Lakhmir Singh Manjit Kaur Class 9 Physics 5th Chapter “Sound” solution
Lakhmir Singh and Manjit Kaur Physics solution: “Sound” Chapter 5. Here you get easy solutions of Lakhmir Singh and Manjit Kaur Physics solution Chapter 5 . Here we have given Chapter 5 all solution of Class 9. Its help you to complete your homework.
- Board – CBSE
- Text Book – Physics
- Class – 9
- Chapter – 05
Lakhmir Singh Manjit Kaur Class 9 Physics 5th Chapter Solution (First Part)
Very Short Answer Type Questions:
l.) Can sound travel through (a) iron, and (b) water?
Answer: sound can travel through iron and water.
2.) Can sound travel through vacuum ?
Answer: No, sound can not travel through vacuum because medium is necessary for propagation of sound.
3.) Name the type of waves which are used by astronauts to communicate with one another on moon (or in outer space).
Answer: Astronauts use radio waves to communicate with one another on moon.
4.) Name one solid, one liquid and one gas through which sound can travel.
Answer: Sound can travel through solid, liquid and gases.
Solid- telephone wire.
Liquid – water.
Gases- mobile communication is through gases.
5.) Which of the following cannot transmit sound ?
Water, Vacuum, Aluminum, Oxygen gas
Answer: Vacuum cannot transmit sound.
6.) Name the physical quantity whose SI unit is ‘hertz’.
Answer: SI unit of frequency is hertz.
7.) What is the SI unit of frequency ?
Answer: SI unit of frequency is hertz.
8.) What type of wave is represented (a) by density-distance graph ?
(b) by displacement-distance graph ?
Answer:
(a) The wave represented by density – time graph is longitudinal wave.
(b) The wave represented by displacement- distance graph is transverse wave.
9.) Is the speed of sound more in water or in steel ?
Answer: The speed of sound is more in steel than water because steel is denser than water.
10.) In which medium sound travels faster : air or iron ?
Answer: The speed of sound is more in iron than air because iron is denser than air.
11.) In which medium sound travels fastest : air, water or steel ?
Answer: The speed of sound is faster in steel than water and steel. Because steel is denser than air and water.
12.) Out of solids, liquids and gases :
(a) in which medium sound travels slowest ?
(b) in which medium sound travels fastest ?
Answer: (a) The speed of sound is slowest in gases.
(b) The speed of sound is fastest in solids.
13.) Which of the following is the speed of sound in copper and which in aluminum ? (a) 5100 m/s (b) 1500 m/s (c) 3750 m/s.
Answer: The speed of sound in copper is 3750 m/s and speed of sound in aluminum is 5100 m/s.
14.) If you want to hear a train approaching from far away, why is it more convenient to put the ear to the track ?
Answer: As we know that speed of sound is faster in denser medium than rare medium.The speed of sound in iron is 15 times greater than speed of sound in air. Therefore it is more convenient to put the ear to the track to listen the sound of train.
15.) What is the speed of sound :
(a) in air ? (b) in water ? (c) in iron ?
Answer:
(a) The speed of sound in air is 344 m/s.
(b) The speed of sound in water is 1498 m/s.
(c) The speed of sound in iron is 5130 m/s.
16.) What name is given to those aircrafts which fly at speeds greater than the speed of sound?
Answer: Supersonic aircrafts fly at speeds greater than the speed of sound.
17.) A jet aircraft flies at a speed of 410 m/s. What is this speed known as ?
Answer: The speed of aircraft at 410 m/s is known as supersonic speed.
18.) What is meant by supersonic speed ?
Answer: If the speed of object is greater than speed of sound then that speed is called as supersonic speed.
19.) State one observation from everyday life which shows that sound travels much more slow than light.
Answer: Lighting in sky proves that speed of sound is much slower than speed of light. When lightning appears in the sky then after some time we hear the sound of lightning.
20.) Name the two types of waves which can be generated in a long flexible spring (or slinky).
Answer: Longitudinal wave and transverse wave can be generated in a long flexible spring.
21.) A stone is dropped on the surface of water in a pond. Name the type of waves produced.
Answer: When a stone is dropped on the surface of water in a pond then transverse waves are generates.
22.) Name the type of waves produced when a tuning fork is struck in air.
Answer: Longitudinal wave are produced by tuning fork.
23.) What is the general name of the waves consisting of :
(a) compressions and rarefactions ? (b) crests and troughs ?
Answer: (a) The wave consisting of compression and rarefactions are called longitudinal wave.
(b) The wave consisting of crests and troughs are called transverse wave.
24.) State the general name of the waves in which the particles of the medium vibrate :
(i) in the same direction as wave.
(ii) at right angles to the direction of wave.
Answer: (i)If the particles vibrates in the same direction as wave then such type of wave are called as longitudinal wave.
(ii) If the particles vibrates at right angles to the direction of wave then such wave are called as transverse wave.
25.) What type of waves are illustrated by the movement of a rope whose one end is fixed to a pole and the other end is moved up and down ?
Answer: The wave produced bymovement of a rope whose one end is fixed to a pole and the other end is moved up and down is called transverse wave.
26.) What should an object do to produce sound ?
Answer: Vibrations are necessary for production of sound.
27.) What is the name of the strings which vibrate in our voice box when we talk ?
Answer: Vocal cord vibrate in our voice box when we talk.
28.) Name the device which is used to produce sound in laboratory experiments.
Answer: Tuning fork is used to produce sound in laboratory experiments.
