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NEET • PHYSICS • WAVES

Waves practice · Mixed difficulty

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91 Medium
91 of 215
Analyzing +4 | -1
Two sound waves are represented by y=a  sin⁡(ωt−kx)y = a\,\,\sin \left( {\omega t - kx} \right) and y=a  cos⁡(ωt−kx)y = a\,\,\cos \left( {\omega t - kx} \right). The wavelength of waves in water are
View answer and explanation
Correct option: A
π2\frac{\pi }{2}
Option A (3) is correct because the given displacement can be expressed as a superposition of three harmonic oscillations using trigonometric identities.
92 Hard
92 of 215
Understanding +4 | -1
In the first overtone position of the resonance tube, the pressure anti-node is present at (L is the length of the tube) which positions from the closed end.
View answer and explanation
Correct option: A
L3,L\frac{\mathrm{L}}{3}, \mathrm{L}
Option A (L3,L\frac{\mathrm{L}}{3}, \mathrm{L}) is correct because in the first overtone (second harmonic) of a closed tube, there are two pressure antinodes, located at L/3 and L from the closed end.
93 Medium
93 of 215
Remembering +4 | -1
If TT is the reverberation time of an auditorium of volume VV then
View answer and explanation
Correct option: D
T∝VT \propto V
Option D (T∝VT \propto V) is correct because the reverberation time is directly proportional to the volume of the auditorium, according to Sabine's formula.
94 Medium
94 of 215
Application +4 | -1
v1 and v2{v_1}\, and \,{v_2} are the velocities of sound at the same temperature in two monoatomic gases of densities ρ1{\rho _1} and ρ2{\rho _2} respectively. If ρ1/ρ2=14{\rho _1}/{\rho _2} = \frac{1}{4} then the ratio of velocities v1   and   v2{v_1}{\mkern 1mu} \,\,and\,\,{\mkern 1mu} {v_2} will be
View answer and explanation
Correct option: C
2:12:1
Option C (2:12:1) is correct because the velocity of sound in a gas is inversely proportional to the square root of the density.
95 Medium
95 of 215
Analyzing +4 | -1
Tuning fork F1{F_1} has a frequency of 256  Hz256\,\,Hz and it is observed to produce 6beats/second6\mathrm{ }beats/second with another tuning fork F2{F_2}. When F2{F_2} is loaded with wax, it still produces 6beats/second6\mathrm{ }beats/second with F1{F_1}. The frequency of F2{F_2} before loading was
View answer and explanation
Correct option: B
262  Hz262\,\,Hz
Option B (262Hz262\mathrm{ }Hz) is correct because adding wax decreases the frequency, and the number of beats remained the same, indicating the original frequency was higher than 256 Hz.
96 Medium
96 of 215
Understanding +4 | -1
Beats are produced by two travelling waves each of loudness II and nearly equal frequencies n1{n_1} and n2{n_2}. The beat frequency will be and maximum loudness hard will be
View answer and explanation
Correct option: B
(n1−n2),4I\left( {{n_1} - {n_2}} \right),4I
Option B ((n1−n2),4I\left( {{n_1} - {n_2}} \right),4I) is correct because the beat frequency is the difference between the two frequencies, and the maximum loudness is proportional to the square of the sum of the amplitudes, resulting in 4I.
97 Medium
97 of 215
Application +4 | -1
The frequency and velocity of sound wave are 600  Hz   and   360  m/s600\,\,Hz\,\,\,and\,\,\,360\,\,m/s respectively. Phase difference between two particles of medium are 60∘60^\circ, the minimum distance between these two particles will be
View answer and explanation
Correct option: A
10cm10\mathrm{ }cm
Option A (10 cm) is correct because the minimum distance is calculated using the phase difference, wavelength, and the relationship between them.
98 Medium
98 of 215
Understanding +4 | -1
A source emits a sound of frequency of 400  Hz400\,\,Hz, but the listener hearts its 390  Hz390\,\,Hz. Then
View answer and explanation
Correct option: C
The listener is moving away from the source
Option C (The listener is moving away from the source) is correct because the observed frequency is lower than the emitted frequency, indicating the listener is moving away from the source.
99 Easy
99 of 215
Application +4 | -1
A sound source of constant frequency resonates with an open end organ pipe of length 30 cm30 \mathrm{~cm} and a close end organ pipe of length 23 cm23 \mathrm{~cm} (both of same diameter). Both pipes are sounding their first overtone. If velocity of sound is 340 ms−1340 \mathrm{~ms}^{-1}, then find the end correction in cm\mathrm{cm}.
View answer and explanation
Correct option: A
1
Option A (1) is correct because the end correction can be found by equating the frequencies of the two pipes and solving for the end correction.
100 Easy
100 of 215
Application +4 | -1
The equation of a transverse wave is given by y=10sin⁡π(0.01x−2t)y = 10\sin \pi (0.01x - 2t) where x and y are in cm and t is in second. Its frequency is
View answer and explanation
Correct option: C
1  sec⁡−11\,\,{\sec ^{ - 1}}
Option C is correct because the frequency is the coefficient of t divided by 2.
101 Medium
101 of 215
Understanding +4 | -1
What is the phase difference between two successive crests in the wave?
View answer and explanation
Correct option: C
2π\pi
Option C (2π\pi) is correct because the phase difference between two successive crests (or any two identical points) in a wave is one full cycle, which corresponds to 2π\pi radians.
102 Medium
102 of 215
Remembering +4 | -1
Velocity of sound is maximum in
View answer and explanation
Correct option: D
Steel
Option D (Steel) is correct because the speed of sound is generally highest in solids due to their high density and elasticity.
103 Medium
103 of 215
Understanding +4 | -1
Statement-I: In longitudinal progressive waves, when the particle is at the mean position, it is a region of maximum compression or rarefaction.
Statement-II: In longitudinal progressive waves, compressions and rarefactions travel forward along the wave.
View answer and explanation
Correct option: A
Statement-I is correct
Option A (Statement-I is correct) is correct because particles at the mean position experience maximum compression or rarefaction in longitudinal waves.
104 Medium
104 of 215
Understanding +4 | -1
If in a resonance tube a oil of density higher than that water is used then at the resonance frequency would
View answer and explanation
Correct option: D
Remain same
Option D (Remain same) is correct because the resonance frequency depends on the length of the air column and the speed of sound in air, which are not affected by the oil's density.
105 Medium
105 of 215
Understanding +4 | -1
Statement I: Sonometer is used to determine frequency of unknown T.F.

