Sound Test

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Sound Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 m $$s^{-1}$$, calculate the time in which one wave is produced ?

A
$$7.3\, \times\, 10^{-3}\, s$$

B
$$8.33\, \times\, 10^{-2}\, s$$

C
$$8.33\, \times\, 10^{-3}\, s$$

D
$$7\, \times\, 10^{-3}\, s$$

Solution

In the question, the wavelength $$\left( \lambda \right) $$ is given as $$20\ cm$$, that is, $$ 0.20\ m$$ and the wave velocity, $$v$$ is given as $$24$$ $$m/s$$.
The frequency $$\nu $$, velocity $$v$$ and wavelength $$\lambda $$ relation is given as:
$$\nu =\dfrac { v }{ \lambda } $$.
Number of waves produced in one second is called as frequency, i.e. $$\nu =\dfrac { v }{ \lambda } $$ $$=\nu =\dfrac { 24 }{ 0.20 } $$ $$= 120$$ waves per second.
Time period of a wave is given as the inverse of frequency.
That is, $$T=\dfrac { 1 }{ \nu } $$ $$=\dfrac { 1 }{ 120 } $$ $$=8.33\times { 10 }^{ -3 }\ seconds$$
Hence, the time period of the wave is given as $$8.33\times { 10 }^{ -3 }\ seconds$$.
Question 2
1 / -0

A man pronounces $$a,\space b, \space c, \space d$$ and $$e$$ and hears last four syllables as an echo. Taking velocity of sound as $$340\space ms^{-1}$$, the distance of reflecting surface, if it takes one-fifth of a second to pronounce or hear one syllable is

A
$$136\space m$$

B
$$316\space m$$

C
$$613\space m$$

D
$$340\space m$$

Solution

Let the distance between the man and the reflecting surface be $$x$$ metres.
Thus the distance travelled by sound to reach back to the man $$d = 2 x$$
The man cannot hear the syllable $$a$$ as at the instant when echo of syllable $$a$$ is about to reach to the man, he pronounces $$e$$. Thus the time taken by man to pronounce all syllables is equal to the time taken by the sound to reach back to the man. i.e $$t = 4 \times \dfrac{1}{5} = \dfrac{4}{5} sec$$
Thus $$d = v t$$
$$2 x = 340 \times \dfrac{4}{5} \implies x = 136 m$$
Question 3
1 / -0

A person is in front of a vertical mountain fires a bullet and hears its echo after a time 3 s. The person walks a distance 'd' towards mountain then fires another bullet and hears its echo after a time 2 s. If velocity of sound in air is 340 ms$$^{-1}$$, the value of 'd' is

A
85 cm

B
170 m

C
255 m

D
340 m

Solution

Given,
Velocity of sound $$v=340$$
Case-1: Echo is heard in 3 sec, let the distance between the mountain and the person is $$x$$.

Total distance travel by sound $$=2x=vt$$
Distance from mountain in first case $$x=\dfrac{v t}{2}$$ $$=\dfrac{340\times 3}{2}$$

Case-2: Echo is heard in 2 sec, let the distance between the mountain and the person is $$y$$.
Total distance travel by sound $$=2y=vt'$$
Distance from mountain in second case $$y=\dfrac{340\times 2}{2}$$

Walking distance $$d=x-y$$

$$\therefore d=\dfrac{340 \times 3}{2}-\dfrac{340\times 2}{2}=\dfrac{340}{2}=170 \ m$$

Option B
Question 4
1 / -0

Identify the longitudinal waves from the following

A
Light waves

B
Radio waves

C
Ultrasonic waves

D
Surface water waves

Solution

Ultrasonic waves are longitudinal waves of frequency greater than $$20,000\ Hz$$. This implies that a human being cannot hear beyond $$20,000\ Hz$$, i.e. the maximum frequency of audible sound is $$20,000\ Hz$$.
Question 5
1 / -0

The loudness of sound decreases with an:
$$(1)$$ Increase in the distance between the ear and the source
$$(2)$$ Decrease in the amplitude of the vibrating body
$$(3)$$ Increase in frequency of sound
$$(4)$$ Decrease in frequency

A
$$1$$ and $$4$$

B
$$2$$ and $$3$$

C
$$1,\ 3$$ and $$4$$

D
$$1 $$ and $$2$$

Solution

The loudness or softness of a sound is determined basically by its amplitude. A sound wave spreads out from its source. As it moves away from the source its amplitude as well as its loudness decreases.
The loudness of sound decreases with an:
(1) Increase in the distance between the ear and the source
(2) Decrease in the amplitude of the vibrating body
Question 6
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Which of the following is/are correct statement(s)?
(1) Besides helping in finding the distance of an object, SONAR also indicates its direction and speed.
(2) Megaphone is a device that works on the ~ principle of reflection of sound.
(3) Ultrasonics are used to detect earthquakes.
(4) Like light, sound waves also obey laws of reflection.

