NCERT Solutions for Class 9 Science Chapter 11 - Sound
You may have noticed the fact that when a ringing phone reaches your ear when you are in another room, but that you hear nothing when the phone is put into a vacuum jar. This is because sound requires a medium to travel, be it air, water, and solids and cannot travel in a vacuum. This concept is discussed in a simplified form in the NCERT Solutions of Class 9 Science Chapter 11- Sound. This aspect can help students to know how vibrations produce sound and how sound waves are generated and reach our ears.
This Story also Contains
- NCERT Solutions Sound Class 9: Exercise Questions
- Sound Class 9 NCERT Solutions: Additional Questions
- Class 9 Science NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions
- NCERT Solutions for Class 9 Physics Chapter 11: Topics
- NCERT Solutions for Class 9 Science Chapter 11 Sound - Important Formulas
- Approach to Solve Questions of Class 9 Chapter 11: Sound
- Benefits of NCERT Solutions Class 9 Chapter 11
- NCERT Solutions for Class 9 Science Chapter-wise
The NCERT solutions for Class 9 Science Chapter 11 give elaborate step-wise explanations involving basics of sound, such as the nature of sound waves, its characteristic features, such as frequency, amplitude, pitch and loudness, and even the fact that sound cannot travel in empty space. In addition to the textbook solutions to exercises, there are additional and Higher Order Thinking Skills (HOTS) problems, carefully composed by experts in the subject, that can improve understanding levels. With these NCERT solutions, not only do these students revisit the essential formulas of sound, but they also get to know how to approach numerical and theoretical questions correctly when they come up in the exams. Important points, applications of sound in life and communication, and technology, and on how to study and prepare to appear in CBSE Class 9 exams are also mentioned in the article.
Also Read,
NCERT Solutions Sound Class 9: Exercise Questions
The NCERT Solutions for Class 9 Science Chapter 11 Sound: Exercise Questions provide clear, step-by-step answers to all the textbook problems. These solutions help students grasp key topics like sound waves, their properties, and applications, making exam preparation easier and more effective.
Q.1. What is sound and how is it produced?
Answer:
A sound is a form of energy that is produced by the vibrations of objects.
It is produced by the to and fro motion of the object. The vibrations are created which causes a disturbance in the adjacent particles of the medium. The disturbance travels in the waveform and creates sound.
Answer:
The compression and rarefactions produced in air near the source of sound (Here the prong) are shown in the diagram below:
Here C is for Compression and R is for Rarefaction.
Answer:
longitudinal waves are formed When oscillation is created parallel to the disturbance of the particles of the medium in the direction of propagation and since the sound waves also create oscillations in the particles of the medium parallel to the disturbance in the direction of propagation.
Therefore, sound waves are called a longitudinal wave .
Answer:
The quality of timber of the sound is that characteristic which enables us to distinguish one sound from another having the same pitch and loudness.
Thus, because of the difference in timbre and pitch of the sound wave, we can identify the voices of friends.
Answer:
This happens because of the difference in the velocities of light and sound waves. Light travels much faster than the sound. That is the reason why we hear the thunder a few seconds after the flash of thunder is seen instead of both are produced simultaneously.
Answer:
Taking the speed of the sound in air as $344\ m/s.$
Then,
we know: $Speed = Wavelength\times Frequency$
Or, $v = \lambda \times \nu$
Hence,
For $\nu =20\ Hz$ ,
$\lambda_{1} = \frac{v}{\nu_{1}} = \frac{344}{20} = 17.2\ m$
and for $\nu =20,000\ Hz$
$\lambda_{2} = \frac{v}{\nu_{2}} = \frac{344}{20,000} = 0.0172\ m$
Hence, the hearing wavelength range for humans is $0.0172m\ to\ 17.2m$ .
(Given: Speed of sound in air is $346\ m/s$ ; Speed of sound in aluminium is $6240\ m/s$ )
Answer:
Here, let us assume the length of the rod be $'l'$ .
Then,
Time taken by the sound wave in air:
$\left ( Given: v_{air} = 346\ m/s \right )$
$Time = \frac{Distance}{Speed}$
$t_{air} = \frac{l}{346} seconds$
Time taken by the sound wave in Aluminium to reach from one end to the other end,
$\left ( Given: v_{aluminium} = 6420\ m/s \right )$
$t_{aluminium} = \frac{l}{6420} seconds$
Therefore, the ratio of the time taken by the sound wave in air and in aluminium will be:
$\frac{t_{air}}{t_{aluminium}} = \frac{\frac{l}{346}}{\frac{l}{6420}} = \frac{6420}{346} = \frac{18.55}{1}$
Hence, the ratio is $18.55:1$ .
