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JEE • PHYSICS • WAVES

Waves practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 543 questions

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121 Medium
121 of 543
Application +4 | -1
In a resonance tube, closed at one end by a smooth moving piston and the other end open, exhibits the first three resonance lengths of air column, L1,L2L_1, L_2 and L3L_3 for the same tuning fork. Then, they are related as
View answer and explanation
Correct option: B
L3=53L2=5L1L_3=\frac{5}{3} L_2=5 L_1
Option B (L3=53L2=5L1L_3=\frac{5}{3} L_2=5 L_1) is correct because it correctly relates the first three resonance lengths in a closed-open tube.
122 Medium
122 of 543
Understanding +4 | -1
Walls of auditorium should be
View answer and explanation
Correct option: A
Good absorber
Option A (Good absorber) is correct because walls of an auditorium should absorb sound to reduce reverberation and create clear acoustics.
123 Easy
123 of 543
Remembering +4 | -1
‘SONAR’ emits which of the following waves
View answer and explanation
Correct option: B
Ultrasonic waves
Option D (Ultrasonic waves) is correct because SONAR emits ultrasonic waves.
124 Medium
124 of 543
Understanding +4 | -1
Statement-I: The change in frequency due to Doppler effect does not depend on separation between the source and the observer.
Statement-II: When a listener and sound source are moving with the same velocity in the same direction, the apparent frequency as heard by listener increases.
View answer and explanation
Correct option: A
Statement-I is correct
Option A is correct because the Doppler effect depends on the relative velocities of the source and observer, not the distance between them.
125 Medium
125 of 543
Analyzing +4 | -1
Three successive resonance frequencies in an organ pipe are 1310,1834 and 2358Hz2358 \mathrm{Hz}. The frequency of 1st 1^{\text {st }} overtone (in Hz\mathrm{Hz} ) is (velocity of sound in air is 340m/s340 \mathrm{m} / \mathrm{s} )
View answer and explanation
Correct option: B
786
Option B (786) is correct because it is calculated by finding the fundamental frequency and then multiplying it by 3 to get the first overtone frequency for a closed pipe.
126 Medium
126 of 543
Understanding +4 | -1
For which of the following cases, there will be no Doppler effect?
View answer and explanation
Correct option: D
All of these
Option B (Same in both balloons) is correct because, according to the ideal gas law, at the same pressure and temperature, equal volumes contain the same number of molecules, regardless of the gas.
127 Medium
127 of 543
Application +4 | -1
A closed organ pipe of length LL and open organ pipe contain gases of densities ρ1  and  ρ2{\mathrm{\rho}_1}\,\,\mathrm{and}\,\,{\mathrm{\rho}_2} respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is
View answer and explanation
Correct option: C
4L3ρ1ρ2\frac{{4L}}{3}\sqrt {\frac{{{\rho_1}}}{{{\rho_2}}}}
Option C (4L3ρ1ρ2\frac{{4L}}{3}\sqrt {\frac{{{\rho_1}}}{{{\rho_2}}}}) is correct because it relates the lengths of the pipes to the densities of the gases, considering the first overtone and equal compressibility.
128 Medium
128 of 543
Application +4 | -1
In a resonance tube the first resonance with a tuning fork occurs at 16  cm16\,\,cm and second at 49  cm49\,\,cm. If the velocity of sound is 330  m/s330\,\,m/s, the frequency of tuning fork is
View answer and explanation
Correct option: A
500500
Option A (500500) is correct because the frequency of the tuning fork is calculated using the difference between the second and first resonance lengths in a resonance tube.
129 Medium
129 of 543
Application +4 | -1
If the frequency of human heart beat is 1.25  Hz1.25\,\,\mathrm{Hz}, the number of heart beats in 11 minute is
View answer and explanation
Correct option: D
7575
Option D (7575) is correct because the number of heartbeats in a minute is the frequency multiplied by the number of seconds in a minute (60).
130 Medium
130 of 543
Analyzing +4 | -1
Statement-I: Beats can be observed by two light sources as in sound.
Statement-II: To observe beats by two light sources, the phase difference between the two sources should change regularly.
View answer and explanation
Correct option: C
Both statement-I &\& II are correct
Option C (Both statement-I & II are correct) is correct because beats can be observed with light sources if the phase difference changes regularly.
131 Medium
131 of 543
Understanding +4 | -1
It is possible to hear beats from the two vibrating sources of frequency
View answer and explanation
Correct option: B
  20  Hz   and  25  Hz\;20\,\,Hz\,\,\,and\,\,25\,\,Hz
