Waves: JEE Mains Previous Year Questions (2021-2024) | Physics for JEE Main & Advanced PDF Download

 Page 1


 
2024 
 
 
Q1 - 2024 (01 Feb Shift 1) 
A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 
6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 
beats per second with it. The frequency of the tuning fork is Hz. 
 
Q2 - 2024 (27 Jan Shift 2) 
A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of 
length 350 cm, both vibrating in fundamental mode. The velocity of sound is ____m/s. 
 
Q3 - 2024 (30 Jan Shift 1) 
In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of 
water is now poured in the organ pipe so that the fundamental frequency is increased to 
110 Hz. If the organ pipe has a cross-sectional area of 2 cm
2
, the amount of water poured 
in the organ tube is _____ g. (Take speed of sound in air is 330 m/s) 
 
Q4 - 2024 (30 Jan Shift 2) 
A point source is emitting sound waves of Power 16 × 10
-8
 W at the origin. The ratio of 
intensity (magnitude only) at two points located at a distances of 2 m and 4 m from the 
origin respectively will be Wm
-2
. [We modified question statement, In official NTA 
paper question was incomplete] 
 
Q5 - 2024 (31 Jan Shift 1) 
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency 
of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe 
will be : 
(1) 60 cm 
(2) 45 cm 
(3) 30 cm 
(4) 15 cm 
 
Q6 - 2024 (31 Jan Shift 2) 
The speed of sound in oxygen at S.T.P. will be approximately: 
(Given, R = 8.3JK
-1
,?? = 1.4 ) 
(1) 310 m/s 
(2) 333 m/s 
Page 2


 
2024 
 
 
Q1 - 2024 (01 Feb Shift 1) 
A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 
6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 
beats per second with it. The frequency of the tuning fork is Hz. 
 
Q2 - 2024 (27 Jan Shift 2) 
A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of 
length 350 cm, both vibrating in fundamental mode. The velocity of sound is ____m/s. 
 
Q3 - 2024 (30 Jan Shift 1) 
In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of 
water is now poured in the organ pipe so that the fundamental frequency is increased to 
110 Hz. If the organ pipe has a cross-sectional area of 2 cm
2
, the amount of water poured 
in the organ tube is _____ g. (Take speed of sound in air is 330 m/s) 
 
Q4 - 2024 (30 Jan Shift 2) 
A point source is emitting sound waves of Power 16 × 10
-8
 W at the origin. The ratio of 
intensity (magnitude only) at two points located at a distances of 2 m and 4 m from the 
origin respectively will be Wm
-2
. [We modified question statement, In official NTA 
paper question was incomplete] 
 
Q5 - 2024 (31 Jan Shift 1) 
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency 
of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe 
will be : 
(1) 60 cm 
(2) 45 cm 
(3) 30 cm 
(4) 15 cm 
 
Q6 - 2024 (31 Jan Shift 2) 
The speed of sound in oxygen at S.T.P. will be approximately: 
(Given, R = 8.3JK
-1
,?? = 1.4 ) 
(1) 310 m/s 
(2) 333 m/s 
(3) 341 m/s 
(4) 325 m/s 
 
Answer Key 
Q1 (6) 
Q2 (294) 
Q3 (400) 
Q4 (4) 
Q5 (4) 
Q6 (1) 
 
Solutions 
 
Q1 
f =
1
2 L
 
T
?? 
f
1
=
1
2
 
6
?? f
2
=
1
2
 
54
?? 
f
1
f
2
=
1
3
 f
2
- f
1
= 12 
f
1
= 6HZ 
 
Q2 
 
f
c
=
v
4l
1
 f
o
=
v
2l
2
 
 f
c
- f
0
 = 7 
v
4 × 150
-
v
2 × 350
= 7 
v
600 cm
-
v
700 cm
= 7 
Page 3


 
2024 
 
 
Q1 - 2024 (01 Feb Shift 1) 
A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 
6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 
beats per second with it. The frequency of the tuning fork is Hz. 
 
Q2 - 2024 (27 Jan Shift 2) 
A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of 
length 350 cm, both vibrating in fundamental mode. The velocity of sound is ____m/s. 
 
