NCERT Solutions for Class 9 Science Chapter 12 Sound
|1. Exercise Questions|
|2. Intext Questions|
NCERT Solutions for Class 9 Science Chapter 12 Sound on this step-by-step Sound answer guide. In some of State Boards and CBSE schools, students are taught thru NCERT books. As the chapter comes to an end, students are requested few questions in an exercising to evaluate their knowledge of the chapter. Students regularly want guidance dealing with those NCERT Solutions for Class 9 Science Chapter 12 Sound.
It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students rating better marks, we have provided grade by grade NCERT answers for all exercises of Class 9 Science Sound so you can be searching for assist from them. Students must solve those exercises carefully as questions withinside the very last exams are requested from those so these exercises at once have an impact on students’ final score. Find all NCERT Solutions for Class nine Science Sound below and prepare for your tests easily.
NCERT Solutions for Class 9 Science
Chapter 12 Sound
Q. 1. What is sound and how is it produced ?
Ans . Sound is a kind of energy produced in the form of waves .
Q. 2. Describe with the help of a diagram , how compressions and rarefactions are produced in air near a source of sound .
Ans . Compression and rarefaction in air : When an object starts vibrating , it creates disturbance in medium . The region where particles come closer to each other is called compression and region where particles go farther to each other is called rarefaction .
In the figure given below , straight lines are showing the normal position of air particles . Dense lines are showing the region of compression and less dense lines are showing region of rarefaction of air particles .
Q. 3. Cite an experiment to show that sound needs a material medium for its propagation .
Ans . Take a glass bell jar , connect it with vacuum pump and suspend an electric bell in it . Connect electric bellwith a battery . Switch on the electric bell and hear the sound of bell . Now remove the air completely from the bell jar using vacuum pump and observe the sound of electric bell .
It is observed that sound of electric bell does not come out after pumping out air from the bell jar . This happens because after creating vacuum in the bell jar there were no air present through which sound wave can propagate .
Q. 4. Why is sound wave called a longitudinal wave ?
Ans . When oscillation is created parallel to the disturbance of the particles of medium in the direction of propagation it is known as longitudinal wave .
Since sound wave creates oscillation in the particles of the medium parallel to the disturbance in the direction of propagation , thus sound waves are called longitudinal waves .
Q. 5. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room ?
Ans . The quality of sound or timber is that characteristic which helps us identify a particular person . Sound produced by two persons may have the same pitch and loudness , but the quality of the two sounds will be different .
Thus , because of difference in timber of the sound wave , one can identify the voice of his friend sitting with others even in dark room .
Q. 6. Flash and thunder are produced simultaneously . But thunder is heard a few seconds after the flash is seen , why ?
Ans . This happens because of the difference in the velocity of light and sound waves . The speed of sound is 330 m / sec in air medium whereas the speed of light is 3 x 108 m / sec .
So , light travels faster than sound . That is why thunder is heard a few seconds after the flash of thunder is seen .
Q. 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 s– ¹ .
Ans . Given :
Velocity of sound = 344 m / s
v₁ = 20Hz
v₂ = 20Hz
We know that :
Case 1 : Frequency ( v₁ ) = 20 kHz = 20,000 Hz
Case 2 : Frequency ( v₂ ) = 20 Hz
Q. 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 .
Ans . We know that ,
The speed of sound in air = 344 m / s
The speed of sound in aluminium = 6420 m / s
Hence , the ratio of time taken by the sound to travel through air and through aluminium .
So , the sound will take 18.06 times more time through air than in aluminium in reaching other boy and the ratio of time taken by the sound to travel through air and through aluminium will be 18.66 : 1
Q. 9. The frequency of a source of sound is 100 Hz . How many times does it vibrate in a minute ?
Ans . Given that ,
Frequency , v = 100 Hz
As we know that the frequency is vibrations per second .
That means , vibrations in 1 sec is = 100
So sound source will vibrate 100 times in a second .