Option B is correct. Beats are audible when the frequency difference between two sources is less than approximately 10 Hz. Only the frequencies 20 Hz and 25 Hz satisfy this condition.
132 Medium
132 of 543
Application +4 | -1
Two trains are moving towards each other with speeds of 20  m/s   and   15  m/s20\,\,m/s\,\,\,and\,\,\,15\,\,m/s relative to the ground. The first train sounds whistle of frequency 600 Hz600\,\mathrm{ }Hz, the frequency of the whistle heard by a passenger in the second train before the meets is (the speed of sound in air is 340m/s340\mathrm{ }m/s)
View answer and explanation
Correct option: D
666Hz666\mathrm{ }Hz
Option D (666Hz666\mathrm{ }Hz) is correct because it is calculated using the Doppler effect formula for approaching source and observer.
133 Medium
133 of 543
Application +4 | -1
When a guitar string is sounded with a 440  Hz440\,\,Hz tuning fork, a beat frequency of 5  Hz5\,\,Hz is heard. If the experiment is repeated with a tuning fork of 437  Hz437\,\,Hz, the beat frequency is 8  Hz8\,\,Hz. The string frequency (Hz)\left( {Hz} \right) is
View answer and explanation
Correct option: A
445
Option A (445) is correct because the string frequency is either 445 Hz or 435 Hz based on the first tuning fork, and the second tuning fork confirms it is 445 Hz.
134 Medium
134 of 543
Understanding +4 | -1
Standing waves can be obtained in an air column even if interfering waves are
View answer and explanation
Correct option: B
Of different amplitudes
Option B (Of different amplitudes) is correct because standing waves can still form even if the interfering waves have different amplitudes, although the nodes will not be perfect nulls.
135 Medium
135 of 543
Understanding +4 | -1
Velocity of sound in air is 332 m/s332 \mathrm{~m} / \mathrm{s}. Velocity in vacuum will be
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because sound waves require a medium to propagate, and a vacuum has no medium.
136 Medium
136 of 543
Application +4 | -1
Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between successive maximum intensities will be
View answer and explanation
Correct option: A
14  sec⁡\frac{1}{4}\,\,\sec
Option A (14  sec⁡\frac{1}{4}\,\,\sec) is correct because the time interval between successive maximum intensities is the reciprocal of the beat frequency, which is the difference between the two frequencies.
137 Medium
137 of 543
Understanding +4 | -1
Assertion: In longitudinal wave propagation the distance between two consecutive compression is equal to wavelength of wave.
Reason: Standing wave does not get transfered.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct because both the assertion and the reason are true statements, but the reason does not explain the assertion.
138 Medium
138 of 543
Application +4 | -1
There are three  sources of sound of equal intensity with frequencies 400,401and 402  vib/sec.400,\mathrm{ }401\mathrm{ }and\,402\,\,vib/sec.. The number of beats heard per second is
View answer and explanation
Correct option: C
22
Option C (2) is correct because the number of beats heard per second is equal to the difference between the highest and lowest frequencies when multiple sound sources of equal intensity are present.
139 Medium
139 of 543
Understanding +4 | -1
The angle between wave velocity and particle velocity in a travelling wave may be
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because the angle between wave velocity and particle velocity can be 0, π2\frac{\pi}{2}, or π\pi depending on the type of wave and the point in the wave.
140 Medium
140 of 543
Application +4 | -1
Two tuning fork P  and  QP\,\,and\,\,\mathrm{Q} when set vibrating give 4   beats/s4{\mkern 1mu} {\mkern 1mu} \,beats/s. if a prong of the fork PP is filed the beats are reduced to 2  s−12\,\,{s^{ - 1}}. What is frequency of PP, if that of QQ is 250  Hz250\,\,Hz?
View answer and explanation
Correct option: A
246  Hz\begin{array}{llllllllllllllllllll} {246\,\,Hz} \end{array}
The correct answer is A (246 Hz) because filing a prong increases the frequency, and the beat frequency decreasing to 2 Hz implies the original frequency of P was lower than Q.
141 Medium
141 of 543
Understanding +4 | -1
Which of the following statement is true about stationary waves?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because in stationary waves, all particles except those at nodes execute SHM, they all reach their mean positions simultaneously, and particles at antinodes have the same maximum velocity at the mean position.
142 Medium
142 of 543
Application +4 | -1
