Sound Test

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Sound Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

A metallic spoon falls on a metal plate making a huge sound and upon bouncing from the plate, it falls on to the pile of sand below it with no noise being heard. Sound is not heard on sand because

A
Sound is absorbed by sand particles

B
Sound generated by the plate is more than the sand particles

C
Vibration of metal plate is more than sand particles

D
Sand does not vibrate

Solution
Question 2
1 / -0

Match the following and select the correct answer from the codes given below.
(i) Pitch (p) Wave form
(ii) Quality (q) Frequency
(iii) Loudness (r) Intensity

A
$$(i) - (q), (ii) - (p), (iii) - (r)$$

B
$$(i) - (p), (ii) - (r), (iii) - (q)$$

C
$$(i) - (r), (ii) - (p), (iii) - (q)$$

D
$$(i) - (q), (ii) - (r), (iii) - (p)$$
Question 3
1 / -0

A long spring whose one end is fixed is stretched from the other end and then left longitudinal waves of frequency $$500\ Hz$$ are produced. If the velocity of the wave is $$250\ m/s$$. Find the distance between two consecutive compressions and rarefaction.

A
$$1.25\ m$$

B
$$0.5\ m$$

C
$$2.5\ m$$

D
$$2\ m$$

Solution

The distance between two consecutive compressions is the wavelength $$\lambda$$.
Given,
Frequency, $$f=500\,Hz$$
Velocity, $$v=250\,m/s$$

We know,
$$v=\lambda f$$
$$250=\lambda \times 500$$

$$\lambda=\dfrac{250}{500}=0.5\,m$$
Question 4
1 / -0

Three different vibrating bodies produce three types of sounds X, Y, and Z.
Sounds X and Y cannot be heard by a man having a normal range of hearing but sound Z can be heard easily.
The sound X is used to find faults and cracks in metals. The sound is similar to that which is produced during an earthquake before the main shock wave is generated.
Name one device which can produce sound like X.

A
Megaphone

B
SONAR

C
Bulb horn

D
Ear mufflers

Solution

SONAR is a technique that uses ultrasound to measure the distance, direction, and speed of underwater objects, such as hills, valleys, etc.
Ultrasound is used to find faults and cracks in metals.
Ultrasonic waves are thrown on metal under investigation. Ultrasounds and infrasounds are not audible to man. Ultrasound has a frequency of more than 20kHz.
So, X is ultrasound and it can be produced by SONAR.
Question 5
1 / -0

Which one of the following material will reflect sound better?

A
Thermocole

B
Curtain made from cloth

C
Steel

D
Paper

Solution

Among the given substances steel is the most rigid material. When sound wave hits the steel, due to rigidity the atoms swing reflecting the sound wave.
Question 6
1 / -0

A stone is dropped into a pond from the top of the tower of height $$h$$. If $$v$$ is the speed of sound in air, then the sound of splash will be heard at the top of the tower after a time:

A
$$\sqrt{\dfrac{2h}{g}} + \dfrac{h}{v}$$

B
$$\sqrt{\dfrac{2h}{g}} - \dfrac{h}{v}$$

C
$$\sqrt{\dfrac{2h}{g}}$$

D
$$\sqrt{\dfrac{2h}{g}} + \dfrac{2h}{v}$$

Solution

Let the time taken by the stone to strike the surface of the water in the pond= $$t_1$$
Using Newton's second equation of motion,
$$s = ut + \dfrac{1}{2}at^2$$ $$(\text{where}, s=\text{displacement},\ t= \text{time},\ a= g\ \text{acceleration due to gravity})$$

$$ \therefore h = ut + \dfrac{1}{2}gt^2_1$$
initial velocityu, $$u = 0$$
$$\implies t_1 = \sqrt{\dfrac{2h}{g}}$$
Let the time taken by the sound to reach the top of the tower, $$t_2$$
$$t_2= \dfrac{h}{v}$$
Total time after which splash of sound is heard, $$t$$

$$t = t_1 + t_2 = \sqrt{\dfrac{2h}{g}} + \dfrac{h}{\nu}$$
Question 7
1 / -0

An underwater sonar source operating at a frequency of $$60\ kHz$$ directs its beam towards the surface. If the velocity of sound in air is $$330\ m/s$$, the wavelength and frequency of waves in air are:

A
$$5.5\ mm,\ 60\ kHz$$

B
$$330\ m,\ 60\ kHz$$

C
$$5.5\ mm,\ 20\ kHz$$

D
$$5.5\ mm,\ 80\ kHz$$

Solution

$$\textbf{Step 1 - Frequency of waves in air}$$
As the frequency of sound does not depends upon medium it will remain unchanged.
so $$f = 60\ kHz$$

$$\textbf{Step 2 - Find wavelength of waves in air}$$
The expression of wavelength is written as,

$$\lambda = \dfrac {v}{f}$$

Here, $$v$$ is speed and $$f$$ is the frequency

$$\lambda = \dfrac {v}{f}$$

$$\lambda = \dfrac {330\ m/s}{60\times 10^{3} Hz} = 5.5\times 10^{-3} m$$

$$\lambda = 5.5\ mm$$
Question 8
1 / -0

A person standing between a pair of tall and wide cliffs claps his hands. He hears the first two echoes at 2 sec and 3 sec respectively. If the speed of sound is 330 m/sec, then

A
Separation between the cliffs is 825m

B
The time of third echo is 5 second

C
every third echo is the loudest

D
after the second echo,none is formed

Solution

Given that,

Speed of sound $$v = 330\ m/s$$

Time

$$ {{t}_{1}}=2s $$

$$ {{t}_{2}}=3s $$
Let $$d_1$$= distance to near wall and $$d_2$$ distance to far wall

Now, the distance to the near wall and back is
$$ {2{d}_{1}}=Speed\times time=330\times 2 $$

$$ {2{d}_{1}}=660\,m $$

Now, the distance is half
$${{d}_{1}}=330\,m$$

Now, the distance to the far wall and back is
$$ {2{d}_{2}}=330\times 3 $$

$$ {2{d}_{2}}=990\ m $$

The distance is half

$${{d}_{2}}=495\,m$$

Now, the total distance is
$$ d={{d}_{1}}+{{d}_{2}} $$

$$ d=330+495 $$

$$ d=825\,m $$
Hence, the Separation between the cliffs is $$825\ m$$
Question 9
1 / -0

Radio waves of speed $$3 \times 10^8 $$ m/s are reflected off the moon and received back on earth. The time elapsed between the sending of the signal and receiving it back at the earth station is 2.5 s. What is the distance of the moon from the earth(in km)?

A
$$8.75 \times 10^5 $$

B
$$3.75 \times 10^4 $$

C
$$3.75 \times 10^5 $$

D
$$8.75 \times 10^4 $$

Solution

During the trip the distance traveled by the signal is $$2\times \text{distance between earth and moon}=2x$$

We know $$time=2.5s=\dfrac{distance}{speed}=2x/(3\times10^8)$$
so $$2x=7.5\times10^8$$ so distance $$x=3.75\times10^8meter=3.75\times10^5Km$$
Question 10
1 / -0

An echo is heard because of the persistence of _____________

A
hearing is 0.1 s

B
vision is 0.8 s

C
hearing is 0.5 s

D
vision is 0.1 s

Solution

An echo is heard because of persistence of hearing of our ear is 0.1 s

The correct option is (a)

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Re-Attempt Weekly Quiz Competition

(i) Pitch	(p) Wave form
(ii) Quality	(q) Frequency
(iii) Loudness	(r) Intensity

Sound Test - 70

Result

Sound Test - 70

Score

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Rank

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SHARING IS CARING

Question 1

Sound is absorbed by sand particles

Sound generated by the plate is more than the sand particles

Vibration of metal plate is more than sand particles

Sand does not vibrate

Solution

Question 2

$$(i) - (q), (ii) - (p), (iii) - (r)$$

$$(i) - (p), (ii) - (r), (iii) - (q)$$

$$(i) - (r), (ii) - (p), (iii) - (q)$$

$$(i) - (q), (ii) - (r), (iii) - (p)$$

Question 3

$$1.25\ m$$

$$0.5\ m$$

$$2.5\ m$$

$$2\ m$$

Solution

Question 4

Megaphone

SONAR

Bulb horn

Ear mufflers

Solution

Question 5

Thermocole

Curtain made from cloth

Steel

Paper

Solution

Question 6

$$\sqrt{\dfrac{2h}{g}} + \dfrac{h}{v}$$

$$\sqrt{\dfrac{2h}{g}} - \dfrac{h}{v}$$

$$\sqrt{\dfrac{2h}{g}}$$

$$\sqrt{\dfrac{2h}{g}} + \dfrac{2h}{v}$$

Solution

Question 7

$$5.5\ mm,\ 60\ kHz$$

$$330\ m,\ 60\ kHz$$

$$5.5\ mm,\ 20\ kHz$$

$$5.5\ mm,\ 80\ kHz$$

Solution

Question 8

Separation between the cliffs is 825m

The time of third echo is 5 second

every third echo is the loudest

after the second echo,none is formed

Solution

Question 9

$$8.75 \times 10^5 $$

$$3.75 \times 10^4 $$

$$3.75 \times 10^5 $$

$$8.75 \times 10^4 $$

Solution

Question 10

hearing is 0.1 s

vision is 0.8 s

hearing is 0.5 s

vision is 0.1 s

Solution

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