# Sound : NCERT Exercise Questions

Q.1 ¬† ¬† What is sound and how is it produced?
Sol.¬† ¬† ¬† Sound is a form of energy which gives the sensation of hearing. It is produced by the vibrations caused in air by vibrating objects.

Q.2 ¬† ¬† Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound?
Sol. ¬† ¬† ¬†When a body vibrates the air in its neighborhood is alternately compressed and rarefied. The compressed air has higher pressure than surrounding air. It therefore pushes the air particles near it causing compression to move forward. A rarefaction or low pressure is created at the original place.¬†These compressions and rarefaction causes particles in the air to vibrate about their mean position. The energy is carried forward in these vibration. This is how sound travels. ¬†

Q.3 ¬† ¬† Cite an experiment to show that sound needs a material medium for its propagation?
Sol. ¬† ¬† ¬†Sound waves are called mechanical waves as they need a material medium to travel.
Take an electric bell and an air tight glass bell jar connected to a vacuum pump. Suspend the bell inside the jar, and press the switch of the bell. You will be able to hear the bell ring.¬†Now pump out the air from the glass jar. The sound of the bell will become fainter and after some time, the sound will not be heard. This is so because almost all air has been pumped out.This shows that sound needs a material medium to travel.

Q.4 ¬† ¬† Why is sound wave called a longitudinal wave?
Sol. ¬† ¬† ¬†Sound wave is called longitudinal wave because it is produced by compressions and rarefactions in the air. The air particles vibrates parallel to the direction of propagation.

¬†Q.5 ¬† ¬† Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Sol. ¬† ¬† ¬†The quality or timber of sound enables us to identify our friend by his voice.

Q.6 ¬† ¬† Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Sol. ¬† ¬† ¬†Thunder is heard after the flash of light is seen because sound and light travel at different speed. Light travels at the speed of about $3 \times {10^8}\,m/s$ whereas sound travels at about 300 m/s.

Q.7 ¬† ¬† A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m/s.
Sol. ¬† ¬† ¬†${\lambda _1} = {V \over \nu } = {{344} \over {20}} = 17.2\,m$
¬† ¬† ¬† ¬† ¬† ¬† ${\lambda _2} = {V \over \nu } = {{344} \over {20,000}} = .0172\,m$

Q.8 ¬† ¬† 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 wave in air and in aluminium to reach the second child.
Sol. ¬† ¬† Velocity of sound wave in air = 346 m/s
¬† ¬† ¬† ¬† ¬† ¬†Velocity of sound wave in aluminium = 6420 m/s
¬† ¬† ¬† ¬† ¬† ¬†Let length of rod be l.
¬† ¬† ¬† ¬† ¬† ¬†Time taken for sound wave in air ${t_1} = {{l} \over {velocity\,in\,air}}$
¬† ¬† ¬† ¬† ¬† ¬†Time taken for sound wave in $A\ell \,\,{t_2} = {{l} \over {velocity\,in\,Aluminium}}$
¬† ¬† ¬† ¬† ¬† ¬†Therefore ${{{t_1}} \over {{t_2}}} = {{velocity\,in\,A\ell } \over {velocity\,in\,air}} ={{6420} \over {346}} = 18.55\,:1$

¬†Q.9 ¬† ¬† The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Sol.¬† ¬† ¬† $\nu = 100\,{Hz} = 100\,vib/\sec$
¬† ¬† ¬† ¬† ¬† ¬† Therefore vibrations in 1 min = 100 √ó 60 = 6000 vibrations.

Q.10 ¬† ¬† Does sound follow the same laws of reflection as light does? Explain.
Sol. ¬† ¬† ¬† ¬†Sound follows the same laws of refection as light. Sound is reflected by hard surfaces and it can be shown that $\angle of\,incidence\,\, = \angle \,of\,reflection$

¬†Q.11 ¬† ¬†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?
Sol.¬† ¬† ¬† ¬† An echo is heard when the time for the reflected sound is heard after 0.1 s.
¬† ¬† ¬† ¬† ¬† ¬† ¬† $Time\,taken\, = {{Total\,distance} \over {velocity}}$.
¬† ¬† ¬† ¬† ¬† ¬† ¬† On a hotter day the velocity of sound is more. If the time taken by echo is less than 0.1 sec it will not be heard.

Q.12 ¬† ¬† Give two practical applications of reflection of sound waves ?
Sol. ¬† ¬† ¬† ¬†Two practical applications of reflection of sound waves are -
¬† ¬† ¬† ¬† ¬† ¬† ¬† (a) Megaphones or loud hailers are designed to send sound in a particular direction.
¬† ¬† ¬† ¬† ¬† ¬† ¬† (b) Stethoscope are based on the principal of multiple reflection of sound within the stethoscope tube enabling the doctor to hear a patient's heartbeat.

