Sound Question Answers: NCERT Class 9 Science

Welcome to the Chapter 12 - Sound, Class 9 Science NCERT Solutions page. Here, we provide detailed question answers for Chapter 12 - Sound. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Sound and excel in their exams. By going through these Sound question answers, you can strengthen your foundation and improve your performance in Class 9 Science. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1
A:

When a disturbance is created on an object, it starts vibrating and sets the particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. As a result, the adjacent particle is disturbed from its mean position and the original particle comes back to rest. This process continues till the disturbance reaches our ears.


Exercise 2
A:

When the school bell is struck with a hammer, it starts vibrating. This disturbance gives rise to the bell moving forward, it pushes the air in front of it. As a result of these vibrations, sound waves are produced.


A:

Waves which need a material medium for propagation are called mechanical waves. Sound waves propagate through a medium because of the interaction of the particles present in that medium. Mechanical waves are governed by Newton’s laws of motion.


A:

No we will not be able to hear the sound because sound needs a medium to propagate. On the moon is no atmosphere, you cannot hear any sound on the moon.


Exercise 3
A:

Our experts will give the answer soon.


A:

Our experts will give the answer soon.


Exercise 4

A:

The speed is defined as the distance travelled by a wave per unit time. Speed, wavelength, and frequency of a sound wave are related by the following equation:


A:

Frequency of the sound wave, = 220 Hz

Speed of the sound wave,  = 440 m s−1

For a sound wave,

Speed = Wavelength × Frequency

Hence, the wavelength of the sound wave is 2 m.


A:

The time period between successive compressions from the source is equal to the time period of the tone, i.e., (1/500). This time period is reciprocal of the frequency of the wave and is given by the relation:


Exercise 5
A:

Intensity of a sound wave is not a physical quantity which can be accurately measured. It does not depend upon the sensitivity of the ear. Loudness is not an entirely physical quantity. The loudness of a sound is defined by its amplitude. The amplitude of a sound decides its intensity, which in turn is perceived by the ear as loudness.


Exercise 6
A:

The speed of sound depends on the nature of the medium. The speed of sound is fastest in case of solids. Its speed decreases in liquids and it is the slowest in gases.

Therefore, in iron, the speed of sound is the fastest at a given temperature.


Exercise 7
A:

Speed of sound, v = 342 m/s

Time taken by sound to travel from the source to reflecting surface, t = 3/2 = 1.5 s

Distance travelled by sound to reflecting surface from the source = v × t = 342 × 1.5  = 513 m


Exercise 8
A:

Concert halls are very big, so the sound might not reach every corner of the hall. Ceilings of concert halls are curved so that sound after reflection spreads uniformly in all parts of the hall.


Exercise 9
A:

In human beings, the audible range of an average human ear lies between 20 Hz to 20,000 Hz. Humans cannot hear sounds having frequency less than 20 Hz and greater than 20,000 Hz.


A:

(a) Infrasound has frequencies less than 20 Hz.

(b) Ultrasound has frequencies more than 20,000 Hz.


Exercise 10
A:

Time taken by the sonar pulse to return, t = 1.02 s

Speed of sound in salt water, v = 1531 m/s

Total distance covered by the sonar pulse = Speed of sound × Time taken

Total distance covered by the sonar pulse = 1.02 x 1531 = 1561.62 m                 ....(i)

Let d be the distance of the cliff from the submarine.

Total distance covered by the sonar pulse = 2d

⇒ 2d = 1561.62                          [From (i)]

⇒ d = 780.81 m

⇒ 2d = 1561.62

d = 780.81 m


Exercise 11
A:

Sound is defined as vibration that travels through the air or another medium as an audible mechanical wave. This creates a disturbance in the medium. It is produced from a vibrating body. This disturbance, when it reaches the ear, produces sound.


A:

Yes, sound follows the same laws of reflection of light. The incident sound wave, reflected sound wave and normal sound wave all lie on the same plane. Also, the angle of incidence of sound is equal to angle of reflection of sound.


A:

As the temp. increases, the speed of sound also increases. Therefore, the speed of sound on a hotter day is more. An echo is heard when the time interval between the original sound and the reflected sound is at least 0.1 s.


A:

(i) Sound navigation and ranging which is used to find the depth of the ocean. This method is known as SONAR.