n1{n_1} is the frequency of the pipe closed at one end and n2{n_2} is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formed
View answer and explanation
Correct option: A
n1n2n2+2n1\frac{{{\mathrm{n}_1}{\mathrm{n}_2}}}{{{\mathrm{n}_2} + 2{\mathrm{n}_1}}}
Option A (n1n2n2+2n1\frac{{{\mathrm{n}_1}{\mathrm{n}_2}}}{{{\mathrm{n}_2} + 2{\mathrm{n}_1}}}) is correct because it represents the fundamental frequency of the closed pipe formed by joining the two pipes, derived from the individual frequencies and lengths of the open and closed pipes.
143 Medium
143 of 543
Analyzing +4 | -1
For a certain organ pipe, three successive resonance frequencies are observed at 425,595425,595 and 765 Hz765 \mathrm{~Hz}. The speed of sound in air is 340 ms−1340 \mathrm{~ms}^{-1}. The pipe is a
View answer and explanation
Correct option: A
Closed pipe of length 1 m1 \mathrm{~m}
Option A (Closed pipe of length 1 m) is correct because the given frequencies are odd multiples, indicating a closed pipe, and the length is calculated using the fundamental frequency formula for a closed pipe.
144 Medium
144 of 543
Application +4 | -1
The velocity of waves in a string fixed at both ends is 2  m/s2\,\,m/s. The string forms standing waves with nodes 5.0  cm5.0\,\,cm apart. The frequency of vibration of the string in HzHz is
View answer and explanation
Correct option: C
20
Option C (20 Hz) is correct because the frequency is calculated using the wave speed and wavelength, where the wavelength is twice the distance between nodes.
145 Medium
145 of 543
Application +4 | -1
A 1000  Hz1000\,\,Hz sound wave in air strikes the surface of a lake and penetrates into water. If speed of sound in water is 1500  ms−11500\,\,\mathrm{m}{\mathrm{s}^{ - 1}}, the frequency and wavelength of waves in water are
View answer and explanation
Correct option: B
1000  Hz,1.5  m1000\,\,\mathrm{ }Hz,\mathrm{ }1.5\,\,m
Option B (1000 Hz,1.5 m1000 \text{ Hz}, 1.5 \text{ m}) is correct because the frequency of a wave remains the same when it travels from one medium to another, and the wavelength changes according to the speed of sound in the new medium.
146 Medium
146 of 543
Analyzing +4 | -1
3 successive overtones frequencies of an organ pipe is given by 360 Hz,600 Hz,840360 \mathrm{~Hz}, 600 \mathrm{~Hz}, 840 Hz\mathrm{Hz}. Find out nature and fundamental frequency of organ pipe.
View answer and explanation
Correct option: B
120 Hz120 \mathrm{~Hz}
Option B (120 Hz120 \mathrm{~Hz}) is correct because the fundamental frequency is found by determining the common difference between successive overtones in a closed organ pipe.
147 Medium
147 of 543
Application +4 | -1
Determine the change in volume of 6 litres of alcohol if the pressure is decreased from 200 cm200 \mathrm{~cm} of Hg\mathrm{Hg} to 75 cm75 \mathrm{~cm}. [velocity of sound in alcohol is 1280 m/s1280 \mathrm{~m} / \mathrm{s}, density of alcohol =0.81 gm/cc=0.81\,\mathrm{gm} / \mathrm{cc}, density of Hg=13.6 gm/cc\mathrm{Hg}=13.6\,\mathrm{gm} / \mathrm{cc} and g=9.81 m/s2]\left.g=9.81 \mathrm{~m} / \mathrm{s}^2\right].
View answer and explanation
Correct option: A
0.75 cc
Option A (0.75 cc) is correct because it is the change in volume calculated using the bulk modulus and the change in pressure.
148 Medium
148 of 543
Application +4 | -1
A micro-wave and an ultrasonic sound wave have the same wavelength. Their frequencies are in the ratio (approximately)
View answer and explanation
Correct option: A
106:1{10^6}:1
Option A (106:1{10^6}:1) is correct because the ratio of frequencies is approximately equal to the ratio of their speeds, microwave speed divided by ultrasonic sound wave speed.
149 Medium
149 of 543
Analyzing +4 | -1
Assertion: In the case of a stationary wave, a person hear a loud sound at the nodes as compared to the antinodes.
Reason: In a stationary wave all the particles of the medium vibrate in phase.
View answer and explanation
Correct option: C
If assertion is true, but reason is false
Option D (If both assertion and reason are false) is correct because sound is loudest at antinodes, and particles in different segments vibrate out of phase.
150 Medium
150 of 543
Understanding +4 | -1
I. In a closed organ pipe, standing waves can be formed.

II. In a closed organ pipe (closed at one end), only odd harmonics are present.

III. The harmonics which are present in a pipe, open at both ends are odd harmonics only.

Which of the following statement(s) related to organ pipe is/are correct?
View answer and explanation
Correct option: C
Both I and II
Option C (Both I and II) is correct because closed organ pipes form standing waves and only contain odd harmonics.
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