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ICSE Class 10 Physics Question 3 of 27

Solved Sample Paper 1 — Question 8

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8
Question

Question 1(viii)

A submarine produces an ultrasonic wave of velocity 1500 ms\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
Answer

37.5 km

Reason — As we know,

Speed of sound (V)=Total distance travelled (2d)time interval (t)\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=2d50⇒d=1500×502⇒d=1500×25⇒d=37500 m1500 = \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 = 375001000\dfrac{37500}{1000} km = 37.5 km

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