# NCERT Solutions for Class 9 Science Chapter 12 Sound

NCERT Solutions for Class 9 Science Class 9 Science Chapter 12 Sound provides detailed answers for all in-text and exercise Questions. These solutions contain an in-depth explanation of each topic involved in the chapter. Students studying in class 9 can access these solutions for free in PDF format.

All these solutions are prepared by expert teachers and updated for the current academic session. NCERT Solutions for Class 9 Science Chapter 12 Sound help students to understand the fundamental concepts given in class 9 Science textbook. We have prepared the answers to all the questions in an easy and well-structured manner. It helps students to grasp the chapter easily.

## NCERT Class 9 Science Chapter 12 Sound Intext Questions (Solved)

Page no: 162

Question 1: How does the sound produced by a vibrating object in a medium reach your ear?

Answer: When a body vibrates the air in its neighbourhood is alternately compressed and rarefied. The compressed air has a higher pressure than surrounding air. It, therefore, pushes the air particles near it causing compression to move forward. A rarefaction or low pressure is created at the original place. These compressions and rarefaction cause particles in the air to vibrate about their mean position. The energy is carried forward in these vibrations. This is how sound travels.

Page no.163

Question 1: Explain how sound is produced by your school bell?

Answer: When the gong strikes the bell, vibrations are produced in the bell which are transmitted through the air to our ears. These vibrations produce a sensation of sound in our ears.

Question 2:  Why are sound waves called mechanical waves?

Answer:  Sound waves are called mechanical waves because they need a material medium to travel.

Question 3:  Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?

Answer: On the moon, sound cannot travel as there is no atmosphere. Sound cannot travel in vacuum so we will not be able to hear any sound.

Page no.166

Question 1: Which waves property determines
(a) loudness   (b) pitch?

Answer: (a) Loudness is determined by the amplitude of the sound. Greater the amplitude more will be the loudness.

(b) Pitch is determined by frequency. Higher is the frequency, greater will be the pitch.

Question 2:  Guess which sound has a higher pitch: guitar or car horn?

Page no. 166

Question 1:  What are wavelength, frequency, time period and amplitude of a sound wave?

Answer:  Wavelength: The distances between two consecutive compressions or rarefaction of a wave.  Its S.I unit is meter.

Frequency: One compression and one rarefaction constitutes one vibration. The number of vibration in a second is called frequency. Its unit is Hertz.

Amplitude: When waves are produced, the particles vibrate about their mean position. The maximum displacement from its mean position of a particle is called its amplitude. It is measured in meters.

Time period: The time taken by the wave to complete one oscillation i.e., the time between two consecutive compressions or rarefactions is called time period.

Question 2:  How are the wavelength and frequency of a sound wave related to its speed?

Speed = Wavelength x frequency
⇒ V = λ × ν

Question 3: Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.

Answer: Given  f = 220 Hz, V = 440 m/s

Question 4:  A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?

Page no. 166

Question 1:  Distinguish between loudness and intensity of sound?

Answer: Loudness and intensity both depend upon the amplitude of the sound. But loudness is the physiological response of our ears to a particular frequency. Our ears are more sensitive to some frequencies as compared to others. Intensity is the amount of sound energy passing per second per unit area. It is proportional to the square of the amplitude.

Page no. 167

Question 1:  In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?

Answer:  Sound travels faster in iron and slowest in air.

Page no.168

Question 1:  An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms−1 ?

Answer:  Given, Time for echo = 3 s
v = 342 m/s
Therefore distance: d = v × t
= 342×3 m
= 513 m.

Page no. 169

Question 1:  Why are the ceilings of concert halls curved?

Answer: The ceilings of concert halls are curved so that after reflections from the surface, sound can reach each and every part of the hall.

Page no. 170

Question 1: What is the audible range of the average human ear?

Answer:  Audible range 20 Hz − 20,000 Hz

Question 2:  What is the range of frequencies associated with (a) Infra sound? (b) Ultrasound?

(a) Infra-sound is less than 20 Hz
(b) Ultra-sound is greater than 20, 000 Hz.

Page no. 172

Question 1: A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?

t = 1.02 s
v = 1531 m/s
so,  distance  d = v × t
= 1531×1.022 m
=780.81m

## NCERT Class 9 Science Chapter 12 Sound Exercise Questions (Solved)

Question 1: What is sound and how is it produced?

Answer: Sound is a form of energy and it is produced due to vibrations of different types of object.
For example: A vibrating tuning fork, a bell, wires in a sitar and a guitar etc.

Question 2: Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

Answer: If we blow a horn, speak, or produce sound by an object in the air we are pushing the air molecules. These molecules, in turn, push the adjacent molecules which impart their energy to the next ones. After losing energy in the interaction, the molecule is back to its original (mean position. This results in the formation of compressions and rarefactions.

Question 3: Cite an experiment to show that sound needs a material medium for its propagation.

Answer: Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar. The bell jar is connected to a vacuum pump, as shown in the following figure. Now, if we press the switch, we will be able to hear the bell.

