Skip to main content
Back to WAVES

Your Progress

0 / 524 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • WAVES

Waves practice · Medium

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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.

Take a chapter quiz →
Showing 31-60 of 524 questions
31 Medium
31 of 524
Application +4 | -1
5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be :
View answer and explanation
Correct option: A
195 Hz
Option A (195 Hz) is correct because it is the frequency of the fork calculated from the beat frequencies and the lengths of the sonometer wire.
32 Medium
32 of 524
Understanding +4 | -1
Assertion (A) : Stationary waves are so called because particles are at rest in stationary waves.

Reason (R) : They are formed by the superposition of two identical waves travelling in opposite directions.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D (If Assertion is false but the Reason is true) is correct because particles in stationary waves do oscillate, and stationary waves are formed by superposition.
33 Medium
33 of 524
Application +4 | -1
A string is under tension so that its length is increased by 1n\frac{1}{n} times its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be
View answer and explanation
Correct option: C
n:1\sqrt n :1
Option C (n:1\sqrt n :1) is correct because the ratio of longitudinal to transverse wave frequencies is proportional to the square root of the factor by which the length increased.
34 Medium
34 of 524
Application +4 | -1
A gas mixture has 24%24 \% of Argon, 32%32 \% of oxygen, and 44%44 \% of CO2\mathrm{CO}_2 by mass. Find the velocity of sound in the gas mixture at 27∘C27^{\circ} \mathrm{C}. Given R=8.4R=8.4 S.I. units. Molecular weight of (Ar=40),(O2=32),(CO2=44)(\mathrm{Ar}=40),\left(\mathrm{O}_2=32\right),\left(\mathrm{CO}_2=44\right).