29.) What is the nature of sound waves in air ?
Answer: The nature of sound wave is longitudinal wave.
30.) What conclusion can be obtained from the observation’ that when the prongs of a sound making tuning fork touch the surface of water in a beaker, the water gets splashed ?
Answer: When tuning fork touch the surface of water in a beaker then tuning fork vibrates and produces sound.
31.) State whether the following statement is true or false :
Sound produced by a vibrating body travels to our ears by the actual movement of air.
Answer: False, Sound produced by a vibrating body travels to our ears by the moment of particles at that space only.
32.) Which of the following travels slowest in air and which one fastest ?
Supersonic aircraft, Light, Sound
Answer: Sound can travels at slowest speed and light travels at fastest speed.
34.) In which medium sound travels faster : air or hydrogen ?
Answer: Sound travels faster in hydrogen than air because hydrogen is more dense than air.
35.) A tuning fork has a number 256 marked on it. What does this number signify ?
Answer: The number on tuning fork gives the information about frequency. If a tuning fork has a number 256 marked in it then it’s frequency is 256 hertz.
36.) What is the time-period of a tuning fork whose frequency is 200 Hz ?
Answer: Given,
Frequency of tuning fork is 200 Hz,
As we know,
Frequency = 1/ period
Period = 1/frequency
Period = 1/200
Period = 0.005
Period = 5 × 10³ s
The time period of that tuning for is 5 × 10³ s.
37.) Calculate the frequency of a wave whose time-period is 0.02 s.
Answer: Given,
Period = 0.02
As we know,
Frequency = 1/ period
Frequency = 1/0.02
Frequency = 50 Hz
The value of frequency is 50 Hz.
38.) What will be the change in the wavelength of a sound wave in air if its frequency is doubled ?
Answer: The frequency is inversely proportional to time period. If wavelength is doubled then its frequency becomes halved.
39.) If 20 waves are produced per second, what is the frequency in hertz ?
Answer: Given, number of waves = 20, time = 1 second.
As we know,
Frequency = number of waves / time
Frequency = 20/1
Frequency = 20 Hz.
The frequency of that wave is 20 Hz.
Short Answer Type Questions:
41.) What is vacuum ? Explain why, sound cannot travel through vacuum ?
Answer: Vacuum is a medium in which there is no any particle. Absence of gases, liquids and solids are considered as vacuum. Propagation of sound is depend on vibration of molecules in the medium. But in vacuum there is absence of molecules thus Vibrations of molecules do not takes place and sound cannot travel through vacuum.
42.) Explain the term ‘amplitude’ of a wave. Draw the diagram of a wave and mark its amplitude on it.
Answer: The maximum perpendicular displacement of the particles from zero displacement point is called as amplitude. Amplitude of the wave is shown in figure as
43.) (a) Distinguish between longitudinal and transverse waves. (b) Are sound waves longitudinal or transverse ?
Answer: Distinguish between longitudinal wave and transverse wave.
(i) If the direction of propagation of wave is perpendicular to direction of vibration of particles then such wave is called as transverse wave. If the direction of propagation of wave is parallel to direction of vibration of particles then such wave is called as longitudinal wave.
(ii) Longitudinal wave consists of compression and rarefactions. Transverse wave consists of Crest and troughs.
(iii) Transvers wave is represented by displacement-time graph and longitudinal wave is represented by density-time graph.
(iv) Longitudinal wave is one dimensional wave because it moves in same direction on motion of particle. Transverse wave is two dimensional wave because direction of propagation of wave is perpendicular to direction of motion of particles.
(v) Examples
a) Transverse wave- light wave
b) Longitudinal wave- sound wave.
44.) A cricket ball is seen to hit the bat first and the sound of hitting is heard a little later. Why?
Answer: We know that speed of sound in air is 344 m/s and speed of light is 3 × 10⁸ m/s. Thus seeing object is due to light wave and hearing is due to sound wave. A cricket ball is seen to hit the bat first and the sound of hitting is heard a little later because speed of light us greater that speed of sound.
45.) Explain why, the flash of lightning reaches us first and the sound of thunder is heard a little later.
Answer: We know that speed of sound in air is 344 m/s and speed of light is 3 × 10⁸ m/s. The flash of lightning reaches us first and the sound of thunder is heard a little later because speed of light greater than speed of sound.
46.) Explain why, the flash of a gun shot reaches us before the sound of the gun shot.
Answer: Lights travels faster in air than sound so we see the flash of gun shot first and after some time we hear the sound of gun shot.
47.) Which of the following terms apply to sound waves in air and which to water waves ? Transverse, Rarefaction, Trough, Crest, Compression, Longitudinal
Answer: The terms related to following are.
Transverse wave- Crest, Trough
Longitudinal wave- Compression, Rarefaction.
48.) (a) Name four ways in which sound can be produced.
(b) Calculate the speed of a sound wave whose frequency is 2 kHz and wavelength 65 cm.
Answer: (a) Sound can be produced by following ways-
i) Hitting an object on another object.
ii) Friction between two objects.
iii) Collision between objects
iv) Rubbing between two objects.
(b) Given, Frequency = f = 2 kHz = 2 × 10³ Hz, wavelength = λ = 65 cm = 0.65 m. Velocity = v= ?
As we know,
v = f × λ
v = 2 × 10³ × 0.65
v = 1300 m/s
Speed of that sound is 130p m/s.