Now start the vacuum pump. When the air in the jar is pumped out gradually, the sound becomes fainter, although the same current is passing through the bell.

After some time when less air is left inside the bell jar you will hear a very feeble sound. At the end, when all the air is removed completely, we will not be able to hear the sound of bell. This show sou &requires material medium.

Question 4: Why sound wave is called a longitudinal wave?

Answer: A sound wave is called a longitudinal wave because it travels in the form of compressions and rarefactions in the medium, where the particles of the medium vibrate in a direction which is parallel to the direction of propagation of the sound wave.

Question 5: Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Answer: The timbre of sound is that characteristic which enables us to distinguish one sound from the other even when these are of the same pitch and loudness. Each person has its own timbre of sound and this characteristic helps us to identify a person from others even without looking at him (i.e., in a dark room).

Question 6: Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Answer: The flash of light is seen earlier than the thunder of sound even though both are produced simultaneously because the speed of light (c) is greater than the speed of sound (v) by 106 as

Question 7: A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m-1.

Question 8: Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Answer: Speed of sound in air = 346 m/s and
Speed of sound in aluminium = 6420 m/s
Since the time taken by sound to travel a given distance in a medium is inversely proportional to its speed in that medium, therefore

Question 9: The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Answer: Since the frequency of the source of sound is 100 Hz,
Number of vibrations of the source in 1 second = 100
Number of vibrations of the source in 1 minute (i.e., 60 seconds)
= 100 ×6
= 6000

Question 10: Does sound follow the same laws of reflection as light does? Explain.

Answer 10: Yes, sound follows the same laws of reflection as light does.
The laws of reflection of sound are as follows:

• The incident sound wave, the reflected sound wave, and the normal at the point of incident all lie in the plane.
• The angle of incidence of sound wave and angle of reflection of sound wave to the normal are equal.

Question 11: When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?

Answer: In any medium as we increase the temperature the speed of sound increases. For example, the speed of sound in air is 331 m/s at 0°c and 344 m/s at 22°c. So, on a hotter day, we cannot hear the echo between the same distances.

Question 12: Give two practical applications of reflection of sound waves.

Answer 12: The practical applications of sound are as follows:

• Megaphones or loudhailers are based on multiple reflections of sound.
• Stethoscope is a medical instrument used for listening to sounds produced within the body is also based on multiple reflection of sound.

Question 13: A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 ms-2 and speed of sound = 340 m/s

Answer: During the downward motion of the stone
Initial velocity u = 0 m/s
Height h = 500 m
Acceleration g = 10ms-2
Now we have

So, stone takes 10 seconds to pond from the top of the tower. Now a sound of splash is produced.
Now the time taken by the sound from the base of the tower to the top of the tower is given by
Time = distance ÷ speed
= 500 ÷ 340
= 1.47 seconds
So, the total time taken = 10 seconds + 1.47 seconds = 11.47 seconds

Question 14: A sound wave travels at a speed of 339 ms-l . If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Answer: We know that, Frequency = Speed/Wavelength
Here, Speed = 339 m s-1
Wavelength = 1.5 cm = 0.015 m
Therefore, Frequency = 339/0.013 = 22600 Hz
It is not audible (as the audible frequency is 20 Hz to 20000 Hz).

Question 15: What is reverberation? How can it be reduced?

Answer: A sound created in a big hall will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible. The repeated reflection that results in this persistence of sound is called reverberation.

To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound absorbing properties.

Question 16: What is loudness of sound? What factors does it depend on?

Answer: The loudness (or softness) of a sound is determined by its amplitude. If the amplitude is higher, it is a louder sound. It depends upon the force with which an object is made to vibrate.

Question 17: Explain how bats use ultrasound to catch a prey.

Answer: Bats can produce and hear sound of frequency up to 100 kHz. The sound produced by flying bat gets reflected from its prey in front of it. By hearing this reflected sound, it can detect the prey even during nights.

Question 18: How is ultrasound used for cleaning?

Answer: To clean any objects, it is placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. The objects thus get thoroughly cleaned.

Question 19: Explain the working and application of a sonar.

Answer: Sonar consists of a transmitter and a detector and is installed in a boat or a ship, as shown in the following figure.

The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector. The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. 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. If the time interval between transmission and reception of ultrasound signal be t and the speed of sound through seawater be v. The total distance, 201 travelled by the ultrasound is then, 2d = v x t.

The sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc.

Question 20: A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.

Answer: Total distance travelled by sound = 2 x 3625 = 7250 m
Time taken = 5 seconds
Therefore, speed = distance/time
= 7250/5 m/s
= 1450 m/s
So, the speed of sound in water is 1450 m/s.

Question 21: Explain how defects in a metal block can be detected using ultrasound.

Answer: Ultrasounds can be used to detect the defect in metal blocks. The cracks or holes inside the metal blocks, which are invisible from outside reduces the strength of the structure. Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect, as shown in the following figure.

Question 22: Explain how the human ear works.