Also , 1 minute is equal to 60 seconds . Thus , the source of sound vibrates 6000 times in a minute .
Q. 10. Does sound follow the same laws of reflection as light does ? Explain .
Ans . Yes , the sound wave follows the same laws of reflection as the light does . The laws of reflection of sound are as follows :
( i ) The incident sound wave , the reflected sound wave and the normal at the point of incidence , all lie in the same plane .
( ii ) The angle of incidence of sound wave and angle of reflection of sound wave to the normal are equal .
When sound waves reflected from a surface , the angle of incidence is equal to the angle of reflection to the normal and the incident wave , normal and reflected wave are in the same plane . This can be proved by experiment . Thus , sound wave obeys the laws of reflection .
Q. 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 ?
Ans . The sound of echo depends upon the distance from source of sound and reflecting surface . The relation between time , distance and speed is as follows :
Time = Distance / Speed
Time is inversely proportional to the speed . As the temperature increases , the speed of sound in a medium increases . Thus , on a hot day due to high temperature the speed of sound increases and the time interval between the original sound and the reflected sound will decrease and we can hear the echo sooner .
Q. 12. Give two practical applications of reflection of sound waves .
Ans . Two practical applications of reflection of sound waves are following :
1. Bulb horn : In bulb horn sound is amplified and sent to the desired direction because of reflection .
2. Stethoscope : It is based on the principle of multiple reflection of sound within the stethoscope tube enabling the doctor to hear a patient’s heartbeat .
Q. 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 m s – 2 and speed of sound = 340 m s – ¹ .
Ans . Given ,
Height of tower = 500 m
Acceleration due to gravity = 10 m / s²
Speed of sound = 340 m / s
We know that :
Q. 14. A sound wave travels at a speed of 339 m s – ¹ . If its wavelength is 1.5 cm , what is the frequency of the wave ? Will it be audible ?
Ans . Given that :
Speed of sound = 339 m / s
Wavelength = 1.5 cm = 0.015 m
We know that ,
As we know , the frequency range of audible sound for humans lies between 20 Hz to 20,000 Hz . Since the frequency of the given sound is more than 20,000 Hz , it is not audible .
Q. 15. What is reverberation ? How can it be reduced ?
Ans . Reverberation is the collection of reflected sounds from the surfaces in an enclosure like an auditorium .
To reduce reverberations , sound must be absorbed as it reaches the walls and the ceiling of a room . That is done by covering walls and roof of halls by sound absorbent material like compressed fibreboard , heavy curtains , rough plastic or draperies .
Q. 16. What is loudness of sound ? What factors does it depend on ?
Ans . Loudness is the degree of sensation of sound produced in the ear .
It depends on the amplitude of the vibration producing that sound . Greater is the amplitude of vibration , louder is the sound produced by it .
Q.17 . Explain how bats use ultrasound to catch a prey .
Ans . Echolocation is the process where ultrasound waves and echoes are used to determine objects in space . This technique is used by bats to navigate and find their foods in the dark . The ultrasonic waves emitted by the bat are reflected from the prey and are detected by its ears . The nature of reflected waves tells the bat :
( i ) the location
( ii ) the nature of its prey
Q. 18. How is ultrasound used for cleaning ?
Ans . Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution .
The high frequency of these ultrasound waves detaches the dirt from the objects .
Q. 19. Explain the working and application of a sonar .
Ans . SONAR is an acronym for sound navigation and ranging .
Working principle : It is based on the principle of the reflection of sound wave ( i.e. , echo ) .
It is used to measure the depth , direction and speed of tmder – water objects such as submarines and ship wrecks with the help of ultrasound . It is also used to measure the depth of seas and oceans too .
Working : A beam of ultrasonic sound is produced and transmitted by the device that produces ultrasonic sound ( called transducer ) of the SONAR , which travels through sea water . When these pulses are intercepted by a distant object or even the sea bottom , they get reflected . These sounds produced by the reflection of this ultrasonic sound is detected and recorded by the detector , which is converted into electrical signals .