A
$$2,\ 3,\ 4$$

B
$$1,\ 2,\ 3$$

C
$$1,\ 2,\ 4$$

D
$$1,\ 3,\ 4$$

Solution

Sonar is a device that uses ultrasonic waves to measure the distance, direction, and speed of underwater objects.
In a megaphone, a tube followed by a conical opening reflects sound successively to guide most of the sound waves from the source in the forward direction towards the audience.
Ultrasonics are not used to detect earthquakes.
Like light, sound waves also obey laws of reflection.
Question 7
1 / -0

When a tuning fork produces sound waves in air, which one of the following is the same in the material of tuning fork as well as in air ?

A
Wave length

B
Frequency

C
Velocity

D
Amplitude

Solution

Sound waves are produced by vibrating objects.

When the tuning fork is hit with a rubber hammer, the tines begin to vibrate. The back and forth vibration of the tines produce disturbances of surrounding air molecules. Since the vibration of air molecules is solely due to vibration of nearby tine of tuning fork,so it means the number of vibration of air molecules will be equal to number of vibration of tine of the tuning fork.So frequency ,which is number of vibrations per second, will be same for both the tuning fork and the air.
Question 8
1 / -0

Wavelength of the light frequency $$100Hz$$ is _________.

A
$$2\times {10}^{6}m$$

B
$$4\times {10}^{6}m$$

C
$$3\times {10}^{6}m$$

D
$$5\times {10}^{6}m$$

Solution

Given that the light frequency is $$100\,Hz$$

We know that the speed of the light is $$3\times 10^{8}\,m/s$$

We have the equation,

$$V=n\lambda$$

where,

$$V$$ is the Wave velocity

$$n$$ is the Frequency

$$\lambda$$ is the Wavelength

$$\implies$$ $$\lambda=\dfrac{V}{n}$$

That is,

$$\lambda=\dfrac{3\times 10^8}{100}=3 \times 10^6\,m$$

Hence the wavelength is $$3\times10^6\,m$$
Question 9
1 / -0

Which of the following is NOT an application of ultrasound waves?

A

B

C

D

Solution

Stethoscope is an acoustic medical device for listening to the internal sound of a living being. It works on reflection of sound.
Question 10
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When a tuning fork is striked using a rubber prong, the compressions and rarefactions in the sound wave correspond to

A
High pressure and low pressure regions

B
Low pressure and high pressure regions

C
Both are high density regions

D
Both are low density regions

Solution

Sound being a longitudinal wave show areas of compression and rarefaction : compressions are regions of high pressure due to particles being close together. rarefactions are regions of low pressure due to particles being spread further apart.

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Re-Attempt Weekly Quiz Competition

Sound Test - 69

Result

Sound Test - 69

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SHARING IS CARING

Question 1

$$7.3\, \times\, 10^{-3}\, s$$

$$8.33\, \times\, 10^{-2}\, s$$

$$8.33\, \times\, 10^{-3}\, s$$

$$7\, \times\, 10^{-3}\, s$$

Solution

Question 2

$$136\space m$$

$$316\space m$$

$$613\space m$$

$$340\space m$$

Solution

Question 3

85 cm

170 m

255 m

340 m

Solution

Question 4

Light waves

Radio waves

Ultrasonic waves

Surface water waves

Solution

Question 5

$$1$$ and $$4$$

$$2$$ and $$3$$

$$1,\ 3$$ and $$4$$

$$1 $$ and $$2$$

Solution

Question 6

$$2,\ 3,\ 4$$

$$1,\ 2,\ 3$$

$$1,\ 2,\ 4$$

$$1,\ 3,\ 4$$

Solution

Question 7

Wave length

Frequency

Velocity

Amplitude

Solution

Question 8

$$2\times {10}^{6}m$$

$$4\times {10}^{6}m$$

$$3\times {10}^{6}m$$

$$5\times {10}^{6}m$$

Solution

Question 9

Solution

Question 10

High pressure and low pressure regions

Low pressure and high pressure regions

Both are high density regions

Both are low density regions

Solution

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