9th Grade > Physics
SOUND MCQs
Total Questions : 57
| Page 5 of 6 pages
Answer: Option D. ->
Sound cannot propagate in outer space.
:
D
Sound needs a medium to travel, but outer space has no atmosphere. Therefore sound cannot travel in outer space. This is the reason why people cannot talk to each other in vacuum.
:
D
Sound needs a medium to travel, but outer space has no atmosphere. Therefore sound cannot travel in outer space. This is the reason why people cannot talk to each other in vacuum.
Answer: Option D. ->
time period
:
D
The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the wave. The relation between time period 'T' and frequency 'F' is:
T=1F
:
D
The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the wave. The relation between time period 'T' and frequency 'F' is:
T=1F
Answer: Option B. ->
False
:
B
:
B
Particles in a longitudinal wave oscillate along the direction of the wave propagation. Sound wave is an example of the longitudinal wave.
Answer: Option A. ->
1
:
A
Speed of sound in a medium is a characteristic of the medium and temperature.
v∝√T√ρM
where:
v: Speed of sound
T: Temperature
ρ: Density
M: Molecular Mass of the medium
Increasing time period decreases the frequency of the wave.
v=fλ still holds because wavelength increases to maintain the speed of the wave in the medium.
:
A
Speed of sound in a medium is a characteristic of the medium and temperature.
v∝√T√ρM
where:
v: Speed of sound
T: Temperature
ρ: Density
M: Molecular Mass of the medium
Increasing time period decreases the frequency of the wave.
v=fλ still holds because wavelength increases to maintain the speed of the wave in the medium.
Answer: Option C. ->
Frequency in A = Frequency in B
:
C
:
C
The frequency of sound waves depends only on the frequency of vibration of the source of sound. It does not depend on the medium of propogation. Thus, the frequency of sound in both media are same.
Answer: Option C. ->
rarefactions
:
C
Sound waves are a sequence of compressions and rarefactions. A compression is a region in a longitudinal wave where the particles are closer than normal. A rarefaction is a region in a longitudinal wave where the particles are farther than normal.
:
C
Sound waves are a sequence of compressions and rarefactions. A compression is a region in a longitudinal wave where the particles are closer than normal. A rarefaction is a region in a longitudinal wave where the particles are farther than normal.
Answer: Option A. ->
4.86 s
:
A
Given, speed of sound, v=350 ms−1 and distance between the observer and the cliff, d=850 m.
This is a case of an echo, so distance travelled = 2d (because sound travels from the observer to the cliff and back to the observer)
We know, speed=distancetime
v=2dt
⇒t=2dv
=(2×850)350=4.86 s.
:
A
Given, speed of sound, v=350 ms−1 and distance between the observer and the cliff, d=850 m.
This is a case of an echo, so distance travelled = 2d (because sound travels from the observer to the cliff and back to the observer)
We know, speed=distancetime
v=2dt
⇒t=2dv
=(2×850)350=4.86 s.
Answer: Option A. ->
frequency
:
A
:
A
The pitch of the sound depends only on frequency. More the frequency, more will be the pitch or shrillness of the voice.
Answer: Option A. ->
Only statement 1 is correct.
:
A
Sound waves require a medium to travel. Due to the absence of atmosphere on the surface of the Moon, there is no medium for the sound to propagate. The propagation of sound on the Moon is not related to the surface temperature of the Moon. Also, the Moon's surface is very cold.
Thus, the statement 1 is correct, but the statement 2 is incorrect.
:
A
Sound waves require a medium to travel. Due to the absence of atmosphere on the surface of the Moon, there is no medium for the sound to propagate. The propagation of sound on the Moon is not related to the surface temperature of the Moon. Also, the Moon's surface is very cold.
Thus, the statement 1 is correct, but the statement 2 is incorrect.
Answer: Option B. ->
343000 Hz
:
B
Given:
Speed of the sound, v=343 ms−1
Wavelength, λ=0.1 cm=0.001 m
Let frequency be f.
Speed of the wave is given as the product of frequency and wavelength.
v=fλ
⟹f=343 ms−10.001 m
f=343000 Hz
Frequency of the sound wave is 343000 Hz.
:
B
Given:
Speed of the sound, v=343 ms−1
Wavelength, λ=0.1 cm=0.001 m
Let frequency be f.
Speed of the wave is given as the product of frequency and wavelength.
v=fλ
⟹f=343 ms−10.001 m
f=343000 Hz
Frequency of the sound wave is 343000 Hz.