Sound Test

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Sound Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

The frequency wavelength and speed of a sound wave are related as:

A
$$\lambda = vT$$

B
$$\lambda = v/T$$

C
$$T = v/\lambda $$

D
$$\lambda = v^2$$

Solution

Hint :- Wavelength is the distance travelled by the sound wave in one cycle of time period.
Explanation:-
$$\bullet$$ Let wavelength of wave is $$\lambda$$, time period is T and velocity of wave is V
$$\bullet$$ $$\lambda= \frac Vf$$ ; where $$f=\frac 1T$$ = frequency of wave
$$\bullet$$ Distance travelled in one time period is given by ,$$\lambda= v\times T$$

$$\textbf{Hence option A correct}$$
Question 2
1 / -0

Two tuning forks A, B vibrate with frequencies in the ratio 2 : 7 and their wavelength in the ratio 3: 4 respectively. Find the tuning fork producing greater velocity of sound.

A
Tuning fork A

B
Tuning fork B

C
Both A and B

D
Cannot be determined

Solution
Question 3
1 / -0

Loudness of sound increases with:

A
The amplitude of vibrating body

B
The increase in surface area of vibrating body

C
The decrease in frequency of vibrating body

D
The increase in distance from the source of sound

Solution

Loudness of a sound $$L = 10\log_{10} \dfrac{I}{I_o}$$
where intensity $$I$$ is proportional to the square of the amplitude $$A$$ of vibrating body i.e. $$I \propto A^2$$
Hence loudness of sound depends directly on the amplitude of vibrating body.
Thus, loudness of sound increases with increase in amplitude of vibrating body.
Question 4
1 / -0

A tuning fork of frequency 420 Hz completes 70 vibrations. Find the distance travelled by sound in air (v = 360 m $$s^{-1})$$

A
20 m

B
50 m

C
60 m

D
80 m

Solution

Number of vibration in $$1sec=420$$
Time taken for $$70$$ vibrations $$=\cfrac{70}{420}=0.167s$$
Distance=speed$$\times$$ Time
As, speed of sound = $$360\ m/s$$
$$\Rightarrow 360\times 0.167=60m$$
Sound travels $$60m$$ when the turning fork makes $$70$$ vibration.
Question 5
1 / -0

What is the distance between a crest and a trough in a wave?

A
$$\lambda$$

B
$$\frac{\lambda }{2}$$

C
$$\frac{\lambda }{4}$$

D
2$$\lambda$$

Solution

The highest surface part of a wave is called crest, and the lowest part is the trough. The horizontal distance between two adjascent crests or troughs is known as the wavelength $$(\lambda)$$.
Question 6
1 / -0

Ultrasonic waves are used for detecting objects under water. What is the technique/device used for this?

A
Ultrasonography

B
Echocardiography

C
Phakoemulsification

D
SONAR

Solution

Sonar is the device that uses ultrasonic waves to measure the distance, direction and detecting the underwater objects.
Question 7
1 / -0

Which kind of waves are used in sonography?

A
Infrared waves

B
Micro waves

C
Sound waves

D
Ultrasonic waves

Solution

Sonography is a medical imaging technique that uses high-frequency sound waves and their echoes. Since high-frequency sound waves are needed for sonography, ultrasonic waves are used. Ultrasonic waves are high-frequency waves with a frequency greater than $$20\,kHz$$
Question 8
1 / -0

A boy stands facing a high wall at a distance of 85 m from it and then blows a whistle. Calculate the time interval when he hears an echo (Speed of sound = 340 m $$s^{-1}$$).

A
1s

B
2s

C
0.5 s

D
3s

Solution

Given,

Distance, $$d=85\,m$$

Speed of sound, $$v=340\,ms^{-1}$$

We have to find the time period $$t$$
Total distance travel by sound wave to hear echo $$=2d$$

We have the equation,

$$v=\dfrac{Distance\ travel}{Time\ consumed}=\dfrac{2d}{t}$$

Then,

$$t=\dfrac{2d}{v}=\dfrac{2\times 85}{340}=\dfrac{170}{340}=0.5\,s$$
Question 9
1 / -0

Arrange the following media in ascending order of the speed of sound in them.
(i) Water
(ii) Steel
(iii) Nitrogen

A
(iii), (ii), (i)

B
(i), (iii), (ii)

C
(iii), (i), (ii)

D
(ii), (i), (iii)

Solution

$$1.$$The speed of the sound depends upon the density of the medium.
$$2.$$The sound propagates faster through the medium with high density and sound propagates slower through the medium with low density.
$$3.$$So, sound travels fastest through solids, faster through the liquid, and slower through the gas.
The velocity of sound in a medium is inversely proportional to the square root of its density.
$$4.$$Here, steel is a solid, water is liquid and nitrogen is gas. So, the ascending order of speed in them is:
Nitrogen, Water, Steel.
Question 10
1 / -0

Quality of sound is used to distinguish between:

A
Noise and music

B
Two different vibrating bodies

C
Two notes produced by same vibrating body

D
All of the above

Solution

Sound "quality" or "timbre" describes those characteristics of sound which allow the ear to distinguish sounds that have the same pitch and loudness. Timbre is then a general term for the distinguishable characteristics of a tone.
Musical sounds are periodic and somewhat regular. They are pleasing to our ears and minds, and sound has been one of the greatest forms of expression since the beginning of time. The unpleasant sound is often described as noise. Noise is more irregular.

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Re-Attempt Weekly Quiz Competition

Sound Test - 60

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Sound Test - 60

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Rank

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SHARING IS CARING

Question 1

$$\lambda = vT$$

$$\lambda = v/T$$

$$T = v/\lambda $$

$$\lambda = v^2$$

Solution

Question 2

Tuning fork A

Tuning fork B

Both A and B

Cannot be determined

Solution

Question 3

The amplitude of vibrating body

The increase in surface area of vibrating body

The decrease in frequency of vibrating body

The increase in distance from the source of sound

Solution

Question 4

20 m

50 m

60 m

80 m

Solution

Question 5

$$\lambda$$

$$\frac{\lambda }{2}$$

$$\frac{\lambda }{4}$$

2$$\lambda$$

Solution

Question 6

Ultrasonography

Echocardiography

Phakoemulsification

SONAR

Solution

Question 7

Infrared waves

Micro waves

Sound waves

Ultrasonic waves

Solution

Question 8

1s

2s

0.5 s

3s

Solution

Question 9

(iii), (ii), (i)

(i), (iii), (ii)

(iii), (i), (ii)

(ii), (i), (iii)

Solution

Question 10

Noise and music

Two different vibrating bodies

Two notes produced by same vibrating body

All of the above

Solution

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