Q8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer:
Given the frequency of the sound wave is 100Hz.
Frequency is the number of oscillations per second.
So, 100Hz means that 100 oscillations are done in 1 second.
Thus, in a minute, the number of oscillations would be,
$= 100\times60 = 6000$
Q.9. Does sound follow the same laws of reflection as light does? Explain.
Answer:
Yes, the sound wave also follows the same laws of reflection as light does. Which are as follows:
(i) The angle of incidence of the sound wave and the angle of reflection of the sound wave to the normal is equal.
(ii) The incident sound wave, the reflected sound wave and the normal at the point of incidence, all lie in the same plane.
Q10. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
Answer:
Listening to the sound of echo depends upon the time interval between the source of a sound and the reflecting sound which should be at least 0.1 seconds.
The speed of the sound increases with an increase in the temperature of the medium. Hence, on a hotter day, the time interval between the source sound and the reflected sound will decrease.
Therefore, we cannot listen to the echo unless the interval is greater than 0.1 seconds.
Q11. Give two practical applications of reflection of sound waves.
Answer:
Two practical applications of sound waves are :
(i) In the stethoscope, the sound of the patient's heartbeat reaches the doctor's ears by multiple reflections throughout the pipe of the stethoscope of sound.
(ii) Horns, megaphones or loudhailers are designed to send sound in a particular direction without spreading it in all directions. There is a conical opening which reflects the sound waves and guides most of the sound waves from the source.
Q12. 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 = $340\: m\: s^{-1}$ .
Answer:
Given the height of the tower, $s =500\ m$ ,
and the velocity of the sound, $v = 340\ m/s$ ,
Acceleration due to gravity, $g = 10\ ms^{-2}$
Let the initial velocity of the stone, $u = 0$ , as the stone is initially at rest.
Let the time taken by the stone to fall to the base of the tower be $t_{1}$
Now, according to the IInd Equation of Motion:
$s = ut_{1}+\frac{1}{2}gt_{1}^2$
$500 = 0\times t_{1}+\frac{1}{2}\times10\times t_{1}^2$
$t_{1}^2 = 100$
$t_{1} = 10 sec$ .
Now, the time taken by the sound to reach the top from the base of the tower,
$t_{2} = \frac{500}{340}sec =1.47sec$
Therefore, the splash is heard at the top after time, $t$
Calculating $t = t_{1}+t_{2} = 10+1.47 = 11.47sec.$
Answer:
Given,
The speed with which sound travels is $339\: m\: s^{-1}$ .
and the wavelength is 1.5 cm.
Then we know the relation,
$v= \lambda \times \nu$
Where, $v$ is velocity, $\lambda$ is the wavelength, and $\nu$ is the frequency of the wave.
Hence,
$\nu = \frac{v}{\lambda} = \frac{339\ m/s}{0.015m} = 22600Hz.$
Calculated frequency is out of audible range of human ears which is $20Hz\ to \ 20,000Hz.$
Therefore, it is inaudible.
Q.14. What is reverberation? How can it be reduced?
Answer:
Reverberation is the repeated reflection after the source stops producing a sound which is also known as persistence of sound. When the wave reaches the wall of a room, it is partly reflected back from the wall. This reflected sound then 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.
To reduce the reverberation, sound must have to be absorbed as it reaches the walls and the ceilings of a room. Materials like wood, fibreboard, rough plastic, heavy curtains, and some seats can be used to reduce the reverberations.
Q15. What is the loudness of sound? What factors does it depend on?
Answer:
Loudness is a physiological response of the ear to the intensity of sound which enables us to distinguish between a soft sound and a loud sound.
The loudness or softness of a sound is determined basically by its amplitude and loudness is proportional to the square of the amplitude of the vibrations.
Where greater the amplitude of vibrations, the louder the sound is produced.
Q16. How is ultrasound used for cleaning?
Answer:
The object to be cleaned is first put in the cleaning solution and then the ultrasonic waves are passed through that solution. The high-frequency ultrasonic waves are capable to remove the dirt from the objects very easily.
Q17. Explain how defects in a metal block can be detected using ultrasound.
Answer:
Ultrasounds can be used to detect cracks and flaws in metal blocks. Metallic components are generally used in the construction of big structures like buildings, bridges, machines and also scientific equipment. The cracks or holes inside the metal blocks, which are invisible from outside reduces the strength of the structure. Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back, indicating the presence of the flaw or defect, as shown
Sound Class 9 NCERT Solutions: Additional Questions
The Sound Class 9 NCERT Solutions: Additional Questions provide extra problems outside the textbook to reinforce the knowledge of such concepts as vibration, frequencies, pitch, and the speed of sound. These organized meaningful questions will enable students to fix their problem-solving ability and go in for exams confidently.