(ii) Stethoscope is used by doctors to measure heartbeat. Dr. is able to hear the sound due to multiple reflection of sound in the stethoscope.


A:

Height of the tower, s = 500 m

Velocity of sound, v = 340 m/s

Acceleration due to gravity, g = 10 m/s2

Initial velocity of the stone, u = 0 

Time taken by the stone to fall to the base of the tower, t1

According to the second equation of motion:


Now, time taken by the sound to reach the top from the base of the tower,

Hence , the splash is heard at the top after time, t

Where, t = t1 + t2 = 10 + 1.47 = 11.47 s


A:

Speed of sound, = 339 m/s

Wavelength of sound, = 1.5 cm = 1.5 / 100 = 0.015 m

Speed of sound = Wavelength × Frequency

Frequency of of sound is 22600 Hz.

The audible range of the human ear is 20 Hz to 20000 Hz. It would not be audible.


A:

Persistence of sound (after the source stops producing sound) due to repeated reflection is known as reverberation. Reverberation can be reduced by absorbing the sound using some materials as it reaches the wall and ceiling of the room and thus prevents the sound from getting reflected. This reflected sound reaches the other wall and again gets reflected partly. Due to this, sound can be heard even after the source has ceased to produce sound.

Some materials which are used to reduce reverberation are fibreboard, heavy curtains, plastics etc.


A:

Loudness is defined as a measure of the response of the ear to the sound. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.


A:

Sound frequencies greater than 20,000 Hz are called ultrasounds. Bats produce high-pitched ultrasonic squeaks. These squeaks reflect on prey and return back to the bats ear. This allows a bat to know the distance of his prey.


A:

The objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency ultrasonic waves are capable of removing the dirt from the objects very easily.


A:

SONAR is an acronym for Sound Navigation And Ranging .

SONAR is a device that uses ultrasonic waves to measure the distance , direction and speed of underwater objects . It is also used to measure the depth of seas and oceans.

A wave of ultrasonic sound is produced and transmitted by the transducer of the SONAR, which travels through sea water and strikes the object on the seabed, gets reflected back and is sensed by the detector. The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance (d) of the under-water object is calculated from the time (t) taken by the echo to return with speed (v) is given by 2d = v × t. This method of measuring distance is also known as ‘echo-ranging’.

A:

Compressions and rarefactions are produced because the disturbance in the medium is caused by sound waves. When an object vibrates then it moves forward, it pushes and compresses the air in front of it creating a region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction.


A:

Time taken to hear the echo, t = 5 s

Distance of the object from the submarine, d = 3625 m

Total distance travelled by the sonar waves during the transmission and reception in water = 2d

Velocity of sound in water,


A:

Ultrasounds can be used to detect cracks and flaws in metal blocks. Defects in metal blocks occur due to crack in them. Cracks have air in them. The speed of sound in metal is greater than that in air. Thus, air acts as a rarer medium for sound waves. Whenever ultrasound is transmitted through a metal block that has defects, these waves get reflected in encountering such cracks. Therefore, these waves are not detected at the detector placed on the other side of the block, opposite to that of the transmitter. This indicates that there is a defect in the block.


A:

Different sounds produced in our surroundings are collected by pinna. The collected sound passes through the auditory canal. At the end of the auditory canal, there is a thin membrane called the eardrum or tympanic membrane. When compression the medium reaches the eardrum the pressure on the outside of the membrane increases and force the eardrum inward. Similarly, the eardrum moves outward when the rarefaction reaches it. The eardrum starts vibrating back and forth rapidly when the sound waves fall on it. The vibrating eardrum sets the small bone hammer into vibration. The vibrations are passed from the hammer to the second bone anvil, and finally to the third bone stirrup. The vibrating stirrup strikes on the membrane of the oval window and passes its vibration to the liquid in the cochlea. This produces electrical impulses in nerve cells. The auditory nerve carries these electrical impulses to the brain. These electrical impulses are interpreted by the brain as sound and we get a sensation of hearing.


A:

Take an electric bell and connect it to electrical supply. An airtight glass bell jar with a vacuum pump.

When we press the switch, we will be able to hear the bell. One can hear the sound of the ringing bell. Now, take a jar and make a small hole in the bottom and insert a small pipe though it. Connect the other end of the pipe to a vacuum cleaner. When the air in the jar is pumped out gradually, the sound becomes feeble although the same amount of current is flowing through the bell. It will be observed that the sound of the ringing bell decreases. When there is no air present inside, we will not be able to hear the sound of the bell. This shows that sound requires a medium for its propagation.


