NCERT Class 9 Science Chapter 12 Sound Extra Questions and Answers
Class 9 Science Chapter 12 Extra Inside Questions and Answers – Sound. Here in this Page Class IX Students can Learn Extra Questions & Answer 12th Chapter Science fully Inside.
We Provided Here Sound Science Chapter 12 Long Answer Type Question, Multiple Choice Questions & Answer, Short Answer Type Questions (2 or 3 marks), and Very Short answer Type Question (1 marks) Solution.
Class 9 Science Chapter 12 Extra Question with Answer – Sound
Science Chapter 12 Sound Class 9 Inside 5 Marks, 3 marks, 2 Marks & And 1 Marks Important Questions and Answers.
Very Short answer type questions:
1.) Define vibration.
Answer: Rapid to and from motion of an object is called vibration.
2.) Can the sound propagate through benzene solution.
Answer: Yes, medium is necessary for propagation of sound and benzene solution is also a medium.
3.) State two examples of medium for propagation of sound.
Answer: Examples of medium for propagation of sound-
a) Air
b) Water
4.) Define compression.
Answer: The region where particles in medium are crowded is called compression.
5.) Give the relationship between frequency and time period.
Answer: The relationship between frequency and time period is as,
Frequency = 1/period
6.) State true or false
Particles in medium travels to listener from source of sound.
Answer: False, particle does not travels through the medium.
7.) Which is the most common medium from propagation of sound?
Answer: Most common medium for propagation of sound is Air.
8.) In which region does the higher pressure of molecules occur?
Answer: Higher pressure of molecules occurs in compression region.
9.) Stay the two types of mechanical waves.
Answer: Longitudinal wave and transverse wave are the types of mechanical wave.
10.) Give the example of longitudinal wave.
Answer: Example of longitudinal wave-
- Sound wave.
- Wave produced in spring.
11.) Give the example of transverse wave.
Answer: Example of transverse wave-
- Light wave
- Wave produced in water.
12.) State one difference between tone and note.
Answer: Tone contain single frequency while note is made up of mixture of frequencies.
13.) Does intensity and loudness interchangeable?
Answer: No, intensity and loudness are two different terms.
14.) Arrange the speed of sound in increasing order in mediums below.
Methanol, helium, brass.
Answer: The increasing order of sound in medium is as,
Brass < Methanol < Helium.
15.) State the difference between echo and reverberation.
Answer:
Echo is a single reflection of sound but in reverberation there will be multiple reflections of sound.
16.) State the principle on megaphone works.
Answer: Megaphone works on principle of multiple reflections of sound.
Short answer type questions:
1.) How did the sound produce?
Answer: There are so many ways of production of sound.
- Striking.
- Plucking.
- Scratching.
- Rubbing.
- Collision.
- Blowing.
- Shaking.
2.) The swings completes 196 rotation in 2 minutes. Calculate the frequency of that swing.
Answer: Given,
Number of rotations = 196,
Time = 2 minutes = 120 seconds.
As we know that,
Frequency = number of rotation/ time
Frequency = 196/120
Frequency = 0.98 Hz
The frequency of that swing will be 0.98 Hz.
3.) Explain wavelength.
Answer:
- As we studied that sound is type of wave.
- Wavelength is the displacement of one complete oscillations.
- Definition: The distance between consecutive trough and crest is called wavelength.
- Wavelength is nothing but it is a distance so it’s SI unit is meter.
- Wavelength is represented by a Greek letter called lambda. It’s symbol is λ.
4.) A person shout in auditorium and hear echo after 4 seconds. If the speed of sound is 346 m/s then calculate distance between that person and wall of the auditorium.
Answer: Given, time = 4 seconds,
Speed of sound = 346 m/s.
As we know that,
Speed = distance/ time.
Distance = speed × time
Distance = 346 × 4
Distance = 1384 m.
Sound travels 1384 m in 4 seconds.
But the distance between that person and wall is half of the total distance because sound travels two times.
Therefore,
Distance between that person and wall = total distance /2
Distance between that person and wall = 1384/2
Distance between that person and wall = 692 m.
The distance between that person and the wall of the auditorium is 692 m.
5.) Calculate the period of pendulum whose frequency is 170 Hz.
Answer: Given,
Frequency of pendulum = 170 Hz.
As we know that,
Frequency =1/period
Period = 1/ Frequency
Period = 1/170
Period = 0.00589 Seconds
The period of that pendulum is 0.00589 seconds.
