MSBSHSE Class 9 Science Chapter 12 Study of Sound Solutions

MSBSHSE Class 9 Science Chapter 12 Study of Sound Solutions helps students to study well for the exams. These solutions, with detailed answers, also have proper step by step explanations. This helps the students to understand the fundamentals of the concepts discussed in this chapter.

Study of Sound is an important topic of Science and it is discussed at length in this Chapter. The solutions given here include questions from the chapter. Main concepts from the chapter include Sound waves, Reflection of sound, The human ear, audible sound, infrasound and ultrasound and so on. Students can master the subject thoroughly by revising these MSBSHSE Class 9 Solutions of Science Chapter 12 Study of Sound. These solutions are developed after research and will help students to gain a thorough conceptual understanding. The well- structured, clear content will also make it easier for the students to learn the subject. These solutions explain the topics from the chapter of the textbook, as per the updated MSBSHSE Syllabus for Class 9. Solving these solutions and answering the questions listed here are designed to help students ace the exams.

Maharashtra Board Class 9 Science Chapter 12- BYJU’S Important Questions & Answers

1.The roof of a movie theatre and a conference hall is curved. Give a scientific reason.

Answer: The roof of the movie theatre as well as that of the conference hall are intentionally made curved, so that the sound produced can be reflected from the roof parts and also from the walls and can be reached to all parts of the theatre or the conference hall. This works on the principle of the reflection of sound. These sound waves that get reflected from the walls and roof of a room multiple times cause a single sound to be heard not once but continuously. This is called reverberation.

2. Give reason why the intensity of reverberation is higher in a closed and empty house.

Answer: In an empty house, there are no sound absorbers like the furniture present and since it is closed, the sound cannot escape the room. Hence, the sound is reflected multiple times from the walls of the closed room without any of these reflections being absorbed. For this reason, the intensity of the reflections are not likely to reduce as it would have if there were sound absorbers in the room. Thus, the intensity of reverberation is higher in a closed and empty house.

3. We cannot hear the echo produced in a classroom. State the reason.

Answer: If it were to be possible for us to hear a distinct echo, then the sound should take more than 0.1 s after starting from the source to get reflected and come back to us. However, the classrooms are designed in a way that the distance between the walls is less than 17.2m. So, the reflections of the sound from and to walls or echoes reach our ears before 0.1s. Owing to this, it becomes difficult to distinguish the original sound from the echo formed in the room.

4. What is an echo? What factors are important to get a distinct echo?

Answer: When you shout loudly, you might hear back the same sound after a little while, this is known as an echo. Hence, an echo is the repetition of the original sound due to reflection by some surface. Meanwhile, for us to be able to hear a distinct echo, the sound should take more than 0.1 s after starting from the source to get reflected and come back to us. Also, to be able to hear a distinct echo, the reflecting surface should be at a minimum distance of half of the above i.e. 17.2 m. Since the velocity of sound depends on the temperature of air, this distance is dependent on the temperature. Meanwhile, the size of the reflector should be large enough in comparison to the wavelength of the sound wave.

5. Study the construction of the Golghumat at Vijapur and discuss the reasons for the multiple echoes produced there.

Answer: Golghumat at Vijapur is an example of multiple echoes being produced there as a result of continuous and multiple reflections.

6. What should be the dimensions and the shape of classrooms so that no echo can be produced there?

Answer: For there to be no echo in the classroom, it should be square shaped, with a distance of less than 17.2m between the walls. Meanwhile, if the ceiling is also shaped like a curve, then the reflection of the sound wave is uniform throughout the room.

7. Where and why are sound absorbing materials used?

Answer: In places such as the auditorium, concert halls, theatres and so on, excessive reverberation is not desirable. To prevent this, the upholotories of the room, its walls and roof etc, and made of sound absorbing materials. Because of this, sounds that are reflected from the rigid surfaces are absorbed by the materials, the reverberations, thus subdued.

