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JEE • PHYSICS • WAVES

Waves practice · Medium

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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 61-90 of 524 questions
61 Medium
61 of 524
Understanding +4 | -1
Water waves produced by a motor boat sailing in water are
View answer and explanation
Correct option: B
Both longitudinal and transverse
The correct answer is B (Both longitudinal and transverse) because water waves produced by a boat have both longitudinal (compression) and transverse (up and down) components.
62 Medium
62 of 524
Understanding +4 | -1
Light can travel in vacuum but not sound, because
View answer and explanation
Correct option: B
Light waves are electromagnetic in nature
The correct answer is B (Light waves are electromagnetic in nature) because electromagnetic waves do not require a medium to propagate, unlike mechanical waves like sound.
63 Medium
63 of 524
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 in a resonance tube depends on the length of the air column and the speed of sound, not the density of the liquid used.
64 Medium
64 of 524
Understanding +4 | -1
Assertion: Transverse waves can travel in all medium.
Reason: All the gases have some bulk modulus.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because transverse waves require a medium with shear modulus to travel, which gases lack, and the reason is irrelevant.
65 Medium
65 of 524
Application +4 | -1
The equation of two harmonic waves are given by
y1=3sin⁡2π(x−vt)cmy_1=3 \sin 2 \pi(x-v t) \mathrm{cm}
y2=4cos⁡2π(x−vt+1.5)cmy_2=4 \cos 2 \pi(x-v t+1.5) \mathrm{cm}
These waves are simultaneously passing through a string. The amplitude of the resulting wave is:
View answer and explanation
Correct option: D
5 cm5 \mathrm{~cm}
Option D (5 cm5 \mathrm{~cm}) is correct because it is the resultant amplitude calculated by considering the phase difference between the two waves.
66 Medium
66 of 524
Analyzing +4 | -1
Assertion: Two waves moving in a uniform string having uniform tension cannot have different velocities.
Reason: Elastic and inertial properties of string are same for all waves in same string. but speed of wave in a string does not depend on its elastic and inertial properties only.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D is correct because both the assertion and the reason are false. Waves in a uniform string with uniform tension have the same speed but can have different velocities (directions), and the speed of a wave in a string depends on its elastic and inertial properties.
67 Medium
67 of 524
Understanding +4 | -1
Assertion: Only longitudinal mechanical waves can propagate in gases.
Reason: Gases have only bulk modulus.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A is correct because longitudinal waves are the only type that can propagate in gases, and this is because gases only possess bulk modulus, which is related to volume changes.
68 Medium
68 of 524
Remembering +4 | -1
When two sinusoidal waves moving at right angles to each other superimpose, they produce
View answer and explanation
Correct option: D
Lissajous figure
Option D (Lissajous figure) is correct because the superposition of two sinusoidal waves at right angles produces a Lissajous figure, which depends on their frequencies and phase difference.
69 Medium
69 of 524
Application +4 | -1
Let speed of sound waves in hydrogen gas at room temperature is v0{v_0}. What will be the speed of sound waves in a room which contains oxygen at same temperature:-
View answer and explanation
Correct option: A
v04\frac{{{v_0}}}{4}
Option A (v04\frac{{{v_0}}}{4}) is correct because the speed of sound is inversely proportional to the square root of the molar mass.
70 Medium
70 of 524
Application +4 | -1
A student sees a jet plane flying from east to west. When the jet is seen just above his head, the sound of jet appears to reach him making angle 60∘60^{\circ} with the horizontal from east. If the velocity of sound is vv. then that of the jet plane is
View answer and explanation
Correct option: D
v/2{v} / 2
Option D (v/2{v} / 2) is correct because the velocity of the jet plane is half the velocity of sound, derived from the geometry of the sound wave propagation.
71 Medium
71 of 524
Remembering +4 | -1
Water waves are
View answer and explanation
Correct option: C
Both longitudinal and transverse
Option C (Both longitudinal and transverse) is correct because water waves exhibit both longitudinal and transverse motion of water particles.
72 Medium
72 of 524
Application +4 | -1
If a source emitting waves a velocity v/4v/4 and the observer moves away from the source with a velocity v/6v/6, the apparent frequency as heard by the observer will be (vv=velocity of sound)
View answer and explanation
Correct option: C
109v\frac{{10}}{9}v
Option C (20 N) is correct because the magnitude of one of the forces is found using the given resultant and angle, along with the condition that the resultant is perpendicular to one of the forces.
73 Medium
73 of 524
Application +4 | -1
Amplitude of 2 waves are a1a_1 and a2a_2, where a1=a2=aa_1=a_2=a and both are superimposed at PP where path difference between then is λ4\frac{\lambda}{4}. Find amplitude of resultant wave.
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
The correct answer is π2\frac{\pi}{2} because the phase difference is calculated from the path difference, and the resultant amplitude depends on this phase difference.
74 Medium
74 of 524
Understanding +4 | -1
If A{A} is the amplitude of the wave coming from a point source at distance r{r} then
View answer and explanation
Correct option: B
Aαr−1A \alpha r^{-1}
Option B (Aαr−1A \alpha r^{-1}) is correct because the intensity of a wave from a point source decreases with the square of the distance, and amplitude is proportional to the square root of intensity.
75 Medium
75 of 524
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/2A/2) is correct because the amplitude of a wave from a point source decreases inversely with the distance from the source.