6.) Calculate the frequency of pendulum whose period is 0.04 seconds.
Answer:
Given,
Period = 0.04 seconds.
As we know that,
Frequency = 1/period
Frequency = 1/0.04
Frequency = 25 Hz.
The frequency of that pendulum is 25 Hz.
7.) A sound wave of frequency 3 kHz moves with 830 m/s. Calculate it’s wavelength.
Answer:
Given,
Frequency = f = 3 KHz = 3× 10³ Hz,
Velocity = v = 830 m/s.
As we know that,
Velocity = frequency × wavelength
v = f × λ
λ = v/f
λ = 830/3×10³
λ = 0.277 m
The wavelength of that wave is 0.277 m.
8.) At echo point, echo is heard after 6 seconds. Calculate distance of echo points from the source. (Speed of sound = 338 m/s.)
Answer:
Given,
time = 6 seconds,
Speed of sound = 338 m/s.
As we know that,
Speed = distance/ time.
Distance = speed × time
Distance = 338 × 6
Distance = 2028 m.
Sound travels 2028 m in 6 seconds.
But the distance between that source and echo point is half of the total distance because sound travels two times.
Therefore,
Distance between source and echo point = total distance /2
Distance between source and echo point = 2028/2
Distance between source and echo point = 1014 m.
Thus the distance between source and echo point is 1014 m.
9.) The speed of sound is 340 m/s .A train is 2 km far from passenger. How mush time is required to hear the sound of horn of a train to that passenger?
Answer: Given,
Speed of sound = 340 m/s,
Distance = 2 km = 2000 m.
As we know that,
Speed = distance / time
Time = distance/speed
Time = 2000/340
Time = 5.88 S.
The passenger listen horn of a train after 5.88 s.
In case you are missed :- Previous Chapter Extra Questions
10.) Minimum time required for echo is 0.1 s. Of the speed of sound is 332 m/s then calculate minimum distance where the person can listen echo.
Answer=
Given, time = 0.1 s,
Speed of sound = 332 m/s,
As we know that,
Speed = distance / time.
Distance = speed × time
Distance = 332 × 0.1 s
Distance = 33.2 m.
But this is the total distance travelled by sound.
As we know that,
Distance of echo = (total distance/2)
Distance of echo = 33.2 /2
Distance of echo = 16.6
Thus the person can stand at 16.6 m from echo to listen echo.
11) Why can not we listen sound in interstellar space?
Answer:
- Sound is a type of longitudinal wave.
- We know that, medium must required for propagation of longitudinal wave.
- Molecules in the medium vibrates and transfer energy to next one. In such a way, sound can propagate through the medium.
- Medium is absent in interstellar space.
- Therefore we cannot listen sound in interstellar space.
12.) State the device which used for treatment of hearing loss patients. How does it works ?
Answer: The device used for treatment of hearing loss patients is hearing aid. It is an electronic device. It amplify sound wave using electrical energy.
13.) Write two medical applications of ultrasonic waves.
Answer: High frequency sound waves are called as ultrasonic waves. Due to high frequency, It is widely used in medical industries.
Applications:
- Diagnosis:
Sound is a type of wave. It can be used for imaging. Thus by using ultrasound waves, we can capture the image of damaged part in the body. Also we can diagnosis the some diseases using such type of ultrasound waves.
- Treatment:
Ultrasonic waves possesses higher energy because of higher frequency. So this high energy is utilized for treatment of some diseases. Ultrasound waves are also used in physiotherapy.
14.) How many times tuning fork of frequency 230 Hz vibrates in 2 minutes?
Answer: Given,
Frequency = 230 Hz,
Time = 2 minutes = 120 seconds.
As we know that,
Frequency = No of oscillations/time.
Number of oscillations = frequency ×time
Number of oscillations = 230 × 120
Number of oscillations = 27600.
That tuning fork Vibrates 27600 time in two minutes.
15.) A submerged send and receive a signal after 8 seconds. Calculate depth of sea. (Speed of sound in water = 1530 m/s)
Answer:
Given that,
Time = 8 s
Distance = 1530 m
We know that, the sound wave produced travels the distance from point of production and then
to the point of reception.
So total distance covered will be 2d.
Hence,
Speed of sound = (2 × depth)/ t
Depth = (speed of sound × t)/2
Depth = (1530 ×8)/2
Depth = 12240/2
Depth = 6120 m .
