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# Compressional wave impulses are sent to the bottom of sea from a ship and the echo is heard after 4s. If bulk modulus of water is $2\times {{10}^{9}}N{{m}^{-2}}$ and the mean temperature is ${{4}^{\circ }}C$, then depth of the sea is\begin{align} & \left( A \right)2000\times {{10}^{3}}m \\ & \left( B \right)2828m \\ & \left( C \right)1414m \\ & \left( D \right)707m \\ \end{align}

Last updated date: 13th Jun 2024
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Hint: For calculating the depth of the sea , consider the concept that time is the ratio of displacement to the velocity. Let us assume the distance is 2d. That is, the initial distance before echo is d and after echo also the sound waves travels a distance d. Then the total distance becomes 2d. The depth can also be defined as the square root of bulk modulus to the density of the material. hence we will get the depth of the sea.

\begin{align} & \dfrac{2d}{{{V}_{S}}}=2 \\ & \\ \end{align}
\begin{align} & \Rightarrow d={{V}_{S}} \\ & \\ \end{align}
In the equation below d is the depth of the sea, B is the bulk modulus and $\rho$ is the density of water. Thus by substituting the value we will get the depth of the sea.
\begin{align} & \Rightarrow d=\sqrt{\dfrac{B}{\rho }} \\ & \Rightarrow d=\sqrt{\dfrac{2\times {{10}^{9}}}{1000}} \\ \end{align}
$\therefore d=1414m$