Quizrr Chapter-Wise Test for JEE Main - 2025

Q.1

Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other, as
shown in the figure. The speed of each pulse is 2 cm s−1 . After 2 s, the total energy of the pulses will be

(A) zero
(B) purely kinetic

(C) purely potential

(D) partly kinetic and partly potential

Q.2

A point mass is subjected to two simultaneous sinusoidal displacements in the x-directions: x1 (t)
​ =
A sin ωt and x2 (t) = A sin(ωt + 2π/3). Adding a third sinusoidal displacement x(t) =
​

B sin(ωt + ϕ) brings the mass to a complete rest. The values of B and ϕ are respectively
3π
(A) 2A and
​

4
​

4π
(B) A and 3
​

5π
(C) 3A and
​

6
​

π
(D) A and 3
​

Q.3

= 0 and x = L. In one experiment, the displacement

The ends of a stretched wire of length L are fixed at x
of the wire is y1 = A sin(πx/L) sin ωt and the energy is E1 , and in another experiment the displacement
​ ​

is y2 = A sin(2πx/L) sin 2ωt and the energy is E2 . Then

​ ​

(A) E2 = E1
​ ​

(B) E2 = 2E1
​ ​

(C) E2 = 4E1
​ ​

(D) E2 = 16E1
​ ​
Q.4

A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first
overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also
generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of
the string is slightly increased the number of beats reduces 2 per second. Assuming the velocity of sound in air
to be 340 m/s, the frequency n of the tuning fork in Hz is

(A) 344

(B) 336

(C) 117.3
(D) 109.3

Q.5

Two vibrating strings made of the same material but of lengths L and 2L have the radii 2r and r respectively.
They are stretched under the same tension. Both the strings vibrate in their fundamental modes-the one of
length L with frequency v1 and the other with frequency v2 . The ratio v1 /v2 is equal to
​ ​ ​ ​

(A) 2

(B) 4

(C) 8
(D) 1

Q.6

A sonometer wire resonates with a given tuning fork, forming standing waves with five antinodes between the
two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M , the
wire resonates with the same tuning fork, forming three antinodes for the same positions of the bridges. The
value of M is

(A) 25 kg
(B) 5 kg
(C) 12.5 kg
1
(D) 25 ​ kg

Q.7

A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm. The
frequency of the tuning fork is 512 Hz. The air temperature is 38∘ C in which the speed of sound is
336 m/s. The zero of the meter scale coincides with the top end of the resonance column tube. When the first
resonance occurs, the reading of the water level in the column is
(A) 14.0 cm
(B) 15.2 cm
(C) 16.4 cm
(D) 17.6 cm

Q.8

A student is performing an experiment with a resonance column. The diameter of the column tube is 4 cm.
∘
The frequency of the tuning fork is 512 Hz. The air temperature is 38 C in which the speed of sound is
−1
336 m s . The zero of the metre scale coincides with the top end of the resonance column tube. When the
first resonance occurs, the reading of the water level in the column is

(A) 14.0 cm
(B) 15.2 cm
(C) 16.4 cm
(D) 17.6 cm

Q.9

A hollow pipe of length 0.8 m is closed at one end. At its open end, a 0.5-m-long uniform string is vibrating in
its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is
50 N and the speed of sound is 320 m s−1 , the mass of the string is
(A) 5 grams

(B) 10 grams
(C) 20 grams
(D) 40 grams

Q.10

A train moves towards a stationary observer with a speed of 34 ms−1 . The train sounds a whistle and its
frequency registered by the observer is f1 . If the train's speed is reduced to 17 ms−1 , the frequency registered
​

is f2 . If the speed of sound is 340 ms−1 then the ratio f1 /f2 is

​ ​ ​

(A) 18
19
​

(B) 12 ​

(C) 2

(D) 19
18
​

Q.11

A siren placed at a railway platform is emitting sound waves of 5kHz frequency. A passenger sitting in a
moving train A records a frequency of 5.5kHz while the train approaches the siren. During his return journey
by a different train B, he records a frequency of 6.0kHz while approaching the same siren. The ratio of the
velocity of the train B to that of the train A is

(A) 242 : 252

(B) 2:1
(C) 5:1
(D) 11 : 6

Q.12

−1
A police car with a siren of frequency of 8kHz is moving with a uniform velocity of 36 km h towards a
tall building which reflects the sound waves. The speed of sound in air is 320 m s−1 . The frequency of the
siren heard by the car driver is