Formula used : The distance ( d ) of the under – water object is calculated from the time ( t ) taken by the echo to return with speed ( v ) is given by 2d = vt * This method of measuring distance is also known as ‘ echo – ranging ‘ .
Q. 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 .
Ans . Time taken between transmission and reception of signal ‘ t ‘ = 5 sec
Distance of the object from the sub marine ‘ d ‘
= 3,625 m
So , total distance travelled by the sound ‘ 2d ‘
= 3625 x 2 = 7250 m
So the speed of sound in water 1,450 m / s .
Q. 21. Explain how defects in a metal block can be detected using ultrasound .
Ans . Ultrasound can be used to detect cracks and flaws in metal blocks . Defects in metal blocks do not allow ultrasound to pass through them and they are reflected back . Ultrasound is passed through one end of a metal block and detectors are placed on the other end . If there is even a small defect , the ultrasound gets reflected back indicating the presence of the flaw or defect . This principle is used to detect defects in metal blocks .
Q.22 . Explain how the human ear works .
Ans . The human ear is a complex organ . It is a highly sensitive part of the human body that enables us to hear a sound . The human ear can be divided into three main parts :
( 1 ) External ear , ( 2 ) Middle ear and ( 3 ) Internal ear .
External ear : It consists of pinna ( auricle ) and ear canal ( external auditory canal ) .
Middle ear : It consists of inner layer of eardrum ( tympanic membrane ) , three tiny bones ( malleus , incus and stapes ) grouped named as ossicles and oval window .
Inner ear : It is also called labyrinth of the ear , part of the ear that contains organs of the senses of hearing and equilibrium and consist of the vestibule , the semicircular canals and the cochlea .
Working of human ear : The external ear catches sound waves and sends them to the ear drum , by the ear canal or external auditory canal . During compression , the pressure increases outside the ear drum which forces the eardrum to move inwards . During rarefaction , the pressure decreases outside the ear drum which forces the eardrum to move outwards . So that vibrations are produced in the eardrum . After that , the three bones , known as ossicles , amplify the sound wave , by working together . These vibrations reach to the inner ear and converted into electrical signals by the vibrating liquid of cochlea sets up electrical impulse in nerve cell present in it . These signals are transmitted by the auditory nerve to the brain . Finally , the brain interprets those signals as sound .
Q. 1. How does the sound produced by a vibrating object in a medium reach your ear ? [ NCERT Q. 1 , Page 162 ]
Ans . Vibrations in an object create disturbance in the medium and consequently compressions and rarefactions . Because of these compressions and rarefactions , sound reaches to our ear .
Q. 2. Explain how sound is produced by your school bell . [ NCERT Q. 1 , Page 163 ]
Ans . School bell starts vibrating when heated that creates compression and rarefaction in air and sound is produced .
Q. 3. Why are sound waves called mechanical waves ? [ NCERT Q. 2 , Page 163 ]
Ans . Sound waves are known as mechanical waves because they need a material medium for propagation , like air or liquids like water , or metals like silver . That is the defining criterion for a mechanical wave . Hence , sound waves are called mechanical wave .
Q. 4. Suppose you and your friend are on the moon . Will you be able to hear any sound produced by your friend ? [ NCERT Q. 3 , Page 163 ]
Ans . Sound waves need a material medium for their propagation . Since there is no atmosphere on the Moon , one person cannot hear the sound produced by another person .
So , I will not be able to hear the sound of my friend on the moon .
Q. 5. Which wave property determines ( a ) loudness , ( b ) pitch ? [ NCERT Q. 1 , Page 166 – I ]
Ans . ( a ) Amplitude of sound waves determines loudness . Louder sound has greater amplitude and vice versa .
( b ) Frequency of the sound waves determined pitch of the sound .
Q. 6. Guess which sound has a higher pitch : guitar or car horn ? [ NCERT Q. 2 , Page 166 – I ]
Ans . Sound of the car horn has higher pitch . Since the pitch of a sound is proportional to its frequency , the guitar has a lower pitch than a car horn .