49.) If a ringing bicycle bell is held tightly by hand, it stops producing sound. Why ?
Answer: We know that vibrations are necessary for production of sound. If we held a hand on ringing bell then there is disturbance produced. Thus due to disturbance for vibrations ringing stops ringing.
50.) Which object is vibrating when the following sounds are produced ?
(i) The sound of a sitar.
(ii) The sound of a tabla
(iii) The sound of a tuning fork
(iv) The buzzing of a bee or mosquito
(v) The sound of a flute
Answer:
(i) The sound of a sitar- The string of sitar vibrates.
(ii) The sound of a tabla- The diaphragm of the tabla vibrates.
(iii) The sound of a tuning fork- The needles of tuning fork vibrates.
(iv) The buzzing of a bee or mosquito
(v) The sound of a flute- Air molecules in flute vibrates.
51.) Give reason for the following :
In most of the cases, we cannot see the vibrations of a sound producing object with our eyes.
Answer: In most cases vibrations are very fast means value of frequency is high so we can not see vibrations of producing sound.
52.) Describe a simple experiment to show that the prongs of a sound producing tuning fork are vibrating.
Answer: We know that, vibrations are necessary for production of sound. We can understand this with a simple experiment.
Apparatus: beaker, water, tuning fork.
Procedure: Fill the water into the beaker as shown in figure.
Hit the tuning fork on the beaker. Hear the produced sound and observe vibrations on the tuning fork.
Conclusion: When prong produced sound then it also vibrates.
53.) When we open a gas tap for a few seconds, the sound of escaping gas is heard first but the smell of gas comes later. Why ?
Answer: When we open a gas tap for few seconds, the sound of gas escaping gas is heard first but the smell of gas comes later because we know that speed of sound is greater than molecules of gas.
54.) A sound signal of 128 vibrations per second has a wavelength of 2.7 m. Calculate the speed with which the wave travels.
Answer: Given, no of vibrations = t = 128,
Time = t = 1 seconds, wavelength =λ= 2.7 m, speed = v =?
As we know
v = f × λ———1
But, Frequency = f = no of oscillations/ time= n/t
Equation 1 becomes,
v = (n/t) × λ
v = (128/1) × 2.7
v = 128 × 2.7
v = 345.6 m/s.
The speed of that wave is 345.6 m/s.
55.) A wave is moving in air with a velocity of 340 m/s. Calculate the wavelength if its frequency is 512 vibrations/sec.
Answer: Given, velocity = v = 340 m/s,
frequency = f = 512 Hz, wavelength =λ=?
As we know,
v = f × λ
λ = v / f
λ = 350 / 512
λ = 0.664 m
The wavelength of that wave is 0.664 m.
- Define the ‘frequency’ and ‘time-period’ of a wave. What is the relation between the two ?
Answer:
Frequency: Frequency is defined as number of oscillations per unit time. Frequency is mathematically expressed as,
f = n/t
Where,
f – frequency,
n- number of oscillations,
t – time required to comple that oscillations.
Time- period: Time required to complete one oscillations is called as time-period.
The relationship betweenfrequency and time-period is as,
f = 1/T
Where,
f- frequency,
T- Time period.
57.) Explain why, a ringing bell suspended in a vacuum chamber cannot be heard outside.
Answer: We know that medium is necessary for propagation of sound. Sound wave is transferred by molecules. But in vacuum there is absence of molecules so we can not heard sound outside when ringing bell is suspended in vacuum chamber.
58.) The frequency of the sound emitted by the loudspeaker is 1020 Hz. Calculate the wavelength of the sound wave in air in cm where its velocity is 340 m/s.
Answer: Given, Frequency =f = 1020 Hz,
Velocity = v = 340 m/s, wavelength = λ = ?
As we know,
v = f × λ
λ = v / f
λ = 340 / 1020
λ = 0.33 m
The wavelength of loudspeaker is 0.33 m.
59.) What is the difference between acompression and a rarefaction in a sound wave ? Illustrate your answer with a sketch.
Answer: Difference between compression and rarefaction.
(i) Molecules are very close to each other in compression but they are far from each other in rarefaction.
(ii) The density of molecules in compression is high but density of molecules in rarefaction is low.
(iii) Intensity of volume is lower in compression buy intensity of sound is high in rarefaction.
(iv) Pressure of molecules is high in compression but low in rarefaction.
(v) The following diagram illustrated the compression and rarefaction of sound.
Long Answer Type Questions:
60.) (a) What is sound ? What type of waves are sound waves in air ?
(b) Describe an experiment to show that sound cannot pass through vacuum.
Answer: (a) Sound is a type of energy which gives sensation to hear.
Nature of sound wave is longitudinal wave.
(b) We know that medium is necessary for propagation of sound. We understand this by a simple experiment as follows.
Apparatus: Evacuated glass chamber, ringing bell.
Procedure: Remove air in the glass chamber. Put a ringing bell into the glass chamber. Adjust the time of bell such that it must ring automatically after 1 minute. Observe the sound of bell outside the glass chamber.
Now, make a small on glass chamber and hear the sound of bell. In first case you can not hear sound but in second case you can hear the sound of bell. From above activity we conclude that,
Conclusion: Medium is necessary for propagation of sound.
61.) (a) How is sound produced ? Explain with the help of an example.
(b) How does sound from a sound producing body travel through air to reach our ears ? Illustrate your answer with the help of a labelled diagram.