(A) 8.50kHz
(B) 8.25kHz
(C) 7.75kHz
(D) 7.50kHz

Q.13

Two monatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate
​ ​

containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in gas 2 is given by

(A) m1 ​

m2
​ ​

(B) m2 ​

m1
​ ​

(C) m
m
1 ​

2 ​

(D) m
m1
2 ​

​
​

Q.14

A vibrating string of length l under the tension T resonates with a mode corresponding to the first overtone
(third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4
beats per second when excited with a tuning fork of frequency n. When the tension of the string is slightly
increased, the beat frequency reduces to 2 beats per second. Assuming the speed of sound in air to be
340 m s−1 , the frequency n of the tuning fork is
(A) 344 Hz
(B) 336 Hz
(C) 117.3 Hz
(D) 109.3 Hz

Q.15

A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s. The
wavelength of the wave is 0.5 m and its amplitude is 10 cm. Ata particular time t, the snap-shot of the wave
is shown in figure. The velocity of point P when its displacement is 5 cm is -

3π ^
(A)
50
jm/s ​

​ ​

3π ^
(B) − 50
jm/s
​

​ ​

3π ^
(C)
50 i
m/s ​

3π ^
− 50 im/s
(D) ​

Q.16

A massless rod is suspended by two identical strings, AB and CD , of equal lengths. A block of mass m is
suspended from the point C such that BO = x, as shown in the figure. Further, it is observed that the
frequency of the first harmonic (fundamental frequency) in AB is equal to the 2nd harmonic in CD. Then the
length of BO is

L
(A) 5 ​

(B) 4L
5
​
 3L
(C) 4 ​

(D) L
4
​

Q.17

Two vibrating strings of the same material but lengths L and 2L have radii 2r and r respectively. They are
stretched under the same tension. Both the strings vibrate in their fundamental nodes, the one of length L with
frequency v1 and the other with frequency v2 . The raio v1 /v2 is given by
​ ​ ​ ​

(A) 2
(B) 4
(C) 8
(D) 1

Q.18

In the experiment to determine the speed of sound using a resonance column,

(A) the prongs of the tuning fork are kept in a vertical plane
(B) the prongs of the tuning fork are kept in a horizontal plane
(C) in one of the two resonances observed, the length of the

(D) in one of the two resonances observed, the length of the resonating air column is close to half the
wavelength of sound in air

Q.19

A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency
registered by the observer is f1 . If the train's speed is reduced to 17 m/s, the frequency registered is f2 . If the
​ ​

speed of sound is 340 m/s, then the ratio f1 /f2 is ​ ​

(A) 18/19
(B) 1/2
(C) 2
(D) 19/18

Q.20

The driver of a car approaching a vertical wall notices that the frequency of the horn of his car changes from
400 Hz to 450 Hz after being reflected from the wall. Assuming speed of sound to be 340 m/s, the speed
of approach of car towards the wall is

(A) 10 m/s
(B) 20 m/s
(C) 30 m/s
(D) 40 m/s

Q.21

An α-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de
λ
Broglie wavelengths are λα and λp respectively. The ratio λp , to the nearest integer, is
​

​ ​ ​

α ​

Q.22

The equation of a transverse wave travelling on a rope is given by 𝑦 = 10 sin𝜋 ( 0.01𝑥 - 2𝑡 ) where 𝑦 and 𝑥
are in cm and 𝑡 is in seconds. The maximum transverse speed of a particle in the rope is about ____ cm/s
(Mark your answer to the nearest integer)

Q.23

A silver sphere of radius 1 cm and work function 4.7eV is suspended from an insulating thread in a free
space. It is under continuous illumination of a 200-nm-wavelength light. As photoelectrons are emitted, the
sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere
is A × 10n (where 1 < A < 10). The value of n is

Q.24

The work function of silver and sodium are 4.6eV and 2.3eV respectively. The ratio of the slope of the
stopping potential versus frequency plot for silver to that of sodium is

Q.25

A police car moving at 22m/s, chases a motorcyclist. The police man sounds his horn at 176 Hz, while
both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, if it
is given that he does not observes any beats.