8. The speed of sound in air at 00 C is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?

Answer: Here, T0 C is denoted as the temperature at which the velocity of air becomes 344m/s,

And increase in velocity of sound in air from 00 C to T0 C= 344 m/s – 332 m/s= 12 m/s

Meanwhile, the velocity of sound in air increases by 0.6 m/s for 10 C rise in temperature

Hence, for 1m/s increase in velocity of sound in air, the temperature should increase by (1/ 0.6)0 C

Thus, if 12 m/s increase in velocity of sound in air, then the temperature will rise by (12 / 0.6)0 C

That is 200 C.

So, the temperature at which the velocity of air becomes 344m/s = T0 C = 200 C – 00 C = 200 C.

9. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her? (The velocity of sound in air is 340 m/s)

Answer: The distance (d) between the lightning and nita is given as X

Meanwhile, speed of sound in air (v) = 340 m//s

Time taken for the sound of lightning to reach Nita (t) = 4 seconds

Here, v = X / t

That is 340 = X/ 4

Hence, X = 340 × 4 = 1360 m.

10. Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.

1. What is the velocity of sound in air?

2. What is the distance between the two walls?

Answer: Given here is time to hear first echo (t) = 4 sec and if

Distance travelled by the sound in this 4 s (d) = 360 × 2 m

Then,velocity of the sound = d / t = (360 × 2) / 4 = 180 m/s

Therefore, velocity of sound in air is 180 m/s

Now, Imagine time taken to hear the second echo = 4 s + 2 s = 6 s

Also, the distance travelled in this 6 s is given as 2x.

Meanwhile, the velocity of sound is 180 m/s

Hence, 2x = 180 × 6

Thus, x = 540 m

Therefore, the distance between the two walls = 540 + 360 = 900m.

11. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How many times as fast as the other?

Answer: Given here is Mass of Hydrogen in Bottle A, W1= 12gm and

Mass of Hydrogen in Bottle B, W2= 48gm

Here, the density ∝ mass ( volume is taken as same for the identical bottes)

Now, p1 / p2 = W1 / W2 and for the

Velocity = v ∝ 1/√p ∝ 1 / √W (T is given as constant)

Therefore, v1 / v2 = √(p1/p2) =√(W1/W2)

Replacing with values, v1 / v2 = √(48/12) = 2/1 =2

Hence, v1=2v2

So, sound will travel faster in bottle A and the speed will be twice as that of bottle B.

12. What are longitudinal waves?

Answer: Sound, a form of energy is a form of waves that creates the sensation of hearing in our ears. Now, a medium is necessary for the propagation of the sound waves. These sound waves produce a chain of compression (place of higher density) and rarefaction, (that is place of lower density) in the medium.The particles of the medium oscillate about their central or mean positions, in a direction parallel to the propagation of the wave known as longitudinal waves.

13. Explain transverse waves.

Answer: Transverse waves are formed by dropping a stone in still water, which causes the particles of water to oscillate up and down. These oscillations are perpendicular to the direction of propagation of the wave.

14. How are the frequency and the wavelength of the sound waves indicated?

Answer: Greek letter lambda (λ) is used to indicate the wavelength of sound waves, while its frequency is indicated by nu (ν).

15. What is amplitude?

Answer: Amplitude is the maximum value of pressure or density.

16. What is the relation between velocity of sound wave and distance?

Answer: The distance covered by a point on the wave (for example the point of highest density or lowest density) in unit time is the velocity of the sound wave.Therefore, the velocity= density / time. At the same time, given that the sound wave covers a distance equal to wavelength( λ) in time period (T), then velocity of the sound wave = wavelength/ time period. Hence, v = λ / T.

17. Who first calculated the speed of sound? How did they do it?

Answer: Borelli and Viviani, the two Italian physicists first calculated the speed of sound in the 1660s. They calculated the time between seeing the flash of a gun and hearing its sound at a long distance. The value they arrive at 350 m/s is near to the value of the currently accepted value of 346 m/s.

18. The velocity of sound in a gaseous medium depends on these physical conditions. Which are they and how are they dependent?

Answer: The velocity of sound in a gaseous medium is dependent upon the physical conditions that is the temperature, density of the gas and its molecular weight.