Thus, the depth of sea is 6120 m .
Long answer type questions:
1.) How does sound propagate through the medium?
Answer:
- Sound is a type of energy. This energy disturb the particles in medium. This disturbance produces vibrations.
- Sound energy vibrates the particle in the medium.
- The particles in the medium vibrates from mean position. They do not make displacement from their mean position.
- First particle displaced from mean position and collides on second particle and come back to it’s mean position.
- Similarly second particles gives the energy to third particle and come back to its mean position.
- Similarly the particles in the medium work as a medium for propagation of sound.
- Thus the net displacement of the particles remains zero.
- Finally this vibrating particles collides on the ear of listener and make sense of sound.
- The net displacement of particles is zero but sound energy will propagate through the medium.
- So we will understand this by considering a wave.
- Sound is mechanical wave.
2.) A) Explain Frequency with example.
B) If a pendulum completes 28 oscillations in a minute then calculate its frequency.
Answer:
A)
Frequency –
- Frequency gives us information about number of oscillations performed in unit time.
- Frequency is a rate of oscillations.
- Definition: Frequency is defined as number of oscillations performed in one second.
- Formula
Frequency = (number of oscillations)/unit time
- SI unit of frequency is per second or Hz.
- If number of oscillations are greater than the value of frequency becomes greater and vice versa.
Examples-
- If a fan completes 1 round in one second the frequency of fan is 1 Hz.
- If a swing completes 10 oscillations in one minute then its frequency is 1/6 Hz.
B) Given,
Pendulum completes 28 oscillations,
Number of oscillations = 28,
Time = 1 minute = 60 seconds,
As,
Frequency = (number of oscillations)/unit time
Frequency = 28/60
Frequency = 0.4667 Hz.
The frequency of that pendulum is 0.4667 Hz.
3) A) How does stethoscope work?
B) The length of tube is 1.2 m. If doctor hear sound of heartbeat after 0.0002 seconds then calculate the speed of sound in tube.
Answer:
A) Stethoscope-
- Stethoscope is device used for listening heartbeats.
- It is also used for listening sound produced in various organs.
- It is based on the principle of multiple reflections of sound.
- Stethoscope has three parts.
a.) Diaphragm- Doctor put the Diaphragm of stethoscope on heart of patient’s heart. Diaphragm vibrates according to the heart beatings.
b.) Tube- Mostly tubes use in stethoscope are made up from rubber. Tube works as a medium for propagation of sound. Multiple reflections of takes place in this tube.
c.) Ear canal- There is a pair of ear canal. It magnify the sound thus doctor listen heartbeat properly.
B) Given,
Distance = 1.2 m,
Time = 0.0002 seconds,
As we know that,
Speed = distance /time
Speed = 1.2/ 0.0002
Speed = 6000 m/s.
The speed of sound in the tube is 6000 m/s.
4.) A) Explain the variation of Speed of sound with the state of medium.
B) A sound of wavelength 18 cm and frequency 4.2 kHz. Calculate time required to travels 2.8 km for this sound ?
Answer:
A) Variation of speed of sound with state of medium.
- As we know that, sound is a longitudinal wave.
- Longitudinal wave must required medium for propagation.
- The speed of sound depends on the nature of medium.
- The sound can propagate through medium by molecules to molecules.
- If the density of molecules is greater than sound can easily propagate. But if the density of molecules in medium is smaller than some distraction arises for propagation of sound.
- In solid medium, the density of molecules is greater than liquid and gases. Therefore the speed of sound is greater in solid.
- The density of molecules in solid, liquids and gases are as
Solid > liquid > gases.
Therefore speed of sound is also varies with such a way.
- Variation of speed of sound.
Solid > liquid > gases.
B)
Given,
Wavelength = λ = 18 cm = 0.18 m,
Frequency = f= 4.2 kHz = 4200 Hz,
Distance = 2.8 km = 2800 m.
Firstly we calculate speed of sound,
As we know that,
Speed = frequency × wavelength.
Speed = f × λ
Speed = 4200 × 0.18
Speed = 756 m/s.
The speed of sound is found to be 756 m/s.
Now we calculate time required for traveling 2.8 km.
As we know that,
Speed = distance / time
Time = distance / speed.
Time = 2800/756
Time = 3.70 seconds.
Time required to travels 2.8 km I
In case you are missed :- Next Chapter Extra Questions