Q.26

When two progressive waves y1 ​ = 4 sin(2x − 6t) and y2 = 3 sin (2x − 6t − π2 ) are superimposed,
​ ​

the amplitude of the resultant wave is

Q.27

A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The
string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm)
between the successive nodes on the string.

Q.28

A stationary source emits sound of frequency f0 ​ = 492 Hz. The sound is reflected by a large car
−1
approaching the source with a speed of 2 m s . The reflected signal is received by the source and superposed
with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound
in air is 330 m s−1 and the car reflects the sound at the frequency it has received.)

Q.29

Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π3 , 2π
​

3
and π . When
​ ​

they are superposed, the intensity of the resulting wave is nI0 . The value of n is
​

Q.30

A stationary source is emitting sound at a fixed frequency f0 , which is reflected by two cars approaching the
​

source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0 . What is the
​

difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant
speeds much smaller than the speed of sound which is 330 ms−1 .

Answers & Solutions

Q.1 Answer:
purely kinetic
Solution:

After 2 s, the pulses will overlap completely. The string becomes straight and therefore does not have
any potential energy. Its entire energy must be kinetic.

Q.2 Answer:
4π
A and 3
​

Solution:
 Conclude from the vector triangle (equilateral).

Q.3 Answer:
E2 = 4E1
​ ​

Solution:

A stationary wave has the equation of the form

y = A sin kx sin ωt
Here, for y1 , we have k1 = L
π
and ω1 = ω
​ ​ ​ ​

∴ v1 = ωk11 = ωL
​

π ​
​

​ ​

2π
For y2 , we have k2 = L and ω2 = 2ω
​ ​ ​ ​

∴ v2 = ωk22 = ωL
​

π
= v1 ​
​

​ ​ ​

Thus, the wave velocities are the same in both the cases. Also, they have the same amplitude. The
frequency for y2 is twice the frequency for y1 ​ ​

Now, since energy ∝ (frequency) 2

∴ E2 = 4E1 ​ ​

Q.4 Answer:
344
Solution:

The frequency (v) produced by the air column is given by

λ × v = v ⇒ v = λv ​

Also, 3λ
4
= ℓ = 75 cm = 0.75 m
​

Q.5 Answer:
1
Solution:

1 1 1 1
v= 2l
​
T
m
​ ​ = 2l
​
T
Aρ
​ ​
= 2l
​
T
πR2 ρ
​ ​
= 2lR
​
T
πρ
​

As T and ρ are the same for both the wires, v ∝ 1

lR
​. Here, LR = L × 2r = 2L × r. ∴ v1 =
​
 v2 ​

Q.6 Answer:
25 kg
Solution:

The frequency of vibration of a string is given by

n= N
2l
​
T
m
​

where N = number of loops (or segments) = number of antinodes. The other symbols have their
usual meanings.
5 9g 3 Mg
Here, n= 2l
​

m
​ ​ = 2l
​

m
​ ​ or M = 25 kg.

Q.7 Answer:
15.2 cm
Solution:

Considering the end correction [e = 0.3D], we get

λ λ
L+e= ⇒L= −e
4 4
​ ​

336 × 100 v
− 0.3 × 4 = 15.2 cm [∵ λ = ]
​ ​

=
512 × 4
​ ​

Q.8 Answer:
15.2 cm
Solution:

=L+e
λ
4
​

⇒ L = λ4 − e = ​
v
4f
​ − 0.6R = 16.4 cm − 1.2 cm = 15.2 cm

Q.9 Answer:
10 grams
Solution:

Vsound
2lpipe ​
​

​ = 2 ( 2lstring
1
) ​
​
T
m
​ ​

mass of the string = mls = 10 g ​

Q.10 Answer:
19
18
​

Solution:

no = ​
ns V
V −vs
​

​
​

ns V ns V
Here, f1 = ​

340 m s−1 −34 m s−1

and f2 ​

​ ​ = ​

340 m s−1 −17 m s−1

​
 340 m s−1 −17 m s−1 19
∴ f1
f2
​

​
​ = 340 m s−1 −34 m s−1 ​ = 18 ​

Q.11 Answer:
2:1
Solution:

When an observer approaches a stationary source, which is emitting sound waves of frequency ns , ​

he hears a frequency n, given by

n = ns (1 + Vv ) ​ ​

where v = velocity of the observer and V = velocity of sound. Here, let vA and vB be the velocities ​ ​

of the trains A and B.