Q3 - 2024 (30 Jan Shift 1) 
In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of 
water is now poured in the organ pipe so that the fundamental frequency is increased to 
110 Hz. If the organ pipe has a cross-sectional area of 2 cm
2
, the amount of water poured 
in the organ tube is _____ g. (Take speed of sound in air is 330 m/s) 
 
Q4 - 2024 (30 Jan Shift 2) 
A point source is emitting sound waves of Power 16 × 10
-8
 W at the origin. The ratio of 
intensity (magnitude only) at two points located at a distances of 2 m and 4 m from the 
origin respectively will be Wm
-2
. [We modified question statement, In official NTA 
paper question was incomplete] 
 
Q5 - 2024 (31 Jan Shift 1) 
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency 
of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe 
will be : 
(1) 60 cm 
(2) 45 cm 
(3) 30 cm 
(4) 15 cm 
 
Q6 - 2024 (31 Jan Shift 2) 
The speed of sound in oxygen at S.T.P. will be approximately: 
(Given, R = 8.3JK
-1
,?? = 1.4 ) 
(1) 310 m/s 
(2) 333 m/s 
(3) 341 m/s 
(4) 325 m/s 
 
Answer Key 
Q1 (6) 
Q2 (294) 
Q3 (400) 
Q4 (4) 
Q5 (4) 
Q6 (1) 
 
Solutions 
 
Q1 
f =
1
2 L
 
T
?? 
f
1
=
1
2
 
6
?? f
2
=
1
2
 
54
?? 
f
1
f
2
=
1
3
 f
2
- f
1
= 12 
f
1
= 6HZ 
 
Q2 
 
f
c
=
v
4l
1
 f
o
=
v
2l
2
 
 f
c
- f
0
 = 7 
v
4 × 150
-
v
2 × 350
= 7 
v
600 cm
-
v
700 cm
= 7 
v
6 m
-
v
7 m
= 7 
v 
1
42
 = 7 
v = 42 × 7 
= 294 m/s 
 
Q3 
?? 4l
1
= 30 ? l
1
=
11
4
?? 
?? 4l
2
= 110 ? l
2
=
3
4
?? 
?l = 2?? 
Change in volume = ?? ?l = 400 cm
3
 
?? = ?? ?? ?? g; ? ?? = 1 g/cm
3
  
 
Q4 
Intensity = Power/Area 
Ratio of intensity will be inverse square ratio of distances 
Ratio =
4
2
2
2
= 4 
 
Q5 
 
 
?? 4
= L
1
 2 
?? 2
 = ?? 
Page 4


 
2024 
 
 
Q1 - 2024 (01 Feb Shift 1) 
A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 
6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 
beats per second with it. The frequency of the tuning fork is Hz. 
 
Q2 - 2024 (27 Jan Shift 2) 
A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of 
length 350 cm, both vibrating in fundamental mode. The velocity of sound is ____m/s. 
 
Q3 - 2024 (30 Jan Shift 1) 
In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of 
water is now poured in the organ pipe so that the fundamental frequency is increased to 
110 Hz. If the organ pipe has a cross-sectional area of 2 cm
2
, the amount of water poured 
in the organ tube is _____ g. (Take speed of sound in air is 330 m/s) 
 
Q4 - 2024 (30 Jan Shift 2) 
A point source is emitting sound waves of Power 16 × 10
-8
 W at the origin. The ratio of 
intensity (magnitude only) at two points located at a distances of 2 m and 4 m from the 
origin respectively will be Wm
-2
. [We modified question statement, In official NTA 
paper question was incomplete] 
 
Q5 - 2024 (31 Jan Shift 1) 
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency 
of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe 
will be : 
(1) 60 cm 
(2) 45 cm 
(3) 30 cm 
(4) 15 cm 
 
Q6 - 2024 (31 Jan Shift 2) 
The speed of sound in oxygen at S.T.P. will be approximately: 
(Given, R = 8.3JK
-1
,?? = 1.4 ) 
(1) 310 m/s 
(2) 333 m/s 
(3) 341 m/s 
(4) 325 m/s 
 
Answer Key 
Q1 (6) 
Q2 (294) 
Q3 (400) 
Q4 (4) 
Q5 (4) 
Q6 (1) 
 
Solutions 
 
Q1 
f =
1
2 L
 
T
?? 
f
1
=
1
2
 
6
?? f
2
=
1
2
 
54
?? 
f
1
f
2
=
1
3
 f
2
- f
1
= 12 
f
1
= 6HZ 
 
Q2 
 
f
c
=
v
4l
1
 f
o
=
v
2l
2
 
 f
c
- f
0
 = 7 
v
4 × 150
-
v
2 × 350
= 7 
v
600 cm
-
v
700 cm
= 7 
v
6 m
-
v
7 m
= 7 
v 
1
42
 = 7 
v = 42 × 7 
= 294 m/s 
 
Q3 
?? 4l
1
= 30 ? l
1
=
11
4
?? 
?? 4l
2
= 110 ? l
2
=
3
4
?? 
?l = 2?? 
Change in volume = ?? ?l = 400 cm
3
 
?? = ?? ?? ?? g; ? ?? = 1 g/cm
3
  
 
Q4 
Intensity = Power/Area 
Ratio of intensity will be inverse square ratio of distances 
Ratio =
4
2
2
2
= 4 
 
Q5 
 
 
?? 4
= L
1
 2 
?? 2
 = ?? 
v = f?? f
2
=
2v
2 L
2
 
v = f
1
 4 L
1
  f
2
=
v
L
2
 
f
1
=
v
4 L
1
 
f
1
= f
2
 
v
4 L
1
=
v
L
2
 
? L
2
= 4 L
1
 
60 = 4 × L
1
 
L
1
= 15 cm 
 
Q6 
v =
 
?? RT
M
=
 
1.4 × 8.3 × 273
32 × 10
-3
 
= 314.8541 ? 315 m/s 
Page 5


 
2024 
 
 
Q1 - 2024 (01 Feb Shift 1) 
A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 
6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 
beats per second with it. The frequency of the tuning fork is Hz. 
 