Answer: (a) As we know that, vibrations are necessary for production of sound. When we hit an object on another then sound will be produced. Also friction between two objects, collision, rubbing between two objects produces sound. In each above activity sound will be produced by Vibrations. Sometime these Vibrations are visible but sometimes not.
Example: Went you hit a stone on another then sound will produce.
(b) Medium is necessary for propagation of sound. Sound is a longitudinal wave. When sound is released in air then it redistributed the density of molecules in air.By this redistribution of molecules energy is transferred from one molecule to another and reaches at destination
The propagation of sound is as
62.) (a) An electric bell is suspended by thin wires in a glass vessel and set ringing. Describe and explain what happens if the air is gradually pumped out of the glass vessel.
(b) Why cannot a sound be heard on the moon ? How do astronauts talk to one another on the surface of moon?
Answer: (a) When we pumped out air from glass vessel and a ring bell set ringing then due to absence of air or any medium we can not hear sound of ringing bell.
(b) There is no air or medium on surface of moon. So we can not heard sound on moon same as earth directly. Thus astronauts use wireless set of radio waves to communicate with each other. Radio wave does not require medium for propagation.
63.) (a) Define the terms ‘frequency’, ‘wavelength’ and ‘velocity’ of a sound wave. What is the relation between them ?
(b) A body vibrating with a time-period of 1/256 s produces a sound wave which travels in air with a velocity of 350 m/s. Calculate the wavelength.
Answer: (a) Frequency: No of oscillations per unit time is called as frequency.
Or
Number of waves completed in unit time is credited frequency of that wave.
Wavelength: The difference between two successive crest or trough is called wavelength of that wave. Also for sound wave, wavelength is a distance of one compressions and rarefactions is a wavelength.
Wavelength is denoted by a Greek letter ‘λ’ called lambda.
Velocity: The distance travelled by sound wave in one second is called as velocity of that sound wave.
The relationship between velocity, frequency and velocity is given as,
v = f × λ
Where,
v- velocity of sound wave,
f – frequency of sound wave,
λ- wavelength of sound wave.
(b) Given, Period = 1/256 seconds, velocity = v = 350 m/s, λ =?
As we know,
v = f × λ
But,
Frequency= 1/ period,
v = (l/period) × λ
λ = v × period
λ = 350×(1//256)
λ = 1.367 m
The wavelength of that wave is 1.367 m.
64.) (a) What are longitudinal waves and transverse waves ? Explain with the help of labelled diagrams.
(b) Give two examples each of longitudinal waves and transverse waves.
Answer: (a)
Longitudinal wave: When direction of displacement of particles is parallel to direction of flow of wave or energy then such wave is called as longitudinal wave. It is illustrated as,
In longitudinal wave, density of particles is as above. We draw a wave diagram of longitudinal wave.
Transverse wave: When direction of displacement of particles is perpendicular to direction of flow of wave or energy then such wave is called as longitudinal wave. The wave diagram of transverse wave is as,
(b) Examples of longitudinal wave- sound waves, seismic wave.
Examples Transverse wave- Light wave, wave produced on surface of water.
65.) (a) Explain the terms ‘compressions’ and ‘rarefactions’ of a wave. What type of waves consist of compressions and rarefactions ?
(b) A worker lives at a distance of 1.32 km from the factory. If the speed of sound in air be 330 m/s, how much time will the sound of factory siren take to reach the worker ?
Answer: (a) Compressions: When wave propagated in the medium, it redistributed the density of molecules in the medium. The density of molecules is as
The space in the medium in which density of molecules is high is called compressions.
And the space in the medium in which density of molecules is low is called rarefactions.
The wave consists of compression and rarefactions is longitudinal wave.
(b) Given, Distance = 1.32 km = 1320 m, speed = 330 m/s.,
As we know,
Speed = distance/time,
Time = distance/time
Time = 1320/330
Time = 4 seconds.
Time required to reach upto worker is 4 seconds.
66.) (a) Explain the terms ‘crests’ and ‘troughs’ of a wave ? What type of waves consist of crests and troughs ?
(b) The flash of a gun is seen by a man 3 seconds before the sound is heard. Calculate the distance of the gun from the man (Speed of sound in air is 332 m/ s).
Answer:
(a) The maximum positive displacement of the particles is called as crest. When wave travels then the particle vibrated from mean position. The displacement of particle is perpendicular to direction of flow of energy.
If the particle displaced above the mean position then such maximum positive displacement is called as crest.
Similarly the negative maximum displacement of the particle from mean position is called as troughs.
The crest and troughs are shown in figure as,
The wave consists of Crest and troughs is called as transverse wave.
(b) Given, time= 3 seconds, speed of sound in air is 332 m/ s.
As we know,
Speed = distance/time
Distance = speed ×time
Distance = 332× 3
Distance = 996 m.
The distance of gun from man is 996 m.
67.) (a) When we put our ear to a railway line, we can hear the sound of an approaching train even when the train is far off but its sound cannot be heard through the air. Why ?
(b) How could you convince a small child that when you speak, it is not necessary for air to travel from your mouth to the ear of a listener ?
Answer: (a) Medium is necessary for propagation of sound. Also speed of sound is depend on medium. If the medium is denser then speed of sound is also greater. So speed of is as
Solid > liquid > gases.
Thus speed of sound is greater in railway lines than air. Therefore When we put our ear to a railway line, we can hear the sound of an approaching train even when the train is far off but its sound cannot be heard through the air.