76 Medium
76 of 524
Application +4 | -1
Two sound waves having a phase difference of 600{60^0} have path difference of
View answer and explanation
Correct option: C
λ/6\lambda /6
Option C (λ/6\lambda /6) is correct because the path difference is calculated using the phase difference and the wavelength of the sound wave.
77 Medium
77 of 524
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 when the tension is increased, the frequency increases, and the beat frequency decreases, indicating the original frequency was lower than the tuning fork's.
78 Medium
78 of 524
Application +4 | -1
In two similar wires of tension 16  N16\,\,N and TT, 3  beats3\,\,beats are heard, then T=T =
View answer and explanation
Correct option: A
49 N49\,N
Option A (49 N49\,N) is correct because the tension is proportional to the square of the frequency, and the beat frequency is the difference between the two frequencies.
79 Medium
79 of 524
Understanding +4 | -1
When two wave of almost equal frequencies n1  and  n2{n_1}\,\,and\,\,{n_2} are produced simultaneously, then the time interval between successive maxima is
View answer and explanation
Correct option: A
1n1−n2\frac{1}{{{n_1} - {n_2}}}
Option A is correct because the time interval between successive maxima is the reciprocal of the beat frequency, which is the difference between the two frequencies.
80 Medium
80 of 524
Remembering +4 | -1
Ultrasonic waves can be used to
View answer and explanation
Correct option: D
All of the above
Option D is correct. Ultrasonic waves have various applications, including detecting submarines, cleaning machinery, and pest control.
81 Medium
81 of 524
Understanding +4 | -1
Stationary waves are formed when
View answer and explanation
Correct option: A
Two waves of equal amplitude and equal frequency travel along the same path in opposite directions
Option A (Two waves of equal amplitude and equal frequency travel along the same path in opposite directions) is correct because this is the condition required for the formation of stationary waves due to superposition.
82 Medium
82 of 524
Analyzing +4 | -1
For a particular tube, three harmonic are 300 Hz300 \mathrm{~Hz}, 600 Hz600 \mathrm{~Hz} and 750 Hz750 \mathrm{~Hz}. The other possible frequencies of oscillations in the tube are
View answer and explanation
Correct option: C
150,450 Hz150,450 \mathrm{~Hz}
Option C (150,450 Hz150,450 \mathrm{~Hz}) is correct because the given frequencies suggest a fundamental frequency of 150 Hz, and 450 Hz is a multiple of this fundamental frequency.
83 Medium
83 of 524
Analyzing +4 | -1
When two tuning forks AA and BB are sounded together, xx beat/s are heard. Frequency of AA is nn. Now, when one prong of fork BB is loaded with a little wax, the number of beats decreases. The frequency of fork BB is
View answer and explanation
Correct option: A
n+xn+x
Option A (n+x) is correct because loading the fork decreases its frequency, and since the beat frequency decreased, the original frequency of B must have been higher than A.
84 Medium
84 of 524
Understanding +4 | -1
The Doppler’s effect is applicable for
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because the Doppler effect applies to both light and sound waves.
85 Medium
85 of 524
Analyzing +4 | -1
Two tuning fork, AA and BB produce notes of frequencies 258  Hz258\,\,Hz and 262  Hz262\,\,Hz. An unknown note sounded with a produces certain beats. When the same note is sounded with BB, the beat frequency gets doubled, the unknown frequency is
View answer and explanation
Correct option: B
254 Hz254\mathrm{ }\,Hz
Option B (254 Hz254\mathrm{ }\,Hz) is correct because the unknown frequency is determined by analyzing the beat frequencies produced with two known frequencies.
86 Medium
86 of 524
Application +4 | -1
A longitudinal progressive wave is given by the equation y=5×10−2sin⁡π(400t+x)my=5 \times 10^{-2} \sin \pi(400 {t}+{x}) {m}. Find (i) amplitude (ii) frequency (iii) wave length and (iv) velocity of the wave. (v) velocity and acceleration of particle at x=16 m{x}=\frac{1}{6} {~m} at t=0.01{t}=0.01 s (vi) maximum particle velocity and acceleration.
View answer and explanation
Correct option: A
8×104 ms−28 \times 10^4 {~ms}^{-2}
Option A (8×104 ms−28 \times 10^4 {~ms}^{-2}) is not the complete answer, but it contains the correct maximum acceleration. The question asks for multiple values, but this option only provides the maximum acceleration, and the other options are incorrect. The complete solution is in the steps below.
87 Medium
87 of 524
Application +4 | -1
A tuning fork of frequency 392  Hz,392\,\,Hz,, resonates with 50  cm50\,\,cm length of a string under tension (T)\left( T \right). If length of the string is decreased by 2%2\%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is
View answer and explanation
Correct option: C
88
Option C (8) is correct because the number of beats is approximately 8, calculated from the change in frequency due to the change in length of the string.
88 Medium
88 of 524
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.
89 Medium
89 of 524
Understanding +4 | -1
Speed of sound waves in a fluid depends upon
View answer and explanation
Correct option: C
Inversely of the square root of density
Option C (Inversely of the square root of density) is correct because the speed of sound in a fluid is inversely proportional to the square root of its density.
90 Medium
90 of 524
Application +4 | -1
A tuning fork vibrates with 2 beats in 0.040.04 second. The frequency of the fork is
View answer and explanation
Correct option: A
50  Hz50\,\,Hz
Option A (50 Hz50 \,Hz) is correct because the beat frequency is calculated by dividing the number of beats by the time interval.
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