CBSE Class 9

CBSE Class 9

Value Based Questions in Science for Class 9 Chapter 12 Sound

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Value Based Questions in Science for Class 9 Chapter 12 Sound

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 12 Sound

VALUE BASED QUESTIONS

Question 1.
Mr. Ravi Bushan, father of Mr. Atul was hard of hearing. Doctor advised him to use hearing aid after diagnosing his ears. But he was not ready to listen to the advice of doctor because he felt that hearing aid is a machine and would cause harm to him. Mr. Atul told his father that hearing aid is not harmful, rather it would help him. Finally, Mr. Ravi decided to have the hearing aid.

  1. What values are shown by Mr. Atul ?
  2. On what principle, hearing aid works ?

Answer:

    1. Concern for his father and
    2. high degree of general awareness.
  2. Hearing aid works on the principle of multiple relfection of sound.

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Question 2.
Ajay’s uncle was advised by his doctor to have echocardiography. His uncle did not know anything regarding the echocardiography. He thought that this test is sensitive and hence he was not ready for it. When Ajay came to know about this, he decided to prepare his uncle for the test. He told his uncle that this test would help the doctor to know the condition of his heart and moreover, this test is very simple. Finally his uncle was convinced and had the echocardiography. The information obtained by the test helped his doctor to treat him well.

  1. What is echocardiography ?
  2. What values are shown by Ajay ?

Answer:

  1. Echocardiography is medical diagnostic technique to construct the image of heart using ultrasonice waves.
    1. Helpful,
    2. concern for his uncle, and
    3. high degree of general awareness.

Question 3.
Harsha was watching a programme based on ships on television. She saw a device attached to a ship through which the man on the ship located the enemy submarines and sent the message to the headquarters.

  1. Name the device fitted in the ship.
  2. On which principle does the device work ?
  3. What values are shown by Harsha ? (CBSE 2015)

Answer:

  1. SONAR
  2. SONAR works on the principle of reflection of sound waves (i.e. echo).
  3. Harsha is inquisitive. She has scientific temperament and takes interest in understanding scientific phenomena.

Question 4.
David while watching ‘National Geographic’ channel on television observed that Bats were easily flying during the night. He did not understand the concept and for this he surfed on internet, and finally found the answer that bats use ultrasound to fly and search their prey at night.

  1. What is ultrasound ? State its one application.
  2. State the principle used by bats.
  3. What value of David’s Nature is depicted from this context ? (CBSE 2015)

Answer:

  1. The sound waves having frequency greater than 20000 Hz are called ultrasound. Ultrasound is used to
    determine the depth of a sea.
  2. Bats use the principle of reflection of ultrasound (i.e. echo).
  3. David is inquisitive. He has scientific temperament and takes interest in understanding the natural phenomena.

Hope given Value Based Questions in Science for Class 9 Chapter 12 Sound are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

HOTS Questions for Class 9 Science Chapter 12 Sound

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HOTS Questions for Class 9 Science Chapter 12 Sound

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 12 Sound

Question 1.
Sound of explosions taking place on other planets are not heard by a person on the earth. Explain, why ?
(CBSE 2011)
Answer:
Sound needs material medium for its propagation from one place to another place. In other words, sound cannot travel through vacuum. Since there is a region in between the planets and the earth, where there is a vacuum, so the sound of explosions taking place on other planets cannot pass through this vaccum. Hence cannot reach the earth.

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Question 2.
Two astronauts on the surface of the moon cannot talk to each other. Explain, why ?
Answer:
Since there is no atmosphere on the surface of the moon (i.e. no medium for the propagation of sound), so the sound cannot travel from one astronaut to another astronaut on the surface of the moon.

Question 3.
A loud sound can be heard at a large distance but a feeble or soft sound cannot be heard at a large distance. Explain, why ?
Answer:
Sound is a form of energy which is transferred from one place to another place. As sound energy is directly proportional to the square of the amplitude of a vibrating body, so loud sound has large energy, whereas soft sound has small energy. As the sound travels through a medium, sound with small energy is absorbed after travelling a small distance in the medium but sound with large energy will be absorbed after travelling a large distance in the medium. Therefore, loud sound can be heard at a large distance but feeble sound cannot be heard at a large distance.

