JEE Mains 26 Feb 2021 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


 
 
 
 
SECTION – A 
 
1. A tuning fork A of unknown frequency produces 5beats/s with a fork of known frequency 340 
HZ. When fork A filed, the beat frequency decreases to 2beats/s. What is the frequency of fork 
A? 
(1) 342 Hz 
(2) 335 Hz 
(3) 338 Hz 
(4) 345 Hz 
Sol. (2) 
 Given 
 BeforeFiled: 
  
 So answer should be 335 Hz or 345 Hz.  
 After Filed : 
  
  
 After filed beat/sec decreases only in case of 335 Hz. 
 
340 
335   
345 
Filed  
Filed  
336, 337, 338………   
346, 347, 348………   
unknown 
frequency ( ?)
340Hz
2 beats/sec
unknown 
frequency
known 
frequency 
340Hz
5 beats/sec
26
th
 Feb. 2021 | Shift - 2
PHYSICS
Page 2


 
 
 
 
SECTION – A 
 
1. A tuning fork A of unknown frequency produces 5beats/s with a fork of known frequency 340 
HZ. When fork A filed, the beat frequency decreases to 2beats/s. What is the frequency of fork 
A? 
(1) 342 Hz 
(2) 335 Hz 
(3) 338 Hz 
(4) 345 Hz 
Sol. (2) 
 Given 
 BeforeFiled: 
  
 So answer should be 335 Hz or 345 Hz.  
 After Filed : 
  
  
 After filed beat/sec decreases only in case of 335 Hz. 
 
340 
335   
345 
Filed  
Filed  
336, 337, 338………   
346, 347, 348………   
unknown 
frequency ( ?)
340Hz
2 beats/sec
unknown 
frequency
known 
frequency 
340Hz
5 beats/sec
26
th
 Feb. 2021 | Shift - 2
PHYSICS
 
 
2. The trajectory a projectile in a vertical plane is y = ?x – ?x
2
, where ? and ? are constants and x 
& y are respectively the horizontal and vertical distance 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
tan ,
4
?
?
?
?
 
(2) 
2
1
tan ,
2
?
?
?
?
 
(3)
2
1
tan ,
? ?
? ?
?
? ?
? ?
? ?
 
 
(4)
2
1
4
tan ,
?
?
?
?
 
Sol. (1) 
 Given : 
 y = ?x – ?x
2
 ....(1) 
 for maximum height, we should find out maximum value of y from equation (1) 
 so, for maximum value of y 
 
dy
dx
 = 0 ? ? – 2 ?x = 0  
 x = 
2
?
?
 ....(2) 
 Now, put value of x from equation (2) in quation (1) 
 y = ?
2
? ? ?
? ?
?
? ?
 - ?
2
2
4
? ? ?
? ?
?
? ?
 
 ? ?
2
2
? ? ?
? ?
?
? ?
 - 
2
4
? ? ?
? ?
?
? ?
?
2
4
?
?
 
 So, H
max
 = 
2
4
?
?
 ....(3) 
 As we know maximum height H
max
 = 
2 2
u sin
2g
?
 ...(4) 
 from (3) and (4) u
2
 = 
2
4
? ? ?
? ?
?
? ?
2
2g
sin
? ?
? ?
?
? ?
 
 and range (R) = 2x = 
2
u 2sin cos
g
? ? ?
 
 2 
2
? ? ?
? ?
?
? ?
 = 
2
2
2g
2sin cos
4 sin
g
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ? ? ?
 
 tan ? = ? ? ? = tan
-1
 ( ?) 
 
Page 3


 
 
 
 
SECTION – A 
 
1. A tuning fork A of unknown frequency produces 5beats/s with a fork of known frequency 340 
HZ. When fork A filed, the beat frequency decreases to 2beats/s. What is the frequency of fork 
A? 
(1) 342 Hz 
(2) 335 Hz 
(3) 338 Hz 
(4) 345 Hz 
Sol. (2) 
 Given 
 BeforeFiled: 
  
 So answer should be 335 Hz or 345 Hz.  
 After Filed : 
  
  
 After filed beat/sec decreases only in case of 335 Hz. 
 