γAr=5/3,γO2=7/5,γCO2=4/3. \gamma_{\mathrm{Ar}}=5 / 3, \gamma_{\mathrm{O}_2}=7 / 5, \gamma_{\mathrm{CO}_2}=4 / 3 \text {. }
View answer and explanation
Correct option: A
303.5 m/s303.5 \mathrm{~m} / \mathrm{s}
Option A (303.5 m/s303.5 \mathrm{~m} / \mathrm{s}) is correct because it is the calculated velocity of sound in the gas mixture based on the given composition and properties.
35 Medium
35 of 524
Understanding +4 | -1
If source and observer both are relatively at rest and if speed of sound is increased then frequency heard by observer will
View answer and explanation
Correct option: D
Will not change
Option D (Will not change) is correct because the frequency heard by the observer depends on the relative motion of the source and observer, not the speed of sound itself when both are at rest.
36 Medium
36 of 524
Application +4 | -1
Equation of a plane progressive given by y=0.6sin⁡2π(t−x2)y=0.6 \sin2 \pi\left(t-\frac{x}{2}\right). On reflection from a denser medium its amplitude becomes 2/32 / 3 of the amplitude of the incident wave. The equation of the reflected wave is
View answer and explanation
Correct option: B
y=−0.4sin⁡2π(t+x2)y=-0.4 \sin 2 \pi\left(t+\frac{x}{2}\right)
Option B (y=−0.4sin⁡2π(t+x2)y=-0.4 \sin 2 \pi(t+\frac{x}{2})) is correct because the amplitude becomes 2/3 and there is a phase change of π\pi upon reflection from a denser medium.
37 Medium
37 of 524
Application +4 | -1
A 20  cm20\,\,cm long string, having a mass of 1.0  g1.0\,\,g, is fixed at both the ends. The tension in the string is 0.5  N0.5\,\,N. the string is set into vibration using an external vibrator of frequency 100  Hz100\,\,Hz. Find the separation (in cmcm) between the successive nodes on the string
View answer and explanation
Correct option: A
5
Option A (5 cm) is correct because it is the calculated separation between successive nodes using the given parameters and the formula for wave speed on a string.
38 Medium
38 of 524
Application +4 | -1
A closed organ pipe and an open pipe of same length produce 22 beats s−1{\mathrm{s}^{ - 1}} when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced will be
View answer and explanation
Correct option: A
7
Option A (7) is correct because the number of beats is the difference between the frequencies of the open and closed pipes, and halving the open pipe's length and doubling the closed pipe's length changes the frequencies, resulting in 7 beats.
39 Medium
39 of 524
Application +4 | -1
A train approaching a platform at a speed of 54 km/h54 \mathrm{~km} / \mathrm{h} sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz1620 \mathrm{~Hz}. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platfrom? The speed of sound in air =332 m/s=332 \mathrm{~m} / \mathrm{s}.
View answer and explanation
Correct option: A
1480 Hz
Option A (1480 Hz) is correct because it is calculated using the Doppler effect formula for the frequency heard after the train has passed.
40 Medium
40 of 524
Application +4 | -1
An observer is approaching a stationary source with a velocity 1/41/4 th of the velocity of sound. Then the ratio of the apparent frequency to actual frequency of source is
View answer and explanation
Correct option: B
5:45:4
Option B (5:45:4) is correct because the apparent frequency increases when the observer approaches the source, and the ratio is calculated using the Doppler effect formula.
41 Medium
41 of 524
Application +4 | -1
The frequency of a tuning fork with amplitude A=1 cmA=1 \mathrm{~cm} is 250 Hz250 \mathrm{~Hz}. The maximum velocity of any particle in air is equal to
View answer and explanation
Correct option: B
5πms −15 \mathrm{\pi} \mathrm{\textrm {ms } ^ { - 1 }}
Option B (5 \pi \textrm{{ ms } ^ { - 1 }}5 \pi \textrm{{ ms } ^ { - 1 }}) is correct because it is calculated using the formula for maximum particle velocity in simple harmonic motion: vmax=ωA=2πfAv_{max} = \omega A = 2\pi fA.
42 Medium
42 of 524
Understanding +4 | -1
The stationary wave y=2a  sin⁡   k  x  cos⁡   ωty = 2a\,\,\sin \,\,\,k\,\,x\,\,\cos \,\,\,\omega t in a closed organ pipe is the result of the superposition of y=a    sin⁡(ωt−kx)y = a\,\,\,\,\sin \left( {\omega t - kx} \right) and
View answer and explanation
Correct option: B
y=−a   sin⁡(ωt+kx)y = - a\,\,\,\sin \left( {\omega t + kx} \right)
Option B (y=−a,,sin⁡(ωt+kx)y = - a\\,\\,\sin ( {\omega t + kx} )) is correct because the superposition of the two waves results in the given stationary wave equation.
43 Medium
43 of 524
Application +4 | -1
A tuning fork of frequency 340 Hz340 \mathrm{~Hz} is sounded above an organ pipe of length 120 cm120 \mathrm{~cm}. Water is now slowly poured in it. The minimum height of water column required for resonance is (speed of sound in air =340 m/s=340 \mathrm{~m} / \mathrm{s} )
View answer and explanation
Correct option: D
45 cm45 \mathrm{~cm}
Option D (45 cm45 \mathrm{~cm}) is correct because it is the minimum height of the water column required for resonance, calculated using the properties of standing waves in a closed pipe.
44 Medium
44 of 524
Application +4 | -1
Equation of motion of wave in the same direction is given by y1=Asin⁡(ωt−Kx),y2=Asin⁡(ωt−kx+ϕ){y}_1={A} \sin (\omega {t}-{Kx}), {y}_2={A} \sin (\omega {t}-{kx}+\phi). The resultant amplitude of the medium particle will be
View answer and explanation
Correct option: B
2Acos⁡ϕ22 A \cos \frac{\phi}{2}
Option B (2Acos⁡ϕ22 A \cos \frac{\phi}{2}) is correct because it represents the resultant amplitude of two waves interfering with a phase difference of ϕ\phi.
45 Medium
45 of 524
Understanding +4 | -1
The relation between velocity of sound in a gas (v)(v) and rms velocity of molecules of gas (vrms)\left(v_{\mathrm{rms}}\right) is:
View answer and explanation
Correct option: A
v=vrms(γ/3)1/2v=v_{\text {rms}}(\gamma / 3)^{1 / 2}
Option A (v=vrms(γ/3)1/2v=v_{\text {rms}}(\gamma / 3)^{1 / 2}) is correct because it represents the correct relationship between the velocity of sound in a gas and the RMS velocity of the gas molecules.
46 Medium
46 of 524
Remembering +4 | -1
The loudness and pitch of a sound depends on
View answer and explanation
Correct option: C
Intensity and frequency
Option C (Intensity and frequency) is correct because loudness is related to the intensity of the sound wave, and pitch is related to its frequency.
