121 past paper MCQs from Physics unit 6,
Waves, across 11 topics. They are arranged topic by topic; bigger topics show a
sample here and link to their full set and to study notes. Tap “Show answer” for the correct option and a short solution.
4.In a ripple tank 40 waves pass through a certain point in one second. If the wavelength of the waves is 5 cm, then find the speed of the wave. (MDCAT 2020)
$2\ \mathrm{m\,s^{-1}}$
$8\ \mathrm{m\,s^{-1}}$
$20\ \mathrm{m\,s^{-1}}$
$200\ \mathrm{m\,s^{-1}}$
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Correct answer: (a) $2\ \mathrm{m\,s^{-1}}$
$v=f\lambda=40\times0.05=2\ \mathrm{m\,s^{-1}}$.
5.The product of frequency and time period is equal to: (MDCAT 2020)
1
2
0
$1/2$
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Correct answer: (a) 1
$T=1/f$, so $fT=1$.
6.The shortest distance between any two points in phase on a wave is called the: (MDCAT 2022)
Speed of Sound: Newton’s Formula and Laplace’s Correction
10.Newton's original formula underestimated speed of sound in air because he: (SIBA MDCAT 2025)
Ignored viscosity
Assumed isothermal
Considered vacuum conditions
Assumed adiabatic
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Correct answer: (b) Assumed isothermal
Newton took the compressions to be isothermal; Laplace corrected them to adiabatic and got the right speed.
11.The option that shows the conditions used by Laplace for connecting the velocity of sound is options medium used thermodynamics process 1 solid adiabatic 2 liquids isobaric 3 gas adiabatic 4 gas isothermal (SZABMU MDCAT 2025)
1
2
3
4
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Correct answer: (c) 3
Laplace took the compressions and rarefactions in a gas to be too quick for heat to escape, that is adiabatic.
12.Speed of sound in solid is greater than in air because ratio: (NUMS MDCAT 2024)
(√E/ρ) solid < (√E/ρ) air
(√E/ρ) solid > (√E/ρ) air
(√E/ρ) solid = (√E/ρ) air
(√E/ρ) solid ≤ (√E/ρ) air
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Correct answer: (b) (√E/ρ) solid > (√E/ρ) air
$v=\sqrt{E/\rho}$, and a solid gains far more in elasticity than it loses to density.
16.If two speakers emit sound at same frequency and phase, maximum loudness occurs when: (UHS MDCAT 2025)
Path difference = √2
Path difference = λ
Path difference = λ/4
Path difference = 3λ/43
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Correct answer: (b) Path difference = λ
Constructive interference needs a path difference of a whole number of wavelengths.
17.Two waves identical traveling in the same medium are superposed: (NUMS MDCAT 2025)
Interference
Diffraction
Reflection
Refraction
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Correct answer: (a) Interference
Two waves of the same kind crossing in one medium add displacement to displacement, which is interference.
18.The path difference between two sound waves coming from a coherent source of wavelength 50 cm at a point is 100 cm. The superposition of the waves at that point produces: (SIBA MDCAT 2025)
Beats
Echo
Loudness
Silence
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Correct answer: (c) Loudness
A path difference of $100\,\mathrm{cm}$ is exactly two wavelengths, so the waves arrive in step and reinforce.
22.Stationary waves are formed in a stretched string of 2m length, such that two vibrating loops are formed. The distance between consecutive nodes formed is: (SZABMU MDCAT 2025)
0.5 m
1 m
2 m
3 m
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Correct answer: (b) 1 m
Two loops in $2\,\mathrm{m}$ make each loop $1\,\mathrm{m}$, and a loop runs from one node to the next.
23.Which one of the following is INCORRECT about the nodes when the string is plucked; (UHS MDCAT 2024)
Amplitude of vibration is zero
Do not move along the string
Produced at the fixed ends of strings
Distance between consecutive nodes is 1 wavelength
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Correct answer: (d) Distance between consecutive nodes is 1 wavelength
Consecutive nodes are half a wavelength apart, not a whole one.
24.When a standing wave is set up on a string fixed at both ends, which of the following statements is true? (PMC Practice 2021)
Sum of the number of antinodes and the number of nodes is always even
Wavelength = length string / number of nodes
The shape of the string at any instant shows a symmetry about the midpoint of the string
Frequency = number of nodes x fundamental frequency
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Correct answer: (c) The shape of the string at any instant shows a symmetry about the midpoint of the string
The pattern of loops between the fixed ends is symmetrical about the middle of the string.
25.The speed ‘v’ of the waves in the string depends upon the tension, F, of the string and m, the mass per unit length of the string. It is given by: (PMC Practice 2021)
v2 = F/m
v = F/m
v x m = F
v = F x m
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Correct answer: (a) v2 = F/m
The speed of a wave on a stretched string is $\sqrt{F/m}$.
26.For a certain organ pipe three successive resonance frequencies are \$425\$, \$595\$ and \$765\,\mathrm{Hz}\$. Taking the speed of sound as \$340\,\mathrm{m\,s^{-1}}\$, the pipe is: (UHS MDCAT 2017)
a closed pipe of length $1\,\mathrm{m}$
a closed pipe of length $2\,\mathrm{m}$
an open pipe of length $1\,\mathrm{m}$
an open pipe of length $2\,\mathrm{m}$
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Correct answer: (a) a closed pipe of length $1\,\mathrm{m}$
The spacing is $170\,\mathrm{Hz}$, and the three are $5$, $7$ and $9$ times $85\,\mathrm{Hz}$. Only odd harmonics appear, so the pipe is closed, and $85=v/4L$ gives $L=1\,\mathrm{m}$.
27.A pipe open at both ends resonates at a fundamental frequency fo. When one end is covered and the pipe is again made to resonate, the fundamental frequency is fc. Which of the following expressions describes the relationship between these two resonants? (PMC MDCAT 2022)
fc = fo
2xfc = fo
fc = 2fo
none
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Correct answer: (b) 2xfc = fo
An open pipe sounds $v/2L$ and a closed one $v/4L$, so the open note is twice the closed.
28.For a certain organ pipe, three successive resonance frequencies are observed at 425, 595, and 765 Hz. The speed of the sound in air is 340 m/s. The pipe is: (UHS MDCAT 2017)
Closed pipe of length 1 m
Closed pipe of length 2 m
Open pipe of length 1 m
Open pipe of length 2 m
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Correct answer: (a) Closed pipe of length 1 m
The frequencies differ by $170\,\mathrm{Hz}$ and are odd multiples of $85\,\mathrm{Hz}$, so the pipe is closed and $340/(4\times85)$ is $1\,\mathrm{m}$.
32.A particle is moving in a uniform circular path whose projection is executing simple harmonic motion on horizontal diameter. The ratio of instantaneous velocity to the maximum velocity of the projection while passing through the center is: (SZABMU MDCAT 2025)
1 : 1
1 : 2
2 : 1
1 : 4
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Correct answer: (a) 1 : 1
The projection is fastest as it passes the centre, so the instantaneous speed there is the maximum speed.