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Question

Physics Question on Electromagnetic waves

Compressional wave pulses are sent to the bottom of sea from a ship and the echo is heard after 2s2 s. If bulk modulus of elasticity of water is 2×109N/m22 \times 10^{9} \,N / m ^{2} and mean temperature is 4C4^{\circ} C, the depth of the sea will be

A

1014 m

B

1414 m

C

2828 m

D

none of these

Answer

1414 m

Explanation

Solution

The speed of sound (longitudinal waves) in water is given by v=Bdv=\sqrt{\frac{B}{d}}
where BB is bulk modulus of water and dd is density.
Given, B=2×109N/m2,d=103kg/m3B=2 \times 10^{9} \,N / m ^{2}, d=10^{3} \,kg / m ^{3}
v=2×109103=1.414×103\therefore v=\sqrt{\frac{2 \times 10^{9}}{10^{3}}}=1.414 \times 10^{3}
=1414m/s=1414\, m / s
When sound travels back to the observer, it covers twice the distance.
So, time of echo. t=2dvt=\frac{2 d}{v}
d=tv2=1414×22=1414m\therefore d=\frac{t v}{2}=\frac{1414 \times 2}{2}=1414\, m