Q. Compressional wave pulse are sent to the bottom of sea from a ship and the echo is heard after 2s. if bulk modulus of elasticity of water is $2 \times 10^{9}Nm^{- 2}$and mean temperature is 4°C, the depth of the sea will be

A) 1014 m
B) 1414 m
C) 2828 m
D) None of these
Explanation

(b) The speed of sound (longitudinal waves) in water is given by
$v = \sqrt{\frac{B}{d}}$
Where B is bulk modulus of water and d is density
Given, $B = 2 \times 10^{9}Nm^{- 2},\ \ \ d = 10^{3}kgm^{- 3}$
$v = \sqrt{\frac{2 \times 10^{9}}{10^{3}}} = 1.414 \times 10^{3}$
$= 1414\ {ms}^{- 1}$
When sound travels back to the observer, it cover’s twice the distance. So, time of echo.
$t = \frac{2d}{v}\ \ \ $
$\therefore\frac{tv}{2} = \frac{1414 \times 2}{2} = 1414\ m$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

Want more practice?
Practice Waves MCQs