Question

An ultrasonic wave is sent from a ship towards the bottom of the sea. It is found that the time interval between the sending and the receiving is $1\cdot 6\text{s}$. Find the depth of the sea if the velocity of sound in seawater is $1400\text{m}{\text{s}}^{-1}$ .
A) $1120\text{m}$
B) $560\text{m}$
C) $1400\text{m}$
D) $112\text{m}$

Answer 1

Hint:When the ultrasonic wave is sent from the ship it travels a distance equal to the depth of the sea to reach the bottom of the sea. As the wave is received back, the wave again travels the same distance. Since the time interval for this action and the speed of the sound wave are given, we can use the depth of the sea to obtain it easily.

Formula used:
The distance travelled by an object is given by, $s=v\times t$ where $v$ is the speed of the object and $t$ is the time taken for the object to cover the distance.

Complete step by step solution.
Step 1: List the parameters given in the question.
An ultrasonic wave travels back and forth from a ship to the bottom of the sea.
The speed of the sound wave is given to be $v=1400\text{m}{\text{s}}^{-1}$ .
The time taken for the sound wave to cover the depth of the sea twice is given to be $t=1\cdot 6\text{s}$ .
Let $d$ be the depth of the sea, then the total distance covered by the wave as it reaches back to the receiver of the ship will be $2d$.

Step 2: Express the relation for the distance covered by the sound wave.
The total distance covered by the sound wave as it reaches the ship can be expressed as $2d=v\times t$ .
$\Rightarrow d={\displaystyle \frac{v\times t}{2}}$ --------- (1)
Substituting for $v=1400\text{m}{\text{s}}^{-1}$ and $t=1\cdot 6\text{s}$ in equation (1) we get, $d={\displaystyle \frac{1400\times 1\cdot 6}{2}}=1120\text{m}$
Thus the depth of the sea is $d=1120\text{m}$ .

So the correct option is A.

Note:Here the given time interval is the time taken for the sound wave to reach the bottom of the sea and then to get back to the ship. Ultrasonic sound waves refer to sound waves which are inaudible to the human ear. They have frequencies greater than the upper limit of the audible frequencies of the human ear.