A:

In longitudinal waves, the motion of the individual particles of the medium is in a direction that is parallel to the direction of energy transport. A longitudinal wave can be created in a slinky if the slinky is stretched out in a horizontal direction and the first coils of the slinky are vibrated horizontally. This is known as longitudinal wave.


A:

Quality of sound is that characteristic which helps us to distinguish one sound from another. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.


A:

The speed of light is greater than the speed of sound. Sound of thunder takes more time to reach the earth as compared to light. Hence, a flash is seen before we hear a thunder.

The speed of light is (3x108 m/s) and the speed of sound is 330 m/s.

A:

Hearing range for humans = 20 Hz to 20 kHz

Speed of sound = 344 m/s

For a sound wave,

Speed = Wavelength × Frequency

i. For = 20 Hz
 

ii. For  = 20,000 Hz
Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

A:

Speed of sound in air = 346 m/s

Speed of sound in Aluminium = 6420 m/s

Let the length of the aluminium rod = d.

Therefore, time taken by the sound wave to reach the other end,

Therefore, time taken by sound wave to reach the other end,

The ratio of time taken by the sound wave in air and aluminium:

A:

The no. of oscillations produced in one second is called frequency .

Frequency of sound = 100 Hz

Total time = 1 min = 60 s

We know that,

Number of oscillations = Frequency × Total time

Number of oscillations/Vibrations = 100 × 60 = 6000

Therefore , the source vibrates 6000 times in a minute, producing a frequency of 100 Hz.


Exercise 0
A:

From the given graph,
Time period of the disturbance,
T = 2 µ s = 2 x 10-6 sec
Velocity of disturbance, v = 1500 m/s
Wavelength of the disturbance,
λ = vT = 1500 x 2 x 10-6 = 3 x 10-3 m


A:

Graph (a) represents the male voice because usually the male voice has less pitch as compared to female voice.


A:

If the time gap between the original sound and reflected sound received by the listener is around 0.1 s, only then the echo can be heard.
Speed of sound in air, v=344 m/s
The minimum distance travelled by the reflected sound wave for the distinctly listening the echo = velocity of the sound x time interval 
= 344 x 0.1 => 34.4 m
But in this situation, the distance travelled by the sound reflected from the building and reaching to the girl will be (6 + 6 = 12 m). Therefore no echo will be heard by the girl.


A:

Because the frequencies of vibrations of a pendulum is below 20 Hz and not in the audible range. But the sound of humming bees in the audible range (20 - 20000 Hz).


A:

If an explosion takes place at the bottom of the lack. The longitudinal wave type of shock waves will take place.


A:

Speed of sound = 340 m/s
Time, t = 10 sec
Distance = Speed x time = 340 x 10 = 3400 m => 3400 / 1000 = 3.4 km


A:

The angle of incidence is always equal to the angle of reflection.
Now, angle made by incident sound with normal
= 90° – 50° = 40°

∴ x = 40°


A:

Ceiling and walls are made curved so that sound after reflection reaches the target audience.


A:

(i) Two sound waves having the same amplitude but different frequencies.

(ii) Two sound waves having the same frequency but different amplitudes.

(iii) Two sound waves having different amplitudes and also different wavelengths.


A:

The distance travelled by sound wave per unit time is known as the speed of time.
Distance travelled by sound wave in periodic time (T) = wavelength (λ) of the sound wave.
Velocity of sound in air, v = 340 m/s

(i) Frequency, u = 256 Hz
    ∴ Wavelength, λ = v / T
    λ = 340 / 256 = 1.33 m

Therefore , the wavelength of the wave is 1.33 m .

(ii) Wavelength, λ = 0.85 m
     ∴ Frequency v =  λ x T
     T =  v / λ = 340 / 0.85 = 400 Hz.


A:

The points of maximum density and minimum density are also called crests and troughs respectively.

Wavelength (A): It is defined as the distance between two successive crests or troughs of a wave is called wavelength. It is measured in the direction of the wave. Time period is the time taken for one complete cycle.