Statement II: In sonometer riders used as indicator.
View answer and explanation
Correct option: B
If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement - I.
Option B is correct because both statements are true, but Statement II (riders as indicators) doesn't explain how a sonometer determines frequency.
106 Medium
106 of 215
Analyzing +4 | -1
The displacement of a particle executing periodic motion is given by y=4  cos⁡2(t/2)   sin⁡  (1000 t)y = 4\,\,{\cos ^2}\left( {t/2} \right)\,\,\,\sin \,\,\left( {1000\,t} \right). This expression may be considered to be a result of superposition of
View answer and explanation
Correct option: B
Three waves
Option B (Three waves) is correct because the given expression can be decomposed into three sinusoidal waves using trigonometric identities.
107 Medium
107 of 215
Application +4 | -1
The fourth overtone of an open organ pipe has the same frequency as the second overtone of a close pipe of length L. The length of open pipe will be.
View answer and explanation
Correct option: B
2L
Option B (2L) is correct because the length of the open pipe is twice the length of the closed pipe when their specified overtones have the same frequency.
108 Easy
108 of 215
Application +4 | -1
A wave length 0.6 cm is produced in air and it travels at a speed of 300 m/s. it will be an
View answer and explanation
Correct option: C
ultrasonic wave
Option C is correct because ultrasonic waves have frequencies above the human hearing range.
109 Easy
109 of 215
Analyzing +4 | -1
A closed organ pipe and an open organ pipe of the same length produce four beats, when sounded together. If the length of the closed organ pipe is increased then the number of beats will
View answer and explanation
Correct option: B
Increase
Option B (Increase) is correct because increasing the length of the closed pipe decreases its frequency, increasing the difference with the open pipe's frequency and thus the number of beats.
110 Medium
110 of 215
Understanding +4 | -1
A wave travelling along a string is described by the equation y=asin⁡(ωt−kx)y = a\sin \left( {\omega t - kx} \right) the maximum particle velocity is
View answer and explanation
Correct option: A
aωa\omega
Option A (aωa\omega) is correct because the maximum particle velocity in a wave is given by the product of the amplitude and the angular frequency.
111 Medium
111 of 215
Analyzing +4 | -1
A tuning fork of known frequency 256  Hz256\,\,Hz makes 5  beats/s5\,\,beats/s with the vibrating string of a piano. The beat frequency decreases to 2  beats/s2\,\,beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was
View answer and explanation
Correct option: C
(256−5)Hz\begin{array}{llllllllllllllllllll} {\left( {256 - 5} \right)Hz} \end{array}
Option C ((256−5)Hz\begin{array}{llllllllllllllllllll}{\left( {256 - 5} \right)Hz}\end{array}) is correct because increasing the tension increases the string's frequency, and the beat frequency decreased, meaning the original frequency was lower than the tuning fork's.
112 Medium
112 of 215
Analyzing +4 | -1
The mathematical form of three travelling waves are given by

