Chhattisgarh State Board Class 9 Science Chapter 14 Sound Exercise Multiple Choice, Fill in the Blanks, Questions and Answers here.
Chhattisgarh State Class 9 Science Chapter 14 Solution
1) Choose the correct option-
(i) When sound waves propagate through a medium, then from one position to another-
(a) Particles of the medium travel
(b) Transfer of energy from one particle to another takes place.
(c) Transformation of energy takes place.
(d) None of these.
Ans: – option (c).
(ii) Sound below the frequency of 20 Hz is called-
(a) amplitude (b) inaudible/infrasound
(c) ultrasound (d) none of these.
Ans: – option (b).
(iii) The distance between consecutive compressions or rarefactions is called-
(a) amplitude (b) frequency
(c) speed of wave (d) wavelength.
Ans: – option (d) wavelength.
(iv) Establish a relation between velocity of transferred wave (v), its frequency and wavelength.
(a) v=u/y. (b) v =uy
(c) v=y/u. (d) none of these.
Ans: – option (b).
(v) Sound waves do not travel in-
(a) solids (b) liquids
(c) gases (d) vacuum.
Ans: – option (d) vacuum.
(vi) If the time period of a vibrating object is 0.05 s, the frequency of waves produced will be-
(a) 5 Hz (b) 20 Hz
(c) 200 Hz (d) 2 Hz.
Ans: – option (b) 20Hz.
2) Fill the following blanks-
(i) The hearing range of sound in humans is ……20Hz-20KHz………..
(ii) SI unit of wavelength is …….. meter……….
(iii) Distance between consecutive compression or consecutive rarefactions is called… wavelength…………
(iv) Velocity of sound depends on … temperature, density,Molecular weight……………
3) What happens to the velocity of sound in a medium if the sound wave frequency is doubled?
Ans: – There will be no change in velocity for changing of the frequency as the velocity is directly proportional to the wavelength the so when wavelength is changing velocity will be varying. The frequency only depends on the source of the sound.
4) The frequency of sound produced by source A is twice to the frequency of sound produced bysource B. Compare the wavelengths of both the sounds.
Ans: – As the frequency is inversely proportional to the wavelength so the source A has twice the frequency of source B so the wavelength of the source A will be half than the source B.
5) Which characteristic of sound helps you to identify your friends by his voice while sitting in anotherroom?
Ans: – The quality of sound is determined by the timber and the pitch of the sound. As everyone has different pitch and timbre so they are identified by their unique voice. In this case by this characteristic, we can differentiate the voice of friends.
6) Sound is produced by colliding two metals, first in air, then in water. Say which medium will producelouder sound when both metals are collided from same distance.
Ans: – The sound depends on medium for the loudness. The sounds travel faster in water as compare to the air because the density of the water is maximum. So, the sound will travel faster in water.
7) What is sound and how is it produced?
Ans: – The vibrating of two particles or many particles creates the sound. Because of vibrating the sound is produced, this vibration creates the flow of sound for this reason which medium has more density sound travels more in that.
8) Explain compressions and rarefaction with help of diagrams.
Ans: –
9) Explain with help of an activity that sound propagation requires a medium.
Ans: – The bell jar experiment is the clear example of the propagation of the sound in a medium. When we take out the air from the jar after that if we ring the bell then we will see that there is no sound at all outside. As the air is not present in there so there is no such medium to propagate the sound from the jar.
10) Explain that sound waves are longitudinal waves.
Ans: – The sound waves are longitudinal in nature. Because of the flowing parallel to the direction of the propagation the sound wave vibrates in that direction. The compression and refraction are the clear explanation of the propagation of the sound wave.
11) Write notes on (i) ECG (ii) ultrasonography (iii) Sonar.
Ans: –
| ECG | Ultrasonography | Sonar |
| The ECG is measurements of the electrical activity of the heart. The depolarization and the repolarization of the sound is creating the images by which we can measure the activity of heart. | In this technique the ultrasonic waves travel through the body and find the damage tissue. These waves are then converted into electrical
signals to generate images of the organ. These images are then displayed on a monitor.
|
This sonar method is used in the submarine to find out the object which float on the sea from the ground. In that case two plane mirror are used for the reflection of the image of the object. |
12) Explain with diagrams velocity, frequency and wavelength of waves. Also establish a relation between them.
Ans: –
13) A sound wave travels at a speed of 3329 m/s. If its wavelength is 1.5 cm, what is the frequency of this wave? Will it be audible?
Ans: – As we know that, speed of the sound=wavelength× frequency.
Here wavelength= 1.5 cm= 0.015m and speed of sound is 3329 m/s.
So, frequency=3329/0.015 = 22600 Hz.
14) The speed of sound in air at a certain temperature is 340 m/s and its wavelength is 0.017 m. If the same sound source is dipped in water what effect will be produced on its wavelength? If the speedof sound in water is 1480 m/s, calculate.
Ans: – The speed of sound= wavelength × frequency;
Here speed of sound= 340 m/s, and wavelength=0.017m.
So, frequency= 340/0.017= 20000 Hz.
In this case, frequency is same but speed of sound is 1480 m/s.
So, wavelength= 1480/20000 = 0.074m.
15) What steps will you take to reduce the pitch of the sound produced by a ‘sitar’?
Ans: – The vibration when will be more than the pitch of the sound will be more. As the sound is produce because of the vibration so when we increase the vibration by using strong pluck, we will see that the sitar is producing more pitch of sounds.
16) A longitudinal wave whose wavelength is 1 cm, propagates in air at a velocity of 330 m/s. Calculate the frequency of wave. Can a normal human being hear this sound wave?
Ans: – The speed of the sound = wavelength × frequency.
Here speed of sound=330 m/s, and wavelength= 1cm=0.01m.
So, frequency= 33000 Hz.
As the range of audible sound of humans are between the 20Hz to 20KHz so this sounds would not be seen by the humans.