For A, we have n = 5.5kHz = (5kHz) (1 + vVA ) or vVA = 0.1 ​

​
​

v v
For B, we have n = 6kHz = (5kHz) (1 + VB ) or VB = 0.2
​ ​

​ ​

∴ vvAB = 2 or vB : vA = 2 : 1. ​

​
​ ​ ​

Q.12 Answer:
8.50kHz
Solution:

f ′ = ( v+v
v−vs
0
) f = ( 320+10
320−10
) (8 × 103 Hz) = 8.5kHz ​
​

​ ​

Q.13 Answer:
m2 ​

m1
​ ​

Solution:

KEY CONCEPT : Cms = ​ ( γRT

M ) Here Cmms ∝
​ ​
1
m; ​ ​

(m )
Cms1
∴ = 2
​
​
​

Crms2 m1
​ ​ ​

​ ​
​

Q.14 Answer:
344 Hz
Solution:

The frequency of the third harmonic of the closed pipe is

3×340 m s−1
f = 3 ( 4lv ) = ​

4×0.75 m
​
= 340 Hz
Since the increase in tension reduces the beat frequency from 4 s−1 to 2 s−1 ,
∴ n − 340 Hz = 4 Hz ⇒ n = 344 Hz

Q.15 Answer:
3π ^
50
​

jm/s
​ ​
Solution:

Since the wave is sinusoidal moving in positive x-axis the point will move parallel to y-axis therefore
options (c) and (d) are ruled out. As the wave moves forward in positive X-direction, the point should
move upwards i.e. in the positive Y-direction. Therefore correct option is a.

Q.16 Answer:
L
5
​

Solution:

T1 + T2 = mg, T1 x = T2 (L − x)
​ ​ ​ ​

1 T1 2 T2
f1 = and f2 =
​

​ ​

2l μ 2l μ
​ ​ ​ ​ ​ ​ ​ ​

where μ = linear mass density and f1 = f2 ​ ​

∴ T1 = 4T2 or T1 x = T41 (L − x)
​ ​ ​
​

​ or x= L
5
​

Q.17 Answer:
1
Solution:

(n 1 ​
= 1
2ℓ
​
( 4πrT 2 ρ ) and n2 = ​ ​ ​
1
4ℓ
​
( πrT2 ρ ) ​ ​

1
n= v
λ
​ = λ
T
m
​ ​ [where λ2 ​ = length of string]
∴ n1
n2
​

​
​ =2× 1
2
​ = 1 [∵ m = mass
length
​ = ρ×A× length
length
​ = ρA])

Q.18 Answer:
the prongs of the tuning fork are kept in a vertical plane
Solution:

The prongs of the tuning fork are kept in a vertical plane in order to set up longitudinal standing
waves in the air column.

Q.19 Answer:
19/18
Solution:
 340 10
n1 = n0 340−34 = ​ ​ ​ n
9 0
​ ​

340 20 n1 10 19 19
n2 = n0 340−17 = ​ ​ ​ n;
19 0
​ ​

n2
​

​
​ = 9
​ × 20
​ = 18
​

Q.20 Answer:
20 m/s
Solution:

450 = 400 ( 340+v

340−vs
s
) ​

​
​

⇒ 98 = 340+v
340−vs
s
​
​

​
​

⇒ 9(340) − 9vs = 8(340) + 8vs ​

​

⇒ 17vs = 340 ​

⇒ vs = 20 m/s ​

Q.21 Answer:
—
Solution:

Kinetic energy = K ⋅ E = qV
h h
De Broglie wavelength = = 2mqV
2mK.E
​ ​

m α qα
​

λ
Therefore ratio λp =
​ ​ ​
​

m p qp
​ ​

α ​ ​ ​ ​

Alpha particle has 4 times the mass and twice the charge of a proton.
λp
= 8≈3
​

λα
​ ​

Q.22 Answer:
—
Solution:

The given equation is 𝑦 = 10sin⁡( 0.01𝜋𝑥 - 2𝜋𝑡 )

