Sound Test

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Sound Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

The equipment based on properties of sound used for medical purposes is based on:

A
frequency ranging from $$20Hz - 20 KHz$$

B
Infrasound

C
Ultrasound

D
None of these

Solution

An ultrasound scanner is an instrument that uses ultrasonic waves for getting images of internal organs of the human body because ultrasonic waves have high frequencies.
Question 2
1 / -0

When the string on a guitar is tightened, its pitch

A
Increases

B
Decreases

C
Remains same

D
May increase or decrease

Solution

The string of a guitar is like a stretched string fixed at both ends, and the frequency of a stretched string is directly proportional to the tension in the string,

$$f\propto \sqrt T$$ ,

When the string is tightened, tension $$T$$ in the string increases, hence frequency $$f$$ also increases. As the pitch of a sound depends upon its frequency, therefore, its pitch also increases.
Question 3
1 / -0

Sonar is a device for:

A
Location and ranging of aircrafts

B
Location and ranging of submarines

C
Producing a musical note of high quality

D
Measuring frequency of musical notes

Solution

Sonar is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. Sonar consists of a transmitter and a detector and is installed in a boat or a ship.
The transmitter produces and transmits ultrasonic waves. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound.
Question 4
1 / -0

Frequency of a sound wave if it travels with a speed of 330 ms$$^{-1}$$ in air and the wavelength of the wave is 330 cm is :

A
100 Hz

B
200 Hz

C
150 Hz

D
none of the above

Solution

We have , a relation in frequency $$f$$ and wavelength $$\lambda$$ of a wave having the velocity $$v$$ as ,
$$v=f\lambda$$ ,
or $$f=v/\lambda$$
Given $$v=330m/s$$ , $$\lambda=330cm=3.3m$$ ,
Therefore velocity of wave, $$f=330/3.3=100Hz$$
Question 5
1 / -0

A longitudinal wave is travelling along a spiral spring with a velocity of $$2 \ m/s$$. If the period of the wave is $$0.2 \ s$$, find the wavelength.

A
$$1 m $$

B
$$0.4 m $$

C
$$ 2 m $$

D
$$3 m $$

Solution

We have , $$v=f\lambda$$ ,
or $$\lambda=v/f$$
So $$\lambda=vT$$ ,
where, $$T= $$ time period and $$f=1/T$$ ,
given
$$T=0.2s , v=2m/s$$ ,
Therefore , $$\lambda=2\times0.2=0.4m$$
Question 6
1 / -0

The characteristics of sound which enables us to distinguish two musical sounds coming from different sources but having the same frequency and loudness is :

A
pitch

B
loudness

C
timber

D
none of these

Solution

Sounds may be generally characterized by pitch, loudness, and quality. Sound "quality" or "timbre" describes those characteristics of sound which allow the ear to distinguish sounds which have the same pitch and loudness. Timbre is then a general term for the distinguishable characteristics of a tone.
Hence Timber is the answer.
Question 7
1 / -0

Transverse waves cannot travel through

A
An iron rod

B
Hydrogen gas

C
A stretched nylon string

D
Lubricating oil

Solution

Transverse waves cannot propagate in a gas because there is no mechanism for driving motion perpendicular to the propagation of the wave.
so the answer is B.
Question 8
1 / -0

A body shouts inside 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 the water level in the well :

A
78 m

B
88 m

C
68 m

D
58 m

Solution

Let d be the distance of body from water level in well (depth of water level) , therefore total distance travelled by sound wave , $$=2d$$ , as the wave comes back after reflection so the distance will be doubled .
Now given , $$v=340m/s , t=0.4s$$ ,
therefore , by , total distance =velocity $$\times$$ total time
$$2d=v\times t$$ ,
or $$d=v\times t/2=340\times0.4/2=68m$$
Question 9
1 / -0

What is the wavelength of a longitudinal wave if the distance between the 5th rarefaction and the 6th rarefaction is 25 cm?

A
25 cm

B
50 cm

C
40 cm

D
20 cm

Solution

The wavelength of a longitudinal wave is defined as the distance between two successive compressions or rarefactions. As given, the distance between the 5th and 6th rarefaction is 25 cm, therefore the wavelength of the wave will be 25 cm.
Question 10
1 / -0

A boy standing between two cliffs claps and hears 2 echoes after 2s and 2.5 s respectively. If the velocity of sound is 330 $$ms^{-1}$$, find the distance between he two cliffs :

A
742.50 m

B
274.50 m

C
472.50 m

D
442.50 m

Solution

Let $$d_{1}$$ and $$d_{2}$$ be the distances of cliffs from the boy .
for first echo , which is heard after $$2s$$ ,
total distance travelled by sound $$=2d_{1}$$ ,
given $$v=330m/s$$ ,
we have , distance=speed$$\times$$time ,
$$2d_{1}=330\times2$$ ,
or $$d_{1}=660/2=330m$$ ,
for second echo , which is heard after $$2.5s$$ ,
total distance travelled by sound $$=2d_{2}$$ ,
given $$v=330m/s$$ ,
we have , distance=speed$$\times$$time ,
$$2d_{2}=330\times2.5$$ ,
or $$d_{2}=825/2=412.5m$$ ,
therefore , distance between cliffs , $$d=d_{1}+d_{2}=330+412.5=742.5m$$

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Sound Test - 40

Result

Sound Test - 40

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SHARING IS CARING

Question 1

frequency ranging from $$20Hz - 20 KHz$$

Infrasound

Ultrasound

None of these

Solution

Question 2

Increases

Decreases

Remains same

May increase or decrease

Solution

Question 3

Location and ranging of aircrafts

Location and ranging of submarines

Producing a musical note of high quality

Measuring frequency of musical notes

Solution

Question 4

100 Hz

200 Hz

150 Hz

none of the above

Solution

Question 5

$$1 m $$

$$0.4 m $$

$$ 2 m $$

$$3 m $$

Solution

Question 6

pitch

loudness

timber

none of these

Solution

Question 7

An iron rod

Hydrogen gas

A stretched nylon string

Lubricating oil

Solution

Question 8

78 m

88 m

68 m

58 m

Solution

Question 9

25 cm

50 cm

40 cm

20 cm

Solution

Question 10

742.50 m

274.50 m

472.50 m

442.50 m

Solution

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