(b) It is not necessary to travel air from your mouth to the ear of a listener for propagation of sound. Only density of the molecules is redistributed in air. First molecule gain the energy from sound and it gives that energy to another molecule and moves back to its initial position. In such way the molecules propagate the sound from mouth of speaker to ear of listener.
Lakhmir Singh Manjit Kaur Class 9 Physics 5th Chapter Solution (Second Part)
Very Short Answer Type Questions:
1.) Which property of sound leads of the formation of echoes ?
Answer: Echoes are formed because sound also reflects from the surface like light.
2.) What name is given to the repetition of sound caused by the reflection of sound waves ?
Answer: The repetition of sound caused by the reflection of sound wave is called echo.
3.) What name is given to the persistence of sound in a big hall or auditorium ?
Answer: Persistence of sound in a big hall or auditorium is called reverberation.
4.) Name three devices which work on the reflection of sound.
Answer: The devices which work on the reflection of sound are:
I.) Megaphone.
II.) Bulb horn.
III.) Stethoscope .
IV.) Soundboard.
5.) What is the other name of a loud-hailer ?
Answer: The other name of a loud-hailer is megaphone.
6.) Name the three characteristics of sound.
Answer: Characteristics of sound:
I.) Loudness
II.) Pitch
III.) Quality or timbre.
7.) Name the unit used to measure the loudness of sound. Also write its symbol.
Answer: The loudness of sound is measured in decibel and it’s symbol is dB.
8.) Name the characteristic which helps us distinguish between a man’s voice and a woman’s voice, even without seeing them.
Answer: We can distinguish between man’s voice and woman’s voice because of pitch. Man’s voice and woman’s voice having same loudness but different pitch.
9.) How does the pitch of a sound depend on frequency ?
Answer: The pitch of sound is directly proportional to frequency of that sound.
10.) Name the characteristic of sound which depends on (a) amplitude (b) frequency, and (c) waveform.
Answer: (a) The loudness of sound is depend on amplitude of that sound.
(b) Pitch of sound is depend on frequency of that sound.
(c) The quality of sound is depend on waveform of that sound.
11.) Name the characteristic of sound which can distinguish between the ‘notes’ (musical sounds) played on a flute and a sitar (both the notes having the same pitch and loudness).
Answer: Quality of sound distinguish between notes played on flute and a sitar having same pitch and loudness.
12.) Name the organs of hearing in our body.
Answer: The ear is the organ of hearing in our body.
13.) Name that part of ear which vibrates when outside sound falls on it.
Answer: Ear-drum of ear vibrates when outside sound falls on it.
14.) Name the three tiny bones present in the middle part of ear.
Answer: Middle ear consist ofhammer, anvil and stirrup.
15.) There are three small bones in the middle ear — anvil, hammer and stirrup :
(a) Which of these bones is in touch with ear-drum ?
(b) Which of these bones is in touch with oval window ?
Answer: (a) Hammer bone is in touch with eardrum.
(b) Stirrup bone is in touch with oval window.
16.) What is the function of three tiny bones in the ear ?
Answer: Three tiny bones in the ear passes on sound vibrations from the ear drum to the liquid in cochlea.
17.) Name the tube which connects the middle ear to throat.
Answer: Eustachian tube connects the middle ear to through.
18.) Name the nerve which carries electrical impulses from the cochlea of ear to the brain.
Answer: Auditory nerve carries electrical impulse to brain.
19.) What is the name of passage in outer ear which carries sound waves to the ear-drum ?
Answer: Ear canal in outer ear which carries sound waves to ear drum.
20.) Why should we not put a pin or pencil in our ears ?
Answer: When we put a pin or pencil in our ears then it tear the ear drum therefore we do not put pin or pencil in out ears.
21.) What type of scans are used these days to monitor the growth of developing baby in the uterus of the mother ?
Answer: Doctor detect the growth of developing baby in the uterus of the mother by makingultrasound scan (sonography).
22.) How is an ultrasound scan for fetus (unborn baby) better than X-ray ?
Answer: X-rays can damage the delicate body cells of unborn baby therefore ultrasound scan for fetus better than x-ray.
23.) What is the name of the device which is used to find the depth of sea (or ocean) by using ultrasonic sound waves ?
Answer: SONAR is used to find the depth of sea.
24.) Write the full name of ‘SONAR’.
Answer: The full name of SONAR is-
SO-sound
N– Navigation
A– And
R– Ranging.
25.) Name the principle on which a soundboard works.
Answer: Soundboard works on principle of reflection of sound.
26.) Name the device which is used to address a small gathering of people.
Answer: Megaphone is used to address a small gathering of people.
27.) Name the device used by doctors to listen to our heartbeats.
Answer: Doctors uses stethoscope to listen to our heartbeats.
28.) What is the shape of a soundboard kept behind the speaker on the stage of a big hall?
Answer: Soundboard are concave in shape.
29.) Name two sound absorbing materials (or objects) which can make our big room less echoey.
Answer: Carpets and fibreboards are the examples of sound absorbing materials.
30.) Can we hear (a) infrasonic waves (b) ultrasonic waves?
Answer: (a) The frequency of sound less than 20 hertz is called infrasonic sound wave. We can not hear this type of sound.
(b) The frequency of sound is greater than twenty thousand hertz is known as ultrasonic wave and we can not hear this type of sound also.
31.) What name is given to the sound waves of frequency too low for humans to hear ?
Answer: The sound waves of frequency too low for humans to hear is known as infrasonic waves.
32.) What name is given to the sound waves of frequency too high for humans to hear ?