Question 4.
A sound wave travelling in a medium is represented as shown in figure,

  1. Which letter represents the amplitude of the sound wave ?
  2. Which letter represents the wavelength of the wave ?
  3. What is the frequency of the source of sound if the vibrating source of sound makes 360 oscillations in 2 minutes ?
    HOTS Questions for Class 9 Science Chapter 12 Sound image - 1

Answer:

  1. Letter X represents the amplitude of the sound wave.
  2. Letter Y represents the wavelength of the sound wave.
  3. Number of oscillations made in 2 minutes (120 s) = 360
    HOTS Questions for Class 9 Science Chapter 12 Sound image - 2
    Hence, frequency of the source of sound = 3 Hz

Question 5.
Represent the following sound waves,
(i) Waves having same amplitude but different frequencies
(ii) Waves having same frequency but different amplitudes
(iii) Waves having different amplitudes and different wave lengths.
(CBSE 2011, 2012)
Answer:
HOTS Questions for Class 9 Science Chapter 12 Sound image - 3

Question 6.
An echo is heard on a day when temperature is about 22° C. Will the echo be heard sooner or later if the temperature falls to 4°C ? (Similar CBSE 2014)
Distance
Answer:
HOTS Questions for Class 9 Science Chapter 12 Sound image - 4
Since, speed of sound in air decreases with the decrease in temperature, so the time after which the echo will be heard increases. Hence, the echo will be heard later than the echo heard when temperature was 22° C.

Question 7.
An echo is heard on a day when temperature is about 22° C. Will the echo be heard sooner or later if the temperature increases to 40° C ? (Similar CBSE 2014)
Answer:
HOTS Questions for Class 9 Science Chapter 12 Sound image - 5
Since, speed of sound in air increases with the increase in temperature, so the time after which the echo will be heard decreases. Hence, echo will be heard sooner than the echo heard when temperature was 22° C.

Question 8.
When we put our ear on a railway track, we can hear the sound of an approaching train even when the train is not visible but its sound cannot be heard through air. Why ? (CBSE 2015, 2016)
Answer:
Sound travels faster in solids than in gases. Therefore, we can hear the sound of an approaching train by putting our ear on a railway track even when the train is not visible.

Hope given HOTS Questions for Class 9 Science Chapter 12 Sound are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

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RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS

Other Exercises

Mark correct alternative in each of the following:
Question 1.
In a cylinder, if radius is doubled and height is halved, curved surface area will be
(a) halved
(b) doubled
(c) same
(d) four times
Solution:
Let radius of the first cylinder (r1) = r
and height (h1) = h
Surface area = 2πrh
If radius is doubled and height is halved
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q1.1
∴ Their surface area remain same (c)

Question 2.
Two cylindrical jars have their diameters in the ratio 3:1, but height 1:3. Then the ratio of their volumes is
(a) 1 : 4
(b) 1 : 3
(c) 3 : 1
(d) 2 : 5
Solution:
Sol. Ratio in the diameters of two cylinder = 3:1
and ratio in their heights = 1:3
Let radius of the first cylinder (r1) = 3x
and radius of second (r2) = x
and height of the first (h1) = y
and height of the second (h2) = 3y
Now volume of the first cylinder = πr2h
= π(3x)2 x y = 9πx2y
and of second cylinder = π(x2) (3y)
∴ Ratio between then = 9πx2y : 3πx2y
= 3 : 1 (c)

Question 3.
The number of surfaces in right cylinder is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
The number of surfaces of a right cylinder is three. (c)

Question 4.
Vertical cross-section of a right circular cylinder is always a
(a) square
(b) rectangle
(c) rhombus
(d) trapezium
Solution:
The vertical cross-section of a right circular cylinder is always a rectangle. (b)

Question 5.
If r is the radius and h is height of the cylinder the volume will be
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q5.1
Solution:
Volume of a cylinder = πr2h (b)

Question 6.
The number of surfaces of a hollow cylindrical object is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
The number of surfaces of a hollow cylindrical object is 4. (d)

Question 7.
If the radius of a cylinder is doubled and the height remains same, the volume will be
(a) doubled
(b) halved
(c) same
(d) four times
Solution:
If r be the radius and h be the height, then volume = πr2h
If radius is doubled and height remain same,
the volume will be
= π(2r)2h = π x 4r2h
= 4πr2h = 4 x Volume
The volume is four times (d)

Question 8.
If the height of a cylinder is doubled and radius remains the same, then volume will be
(a) doubled
(b) halved
(c) same
(d) four times
Solution:
If r be the radius and h be the height, then volume of a cylinder = πr2h
If height is doubled and radius remain same, then volume = πr2(2h) = 2πr2h
∴ Its doubled (a)