Temperature(T) = Velocity of the sound is directly proportional to the square root of the temperature of the medium. Hence, on increasing the temperature four times the velocity becomes double.

Density (p) = The velocity of sound is inversely proportional to the square root of density. So, on increasing the density four times, the velocity can be reduced to half its value.

Molecular weight (M) = The velocity sound is inversely proportional to the square root of molecular weight of the gas. Hence, on increasing the molecular weight four times, the velocity is reduced to half its value.

19. What are audible sounds?

Answer: Audible sounds are the sounds of frequencies in the range, 20 Hz to 20,000 Hz, which the human ear can hear.

20. What are infra sound and ultrasound?

Answer: Sounds of frequencies lower than 20 Hz and higher than 20,000 Hz (20 kHz) cannot be heard by the human ear. The sound with frequency smaller than 20 Hz is called infra sound. Meanwhile, ultrasound are sound waves with frequency greater than 20 kHz.

21. Who has the special ability to hear ultrasonic sounds?

Answer: Some of the animals such as the dog, mouse, bat, dolphin and so on have the special ability to hear ultrasonic sounds. So, they are able to sense some noise that is inaudible to human ears. Meanwhile, children under 5 years of age and some creatures and insects can hear waves with frequency up to 25 kHz. Bats, mice, dolphins, etc,

can also produce ultrasound.

22. What are the uses of ultrasonic sounds?

Answer: Given below are some of the main uses of ultrasonic sounds:

1. To communicate between ships at sea.

2. Used to join plastic surfaces together.

3. To sterilize liquids such as milk by killing the bacteria in it so that it will not spoil for a longer period

4. Ultrasonic waves or the Sonography technology is used in Echocardiography that studies heartbeats

5. Used to obtain images of internal organs in a human body

6. In industry, to clean intricate parts of machines where hands cannot reach

7. Used to identify the cracks and faults in metals blocks

23. Which are the good and bad reflectors of sound?

Answer: A hard and flat surface is a good reflector of sounds, whereas items such as clothes, paper, curtains, carpet, furniture and so on absorb sound instead of reflecting it. Hence, these items are known as bad reflectors.

24. What is called reverberation?

Answer: Sound waves that get reflected from the walls and roof of a room multiple times, cause a single sound to be heard not once but continuously. This is known as reverberation.

25. What is a SONAR and what is it used for?

Answer: The short form for Sound Navigation and Ranging is called SONAR. It is used to locate the direction, distance and speed of an underwater object with the help of ultrasonic sound waves. The SONAR technology first developed during World War I was used to detect enemy submarines. This technology was used in the air also. The bats use this technique to detect obstacles in their path so that they can avoid them and fly freely even in the dark. SONAR, equipped with a transmitter and a receiver that are fitted on ships or boats is mainly used to identify the depth of the sea. It also helps to search underwater hills, valleys, submarines, icebergs, sunken ships and more.

26. What is Sonography?

Answer: Sonography technology makes use of ultrasonic sound waves to generate images of internal organs of the human body. This technology helps to determine the reason of swelling, infection, pain and so on. This technique is also used to study the condition of the heart, the state of the heart after a heart attack as well as the growth of foetus inside the womb of a pregnant woman.

27. Illustrate the structure of the human ear and label the parts.

Answer:

MSBSHSE Class 9 Science Chapter 12 Question 27 Solution

28. What is Outer ear or Pinna? What is its function?

Answer: A peculiar funnel-like shaped outer ear collects the sound waves and passes them via a tube to a cavity in the middle ear. It helps to collect and pass sounds into the middle ear.

29. What is the eardrum?

Answer: The eardrum is a thin membrane in the cavity of the middle ear.

30. What is cochlea?

Answer: A structure of the inner ear that resembles the shell of a snail is the cochlea. It receives the vibrations transmitted from the membrane and converts them into electrical signals that are sent to the brain via the nerve.

Students preparing for the exams, will find these chapter wise MSBSHSE Class 9 Solutions very useful. They can also access other resources such as textbooks and question papers to revise the chapter and practice well.

 

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