Waves And Sound Test

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Waves And Sound Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

In an orchestra, the musical sounds of different instruments are distinguished from one another on the basis of

A
pitch

B
loudness

C
quality

D
overtones

Solution

The musical sounds of different instruments are distinguished from one another on the basis of quality of the sound. Quality of a sound depends upon the overtones, which are present in the sound produced by the musical instrument.
Question 2
1 / -0

If the intensity ratio of two waves is 4 : 1, the ratio of their amplitudes is

A
2 : 1

B
1 : 2

C
4 : 1

D
1 : 4

Solution

Intensity of the sound is directly proportional to the square of the amplitude.

Hence,
Question 3
1 / -0

Bells are made of metal and not of wood because

A
thermal conductivity of metal is greater than that of wood

B
density of the metal is greater than that of wood

C
sound is not conducted by metals but is radiated

D
wood dampens the vibration while metals are elastic

Solution

The greater the elasticity of the medium, the faster the sound travels in the given medium. Sound travels faster in the metals as compared to wood as the metals are more elastic than wood.
Question 4
1 / -0

For each degree centigrade rise of temperature, the velocity of sound in air increases by

A
0.61 m/sec

B
0.61 cm/sec

C
0.61 km/sec

D
0.61 ft/sec

Solution

Let the velocity of sound at 0°C beand velocity at t°C be .

Hence, we can say that the velocity of sound increases by 0.61 m/s for each degree centigrade rise of the temperature.
Question 5
1 / -0

A boy shouts into a deep well and hears the echo 0.4 s after shouting. If the speed of sound is 340 ms^-1, find the depth of water surface from the top of the well.

A
68 m

B
78 m

C
88 m

D
98 m

Solution

Speed is given by,
Question 6
1 / -0

Speed of sound in air is

A
directly proportional to pressure of air

B
directly proportional to square of pressure

C
directly proportional to

D
independent of pressure

Solution

Speed of sound is independent of pressure of the medium. Speed of sound depends on density of the medium.
Question 7
1 / -0

Which of the following options is not TRUE about reverberations?

A
It is the lingering effect of the sound due to repeated reflections.

B
Asbestos and curtains decrease reverberation.

C
It reduces the clarity of sound.

D
None of the above

Solution

The lingering effect of sound due to repeated reflections is called reverberations or persistence of hearing.
This is an undesirable effect as it reduces the clarity of sound. It can be reduced by using sound-absorbing materials on the walls, for example, asbestos, curtains and upholstery.
Question 8
1 / -0

At what temperature, the speed of sound in the air becomes double of its value at 0°C?

A
758°C

B
802°C

C
819°C

D
945°C

Solution

Velocity of sound is directly proportional to the square root of the absolute temperature. Thus,
Question 9
1 / -0

This diagram represents displacement- distance graph of a wave travelling in a straight line. The amplitude and the wavelength of this wave respectively, are

A
2 metres and 1 metre

B
1 metre and 2 metres

C
1 metre and 1 metre

D
2 metres and 2 metres

Solution

Amplitude is defined as the maximum displacement of a particle from the equilibrium position. Here, it is 2 metres.
Wavelength is defined as the distance between the two successive crests or troughs. Here, it is 2 metres.
Question 10
1 / -0

A man standing between two parallel cliffs fires a gun. If he hears first and second echoes after 1.5 seconds and 3.5 seconds respectively, then the distance between the two cliffs is (Take velocity of sound in air as 340 m/s).

A
1190 m

B
850 m

C
595 m

D
510 m

Solution

Let the distance between the nearest cliff and the man = x m
And, the distance between the distant cliff and the man = y m
Distance between two cliffs = (x + y) m
Total distance covered by sound to produce first echo = 2x m and time = 1.5 s
2x = 340 1.5

Hence,

Total distance covered by sound to produce second echo = 2y m, and time = 3.5 s
2y = 340 3.5

Hence

So, the distance between two cliffs = (255 + 595) = 850 m
Question 11
1 / -0

A man fires a gun and hears an echo from the cliff after 2 seconds. He walks towards the cliff about a distance of 85 m and fires the gun again. He hears an echo after 1.5 seconds. In this case, the velocity of sound in air is

A
85 m/s

B
170 m/s

C
340 m/s

D
680 m/s

Solution

Let the speed of sound in air be v m/s.
Let the distance between the cliff and the man be d m.
Total distance covered by sound to produce first echo = 2d m and time = 2 s
Distance = speed x time
2d = 2v
d = v
After moving 85 m, distance = 2d - 170 = 2v - 170

Hence, 0.5v = 170 v = 340 m/s
Question 12
1 / -0

A particle is executing simple harmonic motion. If its amplitude of vibration increases by 20%, then what will be the increase in its total mechanical energy?