y1=(2 cm)sin⁡(3x−6t),y2=(3 cm)sin⁡(4x−12t),  and y3=(4 cm)sin⁡(5x−11t)\begin{aligned} & y_1=(2 {~cm}) \sin (3 x-6 t), \\ & y_2=(3 {~cm}) \sin (4 x-12 t) \text {, } \\ & \text { and } y_3=(4 {~cm}) \sin (5 x-11 t) \\ \end{aligned}

of these waves,
View answer and explanation
Correct option: D
Wave 2 has greatest wave speed and wave 3 has maximum transverse string speed
Option D is correct. Wave speed is ω/k, and maximum transverse speed is Aω. Wave 2 has the greatest wave speed, and wave 3 has the greatest maximum transverse string speed.
113 Medium
113 of 215
Application +4 | -1
Two open pipes of length L are vibrated simultaneously. If length of one of the pipe is reduced by y, then the number of beats heard per second will be. (if the velocity of sound is V and y < < L)
View answer and explanation
Correct option: C
Vy2L\frac{{Vy}}{{2L}}
Option C (Vy2L\frac{{Vy}}{{2L}}) is correct because it represents the beat frequency calculated from the difference in frequencies of the two pipes.
114 Medium
114 of 215
Understanding +4 | -1
From a point source, if amplitude of waves at a distance rr is A, its amplitude at a distance 2r2r will be
View answer and explanation
Correct option: C
A/2A/2
Option C (A/2) is correct because the amplitude of a wave from a point source decreases inversely with the distance from the source.
115 Medium
115 of 215
Application +4 | -1
An open tube is in resonance with a string. If tube is dipped in water, so that 75% of length of the tube is inside water, then ratio of frequency of tube to string is
View answer and explanation
Correct option: A
2:12\mathrm{ }:\mathrm{ }1
Option A (2:12\mathrm{ }:\mathrm{ }1) is correct because the frequency of an open tube is inversely proportional to its length, and the frequency of a string is independent of the tube's length.
116 Medium
116 of 215
Understanding +4 | -1
The equation of a spherical progressive wave is
View answer and explanation
Correct option: D
y=arsin⁡(ωt−kr)y = \frac{a}{r}\sin \left( {\omega t - kr} \right)
Option D is correct because the amplitude of a spherical wave decreases linearly with the distance.
117 Medium
117 of 215
Analyzing +4 | -1
A tuning fork gives 55 beats with another tuning fork of frequency 100  Hz.100\,\,Hz. When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork
View answer and explanation
Correct option: C
105  Hz105\,\,Hz
Option C (105Hz105\mathrm{ }Hz) is correct because adding wax decreases the frequency of the tuning fork, and since the number of beats remained the same, the original frequency must have been higher than 100 Hz.
118 Medium
118 of 215
Application +4 | -1
A wave travelling along positive xx-axis is given by y=Asin⁡(ωt−kx)y = A\sin \left( {\omega t - kx} \right). If it is reflected from rigid boundary such that 80%80\% amplitude is reflected, then equation of reflected wave is
View answer and explanation
Correct option: B
y=−0.8Asin⁡(ωt+kx)y = - 0.8A\sin \left( {\omega t + kx} \right)
Option B (y=−0.8Asin⁡(ωt+kx)y = - 0.8A\sin \left( {\omega t + kx} \right)) is correct because the wave is reflected with 80% amplitude and a phase change of π\pi due to reflection from a rigid boundary.
119 Medium
119 of 215
Analyzing +4 | -1
Statement I Two longitudinal waves given by equation y1(x,t)=2a   sin⁡(ωt−kx){y_1}\left( {x,t} \right) = 2a\,\,\,\sin \left( {\omega t - kx} \right) and y2(x,t)=a  sin⁡(2ωt−2kx){y_2}\left( {x,t} \right) = a\,\,\sin \left( {2\omega t - 2kx} \right) will have equal intensity.
Statement II Intensity of waves of given frequency in same medium is proportional to square of amplitude only
View answer and explanation
Correct option: B
Statement I is true, Statement II is false
Statement I is true, Statement II is false. Statement I is correct because intensity depends on the square of the amplitude and the square of the frequency. Statement II is incorrect because it doesn't account for the frequency's effect on intensity.
120 Medium
120 of 215
Understanding +4 | -1
Two tuning forks of frequency n1{n_1} and n2{n_2} produces n beats per second. If n2{n_2} and n are known, n1{n_1} may be given by
View answer and explanation
Correct option: C
n2±n{\mathrm{n}_2} \pm \mathrm{n}
Option C (n2±n{\mathrm{n}_2} \pm \mathrm{n}) is correct because the beat frequency is the absolute difference between the two frequencies, so n1{n_1} can be either higher or lower than n2{n_2} by n.
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