Question 9.
In a cylinder, if radius is halved and height is doubled, the volume will be
(a) same
(b) doubled
(c) halved
(d) four times
Solution:
Let r be radius and h be height, then Volume = πr2h
If radius is halved and height is doubled
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q9.1

Question 10.
If the diameter of the base of a closed right circular cylinder be equal to its height h, then its whole surface area is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q10.1
Solution:
Let diameter of the base of a cylinder (r) = h
Then its height (h) = h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q10.2

Question 11.
A right circular cylindrical tunnel of diameter 2 m and length 40 m is to be constructed from a sheet of iron. The area of the iron sheet required in m2, is
(a) 40π
(b) 80π
(c) 160π
(d) 200π
Solution:
Diameter of a cylindrical tunnel = 2 m
∴ Radius (r) = \(\frac { 2 }{ 2 }\) = 1m
and length (h) = 40 m
Curved surfae area = 2πrh = 2 x π x 1 x 40 = 80π (b)

Question 12.
Two circular cylinders of equal volume have their heights in the ratio 1 : 2. Ratio of their radii is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q12.1
Solution:
Let r1 and h1 be the radius and height of the
first cylinder, then
Volume = πr12h1
Similarly r1 and h2 are the radius and height of the second cylinder
∴ Volume = πr2h2
But their volumes are equal,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q12.2

Question 13.
The radius of a wire is decreased to one- third. If volume remains the same, the length will become
(a) 3 times
(b) 6 times
(c) 9 times
(d) 27 times
Solution:
In the first case, r and h1, be the radius and height of the cylindrical wire
∴ Volume = πr2h1 …(i)
In second case, radius is decreased to one third
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q13.1
∴ In second case height is 9 times (c)

Question 14.
If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q14.1
Solution:
Let r be the radius and h be the height then volume = πr2h
If height is doubled and volume is same and let x be radius then πr2h = π(x)2 x 2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q14.2

Question 15.
The volume of a cylinder of radius r is 1/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q15.1
Solution:
Let r be the radius and h be the height, then volume = πr2h
This volume is \(\frac { 1 }{ 4 }\) of the volume of a rectangular box
∴ Volume of box = 4πr2h
Let side of base of box = x and height h,
then volume = x2h
∴ 4πr2h = x2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q15.2

Question 16.
The height ft of a cylinder equals the circumference of the cylinder. In terms of ft, what is the volume of the cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q16.1
Solution:
In a cylinder,
h = circumference of the cylinder
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q16.2

Question 17.
A cylinder with radius r and height ft is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?
(a) 2πr(r + h)
(b) πr(r + 2h)
(c) πr(2r + h)
(d) 2πr2 + h
Solution:
r is the radius of the base and ft is the height of a closed cylinder
Then total surface area = 2πr(r + h ) (a)

Question 18.
The height of sand in a cylindrical-shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q18.1
Solution:
Let h be the height and d be the diameter of a cylinder, then
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q18.2

Question 19.
Two steel sheets each of length a1 and breadth a2 are used to prepare the surfaces of two right circular cylinders – one having volume v1 and height a2 and other having volume v2 and height a1. Then,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.1
Solution:
Length of each sheet = a1
and breadth = a2
Volume of cylinder = πr2h
In first case,
v1 is volume and a2 is the height
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.2
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q19.3

Question 20.
The altitude of a circualr cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q20.1
Solution:
In first case,
Let r be the radius and h be the height of the cylinder. Then,
∴ Lateral surface area = 2πrh
In second case,
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS Q20.2

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Solutions for Class 9 Science Chapter 12 Sound

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NCERT Solutions for Class 9 Science Chapter 12 Sound

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 12 Sound. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Physics) Chapter 12 – Sound solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 12 – Sound Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

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NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1.
Explain how sound is produced by your school bell.

                                                            Or

Sound is produced when your school bell is struck with a hammer. Why ? (CBSE 2011)
Answer:
When school bell is struck by a hammer, it starts vibrating. Since the vibrating bodies produce sound, so the vibrating school bell produces the sound.

Question 2.
Why are sound waves called mechanical waves ? (CBSE 2012)
Answer:
Sound waves are characterised by the motion of particles of a medium. Hence sound waves are called mechanical waves.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend ?
Answer:
Sound waves need material medium like air to move from one place to another place. Since there is no air on the moon, so sound cannot travel from one place to another place. Hence, we cannot hear sound on the moon.

Question 4.
Which wave property determines
(a) loudness,
(b) pitch ? (CBSE 2011, 2012)
Answer:
(a) Amplitude of the wave determines loudness.
(b) Frequency of the wave determines pitch.