Q1. Explain the working and application of a sonar.
Answer:
Working of SONAR:
Sonar consists of a transmitter and a detector and is installed in a boat or a ship, as shown in the Figure below:
The transmitter produces and transmits ultrasonic waves. These waves travel through water and, after striking the object on the seabed, are reflected back and are sensed by the detector. The detector converts the ultrasonic waves into electrical signals, which are appropriately interpreted. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between the transmission and reception of the ultrasound.
Applications of SONAR:
The sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarines, icebergs, sunken ships, etc.
Answer:
Given the distance of the object is 3625 m from the submarine and the time taken is 5 seconds.
But the time gave includes of both journey downward and upward,
hence the time taken will be half of this time.
$t = \frac{5}{2}sec$
And as $Speed = \frac{Distance}{Time}$
Putting the values in the equation, we get
$Speed = \frac{3625}{\frac{5}{2} } = 1450\ m/s$
Therefore, the speed of sound is 1450 m/s.
Q3. Explain how bats use ultrasound to catch prey.
Answer:
Bats emits very high frequencies sounds from their mouth which when touches the prey, and gets reflected back to the bat. Then the bat detects these waves and estimates the distance and the direction of the prey.
Q4. Explain how the human ear works.
Answer:
The outer ear is called the ‘pinna’. It collects the sound from the surroundings. 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 of the medium reaches the eardrum the pressure on the outside of the membrane increases and forces the eardrum inward. Similarly, the eardrum moves outward when a rarefaction reaches it. In this way, the eardrum vibrates. The vibrations are amplified several times by three bones (the hammer, anvil, and stirrup) in the middle ear. The middle ear transmits the amplified pressure variations received from the sound wave to the inner ear. In the inner ear, the pressure variations are turned into electrical signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.
Class 9 Science NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions
The Class 9 Science NCERT Chapter 11: Higher Order Thinking Skills (HOTS) Questions are designed to challenge students with advanced problems related to sound waves, vibrations, and their applications. These questions go beyond basic concepts, encouraging logical reasoning, deeper analysis, and exam-oriented preparation.
Q.1 On average, a human heart is found to beat 85 times in a minute, calculate its time period.
Answer:
The beat frequency of the heart $85 / 1 \mathrm{~min}$
$
\begin{aligned}
& =85 / 60 \\
& =1.42 \mathrm{~Hz}
\end{aligned}
$
The period
$
\begin{aligned}
T & =\frac{1}{f} \\
T & =\frac{1}{1.42} \\
T & =\frac{100}{142} \\
T & =0.7043 \mathrm{sec}
\end{aligned}
$
Q.2 The velocity of sound in gas at 51°C is 340m/s. What will be the velocity of sound if pressure is doubled and temperature becomes 127°?
Answer:
Initial temperature:
$
T_1=51^{\circ} \mathrm{C}=(273+51) K=324 K
$
Final temperature:
$
T_2=127^{\circ} \mathrm{C}=(273+127) K=400 K
$
We know that $c \propto \sqrt{T}$
$
\begin{aligned}
& \frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}} \\
& \Longrightarrow c_2=c_1 \sqrt{\frac{T_1}{T_2}}=340 \sqrt{\frac{400}{324}}=340 * \frac{20}{18}=378 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
Q.3 If the temperature of the atmosphere is increased, the following characteristics of the sound wave is affected
(a) Amplitude
(b) Frequency
(c) Velocity
(d) Wavelength
Answer:
$
\begin{aligned}
& \because \frac{P}{\rho}=\frac{R T}{M} \\
& \therefore V=\sqrt{\frac{\gamma R T}{M}}
\end{aligned}
$
wherein
$
V \propto \sqrt{T}
$
$T$ is in Kelvin
So, velocity increases with an increase in temperature.
Hence, the answer is the option (3).
NCERT Solutions for Class 9 Physics Chapter 11: Topics
The NCERT Solutions of Class 9 Physics Chapter 11 Sound includes all the relevant topics connected with production, transmission, and characteristics of sound waves. These solutions feature step-by-step explanation of every concept providing students with opportunity to further consolidate their foundations and adequately prepare to write exams.
