
.
(1)
15
4
(2)
29
15
(3)
111
80
(4)
15
11
Page 5


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
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 ONLY ONE
is correct.
Choose the correct answer :
1. In a Young’s double slit experiment two slits are
separated by 2 mm and the screen is placed
one meter away. When a light of wavelength
500 nm is used, the fringe separation will be :
(1) 0.25 mm
(2) 0.75 mm
(3) 0.50 mm
(4) 1 mm
Answer (1)
Sol.
?
??
D
d
?
?
??
?
?
9
3
500 10 1
210
= 250 × 10
–6
= 0.25 mm
2. A planet revolving in elliptical orbit has :
A. a constant velocity of revolution.
B. has the least velocity when it is nearest to
the sun.
C. its areal velocity is directly proportional to
its velocity.
D. areal velocity is inversely proportional to its
velocity.
E. to follow a trajectory such that the areal
velocity is constant.
Choose the correct answer from the options
given below :
(1) A only
(2) E only
(3) D only
(4) C only
Answer (2)
Sol.
dA
dt
 = constant according to Kepler’s law
3. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : Body ‘P’ having mass M moving
with speed ‘u’ has head-on collision elastically
with another body ‘Q’ having mass ‘m’ initially
at rest. If m << M, body ‘Q’ will have a maximum
speed equal to ‘2u’ after collision.
Reason R : During  elastic collision, the
momentum and kinetic energy are both
conserved.
In the light of the above statements, choose the
most appropriate answer from the options
given below :
(1) Both A and R are correct and R is the
correct explanation of A
(2) A is not correct but R is correct
(3) A is correct but R is not correct
(4) Both A and R are correct but R is NOT the
correct explanation of A
Answer (1)
Sol. V
2
 = 
?
?
2M
u
Mm
?
1
m<<M
?
?? ?
2
2M
Vu2u
M
4. LED is constructed from Ga-As-P
semiconducting material. The energy gap of
this LED is 1.9 eV. Calculate the wavelength of
light emitted and its colour.
[h = 6.63 × 10
–34
 Js and c = 3 × 10
8
 ms
–1
]
(1) 654 nm and red colour
(2) 1046 nm and blue colour
(3) 1046 nm and red colour
(4) 654 nm and orange colour
Answer (1)
Sol. ?E
g
 = 1.9 eV
?
??
1242
nm
1.9
= 654 nm of red colour
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
5. Assume that a tunnel is dug along a chord of
the earth, at a perpendicular distance (R/2)
from the earth’s centre, where ‘R’ is the radius
of the Earth. The wall of the tunnel is
frictionless. If a particle is released in this
tunnel, it will execute a simple harmonic motion
with a time period :
(1)
g
2R ?
(2)
2R
g
?
(3)
R
2
g
?
(4)
1 g
2R ?
Answer (3)
Sol.
x
r
??
?? ? ?
??
??
3
GMr x
Fm
r
R
?
??
2
23
dx GM
x
dt R
? ?? ??
3
RR
T2 2
GM g
6. A particle is moving with uniform speed along
the circumference of a circle of radius R under
the action of a central fictitious force F which
is inversely proportional to R
3
. Its time period
of revolution will be given by :
(1) ?
4
3
TR (2) ?
3
2
TR
(3) ?
2
TR (4) ?
5
2
TR
Answer (3)
Sol.
?
3
K
F
R
??
?
2
3
mv K
R R
??
?
1
v
R
?
?
?
2R
T
v
?
?
2
TR
7. The normal density of a material is ? and its bulk
modulus of elasticity is K. The magnitude of
increase in density of material, when a pressure
P is applied uniformly on all sides, will be :
(1)
?P
K
(2)
?K
P
(3)
?
K
P
(4)
?
PK
Answer (1)
Sol.
?
?
? ??
??
??
P
K
V
–
V
?
???
?
?
V
–
V
?
?