Answer: The sound waves of frequency too high for humans to hear is known as ultrasonic waves.
33.) What type of sound waves are produced by a vibrating simple pendulum ?
Answer : The sound wave produced by simple pendulum is Infrasonic waves.
34.) What happens to the pitch of a sound if its frequency increases ?
Answer: If the frequency of sound increases then it’s pitch also increases and the rate of vibrations also increases.
35.) What happens to the loudness of a sound if its amplitude decreases ?
Answer: If the amplitude of sound decreases then sound becomes soft or faint.
36.) What name is given to sound waves of frequencies higher than 20 kHz ?
Answer: The sound waves having frequency greater than 20 kHz is known as ultrasonic waves.
Short Answer Type Questions:
38.) On which day, a hot day or a cold day, an echo is heard sooner ? Give reason for your answer
Answer: We know that speed of sound is depend on atmospheric temperature. If the temperature increases then speed of sound also increases. Therefore the echo is heard sooner on hoy day.
39.) In which medium, air or water, an echo is heard much sooner ? Why ?
Answer: The speed of sound in air is 344 m/s therefore minimum distance for echo will be 17.2 m. The speed of sound in water is 1500 m/s thus the minimum distance for echo will be 75 m. Therefore echo is heard much sooner in water.
Doctor detect the growth of developing baby in the uterus of the mother by makingultrasound scan Answer: If the multiple reflection of sound takes place in the auditorium then such repeated reflection is called as reverberation. If the reverberation time is too long then sound becomes blurred, distorted and confusing.
41.) How can reverberations in a big hall or auditorium be reduced ?
Answer: The reverberation in a big hall be reduced by following ways-
i) Put sound absorbing material on wall, it reduces reverberation.
ii) Put carpet on floor. It can absorb sound and avoid multiple reflection.
iii) Put heavy veils on windows.
iv) We can also made seats of big hall by using sound absorbing materials to avoid reverberation.
42.) Why do we hear more clearly in a room with curtains than in a room without curtains ?
Answer: The curtains reduced reverberation in the room therefore it reduces reverberation and we can hear more clearly than in a room with curtains.
43.) What is a megaphone ? Name the principle on which a megaphone works.
Answer: The device which made up for send sound in particular direction without spreading is called megaphone.
The megaphone works on the principle of reflection of sound.
44.) What is a bulb horn ? Name the principle on which a bulb horn works.
Answer: The bulb horn is a musical instrument like shehnaiand trumpet which made up for send a sound in particular direction.
The bulb horn is based on the principle of reflection of sound.
45.) What is a stethoscope? Name the principle on which a stethoscope works.
Answer: The device used by doctors to listen the sound of heart beating is called stethoscope. It is also based on the principle of multiple reflection of sound.
46.) What is a soundboard ? Explain the working of a soundboard with the help of a labelled diagram.
Answer: We can not hear sound clearly in auditorium because of much absorption of sound by walls, ceilings, seats etc. The solution for this type of problem is soundboard. The soundboard is consist of concave type reflecting surface placed behind the speaker. Soundboard reflects the sound waves. So listeners hear sound clearly. The structure of soundboard is as,
47.) (a) What is meant by the ‘loudness’ of sound ? On what factor does the loudness of a sound depend ?
(b) Draw labelled diagrams to represent (a) soft sound, and (b) loud sound, of the same frequency.
Answer: (a) The energy reaching on ear per second is called as loudness of sound. The loudness of sound depend on the amplitude of sound. If amplitude is greater then loudness also greater.
(b) The diagram below represents soft and loud sound of same frequency.
The diagram above represents load sound.
The diagram above represents faint sound.
48.) (a) Explain the term ‘pitch’ of a sound. On what factor does the ‘pitch’ of a sound depend ?
(b) Draw labelled diagrams to represent sound of (a) low pitch, and (b) high pitch, of the same loudness
Answer: (a)The property of sound which distinguish the sound of same amplitude or loudness is called pitch. The pitch of sound is depend on frequency of that sound.
(b) The term pitch is related to frequency of sound. The following diagrams show low pitch sound and high pitch sound of same amplitude.
The above diagram represents low pitch of sound.
The above diagram shows high frequency sound.
49.) What is meant by the quality (or timbre) of sound?On what factors does the quality (or timbre ) of a sound depend?
Answer: The characteristic of sound which distinguish sound of same amplitude and frequency. We can distinguish the sound of same singer because of timbre.
Timbre or quality of sound is depend on shape of the sound waves.
50.) Explain why, if we strike a table lightly, we hear a soft sound but if we hit the table hard, a loud sound is heard.
Answer: When we strike a table lightly then the sound produced by this is of low amplitude but if we hit the table hard then sound produced is of high amplitude. We know that loudness of sound is depend on amplitude thus the sound produced by hitting the table is loud than striking a table.
51.) Give one use of ultrasound in industry and one in hospitals.
Answer:
Industrial uses of ultrasound-
I) We can detect the cracks in metal block without damaging it with the help of ultrasound.
II) Ultrasound is used for cleaning the unreachable parts of the machine.
III) Ultrasound is used to identify the defective product in industry.
Use of ultrasound in hospitals-
I) Ultrasound scan is used to monitor the growth of unborn baby in the uterus
II) Ultrasound breaks the stones in kidney.
III) Ultrasound scan is used in medical for diagnosis.
52.) How is it that bats are able to fly at night without colliding with other objects ?