Frequently Asked Questions about Sound - Class 9 Science

    • 1. How many questions are covered in Sound solutions?
    • All questions from Sound are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
    • 2. Are the solutions for Sound helpful for exam preparation?
    • Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
    • 3. Can I find solutions to all exercises in Sound?
    • Yes, we provide solutions to all exercises, examples, and additional questions from Sound with detailed explanations.
    • 4. How do these solutions help in understanding Sound concepts?
    • Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
    • 5. Are there any tips for studying Sound effectively?
    • Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

Exam Preparation Tips for Sound

The Sound is an important chapter of 9 Science. This chapter’s important topics like Sound are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

Latest Blog Posts

Stay updated with our latest educational content and study tips

The Secret to Smarter Learning — Building Strong Critical Thinking Skills

In today’s world of endless information , knowing how to think is more important than knowing what to think . From school projects to real – life decisions , critical thinking helps students question ideas , analyze facts and form logical conclusions . But what exactly does critical thinking mean ? Simply put , it’s […]

Read More

Study Smarter, Not Harder: Build Productive Habits That Stick

Every student dreams of better grades , stronger focus and more study time – but the real challenge isn’t starting, it’s staying consistent . Building productive study habits is not about studying all day , it’s about studying smart . In today’s fast – paced digital world, distractions are everywhere – from endless phone notifications […]

Read More

The Hidden Risks of Online Gaming for Children — Is your child safe while gaming online?

Online gaming has rapidly become one of the most popular pastimes among children. Whether it’s multiplayer mobile games , PC adventures or console challenges , kids are spending more time than ever in the virtual world . On the surface, gaming seems entertaining and even educational – improving hand- eye coordination , teamwork and problem […]

Read More

The Role of Parents in Digital Literacy – Guiding Kids for a Smarter Online Future

Kids today are surrounded by screens from the moment they wake up . Whether it’s smart classrooms, online lessons or video games with friends technology has quietly become a part of everything they do. It’s amazing how much they can learn, explore and create with just a tap or a click. But it also brings […]

Read More

Benefits of Using Our NCERT Solutions for Class

When it comes to excelling in your studies, having a well-structured study guide can make a huge difference. Our NCERT Solutions for Class provide you with a comprehensive, easy-to-understand, and exam-focused resource that is specifically tailored to help you maximize your potential. Here are some of the key benefits of using our NCERT solutions for effective learning and high scores:

NCERT Solutions for Effective Exam Preparation

Preparing for exams requires more than just reading through textbooks. It demands a structured approach to understanding concepts, solving problems, and revising thoroughly. Here’s how our NCERT solutions can enhance your exam preparation:

  • Clear Understanding of Concepts: Our NCERT solutions are designed to break down complex topics into simple, understandable language, making it easier for students to grasp essential concepts in . This helps in building a solid foundation for each chapter, which is crucial for scoring high marks.
  • Step-by-Step Solutions: Each solution is presented in a detailed, step-by-step manner. This approach not only helps you understand how to reach the answer but also equips you with the right techniques to tackle similar questions in exams.
  • Access to Important Questions: We provide a curated list of important questions and commonly asked questions in exams. By practicing these questions, you can familiarize yourself with the types of problems that are likely to appear in the exams and gain confidence in answering them.
  • Quick Revision Tool: Our NCERT solutions serve as an excellent tool for last-minute revision. The solutions cover all key points, definitions, and explanations, ensuring that you have everything you need to quickly review before exams.

Importance of Structured Answers for Scoring Higher Marks

In exams, it's not just about getting the right answer—it's also about presenting it in a well-structured and logical way. Our NCERT solutions for Class are designed to guide you on how to write answers that are organized and effective for scoring high marks.

  • Precise and Concise Answers: Our solutions are crafted to provide answers that are to the point, without unnecessary elaboration. This ensures that you don't waste time during exams and focus on delivering accurate answers that examiners appreciate.
  • Step-Wise Marks Distribution: We understand that exams often allot marks based on specific steps or points. Our NCERT solutions break down each answer into structured steps to ensure you cover all essential points required for full marks.
  • Improved Presentation Skills: By following the format of our NCERT solutions, you learn how to present your answers in a systematic and logical manner. This helps in making your answers easy to read and allows the examiner to quickly identify key points, resulting in better scores.
  • Alignment with NCERT Guidelines: Since exams are often set in alignment with NCERT guidelines, our solutions are tailored to follow the exact format and language that is expected in exams. This can improve your chances of scoring higher by meeting the examiner's expectations.