?
?? ??
??
?
??
P
K
?
?? ?
?
?
P
K
?
?
?? ?
P
K
8. An alternating current is given by the equation
??? ?
12
iisin t icos t . The rms current will be :
(1)
?? ?
2
12
1
ii
2
(2)
?? ?
12
1
ii
2
(3) ??
?
1
22
2
12
1
ii
2
(4) ??
?
1
22
2
12
1
ii
2
Answer (4)
Sol.??? ?
12
iisin t icos t
? ?? ?
?
T
2
2
0
rms
idt
i
T
??
?? ?? ? ? ?
?
?
T
22 2 2
12 12
0
i sin t i cos t 2ii sin t cos t dt
T
?? ?
2 2
1 2
i i
0
22
?
??
??
1
22
2
rms 1 2
1
iii
2
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
9. The temperature ? at the junction of two
insulating sheets, having thermal resistances R
1
and R
2
 as well as top and bottom temperatures
?
1 
and ?
2
 (as shown in figure) is given by :
?
2
?
?
1
R
2
R
1
(1)
??
12 2 1
21
R– R
R–R
(2)
??
22 1 1
21
R– R
R–R
(3)
???
?
11 2 2
12
RR
RR
(4)
???
?
12 2 1
12
RR
RR
Answer (4)
Sol.
R
2
?
2
?
?
1
R
1
?? ??
?
1 2
21
– –
RR
???? ?? ?
21 1 2 1 2
R– R R – R
?
???
??
?
12 2 1
12
RR
RR
10. In a typical combustion engine the work done by
a gas molecule is given by 
?
?? ?
2
x
–
2
kT
We
, where x
is the displacement, k is the Boltzmann constant
and T is the temperature, If ? and ? are
constants, dimensions of ? will be :
(1) [MLT
–1
](2)[M
0
LT
0
]
(3) [MLT
–2
](4)[M
2
LT
–2
]
Answer (2)
Sol.
?
?? ?
2
x
–
2
kT
We
??
??
?? ? ?
??
??
2–2
–2
22
kT ML T
MT
xL
?? ?? ?
??
22–2
MLT
?
?? ??
??
2–2
2
–2
MLT
MT
? ? ? ??
00
MLT
11. Find the gravitational force of attraction
between the ring and sphere as shown in the
diagram, where the plane of the ring is
perpendicular to the line joining the centres. If
8R is the distance between the centres of a
ring (of mass ‘m’) and a sphere (mass ‘M’)
where both have equal radius ‘R’
M
R
Y
m
X
R
8R
(1)
?
2
1GMm
R 38
(2) ?
8GmM
9R
(3)
?
2
8 GmM
27
R
(4)
?
2
22 GMm
3
R
Answer (3)
Sol. Sphere can be supposed to be concentrated at
centre.
m
R
8R
M
3/2
22
GMm 8R
F
R8R
?
??
?
??
2
8GMm
F
27R
?
12. If ?
1
 and ?
2
 are the wavelengths of the third
member of Lyman and first member of the
Paschen series respectively, then the value of
?
1
 : ?
2 
 is
(1) 1 : 3 (2) 7 : 108
(3) 7 : 135 (4) 1 : 9
Answer (3)
Sol.
??
?? ???
??
? ??
1
1
111 16
R
116 15R
??
?? ???
??
? ??
2
2
1 1 1 144
R
916 7R
?
?? ?
?
1
2
16 7R 7
15R 144 135
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
13. Given below are two statements : one is
labelled as Assertion A and the other is labelled
as Reason R.
Assertion A : An electron microscope can
achieve better resolving power than an optical
microscope.
Reason R : The de Broglie’s wavelength of the
electrons emitted from an electron gun is much
less than wavelength of visible light.
In the light of the above statements, choose the
correct answer from the options given below :
(1) A is false but R is true.
(2) Both A and R are true and R is the correct
explanation of A.