Question 5.
Guess which sound has a higher pitch : guitar or car horn ?
Answer:
Guitar, because frequency of sound produced by guitar is higher than the sound produced by car horn.

Question 6.
What are wavelength, frequency, time period and amplitude of a sound wave ? (CBSE 2012)
Answer:
Wavelength (or length of a wave): The distance between two successive regions of high pressure or high density {or compressions) or the distance between two successive regions of low pressure or low density (or rarefactions) is known as wavelength of a sound wave. It is denoted by λ (read as lambda).
In S.I., unit of wavelength is metre (m).
Frequency: The number of compressions or rarefactions crossing a point per unit time is known as the frequency of a sound wave. It is denoted by μ (read as Neu). In S.I., unit of frequency is hertz (Hz).
1 hertz = one oscillation completed by a vibrating body or a vibrating particle in one second.
Time period: Time taken by two consecutive compressions or rarefactions to cross a fixed point. Amplitude. The maximum displacement of a vibrating body from its rest position or mean position.

Question 7.
How are the wavelength and frequency of a sound wave related to its speed ? (CBSE 2011, 2012)
Answer:
V = vλ.

Question 8.
Calculate the wavelength of a sound wave whose frequency is 200 Hz and speed is 440 m/s in a given medium.
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 1

Question 9.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of sound. What is the time interval between successive compressions from the source ?
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 2

Question 10.
Distinguish between loudness and intensity of sound.
Answer:

LoudnessIntensity of a sound
1.Loudness is a subjective quantity. It depends upon the sensitivity of the human ear. A sound may be loud for a person but the same sound may be feeble for another person who is hard of hearing even when both are sitting at the same distance from the source of sound.

 

Intensity of a sound is an objective physical quantity. It does not depend on the sensitivity of a human ear.
 2. Loudness cannot be measured as a physical quantity because it is just sensation which can be felt only.

 

Intensity of a sound can be measured as a physical quantity.

Question 11.
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature ?
Answer:
Sound travels the fastest in iron.

Question 12.
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 m s-1 ? (CBSE 2011)
Answer:
Time taken by sound to travel from the source to the reflecting surface, t = 3/2 = 1.5 s
Speed, v = 342 m s
Distance of reflecting surface from the source, S = vt = 342 x 1.5 = 513 m.

Question 13.
Why are the ceilings of concert halls curved ? (CBSE 2011, 2012)
Answer:
So that the sound after reflection from the ceiling reaches all the corners of the hall.

Question 14.
What is the audible range of the average human ear ? (CBSE, 2011, 2012, 2015)
Answer:
20 Hz to 20,000 Hz.

Question 15.
What is the range of frequencies associated with :
(a) infra sound ?
(b) ultra sound ? (CBSE 2011, 2013)
Answer:
(a) Frequencies less than 20 Hz and greater than zero.
(b) Frequencies greater than 20,000 Hz and equal to 107 Hz.

Question 16.
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 ?
Answer:
Time taken by the pulse to go from submarine to the cliff, t = 1.02/2 =0.51 s
Speed of sound, v = 1531 m/s
Distance of cliff from the submarine, S = vt = 1531 x 0.51 = 780.81 m.

NCERT CHAPTER END EXERCISE

Question 1.
What is sound and how is it produced ? (CBSE2012)
Answer:
Sound is a form of energy which produces the sensation of hearing in our ears. Sound is produced by forcing an object to vibrate. In other words, sound is produced by a vibrating object.

Question 2.
Why is sound wave called a longitudinal wave ? (CBSE 2012)
Answer:
When sound waves travel in medium, the particles of the medium vibrate about their equilibrium positions along the direction of the propagation of the waves.

Question 3.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room ? (CBSE 2011)
Answer:
Timber or quality of sound.

Question 4.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why ?

          Or

There is some time interval between observing a flash and hearing a thunder. Explain. (CBSE 2013)
Answer:
Thunder is heared after some time interval the flash is seen because speed of sound is less than the speed of light.

Question 5.
A person has a hearing range of 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies ?
Answer:
Take the speed of sound.in air as 344 m s-1.
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 3

Question 6.
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 waves in air and in the aluminium to reach the second child. (Speed of sound in aluminium is 6420 m s-1 and in air is 346 m s-1). (CBSE 2013)
Answer:
Let l = length of the rod
Time taken by sound to travel distance / in aluminium rod,
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 4
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 5

Question 7.
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute ?
(CBSE 2011, 2012)
Answer:
Frequency of source = 100 Hz.
∴ Number of times the source of sound vibrates in 1 s = 100
∴ Number of times the source vibrates in a minute or 60 s = 100 x 60 = 6000.