340 
335   
345 
Filed  
Filed  
336, 337, 338………   
346, 347, 348………   
unknown 
frequency ( ?)
340Hz
2 beats/sec
unknown 
frequency
known 
frequency 
340Hz
5 beats/sec
26
th
 Feb. 2021 | Shift - 2
PHYSICS
 
 
2. The trajectory a projectile in a vertical plane is y = ?x – ?x
2
, where ? and ? are constants and x 
& y are respectively the horizontal and vertical distance 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
tan ,
4
?
?
?
?
 
(2) 
2
1
tan ,
2
?
?
?
?
 
(3)
2
1
tan ,
? ?
? ?
?
? ?
? ?
? ?
 
 
(4)
2
1
4
tan ,
?
?
?
?
 
Sol. (1) 
 Given : 
 y = ?x – ?x
2
 ....(1) 
 for maximum height, we should find out maximum value of y from equation (1) 
 so, for maximum value of y 
 
dy
dx
 = 0 ? ? – 2 ?x = 0  
 x = 
2
?
?
 ....(2) 
 Now, put value of x from equation (2) in quation (1) 
 y = ?
2
? ? ?
? ?
?
? ?
 - ?
2
2
4
? ? ?
? ?
?
? ?
 
 ? ?
2
2
? ? ?
? ?
?
? ?
 - 
2
4
? ? ?
? ?
?
? ?
?
2
4
?
?
 
 So, H
max
 = 
2
4
?
?
 ....(3) 
 As we know maximum height H
max
 = 
2 2
u sin
2g
?
 ...(4) 
 from (3) and (4) u
2
 = 
2
4
? ? ?
? ?
?
? ?
2
2g
sin
? ?
? ?
?
? ?
 
 and range (R) = 2x = 
2
u 2sin cos
g
? ? ?
 
 2 
2
? ? ?
? ?
?
? ?
 = 
2
2
2g
2sin cos
4 sin
g
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ? ? ?
 
 tan ? = ? ? ? = tan
-1
 ( ?) 
 
 
 
3. 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) 
2
2gh
I mr ?
 
(2) 2gh 
(3)
2
2
2
mgh
I mr ?
 
(4)
2
2mgh
I mr ?
 
Sol. (4) 
 
 
m
m
h
A
B
R 
?
? using energy conservation between A and B point 
 mgh = 
1
2
 m (wR)
2
 + 
1
2
 I ?
2
 
 2mgh = (MR
2
 + I) ?
2
 
 ?
2
 = 
2
2mgh
I MR ?
 
 
 
 
 
 
 
 
Page 4


 
 
 
 
SECTION – A 
 
1. A tuning fork A of unknown frequency produces 5beats/s with a fork of known frequency 340 
HZ. When fork A filed, the beat frequency decreases to 2beats/s. What is the frequency of fork 
A? 
(1) 342 Hz 
(2) 335 Hz 
(3) 338 Hz 
(4) 345 Hz 
Sol. (2) 
 Given 
 BeforeFiled: 
  
 So answer should be 335 Hz or 345 Hz.  
 After Filed : 
  
  
 After filed beat/sec decreases only in case of 335 Hz. 
 
340 
335   
345 
Filed  
Filed  
336, 337, 338………   
346, 347, 348………   
unknown 
frequency ( ?)
340Hz
2 beats/sec
unknown 
frequency
known 
frequency 
340Hz
5 beats/sec
26
th
 Feb. 2021 | Shift - 2
PHYSICS
 
 
2. The trajectory a projectile in a vertical plane is y = ?x – ?x
2
, where ? and ? are constants and x 
& y are respectively the horizontal and vertical distance 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
tan ,
4
?
?
?
?
 
(2) 
2
1
tan ,
2
?
?
?
?
 
(3)
2
1
tan ,
? ?
? ?
?
? ?
? ?
? ?
 
 
(4)
2
1
4
tan ,
?
?
?
?
 
Sol. (1) 
 Given : 
 y = ?x – ?x
2
 ....(1) 
 for maximum height, we should find out maximum value of y from equation (1) 
 so, for maximum value of y 
 
dy
dx
 = 0 ? ? – 2 ?x = 0  
 x = 
2
?
?
 ....(2) 
 Now, put value of x from equation (2) in quation (1) 
 y = ?
2
? ? ?
? ?
?
? ?
 - ?
2
2
4
? ? ?
? ?
?
? ?
 