(3) Both A and R are true but R is NOT the
correct explanation of A.
(4) A is true but R is false.
Answer (2)
Sol. Resolving power increases on decreasing the
wavelength.
14. Five equal resistances are connected in a
network as shown in figure. The net resistance
between the points A and B is :
D
C
R
R
E
R
R
R
A
B
(1)
R
2
(2)
3R
2
(3) R (4) 2R
Answer (3)
Sol. E and D are at same potential
So, 
??
eq
11 1
R2R2R
R
eq
 = R
15. If two similar springs each of spring constant
K
1
 are joined in series, the new spring constant
and time period would be changed by a
factor :
(1)
1
,2 2
4
(2)
1
,2 2
2
(3)
1
,2
2
(4)
1
,2
4
Answer (3)
Sol.
??
1
11 1
KKK
?
1
K
K
2
?
1
T
K
16. Four identical solid spheres each of mass ‘m’
and radius ‘a’ are placed with their centres on
the four corners of a square of side ‘b’. The
moment of inertia of the system about one side
of square where the axis of rotation is parallel
to the plane of the square is :
(1)
2
4
ma
5
(2)
22
8
ma +mb
5
(3)
22
8
ma +2mb
5
(4)
22
4
ma +2mb
5
Answer (3)
Sol.
b
????
22
2
Ima 42mb
5
??
22
8
ma 2mb
5
17. Consider the combination of 2 capacitors C
1
and C
2
, with C
2
 > C
1
, when connected in
parallel, the equivalent capacitance is 
15
4
times
the equivalent capacitance of the same
connected in series. Calculate the ratio of
capacitors, 
2
1
C
C
.
(1)
15
4
(2)
29
15
(3)
111
80
(4)
15
11
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
Answer (None of the option)
Sol.
?? ??
?
12
12
12
CC 4
CC
CC 15
?? ??
2
12 12
4 C C 15C C
Put 
?
2
1
C
x
C
4 (x + 1)
2
 = 15 x
None of the values of x satisfies the condition.
18. Find the electric field at point P (as shown in
figure) on the perpendicular bisector of a
uniformly charged thin wire of length L carrying
a charge Q. The distance of the point P from
the centre of the rod is ?
3
aL
2
.
a
L
E
P
O
Q
(1)
??
2
0
Q
3L
(2)
??
2
0
Q
4L
(3)
??
2
0
3Q
4L
(4)
??
2
0
Q
23 L
Answer (4)
Sol.
?
?
Q
=
L
?
?
??
0
E= ×2sin
4a
? = 30°
?
??
?? ?
2
0
0
QQ
E=
L
23 L
4L 3
2
19. A large number of water drops, each of radius
r, combine to have a drop of radius R. If the
surface tension is T and mechanical equivalent
of heat is J, the rise in heat energy per unit
volume will be
(1)
??
?
??
??
2T 1 1
Jr R
(2)
3T
rJ
(3)
??
?
??
??
3T 1 1
Jr R
(4)
2T
rJ
Answer (3)
Sol.
nr
3
 = R
3
H = T[4 ?nr
2
 – 4 ? R
2
]
Rise in Heat energy/volume (Q)
Q = 
?
??
?
??
?
22
3
4T
nr R
4
Jn. r
3
Solving we get
??
??
??
??
3T 1 1
Q
Jr R
20. A short straight object of height 100 cm lies
before the central axis of a spherical mirror
whose focal length has absolute value |f| = 40 cm.
The image of object produced by the mirror is
of height 25 cm and has the same orientation
of the object. One may conclude from the
information :
(1) Image is virtual, opposite side of convex
mirror
(2) Image is virtual, opposite side of concave
mirror
(3) Image is real, same side of convex mirror
(4) Image is real, same side of concave mirror
Answer (1)
Sol. Image is diminished  and magnification is
positive.
? It is possible if object is placed in front of
convex mirror.