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Physics MCQs for NEET — Practice Questions with Answers

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A tuning fork makes 256 vibrations per second in air. When the velocity of sound is 330 m/s, then the wavelength of the tone emitted is :

0.56 m

0.89 m

1.11 m

1.29 m

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Explanation

$v = n λ$ ⇒ $λ = \frac{v}{n} = \frac{330}{256} = 1 .29 m$

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Sound waves have the following frequencies that are audible to human beings :

5 c/s

27000 c/s

5000 c/s

50,000 c/s

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Explanation

The audible range of frequency is 20Hz to 20kHz.

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The minimum audible wavelength at room temperature is about

0.2 Å

5 Å

5 cm to 2 metre

20 mm

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Explanation

Since maximum audible frequency is 20,000 Hz,

hence $λ_{\min} = \frac{v}{n_{\max}} = \frac{340}{20, 000} \approx 20 m m$

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The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is

$\sqrt{2 / 7}$

$\sqrt{1 / 7}$

$\sqrt{3} / 5$

$\sqrt{6} / 5$

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Explanation

Velocity of sound in gas $v = \sqrt{\frac{γ R T}{M}}$ ⇒ $v \propto \sqrt{\frac{γ T}{M}}$

⇒ $\frac{v_{N_{2}}}{v_{H e}} = \sqrt{\frac{γ_{N_{2}}}{γ_{H e}} \times \frac{M_{H e}}{M_{H_{2}}}} = \sqrt{\frac{\frac{7}{5} R \times 4}{\frac{5}{3} R \times 28}} = \frac{\sqrt{3}}{5}$

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The frequency of a rod is 200 Hz. If the velocity of sound in air is 340 ms^–1, the wavelength of the sound produced is :

1.7 cm

6.8 cm

1.7 m

6.8 m

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Explanation

$λ = \frac{v}{n} = \frac{340}{200} = 1 .7$ m

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A stone is dropped into a lake from a tower 500 metre high. The sound of the splash will be heard by the man approximately after

[Given : Speed of sound = 340 m/s ]

11.5 seconds

21 seconds

10 seconds

14 seconds

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Explanation

The time taken by the stone to reach the lake

$t_{1} = \sqrt{(\frac{2 h}{g})} = \sqrt{(\frac{2 \times 500}{10})} = 10 s e c$ (Using $h = u t + \frac{1}{2} g t^{2}$ )

Now time taken by sound from lake to the man

$t_{2} = \frac{h}{v} = \frac{500}{340} \approx 1 .5 \sec$

⇒ Total time $= t_{1} + t_{2} = 10 + 1 .5 = 11 .5 \sec .$

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When sound waves travel from air to water, which of the following remains constant

Velocity

Frequency

Wavelength

All the above

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Explanation

When medium changes, velocity and wavelength changes but frequency remains constant.

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At what temperature velocity of sound is double than that of at 0°C

819 K

819°C

600°C

600 K

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Explanation

$v \propto \sqrt{T}$ ⇒ $\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$ ⇒ $2 = \sqrt{\frac{T_{2}}{(273 + 0)}}$

⇒ $T_{2} = 273 \times 4 = 1092 K = 819^{o} C$

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If velocity of sound in a gas is 360 m/s and the distance between a compression and the nearest rarefaction is 1m, then the frequency of sound is

90 Hz

180 Hz

360 Hz

720 Hz

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Explanation

Distance between a compression and the nearest rarefaction is $\frac{λ}{2} = 1 m .$

Hence $n = \frac{v}{λ} = \frac{360}{2} = 180 H z$ .

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If the density of oxygen is 16 times that of hydrogen, what will be the ratio of their corresponding velocities of sound waves :

1 : 4

4 : 1

16 : 1

1 : 16

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Explanation

$v = \sqrt{\frac{γ P}{ρ}} \Rightarrow \frac{v_{O_{2}}}{v_{H_{2}}} = \sqrt{\frac{ρ_{H_{2}}}{ρ_{o_{2}}}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$

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