CBSE Class 10 Physics Waves and Sound MCQs Set 27

Answer: (b) 50

Explanation:
1. In 1 s, 25 compressions pass the point.
2. Also 25 rarefactions pass.
3. Total = 25 + 25 = 50.

Teacher's Note:
1. Each oscillation consists of one compression and one rarefaction.

Answer: (d) mass

Explanation:
1. Pressure, density and volume of the region all change as the wave passes.
2. The mass of the medium in the region remains the same.

Teacher's Note:
1. Mass is conserved; only the distribution of particles changes.

Answer: (b) there is no atmosphere

Explanation:
1. The Moon has practically no atmosphere.
2. Without a medium sound cannot propagate, so no sound reaches the other person.

Teacher's Note:
1. Astronauts talk through radio waves, which need no medium.

Answer: (d) the transmitting medium

Explanation:
1. The speed depends on the elasticity and density of the medium (and on its temperature).
2. It does not depend on the properties of the wave such as amplitude or frequency.

Teacher's Note:
1. The same sound travels at different speeds in air, water and steel.

Answer: (b) ultrasonic waves

Explanation:
1. SONAR sends high-frequency ultrasonic pulses and detects the echo.
2. Ultrasonic waves are used because they are directional and travel as narrow beams.

Teacher's Note:
1. SONAR stands for sound navigation and ranging.

Answer: (b) t = T

Explanation:
1. The wave repeats after one time period.
2. The region returns to the same state, minimum density, after T.

Teacher's Note:
1. At T/2 the region has maximum density.

Answer: (c) the frequency of the sitar string to match other instruments

Explanation:
1. Tension controls the frequency of vibration of the string.
2. The sitarist tunes the string to the required frequency, which sets the pitch.

Teacher's Note:
1. Tuning changes the pitch, not the loudness.

Answer: (c) the disturbance moves

Explanation:
1. Only the disturbance travels.
2. The air particles merely oscillate about their positions.

Teacher's Note:
1. Waves carry energy without carrying matter.

Answer: (d) 0.010 s

Explanation:
1. A crest changes to a trough after half a period.
2. t = T/2 = 0.01 s.

Teacher's Note:
1. After T/4 the displacement is zero.

Answer: (b) the intensity of sound is 10\( ^{-12} \) W/m\( ^2 \)

Explanation:
1. The decibel scale is defined with reference to the faintest audible intensity.
2. That reference is 10\( ^{-12} \) W/m\( ^2 \), which is 0 dB.

Teacher's Note:
1. 0 dB does not mean no sound.

Answer: (c) 330 m/s

Explanation:
1. The sound travels to the mountain and back: 2 \( \times \) 660 = 1320 m.
2. v = 1320/4 = 330 m/s.

Teacher's Note:
1. Remember to double the distance for an echo.

Answer: (d) a tuning fork

Explanation:
1. A tuning fork vibrates at a fixed natural frequency.
2. It gives a pure note of a known frequency.

Teacher's Note:
1. The frequency is stamped on the fork.

Answer: (d) amplitude

Explanation:
1. Loudness depends on the amplitude of vibration.
2. A louder sound has a larger amplitude.

Teacher's Note:
1. Frequency decides the pitch.

Answer: (a) break the step

Explanation:
1. Marching in step can apply periodic force at the bridge's natural frequency.
2. That can cause resonance and a dangerously large amplitude.
3. So soldiers break step.

Teacher's Note:
1. The Broughton bridge in England collapsed in this way in 1831.

Answer: (d) 1800

Explanation:
1. Number of waves = f \( \times \) t.
2. = 3600 \( \times \) 0.50 = 1800.

Teacher's Note:
1. One wavelength is produced per vibration of the source.

Answer: (d) supersonic speed

Explanation:
1. Speed greater than the speed of sound is called supersonic.

Teacher's Note:
1. Ultrasonic and infrasonic refer to frequencies.

Answer: (d) 2000

Explanation:
1. Speed = 600/2 = 300 m/s.
2. \( \lambda \) = 300/1000 = 0.3 m.
3. Number of wavelengths = 600/0.3 = 2000.

Teacher's Note:
1. The same number is f \( \times \) t = 1000 \( \times \) 2 = 2000.

Answer: (c) 21 m/s

Explanation:
1. v = f\( \lambda \) = 0.07 \( \times \) 300.
2. v = 21 m/s.

Teacher's Note:
1. Wave speed = frequency \( \times \) wavelength.

Answer: (c) 17 m or more

Explanation:
1. The echo must return after at least 0.1 s.
2. The sound covers 340 \( \times \) 0.1 = 34 m in this time.
3. The obstacle must be at least 34/2 = 17 m away.

Teacher's Note:
1. Distance = v \( \times \) t/2.

Answer: (d) 7 m

Explanation:
1. Frequency f = 340/1.7 = 200 Hz and it stays constant.
2. \( \lambda \) in water = 1400/200 = 7 m.

Teacher's Note:
1. The frequency is unchanged when the medium changes.

Answer: (d) 0

Explanation:
1. Sound is a mechanical wave and needs a material medium.
2. In vacuum there are no particles to vibrate, so sound cannot travel and its speed is zero.

Teacher's Note:
1. Light, unlike sound, can travel through vacuum.

Answer: (b) a longitudinal wave

Explanation:
1. The coils are displaced along the length of the slinky, parallel to the direction in which the pulse moves.
2. That makes it a longitudinal wave.

Teacher's Note:
1. A sideways flick of the slinky would give a transverse wave.

Answer: (a) the force exerted by air and friction at the support

Explanation:
1. Air resistance and friction at the support remove energy from the pendulum.
2. So the amplitude falls gradually until it stops.

Teacher's Note:
1. Such a loss of amplitude is called damping.