Comparing with the general wave equation 𝑦 = 𝐴sin𝑘𝑥 - 𝜔𝑡
Hence, 𝜔 = coefficient of 𝑡 = 2𝜋
Amplitude, 𝐴 = 10
⇒ Maximum speed of the particle, 𝑣max = 𝐴𝜔
⇒ 𝑣max = 10 × 2𝜋
⇒ 𝑣max = 10 × 2 × 3.14
⇒ 𝑣max = 62.8 ≈ 63 cm s-1

Q.23 Answer:
—
Solution:
 = 1 cm = 1 × 10−2 m
Here, radius of sphere R
Work function, W = 4.7eV
Energy of incident radiation
hc 1240eVnm
= λ
​ = 200 nm
​ = 6.2eV
According to Einsteins photoelectric equation
hc
λ
= ϕ + eVs ​

6.2eV = 4.7eV + eVs ​

Vs = 1.5 V​

The sphere will stop emitting photoelectrons, when the potential on its surface becomes equal to
1.5 V
1 Q
∴ 4πε 0 R
= 1.5 ​
​ ​

1 Ne
⇒ 4πε 0 R
= 1.5 ​
​ ​

where N = Number of photoelectrons emitted

e = charge of each electron
1.5×1×10−2
N = 1.5×R
1
×e
= 9×109 ×1.6×10−19
​ ​

4πτ0 ​
​

N = 15
16
× 19 × 108 = 48
5
× 108 ​ ​ ​

N = 50
48
× 107 = 1.04 × 107 ​

∴Z=7

Q.24 Answer:
—
Solution:
ϕ
For photoelectric effect: hv
e
​ − e
​ = V0 ​

The slope is: tan θ = h

e
​ = constant
∴ The ratio will be 1.

Q.25 Answer:
—
Solution:

f1 = frequency of the police car heard by motorcyclist, f2 = frequency of the siren heard by
​ ​

motorcyclist.
 330−v
f1 = 330−22
​ × 176; f2 = 330+v
330
× 165 ​ ​ ​

∵ f1 − f2 = 0 ⇒ v = 22 m/s
​ ​

Q.26 Answer:
—
Solution:

Resultant amplitude, A = A21 + A22 + 2A1 A2 cos ϕ ​ ​ ​ ​ ​

= 42 + 32 + 2 × 4 × 3 × cos π2 = 16 + 9 + 0 = 5 ​ ​ ​

Q.27 Answer:
—
Solution:

0.5
We know that, v = T
μ
​ ​ = 10−3 /0.2
​ ​ = 10 m/s The wavelength of the wave established
v 10
λ= f
​
= 100
​
= 0.1 m = 10 cm
The gap(separation) between two successive nodes(or antinodes ) = λ/2
= 10/2 cm
= 5 cm
​

Q.28 Answer:
—
Solution:

Frequency received by the approaching car (as observer)

f1 = ( v+v
​

v
0
)f ​

Frequency of the reflected wave as received by the source

f2 = ( v−v
​
v
0
) f1 = ( v+v
v−v0
0
)f ​
​ ​
​

​
​

∴ beat frequency = f2 − f = ( v+v − 1) f = ( 332 −1 −1

​

v−v0
0 ​

​
​

328
​ − 1) 492 s =6s

Q.29 Answer:
—
Solution:

y= I0 [sin O + sin π3 + sin 2π

​

3 + sin π ] ​ ​

y= I0 [ ​

2
3 ​

​ + 2
3
]​

​ = 3 I0 ​ ​

∴ Ir = y = 3I0
​
2 ​
⇒ n=3

Q.30 Answer:
—
Solution:

Let v be the speed of sound and vc and f0 the speed and frequency of car. ​ ​

The frequency of sound reflected by the car is

∴ f1′ = f0 [ v+v
​

v−vc
c
] ​
​

​
​

Differentiating the above equation w.r.t. vc , we get ​

= f0 [ ]
df1′ (v−vc ) dvd ​ ​ (v+vc )−(v+vc ) dvd
​ ​ ​ (v−vc ) ​

c ​
c ​

dvc (v−vc )2
​ ​ ​

​
​

∴ df1′
dvc
​

​ = f0 [ (v−v
2v
​

) 2 ] = f0 v 2 (∵ vc ≪< v )
2v
​ ​ ​

c
​

​
​

df1′
∴ × 100 = v2 × dvc
​

f0
​ ​ ​

∴ 0.012 = 2×dv
330
c
∴ dvc = 0.198 m/s ≈ 7 km/h ​

​ ​