[4 Mark Questions]

Question 14 Marks

Explain echo method to measure velocity’ of sound.

Answer

Apparatus required: A source of sound pulses, a measuring tape, a sound receiver, and a stop watch.
Procedure:
1. Measure the distance ' $d$ ' between the source of sound pulse and the reflecting surface using the measuring tape.
2. The receiver is also placed adjacent to the source. A sound pulse is emitted by the source.
3. The stopwatch is used to note the time interval between the instant at which the sound pulse is sent and the instant at which the echo is received by the receiver. Note the time interval as ' $t$ '.
4. Repeat the experiment for three or four times. The average time taken for the given number of pulses is calculated.
Calculation of speed of sound: The sound pulse emitted by the source travels a total distance of $2 d$ while travelling from the source to the wall and then back to the receiver. The time taken for this has been observed to be ' $t$ '. Hence, the speed of sound wave is given by:
$\text { Speed of sound }=\frac{\text { distenceTravelled }}{\text { TimeTaken }}$
$=\frac{2 d}{t}$
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Question 24 Marks

What is the Doppler effect? Derive the formula for the change in apparent frequency
1. Both source and listener move towards and away from each other.
2. Both source and listener move one behind the other source follows the listener and listener follows the source.
3. A source at rest, listener moves towards and away from the source.
4. A listener at rest, the source moves towards and away from the listener.

Answer

Position of source and listener	Note	Expression for apparent frequency
(i) Both source and listener move (ii) They move towards each other	(a) Distance between source and listener decreases. (b) Apparent frequency is more than actual frequency.	$n^{\prime}=(v-v L v+v S) n$
(i) Both source and listener move (ii) They move away from each other	(a) Distance between source and listener increases. (b) Apparent frequency is less than actual frequency. (c) $v_S$ and $v_L$ become opposite to that in case-1.	$n^{\prime}=(v-v L v+v S) n$

Position of source and listener	Note	Expression for apparent frequency
(i) Both source and listener move (ii) They move one behind the other (iii) Source follows the listener	(a) Apparent frequency depends on the velocities of the source and the listener. (b) $v_S$ becomes opposite to that in case-2.	$n^{\prime}=(v-v L v+v S) n$
(i) Both source and listener move (ii) They move one behind the other (iii) The listener follows the source	(a) Apparent frequency depends on the velocities of the source and the listener. (b) $v_S$ and $v_L$ become opposite to that in case-3.	$n^{\prime}=(v-v L v+v S) n$

Position of source and listener	Note	Expression for apparent frequency
(i) Source at rest (ii) Listener moves towards the source	(a) Distance between source and listener decreases. (b) Apparent frequency is more than actual frequency. (c) $v_S$ = 0 in case-1.	$n^{\prime}=(v+v L v) n$
(i) Source at rest (ii) Listener moves away from the source	(a) Distance between source and listener increases. (b) Apparent frequency is less than actual frequency. (c) $v_s$ = 0 in case-2.	$n^{\prime}=(v-v L v) n$

Position of source and listener	Note	Expression for apparent frequency
(i) Listener at rest (ii) Source moves towards the listener	(a) Distance between source and listener decreases. (b) Apparent frequency is more than actual frequency. (c) $v_L$ = 0 in case-1.	$n^{\prime}=(v v-v S) n$

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Question 34 Marks

Write down the concept of Ear trumpet and Megaphone.

Answer

Ear trumpet:

Ear trumpet is a hearing aid, which is useful by people who have difficulty in hearing.
In this device, one end is wide and the other end is narrow. The sound from the sources fall into the wide end and are reflected by its walls into the narrow part of the device. This helps in concentrating the sound and the sound enters the eardrum with more intensity. This enables a person to hear the sound better.

Megaphone:

A megaphone is a horn – shaped device used to address a small gathering of people.
It’s one end is wide and the other end is narrow. When a person speaks at the narrow end, the sound of his speech is concentrated by the multiple reflections from the walls of the tube. Thus, his voice can be heard loudly over a long distance.

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Question 44 Marks

Classify sound waves based on their frequencies.

Answer

(i) Audible waves : These are sound waves with a frequency ranging between 20 Hz and 20,000 Hz. These are generated by vibrating bodies such as vocal cords, stretched strings etc.

(ii) Infrasonic waves : These are sound waves with a frequency below 20 Hz that cannot be heard by the human ear. eg: waves produced during earth quake, ocean waves, sound produced by whales, etc.

(iii) Ultrasonic waves : These are sound waves with a frequency greater than 20 kHz, Human ear cannot detect these waves, but certain creatures like mosquito, dogs, bats, dolphins can detect these waves, eg: waves produced by bats.

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Question 54 Marks

Write down the applications of echo?

Answer

Applications of echo:

Some animals communicate with each other over long distances and also locate objects by sending the sound signals and receiving the echo as reflected from the targets.
The principle of echo is used in obstetric ultrasonography, which is used to create real – time visual images of the developing embryo or fetus in the mother’s uterus. This is a safe testing tool, as it does not use any harmful radiations.
Echo is used to determine the velocity of sound waves in any medium.

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