Question

Compressional wave pulses 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 $2\times {10}^{9}N{m}^{-2}$ and mean temperature is $4°C$, then the depth of the sea will be

• A. $1014m$
• B. $1414m$
• C. $2828m$
• D. $1825m$

Answer 1

The correct option is B

$1414m$

Step 1. Given Data,

Time of echo $t=2s$

Bulk Modulus $B=2\times {10}^{9}N{m}^{-2}$

Mean Temperature $T=4°C$

Assume,

Density of water $d={10}^{3}kg{m}^{-3}$

We have to calculate the depth of sea.

Step 2. Calculate the speed of sound in water,

We know that,

The speed of sound in water is given by, $v=\frac{\sqrt[]{B}}{d}$ (where, $d$ is density of water)

$$\begin{gathered} \Rightarrow v=\frac{\sqrt{2\times {10}^9}}{{10}^3}m{s}^{-1} \\ \Rightarrow v=1.414\times {10}^{3}m{s}^{-1} \\ \Rightarrow v=1414m{s}^{-1} \end{gathered}$$

Step 3. Calculate the depth of sea that is Distance $D$,

(Note: When sound travels back to the observer, it covers twice the distance.)

So, time of echo $t=\frac{2D}{v}$ (where, $D$ is Distance.)

$$\begin{gathered} \Rightarrow D=\frac{tv}{2}m \\ \Rightarrow D=\frac{2\times 1414}{2}m \\ \Rightarrow D=1414m \end{gathered}$$

Hence, the correct option is $\text{'}B\text{'}$.