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(i) What is the principle on which SONAR is based?
(ii) An observer stands at a certain distance away from a cliff and produces a loud sound. He hears the echo of the sound after $1.8s$. Calculate the distance between the cliff and the observer if the velocity of sound in air is $340m{{s}^{-1}}$.

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Last updated date: 27th Mar 2024
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MVSAT 2024
Answer
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Hint: SONAR stands for Sound Navigation and Ranging. It uses ultrasonic waves to find out the distance of an object from the transmitter or to get a clear picture of the path ahead when it is not possible to get a clear sight of the path. The second part of the question can be solved by relating the speed of sound with the distance travelled in the time period and solving for the required variable.

Formula used: $\text{Distance = Speed}\times \text{Time}$

Complete step by step answer:
(i) SONAR is a technology that utilizes ultrasound waves to find the distance between the object and the transmitter and to know whether there are any obstacles in the path ahead when it is not possible to get a visual on the path ahead. It stands for Sound Navigation and Ranging.
In this method a transmitter sends out an ultrasound wave that travels ahead in the medium and is reflected back from an object and received by the same body that sent out the wave. By calculating the time gap between the emission and reception of the wave and by knowing the speed of sound in the medium, the distance between the transmitting body and the obstacle can be found out.
Submarines especially use this technology since it is not very easy to see underwater. Ultrasound waves are used since they have a higher frequency than normal sound waves, so they do not get dissipated in the water. Also, they can be distinguished from other sounds of the submarine such as the engine.

(ii) The second part of the question can be solved by using the relation between the distance, speed and time of a body.
$\text{Distance = Speed}\times \text{Time}$ --(1)
Hence, let us analyze the question.
Let the distance travelled by the sound be $L$.
The time span in which the echo is heard is $t=1.8s$.
The speed of sound in air is given as $v=340m{{s}^{-1}}$.
Hence, using these values in (1), we get,
$L=340\times 1.8=612m$ --(2)
Now, the sound travels this total distance by travelling the same path twice, that is the path between the person and the cliff, once after the person has clapped and the sound reaches the cliff and one more time while the sound is reflected from the cliff and reaches back to the person.
Hence, the distance $d$ between the cliff and the person must be half of the total distance covered by the sound wave.
$\therefore d=\dfrac{L}{2}=\dfrac{612}{2}=306m$ [Using (2)]

Hence, the required distance between the person and the cliff is $306m$.

Note: Sometimes students forget that the distance that the sound travels is actually twice the distance between the source of sound and the obstacle due to which the echo occurs. Forgetting to factor this in will lead to a completely wrong result. Hence, students must be careful of this point.
Students might also think that why they cannot hear an echo in a very small room. This is because the brain needs at least one-tenth of a second to distinguish between two sounds. Hence, in a small room, the original sound and the echo do not have this time gap between them due to the small distance that the sound has to travel and hence, even if there is an echo, the brain cannot distinguish it from the original sound.
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