 ? ?
2
2
? ? ?
? ?
?
? ?
 - 
2
4
? ? ?
? ?
?
? ?
?
2
4
?
?
 
 So, H
max
 = 
2
4
?
?
 ....(3) 
 As we know maximum height H
max
 = 
2 2
u sin
2g
?
 ...(4) 
 from (3) and (4) u
2
 = 
2
4
? ? ?
? ?
?
? ?
2
2g
sin
? ?
? ?
?
? ?
 
 and range (R) = 2x = 
2
u 2sin cos
g
? ? ?
 
 2 
2
? ? ?
? ?
?
? ?
 = 
2
2
2g
2sin cos
4 sin
g
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ? ? ?
 
 tan ? = ? ? ? = tan
-1
 ( ?) 
 
 
 
3. 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) 
2
2gh
I mr ?
 
(2) 2gh 
(3)
2
2
2
mgh
I mr ?
 
(4)
2
2mgh
I mr ?
 
Sol. (4) 
 
 
m
m
h
A
B
R 
?
? using energy conservation between A and B point 
 mgh = 
1
2
 m (wR)
2
 + 
1
2
 I ?
2
 
 2mgh = (MR
2
 + I) ?
2
 
 ?
2
 = 
2
2mgh
I MR ?
 
 
 
 
 
 
 
 
 
 
4. Find the peak current and resonant frequency of the following circuit (as shown in figure) 
    
(1) 0.2 A and 100 Hz 
(2) 2 A and 50 Hz 
(3) 2 A and 100 Hz 
(4) 0.2 A and 50 Hz 
Sol. (4) 
 Peak current in series LCR  CKT 
 
? ?
0
2
2
30
L C
v
i
z
x x R
? ?
? ?
  
 
? ?
2
2
30
10 100 (120)
i ?
? ?
 
 
30 1
0.2 .
150 5
i Amp ? ? ? 
 ? ? ?
L
X L ? 
 ? ?(100) (100 × 10
–3
) ? 10 
 
1
?
?
L
X
c ?
6
1
100 100 10
?
?
? ?
 
 
6
4
10
100
10
? ? 
 Resonance frequency 
1
?
LC
? 
 
3 6 5
1 1
100 10 100 10 10
? ? ?
? ?
? ? ?
? 
 ? ? = 2 ?F 
 
5
1 1
2
10
F
?
?
? ? 
 
5
1
10
2 ?
?  
100
10
2 ?
? 
? 50Hz 
  
V=30 sin 100t 
120 ? 
100 ?F
100mH 
~
Page 5


 
 
 
 
SECTION – A 
 
1. A tuning fork A of unknown frequency produces 5beats/s with a fork of known frequency 340 
HZ. When fork A filed, the beat frequency decreases to 2beats/s. What is the frequency of fork 
A? 
(1) 342 Hz 
(2) 335 Hz 
(3) 338 Hz 
(4) 345 Hz 
Sol. (2) 
 Given 
 BeforeFiled: 
  
 So answer should be 335 Hz or 345 Hz.  
 After Filed : 
  
  
 After filed beat/sec decreases only in case of 335 Hz. 
 
340 
335   
345 
Filed  
Filed  
336, 337, 338………   
346, 347, 348………   
unknown 
frequency ( ?)
340Hz
2 beats/sec
unknown 
frequency
known 
frequency 
340Hz
5 beats/sec
26
th
 Feb. 2021 | Shift - 2
PHYSICS
 
 
2. The trajectory a projectile in a vertical plane is y = ?x – ?x
2
, where ? and ? are constants and x 
& y are respectively the horizontal and vertical distance 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
tan ,
4
?
?
?
?
 
(2) 
2
1
tan ,
2
?
?
?
?
 
(3)
2
1
tan ,
? ?
? ?
?
? ?
? ?
? ?
 
 
(4)
2
1
4
tan ,
?
?
?
?
 