Q2 - 2024 (27 Jan Shift 2) 
A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of 
length 350 cm, both vibrating in fundamental mode. The velocity of sound is ____m/s. 
 
Q3 - 2024 (30 Jan Shift 1) 
In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of 
water is now poured in the organ pipe so that the fundamental frequency is increased to 
110 Hz. If the organ pipe has a cross-sectional area of 2 cm
2
, the amount of water poured 
in the organ tube is _____ g. (Take speed of sound in air is 330 m/s) 
 
Q4 - 2024 (30 Jan Shift 2) 
A point source is emitting sound waves of Power 16 × 10
-8
 W at the origin. The ratio of 
intensity (magnitude only) at two points located at a distances of 2 m and 4 m from the 
origin respectively will be Wm
-2
. [We modified question statement, In official NTA 
paper question was incomplete] 
 
Q5 - 2024 (31 Jan Shift 1) 
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency 
of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe 
will be : 
(1) 60 cm 
(2) 45 cm 
(3) 30 cm 
(4) 15 cm 
 
Q6 - 2024 (31 Jan Shift 2) 
The speed of sound in oxygen at S.T.P. will be approximately: 
(Given, R = 8.3JK
-1
,?? = 1.4 ) 
(1) 310 m/s 
(2) 333 m/s 
(3) 341 m/s 
(4) 325 m/s 
 
Answer Key 
Q1 (6) 
Q2 (294) 
Q3 (400) 
Q4 (4) 
Q5 (4) 
Q6 (1) 
 
Solutions 
 
Q1 
f =
1
2 L
 
T
?? 
f
1
=
1
2
 
6
?? f
2
=
1
2
 
54
?? 
f
1
f
2
=
1
3
 f
2
- f
1
= 12 
f
1
= 6HZ 
 
Q2 
 
f
c
=
v
4l
1
 f
o
=
v
2l
2
 
 f
c
- f
0
 = 7 
v
4 × 150
-
v
2 × 350
= 7 
v
600 cm
-
v
700 cm
= 7 
v
6 m
-
v
7 m
= 7 
v 
1
42
 = 7 
v = 42 × 7 
= 294 m/s 
 
Q3 
?? 4l
1
= 30 ? l
1
=
11
4
?? 
?? 4l
2
= 110 ? l
2
=
3
4
?? 
?l = 2?? 
Change in volume = ?? ?l = 400 cm
3
 
?? = ?? ?? ?? g; ? ?? = 1 g/cm
3
  
 
Q4 
Intensity = Power/Area 
Ratio of intensity will be inverse square ratio of distances 
Ratio =
4
2
2
2
= 4 
 
Q5 
 
 
?? 4
= L
1
 2 
?? 2
 = ?? 
v = f?? f
2
=
2v
2 L
2
 
v = f
1
 4 L
1
  f
2
=
v
L
2
 
f
1
=
v
4 L
1
 
f
1
= f
2
 
v
4 L
1
=
v
L
2
 
? L
2
= 4 L
1
 
60 = 4 × L
1
 
L
1
= 15 cm 
 
Q6 
v =
 
?? RT
M
=
 
1.4 × 8.3 × 273
32 × 10
-3
 
= 314.8541 ? 315 m/s 
Numerical 
Q.1. The fundamental frequency of vibration of a string 
stretched between two rigid support is  50 Hz. The mass of the 
string is  18 g and its linear mass density is  20 g / m. The speed 
of the transverse waves so produced in the string is 
___________ ms
-1 
Q.2. In an experiment with sonometer when a mass of  180 g is 
attached to the string, it vibrates with fundamental frequency 
of  30 Hz. When a mass m is attached, the string vibrates with 
fundamental frequency of  50 Hz. The value of m is ___________ 
g. 
JEE Main 2023 (Online) 13th April Evening Shift 
 
JEE Main 2023 (Online) 15th April Morning Shift 
 
JEE Main 2023 (Online) 12th April Morning Shift 
 
Q.3. For a certain organ pipe, the first three resonance 
frequencies are in the ratio of 1 :3: 5 respectively. If the 
frequency of fifth harmonic is  4 0 5 Hz and the speed of sound in 
air is  3 2 4 ms -1 the length of the organ pipe is _________ m.