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Physics MCQs for NEET — Practice Questions with Answers

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What is the phase difference between two successive crests in the wave?

π/2

2π

4π

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Explanation

Since distance between two consecutive crests is λ, so

$ϕ = \frac{2 π}{λ} \times λ = 2 π .$

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A wave of the frequency of 500 Hz has a velocity of 360 m/sec. The distance between two nearest points 60° out of phase, is :

0.6 cm

12 cm

60 cm

120 cm

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Explanation

The distance between two points i.e. path difference between them $Δ = \frac{λ}{2 π} \times ϕ = \frac{λ}{2 π} \times \frac{π}{3} = \frac{λ}{6} = \frac{v}{6 n}$ $(∵ v = n λ)$

⇒ $Δ = \frac{360}{6 \times 500} = 0 .12 m = 12 c m$

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Sound waves of wavelength greater than that of audible sound are called :

Seismic waves

Sonic waves

Ultrasonic waves

Infrasonic waves

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Explanation

Infrasonic waves have frequency less than (20 Hz) audible sound and wavelength more than audible sound.

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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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