Q.13 ¬† ¬† 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 =¬†$10\,m/{s^2}$ square and speed of sound = 340 m/s .
Sol. ¬† ¬† ¬† ¬†Height of tower = 500 m
¬† ¬† ¬† ¬† ¬† ¬† ¬† Acceleration due to gravity = $10\,m/{s^2}$
¬† ¬† ¬† ¬† ¬† ¬† ¬† Speed of sound = 340 m/s
¬† ¬† ¬† ¬† ¬† ¬† ¬† Time for the stone to reach the water surface.
¬† ¬† ¬† ¬† ¬† ¬† ¬† ${t_1} = \sqrt {{{2h} \over g}} = \sqrt {{{2 \times 500} \over {10}}} = 10\,s$ $\left[ {h = ut\, + {1 \over 2}g{t^2}\,where\,u = 0} \right]$
¬† ¬† ¬† ¬† ¬† ¬† ¬† Time for sound of splash to reach the top
¬† ¬† ¬† ¬† ¬† ¬† ¬† ${t_2} = {{Distance} \over {Velocity}} = {{500} \over {340}} = 1.47\,s$
¬† ¬† ¬† ¬† ¬† ¬† ¬† Therefore Total time ${t_1} + {t_2} = 11.47\,s$

Q.14 ¬† ¬† A sound wave travels at a speed of 339 m/s . If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Sol. ¬† ¬† ¬† ¬†Velocity = 339 m/s
¬† ¬† ¬† ¬† ¬† ¬† ¬† $\lambda = 1.5\,cm = 1.5\, \times {10^{ - 2}}\,m$
¬† ¬† ¬† ¬† ¬† ¬† ¬† $\nu = {V \over \lambda } = {{339} \over {1.5 \times {{10}^{ - 2}}}} = 22600\,{Hz}$
¬† ¬† ¬† ¬† ¬† ¬† ¬† It will not be audible as human range is up to 20,000 Hz only.

¬†Q.15 ¬† ¬†What is reverberation? How can it be reduced ?
Sol. ¬† ¬† ¬† ¬†The persistence of sound due to its repeated reflection until it is not audible is called reverberation.¬†It can be reduced by covering walls and roof of halls by sound absorbent material like compressed fiberboard, rough plaster or draperies.

Q.16 ¬† ¬† What is loudness of sound? What factors does it depend on?
Sol. ¬† ¬† ¬† ¬†Loudness is the physiological response of our ears to sound. It depends on amplitude and frequency and also age of the person. Human beings can hear sound of frequency between 20 Hz to 20,000 Hz.

Q.17 ¬† ¬† Explain how bats use ultrasound to catch a prey ?
Sol. ¬† ¬† ¬† ¬†Bats emit ultrasound. These are reflected by various obstacles and return to the bat‚Äôs ear. The nature of reflection tells the bat where the obstacles or prey is and accordingly the bat is able to catch its prey.

Q.18 ¬† ¬† ¬†How is ultrasound used for cleaning?
Sol. ¬† ¬† ¬† ¬†¬†Ultrasound are high frequency waves. Objects to be cleaned are placed in a cleansing solution and ultrasonic waves are passed. The continuous high frequency vibration cause the dust, grime etc. to detach from the object and can then be easily washed away.

Q.19 ¬† ¬† Explain the working and application of a sonar ?
Sol.¬† ¬† ¬† ¬† SONAR stands for Sound, Navigation And Ranging. It is a device used to measure distance, direction and speed of underwater objects. It has a transmitter and detector near its base. The transmitter transmits ultrasonic signals which get reflected by various underwater objects. These are received by the detector which can convert these waves into appropriate electrical signals and give us the required information.

Q.20 ¬† ¬† 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.
Sol.¬† ¬† ¬† ¬† Time = 5 s for echo to return
¬† ¬† ¬† ¬† ¬† ¬† ¬† Distance of object = 3625 m
¬† ¬† ¬† ¬† ¬† ¬† ¬† $V = {{2 \times d} \over t} = {{2 \times 3625} \over 5} = 1450\,m/s$

Q.21 ¬† ¬† Explain how defects in a metal block can be detected using ultrasound ?
Sol.¬† ¬† ¬† ¬† Ultrasound is passed through the metal block which has to be tested. In case of flaws ultrasound does not pass through it but is reflected back. A detector on the other side of the block does not receive all the transmitted waves and hence the flow is detected . Q.22 ¬† ¬† Explain how the human ear works ?
Sol. ¬† ¬† ¬† ¬†The human ear consists of three parts ‚Äď the outer ear, middle ear and inner ear.

Outer ear : This is also called ‚Äėpinna‚Äô. It collects the sound from the surrounding and directs it towards auditory canal.
Middle ear : The sound reaches the end of the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane to vibrate. These vibrations are amplified by three small bones- hammer, anvil and stirrup.¬†Inner ear : These vibration reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound. Get Lifetime Subscription at Never Before Price. ApplyNow