Answer: Bat produces high frequency ultrasound while flying. The ultrasound hit on the surface of object which is on the path of bat and reflect back. Bat can detect the object in the path by listening echo. In this way bat can fly at night without colliding with other objects.
53.) Explain how, bats use ultrasound to catch the prey.
Answer: Bat produces high frequency ultrasound while flying. This sound hit on the surface of prey and reflect back. Thus the bat listen this echo and detect the prey on the flying path. In this way bat catch the prey.
54.) Explain how, flaws (or defects) in a metal block can be detected by using ultrasound.
Answer: Detection of cracks in metal block is a challenging task in industries. Ultrasound play a vital role in detecting cracks.
Pass the ultrasound in the metal block. Place a ultrasound detector on another side of that metal block. If the metal is defect free then total ultrasound waves transmit through the metal block as below.
If the metal block has some defect then ultrasound waves reflect back. It does not transmit in defective region. We can illustrate this by diagram below.
55.) Why are the ceilings of concert halls made curved ? Draw a labelled diagram to illustrate your answer.
Answer: As we know that sound reflects when it falls on hard surface. If we make structure of Ceilings are curved then total sound falling on ceiling get reflects and reach to listener. The structure of ceiling must be in concrete hall is,
56.) Draw a labelled diagram to show the multiple reflections of sound in a part of the stethoscope tube.
Answer: Stethoscope is a device which measure the heart beating rate is based on principle of reflection of sound waves. The multiple reflections of sound waves in stethoscope is as,
57.) What is the range of frequencies associated with (a) infrasound (b) audible sound, and (c) ultrasound ?
Answer: The range of frequencies associated with
(a) infrared is lower than 20 Hz.
(b) Audible sound is between 20 Hz to 20,000 Hz.
(c) ultrasound is greater than 20000 Hz.
58.) (a) What is the difference between infrasonic waves and ultrasonic waves ?
(b) Choose the infrasonic waves and ultrasonic waves from the following frequencies :
(i) 10,000 Hz(ii) 30,000 Hz
(iii) 18 Hz (iv) 50,000Hz
(v) 10 Hz.
Answer: (a) Difference between infrasonic and ultrasonic waves.
i) The frequency of infrasonic wave is lower than audible frequency and the frequency of ultrasonic wave is greater than audible range.
ii) The objects which moves very slowly produces infrasonic waves. The object which moves very rapidly produces ultrasonic waves.
iii) Earthquake and elephant produces infrasonic waves. Bats and dolphin produces ultrasonic waves.
(b) Frequency of infrasonic waves- (iii) 18 Hz,
(v) 10 Hz
Frequency of ultrasonic waves- (ii) 30000 Hz,
(iv) 50000 Hz
59.) (a) What is the frequency range of hearing in humans ?
(b) Which of the following sound frequencies cannot be heard by a human ear ?
(i) 10 Hz (ii) 100 Hz
(iii) 10,OOO Hz (iv) 15 Hz
(v) 40,000 Hz
Answer: (a) The frequency range of hearing in humans is between 20 Hz to 20 kHz.
(b) The sound frequency cannot be heard by human ear are-
(i) 10 Hz
(iv) 15 Hz
(v) 40,000 Hz
60.) The echo of a sound is heard after 5 seconds. If the speed of sound in air be 342 m/s, calculate the distance Of the reflecting surface.
Answer: Given, time = t = 5 seconds,
Speed of sound in air = v = 342 m/s.
We have to calculate distance of the reflecting surface = d =?
Firstly we calculate total distance travelled by sound in 5 seconds.
Speed = distance/ time
distance = speed × time
distance = 342 × 5
distance = 1710 m
This is the total distance travelled by sound in 5 seconds. But we know that the distance between reflecting surface and observer is half of the total distance.
Distance of reflecting surface = ½ ×(total distance)
Distance of reflecting surface = ½ ×(1710)
Distance of reflecting surface = 855 m.
Thus the distance of reflecting surface is 855 m.
61.) The speed of sound in water is 1500 metres per second. How far away from an under-sea rock should a deep sea diver be so that he can hear his own echo ?
Answer: Given, Speed of sound in water = 1500 m.
We know that, minimum time required for echo is 0.1 second.
Thus, total distance travelled by sound = speed × time
Distance = 1500 × 0.1
Distance = 150 m
But the distance of echo is ½ time total distance travelled by sound.
The distance of echo = ½ × (total distance)
The distance of echo = ½ × (150)
The distance of echo = 75 m.
The deepness of sea must be 75 m for echo.
Long Answer Type Questions
62.) Solution:
Answer: (a) The nature of sound is longitudinal wave. So the sound also exhibit reflection phenomenon same as light.
Reflection of sound: The bouncing of sound in a same medium is called reflection of sound. Solid and hard continuous surfaces are the best for reflection of sound.
(b) The examples of good reflectors of sound-
Cliff, hard wood, surface of metals etc.
(c) Sound also shows reflection like light. We studied the laws of reflection of light. Similarly sound holds same laws.
The laws of reflection of sound-
- Incident sound, reflected sound and normal lies on same plane.
- Angle of incident is equal to angle of reflection.
- Incident sound waves and reflected sound waves lies in same medium.
63.) Solution:
Answer: (a) As we know that, sound exhibit reflection phenomenon. When sound waves incident on a plane of surface then it reflection in same medium and we hear that same sound.
Echo: When sound waves incident on a surface and reflect back o the observer then this phenomenon is called as echo.
