Question

A depth measuring device emits a signal of $20000$ vibration per second which has speed of $1000\,{\text{m/s}}$ in water. The pulse is reflected from the ocean and returns to the device in $0.5\,{\text{s}}$ after the signal is emitted, the depth of the ocean is:
A. $500\,{\text{m}}$
B. $750\,{\text{m}}$
C. $250\,{\text{m}}$
D. $1000\,{\text{m}}$

Answer 1

First of all, we will use the formula relating velocity, distance and time.
The total distance is twice the depth of the ocean. We will substitute the values and manipulate accordingly.

Complete step by step answer:
In the given problem, we have,

The number of vibrations per second, of the emitted signal is given as $20000$ vibration per second.
The speed of the sound signal when it travels in water is $1000\,{\text{m/s}}$ .
The total time taken by the emitted sound to come back to the device is $0.5\,{\text{s}}$ .
We are needed to calculate the depth of the ocean, which was recorded by this device.

This problem is based on the reflection of sound from a surface. When sound is emitted, it travels in forward direction and hits the surface. After hitting the surface, it retraces the path through which it has travelled. We can easily calculate the total distance travelled by the sound wave. In other words, we can say that, the total distance is the sum of the distance it has travelled from the device to the sea bed and the distance travelled by the same from the sea-bed to the device.
Let’s say the depth of the ocean be $d$ .
We know,
$2d = vt$ …… (1)
Where,
$d$ indicates the one-way distance from the device to the sea bed.
$v$ indicates the speed of the sound wave.
$t$ indicates the total time taken by the emitted sound to come back to the device.
In equation (1), the distance is written as twice, because it travels double distance.

Substituting the required values in equation (1), we get:
2d = vt \\\ 2d = 1000 \times 0.5 \\\ d = \dfrac{{1000 \times 0.5}}{2} \\\ d = 250\,{\text{m}} \\\
Hence, the depth of the ocean is found out to be $250\,{\text{m}}$ .

So, the correct answer is “Option C”.

Note:
This problem is based on reflection of sound. While calculating the depth of the ocean, remember that the depth is half the total distance traversed by the sound. It other words it can be said that the emitted sound hits the sea-bed in $0.25\,{\text{s}}$ and the reflected sound takes another $0.25\,{\text{s}}$ to return to the device.