A submarine produces an ultrasonic wave of velocity 1500 $\dfrac{\text{m}}{\text{s}}$ in water. The officer receives signal after 50 s of emission of ultrasonic waves. Find the distance of object which is present in the bottom of sea.

1. 25.7 km
2. 37.5 km
3. 37.5 m
4. 50 km

37.5 km

Reason — As we know,

$\text {Speed of sound (V)} = \dfrac{\text{Total distance travelled (2d)}}{\text {time interval (t)}} \\[0.5em]$

Total distance travelled by the sound in going and then coming back = 2d

Given,

t = 50 s

V = 1500 ms^-1

Substituting the values in the formula above, we get,

$1500 = \dfrac{2d}{50} \\[0.5em] \Rightarrow d = \dfrac{1500 \times 50}{2} \\[0.5em] \Rightarrow d = 1500 \times 25 \\[0.5em] \Rightarrow d = 37500 \text{ m}$

Converting m to km

1000 m = 1 km

∴ 37500 m = $\dfrac{37500}{1000}$ km = 37.5 km

Hence, distance of the submarine from the object = 37.5 km