When a person shout in the cliff then the sound wave incident on cliff and returned to that person. The sound of shouting of that person reach after reflection and he listen the multiple sound. In this way, echo is formed.
(b) We can calculate minimum distance for echo.
As we know that speed of sound at 20⁰ is 344 m/s, and time required to produce echo is 0.1 second.
As we know that from kinematics,
Total Distance = speed × time
Total Distance = 344 × 0.1
Total Distance = 34.4 m.
But this is the total distance. This is the sum of distance going from us and return back.
Therefore minimum distance =(Total distance)/2
Minimum distance for echo = 34.4/2
Minimum distance for echo = 17.2 m.
The minimum distance for echo at 20⁰ must be 17.2 m.
(c) Given, Speed of sound = 330 m/s, distance between cliff and man is = 825 m. But to listen echo, sound must travel 2 time distance between observer and reflecting surface.
Total Distance travelled by sound = 2 × 825
Total Distance = 1650
As we know that,
Speed = Distance/time
Time = distance/speed
Time = 1650/330
Time = 5 seconds.
The time required for echo is 5 seconds.
64.) Solution:
Answer: (a) We know that, the nature of sound is longitudinal wave. The sound having frequency greater than 20000 Hz and can not heard by human is called ultrasound. The frequency of ordinary sound is between 20 Hz to 20000 Hz and the frequency of ultrasound is greater than 20000 Hz. We can heard ordinary sound but human ears does not ability to listen ultrasound. Some animals like bat produces and listen ultrasound.
(b) Applications of ultrasound:
I) Industrial applications
a) Ultrasound is used to detects the cracks in metals.
b) Ultrasound is used to clean some untouchable parts of machines.
II) Medical use:
a) Ultrasound is used to detect the internal structure of cells.
b) Ultrasound is used to diagnosis the disease.
c) Ultrasound is used to see the growth of unborn baby.
d) Ultrasound is used to measure the depth of sea.
65.) Solution:
Answer: (a) Infrasonic sound: The sound having frequency lower than 20 Hz and can not hear by human is called infrasonic sound. This type of sound is produce by slow Vibrating objects.
Whales, elephants, rhinoceros this type of animals will produces infrasonic sound.
(b) The sound having frequency greater than 20000 Hz and can not heard by human is called ultrasound. This type of sound produces by fast moving objects.
Bat, dolphins, porpoise, rats will produces ultrasonic sound waves.
(c) Given, Frequency of sound i. e.
F1 = 20 Hz, F2 =20 kHz =20000 Hz,
Speed of sound =v = 344 m/s.
As we know the relation between frequency and velocity,
Velocity = frequency × wavelength
v = F × λ.
Firstlywe calculatefrequencyfor F1 = 20 Hz,
V = F1 × λ1
λ1= V/F1
λ1= 344/20
λ1= 17.2 m
Similarly we calculate frequency for F2 = 20000 Hz.
V =F2 × λ2
λ2= V/F2
λ2= 344/20000
λ2= 0.0172 m
Thus, the wavelength range of audible sound is between 17.2 m to 0.0172 m.
66.) Solution:
Answer:(a)
Echolocation- The process of detection of objects in the way of some animals like bats through ultrasonic waves is called echolocation.
Echocardiography- The study of heart using ultrasound is called as Echocardiography.
Ultrasonography- The process of capturing image and video using ultrasound is called Ultrasonography.
67.) Solution:
Answer: (a) We know that, when sound wave incident on the surface then it reflects in same medium as reflection of light. This property of sound wave is used for detection of object or the depth of sea.
SONAR- The devices which measure depth of sea using ultrasound is called SONAR.
The Long form of SONAR is Sound Navigation And Ranging.
(b) Given, Time = 3 seconds, Speed of sound in water =1440 m/s.
Now we calculate total distance travelled by sound waves in 3 seconds.
As,
Speed = distance/time
Distance = speed × time
Distance = 1440 × 3
Distance = 4320 m.
But this is the total distance travelled by sound wave in 3 second. But the distance of reflecting surface is half of the total distance travelled by sound wave.
Distance of reflecting surface= (total distance)/2
Distance of reflecting surface = 4320/2
Distance of reflecting surface = 2160 m
Thus, the depth of sea is 2160 m
68.) Solution:
Answer: Human ear is most sensitive organ in the body. It responds to sound waves in audible range.
Construction: Human ear consist of outer ear, inner ear and middle ear. The outer ear consist of pinna. The pinna is connected to 2 to 3 centimetre longcanal. Canal is connected to ear drum. The middle ear is made up from bones named hammer, anvil and stirrup. Hammer is connected to ear drum and stirrup is connected to inner ear. The anvil is between hammer and stirrup. The middle ear has large empty part and contains air. The middle ear connect to through by a tube known as eustachian tube. The structure of inner ear is like a coil named cochlea. It is made up from liquid. The construction of ear is as,
Working: The sound waves enter in ear by pinna. It reaches at ear drum through canal. As we know that, sound waves consist of compressions and rarefaction. In compressions density of molecules is high but in rarefaction the density of molecules is low. When compressions strike on ear drum then it pushes ear drum inward. Similarly ear drum pushes outward because of rarefaction. The ear drum get continuous vibrations due to sound waves falling on it. The vibrations are passed to cochlea through hammer bone, anvil and stirrup bone.The cochlea converts vibrations into electrical signals. This electrical signals reach to brain via nerves cells. In this way human eye works.
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