Question 8.
Does sound follow the same laws of reflection as light does ? Explain.
Answer:
Yes. Sound waves are reflected just like light waves.

Question 9.
When a sound is reflected from a distant object, an echo is produced. Let the distance of the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day ?
Answer:
Let d = distance between the reflecting surface and the source of sound
v = speed of sound in air.
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 6
On a hotter day, speed of sound increases with increase in temperature. Hence, the time after which echo is heard decreases. If the time taken by the reflected sound is less than 0-1 s after the production of original sound, then echo is not heard. However, if this time is greater than 0-1 s, then echo will be heard.

Question 10.
Give two practical applications of multiple reflection of sound waves. (CBSE 2011, 2012)
Answer:

  1. Megaphone
  2. Hearing aid.

Question 11.
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 .
(CBSE 2011, 2012)
Answer:
Time, after which splash is heard = time taken by the stone to reach the surface of water in a pond + time taken by the sound of splash to reach the top of tower.
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 7
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 8

Question 12.
A sound wave travels at a speed of 399 m s-1. If its wavelength is 1.5 m, what is the frequency of the wave ? Will it be audible ?
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 9
Since audible range of frequencies is 20 Hz to 20,000 Hz. Hence, the given frequency will not be audible.

Question 13.
What is reverberation ? How can it be reduced ? (CBSE 2012, 2014)
Answer:
The phenomenon of prolongation of original sound due to the multiple reflection of sound waves even after the source of sound stops producing sound is called reverberation.
Reverberation can be reduced by covering the roof and walls of a hall by sound absorbing materials.

Question 14.
What is loudness of sound ? What factors does it depend on ? (CBSE 2011)
Answer:
Loudness of a sound is a subjective quantity which causes unpleasant effect in our ear.
Loudness depends upon the amplitude of the vibrating body and the sensitivity of human ear.

Question 15.
Explain how bats use ultrasound to catch a prey. (CBSE 2011, 2012)

                                                  Or

Bats have no eyes, yet they can ascertain distances. (CBSE 2013)
Answer:
Bats can produce ultrasonic waves by flapping their wings. They can also detect these waves. The ultrasonic waves produced by the bats after reflection from the obstacles like buildings guide them to remain away from the obstacles during their flights. Hence, they can fly during night without hitting the obstacles. Bats also catch their prey during night with the help of ultrasonic waves. The ultrasonic waves produced by a bat spread out. These waves after reflecting from a prey say an insect reach the bat. Hence, the bat can easily locate its prey (Figure 24).
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 10

Question 16.
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. (CBSE 2011)
Answer:
NCERT Solutions for Class 9 Science Chapter 12 Sound image - 11

NCERT Solutions for Class 9 Science Chapter 12 Sound

Hope given NCERT Solutions for Class 9 Science Chapter 12 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

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RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS

Other Exercises

Question 1.
Write the number of surface of a right circular cylinder.
Solution:
Three, two circular and one curved.

Question 2.
Write the ratio of total surface area to the curved surface area of a cylinder of radius r and height h.
Solution:
∵ Radius = r
and height = h
∴ Curved surface area = 2πrh
and total surface area = 2πr(h + r)
∴ Ratio = 2πr(h + r) : 2πrh
= h + r : h

Question 3.
The ratio between the radius of the base and height of a cylinder is 2 : 3. If its volume is 1617 cm3, find the total surface area of the cylinder.
Solution:
Ratio in radius and height of the cylinder = 2 : 3
Let radius (r) = 2x
Then height (h) = 3x
∴ Volume = πr2h
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Q3.1
RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS Q3.2

Question 4.
If the radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3, then find the ratio of their volumes.
Solution:
Ratio of radii of two cylinder = 2:3
Let radius of first cylinder (r1) = 2x
and second cylinder (r2) = 3x
and ratio in their heights = 5:3
Let height of first cylinder (h1) = 5y
and height of second (h2) = 3y
∴ Volume of the first cylinder =πr2h
= π x (2x)2 x 5y = 20πx2y
and volume of second cylinder = π(3x)2 x 3y = 27πx2y
Now ratio between then,
= 20πx2y: 21πx2y
= 20 : 27

Hope given RD Sharma Class 9 Solutions Chapter 19 Surface Areas and Volume of a Circular Cylinder VSAQS are helpful to complete your math homework.

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