A
44%

B
21%

C
20%

D
10%

Solution

Energy of a particle executing SHM is directly proportional to the square of the amplitude.

If A is increased by 20%, then new amplitude of the SHM: K' = 1.2A
A'² = 1.44A²
New value of energy, E' = 1 .44kA²
Percentage change in energy is:

Hence, we can say that change in the energy is 44%.
Question 13
1 / -0

A key of a mechanical piano is struck gently and then struck again but much this time. In the second case,

A
sound will be louder but pitch will not be different

B
sound will be louder and pitch will also be higher

C
sound will be louder but pitch will be lower

D
both loudness and pitch will remain unaffected

Solution

Piano is a stringed instrument. When the key is struck harder, the string below it vibrates with higher amplitude, as the intensity of the sound is directly proportional to the square of the amplitude. Hence, sound will be louder. The pitch (frequency) depends upon the tension, length and the mass of the string. Hence, pitch of the sound will remain the same.
Question 14
1 / -0

The intensity of sound wave gets reduced by 20% on passing through a slab. The reduction in intensity on passing through two consecutive slabs is:

A
40%

B
36%

C
30%

D
50%

Solution

Let the intensity of the light be I.
Intensity of the light after passing through first slab is I'.

Intensity of the light after passing through second slab is I''.

Hence, percentage loss of Intensity is
Question 15
1 / -0

A spherical source of power 4 W and frequency 800 Hz is emitting sound waves. The intensity of waves at a distance of 200 m is:

A
2 x 10^–4 W/m²

B
8 x 10^–6 W/m²

C
4 W/m²

D
1 x 10^–4W/m²

Solution

From the formula,
Intensity =

= 7.96 × 10^-6
= 7.96 × 10^-6 W/m²
8 × 10^-6 W/m²
Question 16
1 / -0

The speed of sound in humid air is greater than that in dry air because

A
humid air is less dense

B
humid air is dense

C
humid air is elastic

D
humid air is less elastic

Solution

Speed of sound in different media depends upon some inertia factors, such as molecular mass, density and the elasticity of the medium. Since humid air is less dense than dry air, so the speed of sound in humid air is greater than that in dry air.
Hence, option (1) is correct.
Question 17
1 / -0

Ultrasonic, infrasonic and audible waves travel through a medium with speeds of v_u, v_i and v_a,respectively. Which of the following relations holds true?

A
v_i = v_a = v_u

B
v_u > v_a > v_i

C
v_u < v_a < v_i

D
v_a v_u = v_i

Solution

Velocity of a sound is dependent on the medium. It is independent of the frequency, so v_i = v_a = v_u.
Question 18
1 / -0

What is the change in intensity level when the intensity of a sound increases by a factor of 10⁵?

A
50 dB

B
60 dB

C
70 dB

D
80 dB

Solution

I/I_o= 10⁵
The increase in intensity level is given by:
L = 10 log₁₀(I/I_o) = 10 log₁₀(10⁵)
= 10 × 5 = 50 dB
Question 19
1 / -0

The speed of sound in hydrogen at STP is v. What is the speed of sound in helium at STP?

A

B

C
v

D
2v

Solution

Density of helium at STP = 2 times the density of hydrogen at STP
Since v ∝ 1/, the speed of sound in helium will be v /
Hence, the correct choice is (1).
Question 20
1 / -0

In a sinusoidal wave, the time required for a particular point to move from maximum displacement is 0.17 s. Then the frequency of the wave will be

A
1.47 Hz

B
2.94 Hz

C
4.17 Hz

D
0.92 Hz

Solution

Frequency is reciprocal of time period.
For a sinusoidal wave, time period is
T = 4 t
Time required for maximum displacement is
T = 4 × 0.17
= 0.68 s
Also frequency is inverse of time period,
hence,
n = = 1.47 Hz.