Sol. (1) 
 Given : 
 y = ?x – ?x
2
 ....(1) 
 for maximum height, we should find out maximum value of y from equation (1) 
 so, for maximum value of y 
 
dy
dx
 = 0 ? ? – 2 ?x = 0  
 x = 
2
?
?
 ....(2) 
 Now, put value of x from equation (2) in quation (1) 
 y = ?
2
? ? ?
? ?
?
? ?
 - ?
2
2
4
? ? ?
? ?
?
? ?
 
 ? ?
2
2
? ? ?
? ?
?
? ?
 - 
2
4
? ? ?
? ?
?
? ?
?
2
4
?
?
 
 So, H
max
 = 
2
4
?
?
 ....(3) 
 As we know maximum height H
max
 = 
2 2
u sin
2g
?
 ...(4) 
 from (3) and (4) u
2
 = 
2
4
? ? ?
? ?
?
? ?
2
2g
sin
? ?
? ?
?
? ?
 
 and range (R) = 2x = 
2
u 2sin cos
g
? ? ?
 
 2 
2
? ? ?
? ?
?
? ?
 = 
2
2
2g
2sin cos
4 sin
g
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ? ? ?
 
 tan ? = ? ? ? = tan
-1
 ( ?) 
 
 
 
3. 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) 
2
2gh
I mr ?
 
(2) 2gh 
(3)
2
2
2
mgh
I mr ?
 
(4)
2
2mgh
I mr ?
 
Sol. (4) 
 
 
m
m
h
A
B
R 
?
? using energy conservation between A and B point 
 mgh = 
1
2
 m (wR)
2
 + 
1
2
 I ?
2
 
 2mgh = (MR
2
 + I) ?
2
 
 ?
2
 = 
2
2mgh
I MR ?
 
 
 
 
 
 
 
 
 
 
4. Find the peak current and resonant frequency of the following circuit (as shown in figure) 
    
(1) 0.2 A and 100 Hz 
(2) 2 A and 50 Hz 
(3) 2 A and 100 Hz 
(4) 0.2 A and 50 Hz 
Sol. (4) 
 Peak current in series LCR  CKT 
 
? ?
0
2
2
30
L C
v
i
z
x x R
? ?
? ?
  
 
? ?
2
2
30
10 100 (120)
i ?
? ?
 
 
30 1
0.2 .
150 5
i Amp ? ? ? 
 ? ? ?
L
X L ? 
 ? ?(100) (100 × 10
–3
) ? 10 
 
1
?
?
L
X
c ?
6
1
100 100 10
?
?
? ?
 
 
6
4
10
100
10
? ? 
 Resonance frequency 
1
?
LC
? 
 
3 6 5
1 1
100 10 100 10 10
? ? ?
? ?
? ? ?
? 
 ? ? = 2 ?F 
 
5
1 1
2
10
F
?
?
? ? 
 
5
1
10
2 ?
?  
100
10
2 ?
? 
? 50Hz 
  
V=30 sin 100t 
120 ? 
100 ?F
100mH 
~
 
 
 
5. The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors 
, a b
? ?
and c
?
respectively. Then choose the correct relation for these vectors. 
(1) 2 b a c ? ?
? ? ?
 
(2) b a c ? ?
? ? ?
 
(3) 2 b a c ? ?
? ? ?
 
(4)
? ?
2 b a a c c ? ? ?
? ? ? ? ?
 
Sol. (4) 
  
 
 We see from the diagram that because of the law of reflection, the component of the unit vector 
a
?
along b
?
changes sign on reflection while the component parallel to the mirror remain 
unchanges. 
 
11
a a a
?
? ?
? ? ? ? ? ? ?
 
 and
? ?
. a c a c
?
?
? ? ? ? ? ? ?
 
 we see that the reflected unit vector is 
 
11 ?
? ?
? ? ? ? ? ? ?
b a a 2( ) ? ? ?
? ? ? ?
a a c c 
 
6. A radioactive sample is undergoing ? decay. At any time t
1
, its activity is A and another time t
2
, 
the activity is 
5
A
. What is the average life time for the sample?  
(1)
2 1
1 5
t t
n
?
 
(2) 
2 1
1 ( )
2
n t t ?
 
(3)
1 2
1 5
t t
n
?
 
(4)
2 1
1 5 n
t t ?
 
Reflected ray
Normal
Incident ray
?i
?r b
?
?
c
?
a