Q. 7. What are wavelength , frequency , time period and amplitude of a sound wave ? [ NCERT Q. 1 , Page 166 – II ]
Ans . Amplitude : The maximum displacement of a vibrating particle of the medium from the mean position .
Wavelength : The distance between two consecutive compressions or rarefaction of wave .
Time period : Time for a particle on a medium to make one complete vibrational cycle .
Frequency : The number of vibrations made by the particles in one second .
Q. 8. How are the wavelength and frequency of a sound wave related to its speed ? [ NCERT Q. 2 , Page 166 – II ]
This means the speed is equal to the product of wavelength and frequency of the sound wave . This equation is also called the ‘ wave equation ‘ and is applicable to all types of wave .
Q. 9. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m / s in a given medium . [ NCERT Q. 3 , Page 166 – II ]
Ans . Frequency ( v ) = 220 Hz
Velocity ( v ) = 440 m / s
Wavelength ( A ) = ?
As we know speed ‘ v ‘ , frequency v of the wave and its wavelength ‘ λ ‘ are related as follows :
Speed = Frequency x Wavelength
So , the wavelength of the sound wave will be 2 m .
Q. 10. 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 ? [ NCERT Q. 4 , Page 166 – II ]
Ans . The time interval between compressions is called time interval .
Here , the frequency ( v ) = 500 Hz
T ( Time period ) = ?
We know that :
Thus time interval between two consecutive compressions in the given wave is 0.002s .
Q.11 . Distinguish between loudness and intensity of sound . [ NCERT Q. 1 , Page 166 – III ]
Ans . Intensity is an average energy transported per second per unit area perpendicular to the direction of propagation . The loudness of sound depends on its intensity but the relationship is not linear .
Loudness is the degree of sensation of sound produced in the ear . Loudness is related to the amplitude of sound waves whereas intensity is the rate at which power is transferred across a given area . Loudness of sound is determined by amplitude and intensity of the sound wave is determined by frequency of sound waves .
Q. 12. In which of the three media , air , water or iron , does sound travel the fastest at a particular temperature ? [ NCERT Q. 1 , Page 167 ]
Ans . At particular temperature , sound travels fastest in iron .
Q. 13. An echo is heard in 3 s . What is the distance of the reflecting surface from the source , given that the speed of sound is 342 m s – ¹ ? [ NCERT Q. 1 , Page 168 ]
Ans . Speed of sound = 342 m / s
Echo returns in time = 3s
Total distance travelled by the sound
= Speed x Time – = 342 × 3 × 1,020 m
As we know , to return an echo sound has to cover distance in two ways .
So , sound has travelled twice the distance of the reflecting surface and the source .
Hence , the distance of the reflecting surface from the source will be 1,026 / 2 = 513 m .
Q. 14. Why are the ceilings of concert halls curved ? [ NCERT Q. 1 , Page 169 ]
Ans . Ceilings and walls of concert halls are curved so that the sound after reflection reaches uniformly to every corner of the concert hall and the audience can listen the sound clearly .
Q. 15. What is the audible range of the average human ear ? [ NCERT Q. 1 , Page 170 ]
Ans . Audible range of the average human ear is 20 Hz to 20,000 Hz .
Q. 16. What is the range of frequencies associated with
( a ) Infrasound ?
( b ) Ultrasound ? [ NCERT Q. 2 , Page 170 ]
Ans . ( a ) Infrasound : Less than 20 Hz .
( b ) Ultrasound : More than 20,000 Hz .
Q. 17. 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 ? [ NCERT Q. 1 , Page 172 ]
Ans . Speed of sound wave = 1531 m / s
Time ( t ) = 1.02 s
To return the SONAR pulse back , its wave has to travel two ways .
The total distance travelled by sub marine is 1561.32 .
So , the distance between the source and reflecting surface ( cliff ) will be = ( Distance of the cliff from the sub marine / 2 ) = 1561/2 = 780.31 m .
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