47 Medium
47 of 524
Understanding +4 | -1
Statement-I: Mechanical waves require material medium for their propagation.
Statement-II: Transverse waves cannot be generated with in the liquids.
View answer and explanation
Correct option: C
Both statement-I &\& II are correct
Option C (Both statement-I & II are correct) is correct because mechanical waves require a medium, and transverse waves generally cannot propagate within liquids.
48 Medium
48 of 524
Remembering +4 | -1
Ultrasonic waves are those waves
View answer and explanation
Correct option: B
Man can’t hear
Option B (Man can’t hear) is correct because ultrasonic waves have frequencies beyond the human audible range.
49 Medium
49 of 524
Application +4 | -1
Ten tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tuning forks produces 8 beats/s. The highest frequency is four times that of the lowest. Possible highest and lowest frequencies (in Hz{Hz} ) are
View answer and explanation
Correct option: C
96 and 24
Option C (96 and 24) is correct because the frequencies are in arithmetic progression and the highest frequency is four times the lowest.
50 Medium
50 of 524
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 is correct because the frequency of the first tuning fork is 105 Hz. Since loading the fork doesn't change the number of beats, the original frequency must have been higher than 100 Hz.
51 Medium
51 of 524
Understanding +4 | -1
If n1,n2,n3….{n_1},{n_2},{n_3} \ldots . are the frequencies of segments of a stretched string, the frequency nn of the string is given by
View answer and explanation
Correct option: C
1n=1n1+1n2+1n3+…\frac{1}{n} = \frac{1}{{{n_1}}} + \frac{1}{{{n_2}}} + \frac{1}{{{n_3}}} + \ldots
Option C is correct. The frequency of the entire string is related to the frequencies of its segments by the reciprocal relationship 1n=1n1+1n2+1n3+…\frac{1}{n} = \frac{1}{{{n_1}}} + \frac{1}{{{n_2}}} + \frac{1}{{{n_3}}} + \ldots.
52 Medium
52 of 524
Application +4 | -1
Two waves represented by y1=10sin⁡(2000πt+2x)\mathrm{y}_1=10 \sin (2000 \pi \mathrm{t}+2 \mathrm{x}) and y2=10sin⁡(2000πt+2x+π/2)\mathrm{y}_2=10 \sin (2000 \pi \mathrm{t}+2 \mathrm{x}+\pi / 2) are superposed at any point at a particular instant. The resultant amplitude is:
View answer and explanation
Correct option: C
14.1 units
Option C (14.1 units) is correct because it is the resultant amplitude calculated using the superposition principle with a phase difference of π/2\pi / 2.
53 Medium
53 of 524
Application +4 | -1
The wavelengths of two waves are 50 and    51  cm50\mathrm{ }\,and\;\,\,51\,\,cm respectively. If the temperature of the room is 20 ∘C20\,^\circ C, then what will be the number of beats produced per second by these waves, when the speed of sound at 0 ∘C0\,^\circ C is 332  m/sec⁡332\,\,m/\sec
View answer and explanation
Correct option: A
14
Option A (14) is correct because the number of beats produced per second is calculated using the difference in frequencies of the two waves.
54 Medium
54 of 524
Understanding +4 | -1
It is possible to distinguish between the transverse and longitudinal waves by studying the property of
View answer and explanation
Correct option: D
Polarisation
Option D (Polarisation) is correct because polarization is a property unique to transverse waves, distinguishing them from longitudinal waves.
55 Medium
55 of 524
Application +4 | -1
The fundamental frequency of an open pipe of length 1 m1 \mathrm{~m}, if the speed of sound in air is 340 ms−1340 \mathrm{~ms}^{-1} is
View answer and explanation
Correct option: B
170 Hz170 \mathrm{~Hz}
Option B (170 Hz170 \mathrm{~Hz}) is correct because it is the fundamental frequency calculated using the formula v=v2lv = \frac{v}{{2l}} for an open pipe.
56 Medium
56 of 524
Understanding +4 | -1
Energy is not carried by which of the following waves
View answer and explanation
Correct option: A
Stationary
Option A is correct because stationary waves do not propagate energy; they store energy in the form of potential and kinetic energy.
57 Medium
57 of 524
Analyzing +4 | -1
A sound wave of frequency nn travels horizontally to the right. It is reflected from a large vertical plane surface moving to the left with speed vv. The speed of the sound in the medium is c.c. Then
View answer and explanation
Correct option: C
The number of waves strike the surface per second is n[c+vc]n\left[ {\frac{{c + v}}{c}} \right]
Option C (The number of waves strike the surface per second is n[c+vc]n\left[ {\frac{{c + v}}{c}} \right]) is correct because it represents the apparent frequency observed by the moving surface due to the Doppler effect.
58 Medium
58 of 524
Application +4 | -1
A string on a musical instrument is 50 cm long and its fundamental frequency is 270 Hz. If the desired frequency of 1000 Hz is to be produced, the required length of the string is (in cm)
View answer and explanation
Correct option: A
13.5
Option A (13.5) is correct because the frequency of a string is inversely proportional to its length, so n∝1Ln \propto \frac{1}{L}.
59 Medium
59 of 524
Application +4 | -1
If a length of a closed organ pipe is 1  m1\,\,m and velocity of sound is 400  m/s400\,\,m/s, then the frequency for the second note is
View answer and explanation
Correct option: B
3×3304Hz3 \times \frac{{330}}{4}Hz
Option B (3×3304Hz3 \times \frac{{330}}{4}Hz) is correct because the second note in a closed organ pipe corresponds to the third harmonic, and the frequency is calculated using the formula f3=3v4lf_3 = \frac{{3v}}{{4l}}.
60 Medium
60 of 524
Application +4 | -1
At 16∘C16^{\circ} {C}, two open end organ pipes, when sounded together give 34 beats in 2 s2 {~s}. How many beats per second will be produced, if the temperature is raised to 51∘C51^{\circ} {C} ?
(Neglect increase in length of the pipes)
View answer and explanation
Correct option: A
18 s−118 {~s}^{-1}
Option A (18 s−118 {~s}^{-1}) is correct because it is the beat frequency calculated using the temperature dependence of the speed of sound.
Revision notes and key takeaways

Review WAVES in your textbook or class notes, then use this set to check your understanding.

Showing 31-60 of 524 questions
0 /524
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.