11.1 Production of Sound
11.2 Propagation of Sound
11.2.1 Sound Waves are Longitudinal Waves
11.2.2 Characteristics of a Sound Wave
11.2.3 Speed of Sound in Different Media
11.3 Reflection of Sound
11.3.1 Echo
11.3.2 Reverberation
11.3.3 Uses of Multiple Reflections of Sound
11.4 Range of Hearing
11.5 Applications of Ultrasound
NCERT Solutions for Class 9 Science Chapter 11 Sound - Important Formulas
NCERT Solutions Class 9 Science Chapter 11 Sound Important Formulas will cover all the important formulas, which should be known by students when solving numerical problems and answering questions on different topics such as the speed of waves, frequencies, wavelengths of waves and echoes. These formulas make learning easy and can be used as a writing revision guide.
1. Speed, Frequency, and Wavelength of Sound
$
v=f \times \lambda
$
$v=$ velocity of sound (m/s)
$f=$ frequency (Hz)
$\lambda=$ wavelength (m)
2. Relation between Time Period and Frequency
$
f=\frac{1}{T}
$
$f=$ frequency (Hz)
$T=$ time period (s)
3. Echo Formula
$
d=\frac{v \times t}{2}
$
$d=$ distance of obstacle (m)
$v=$ velocity of sound in air ( $\approx 343 \mathrm{~m} / \mathrm{s}$ at $25^{\circ} \mathrm{C}$ )
$t=$ time taken for echo (s)
Approach to Solve Questions of Class 9 Chapter 11: Sound
Analyze the question in order to find out what quantities (time, distance, frequency, or wavelength) are involved. Identify the nature of the question - is it conceptual (such as why sound cannot travel in a vacuum) or is it a numerical calculation (such as speed, echo, or frequency) based?. Know the question, are they asking, what is the nature of sound waves, the properties of vibration, echo, SONAR etc. Apply the corresponding concepts and formulas like $v=f \times \lambda$ for wave speed, or $d=\frac{v \times t}{2}$ for echo problems. Wherever necessary, use clear and labeled diagrams or wave depictions (such as a compression and rarefaction diagram) as they have a greater impact upon an answer. List all of the given data step-by-step before starting to solve it. ALWAYS convert quantities into proper SI units. Do not neglect the units and directions of the travel of a wave when you examine displacement, velocity, or reflection of sound. Lastly, organize the solution in a logical manner by describing it through logical reasoning and divisive steps.
Benefits of NCERT Solutions Class 9 Chapter 11
- Good Concept Clarity Level - Specifically, assists students in appreciating how sound is produced, travels and heard using a series of easy solutions.
- Exam Preparation- Prepares students with all NCERT exercise questions and additional practice getting them ready to take competitive board exams offered by CBSE.
- Application-Based Learning- Describes the practical applications of sound waves, echo, SONAR so that concepts become easier to relate.
- Formula Support - All the valuable formulas are given topic-wise, and this helps a lot in solving numerical problems within no time.
- Time-Saving - Properly organized solutions save the time spent on revision as direct answers and key points are provided in one place.
- Increases Confidence- HOTS and other questions enhance critical thinking and help students be prepared to take competitive tests like NTSE and Olympiads.
- Step-by-Step Problem Solving - Teaches students to answer numerical and theoretical questions systematically.
NCERT Solutions for Class 9 Science Chapter-wise
NCERT Solutions for Class 9 Science Chapter-wise links provide easy access to all chapters in one place, making learning simple and organized. These solutions cover every exercise question with step-by-step answers, helping students prepare effectively for exams and build strong concepts in Physics, Chemistry, and Biology.
NCERT Solutions for Class 9 - Subject Wise
Also Check NCERT Books and NCERT Syllabus here:
Frequently Asked Questions (FAQs)
Q: How can I download the Class 9 Science Chapter 11 Sound Exercise Question Answer PDF?
A:
The Class 9 Science Chapter 11 Sound Exercise Question Answer PDF is available on Careers360 and includes step-by-step solutions to all textbook questions, making exam revision easy and quick.
Q: Is there a PDF available for Class 9 Science Chapter 11 Exercise Question Answer?
A:
Yes, the Class 9 Science Chapter 12 Exercise Question Answer PDF provides complete solutions to the chapter's problems, designed as per the latest NCERT curriculum.
Q: What does the Sound Chapter Class 9 PDF include?
A:
The Sound Chapter Class 9 PDF includes important definitions, diagrams, solved questions, and summaries, making it a helpful tool for revision and understanding the chapter.
Q: What are the features of NCERT Solutions for Class 9 Science Chapter 11 Sound?
A:
Class 9 NCERT science chapter has 17 exercise questions, solutions of these questions are explained in the NCERT solutions of Sound Class 9 chapter. All the questions are according to the CBSE NCERT syllabus of Class 9.
Q: Where can I find more questions on Class 9 NCERT chapter Sound?
A:
More questions on NCERT Class 9 chapter Sound can be solved from NCERT Exemplar questions for Class 9
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