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Waves MCQs for NEET — Physics Questions with Answers

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The ratio of masses of nitrogen and oxygen is 14:16. The temperature at which the speed of sound in nitrogen will be same at that in oxygen at 55°C is -

35°C

48°C

65°C

14°C

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Explanation

$v = \sqrt{\frac{γ R T}{M}}$ ⇒ $\frac{T_{N}}{T_{0}} = \frac{M_{N}}{M_{0}}$ ⇒ $\frac{T_{N}}{273 + 55} = \frac{14}{16} = \frac{7}{8}$

⇒ $T_{N} = 287 K = 14 ° C$

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A wavelength 0.60 cm is produced in air and it travels at a speed of 300 ms^–1. It will be an

Audible wave

Infrasonic wave

Ultrasonic wave

None of the above

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Explanation

$n = \frac{v}{λ} = \frac{300}{0 .6 \times 10^{- 2}} H z$ $= \frac{3}{6} \times 10^{5} H z = 50, 000 H z$

⇒ Wave is ultrasonic.

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An observer standing near the seashore observes 54 waves per minute. If the wavelength of the water wave is 10m then the velocity of a water wave is :

540 ms^-1

5.4 ms^-1

0.184 ms^-1

9 ms^-1

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Explanation

Number of waves per minute = 54

∴ Number of waves per second = 54/60

Now v = nλ ⇒ $n = \frac{54}{60} \times 10 = 9 m / s .$

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The equation of a wave is $y = 2 \sin π (0 .5 x - 200 t)$ , where x and y are expressed in cm and t in sec. The wave velocity is :

100 cm/sec

200 cm/sec

300 cm/sec

400 cm/sec

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Explanation

Comparing given equation with standard equation of progressive wave. The velocity of wave

$v = \frac{ω (Co-efficient of t)}{k (Co-efficient of x)} = \frac{200 π}{0 .5 π} = 400 c m / s$

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Equation of a progressive wave is given by

$y = 0 .2 \cos π (0 .04 t + . 02 x - \frac{π}{6})$

The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of π/2 :

4 cm

8 cm

25 cm

12.5 cm

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Explanation

Comparing with $y = a \cos (ω t + k x - ϕ)$ ,

We get $k = \frac{2 π}{λ} = 0 .02 π \Rightarrow λ = 100 c m$

Also, it is given that phase difference between particles $Δ ϕ = \frac{π}{2} .$

Hence path difference between them $Δ = \frac{λ}{2 π} \times Δ ϕ = \frac{λ}{2 π} \times \frac{π}{2} = \frac{λ}{4} = \frac{100}{4} = 25 c m$

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Two waves are given by $y_{1} = a \sin (ω t - k x)$ and $y_{2} = a \cos (ω t - k x)$ . The phase difference between the two waves is :

$\frac{π}{4}$

$\frac{π}{8}$

$\frac{π}{2}$

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Explanation

$y_{1} = a \sin (ω t - k x)$

and $y_{2} = a \cos (ω t - k x) = a \sin (ω t - k x + \frac{π}{2})$

Hence phase difference between these two is $\frac{π}{2} .$

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If the amplitude of waves at distance r from a point source is A, the amplitude at a distance 2r will be :

A/2

A/4

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Explanation

$I \propto A^{2} \propto \frac{1}{d^{2}}$ ⇒ $a \propto \frac{1}{d}$

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A wave is reflected from a rigid support. The change in phase on reflection will be :

π/4

π/2

2π

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Explanation

After reflection from rigid support, a wave suffers a phase change of π.

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A plane wave is represented by $x = 1 .2 \sin (314 t + 12 .56 y)$

Where x and y are distances measured along in x and y direction in meters and t is time in seconds. This wave has

A wavelength of 0.25 m and travels in + ve x-direction

A wavelength of 0.25 m and travels in + ve y-direction

A wavelength of 0.5 m and travels in – ve y-direction

A wavelength of 0.5 m and travels in – ve x-direction

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Explanation

The given equation representing a wave travelling along –y direction (because ‘+’ sign is given between t term and x term).

On comparing it with $x = A \sin (ω t + k y)$

We get $k = \frac{2 π}{λ} = 12 .56$ ⇒ $λ = \frac{2 \times 3 .14}{12 .56} = 0 .5 m$

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The equation of a wave traveling in a string can be written as $y = 3 \cos π (100 t - x)$ . Its wavelength is :

100 cm

2 cm

5 cm

None of the above

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Explanation

Comparing the given equation with $y = a \cos (ω t - k x)$

We get $k = \frac{2 π}{λ} = π$ ⇒ λ = 2cm

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