Question

A man is swimming at a depth $d$ in the sea at a distance $(L>>d)$ from a ship $\left(S\right)$. An explosion occurs in the ship and after hearing the sound, the man immediately moves to the surface. It takes $0.8\text{s}$ for the man to rise to the surface after he hears the sound of explosion. $0.2\text{s}$ after reaching the surface, he once again hears the sound of explosion. Then, calculate $L$.


Given: Speed of sound in air = $340{\text{ms}}^{-1}$; Bulk modulus of water $2\times {10}^9\text{Pa}$

• A. $342.8\text{m}$
• B. $374.2\text{m}$
• C. $447.6\text{m}$
• D. $555.5\text{m}$

Answer 1

The correct option is C $447.6\text{m}$

Speed of sound in water

$v_{\omega }=\sqrt{\frac{B}{\rho }}=\sqrt{\frac{2\times {10}^9}{{10}^3}}=1414{\text{ms}}^{-1}$

and given, speed of sound in air $v_{air}=340{\text{ms}}^{-1}$

First sound is heard due to sound wave travelling in water. It takes time $t_1$ for sound to travel from $S$ to $M$ in water. Second sound is heard because of sound travelling through air reaching the man. This sound takes time $t_2$ to travel through $L$ distance in air.

From the data, we can conclude that, time difference between the sound heard in water and sound heard in air $\Delta t=0.8+0.2=1.0\text{s}$

$\because d\ll L$ , we can say that $SM\approx L$

$\therefore \frac{L}{v_{air}}-\frac{L}{v_{\omega }}=1$

$\Rightarrow L=\frac{v_{air}{v}_{\omega }}{v_{\omega }-v_{air}}=\frac{340\times 1414}{1414-340}=447.6\text{m}$