If average depth of an ocean is $4000 \mathrm{~m}$ and the bulk modulus of water is $2 \times 10^9 \mathrm{~Nm}^{-2}$, then fractional compression $\frac{\Delta V}{V}$ of water at the bottom of ocean is $\alpha \times 10^{-2}$. The value of $\alpha$ is _______ (Given, $\mathrm{g}=10 \mathrm{~ms}^{-2}, \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}$)
Answer (integer)
2
Solution
<p>$$\begin{aligned}
& \mathrm{B}=-\frac{\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)} \\
& -\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)=\frac{\rho \mathrm{gh}}{\mathrm{B}}=\frac{1000 \times 10 \times 4000}{2 \times 10^9} \\
& =2 \times 10^{-2}[-\mathrm{ve} \text { sign represent compression }]
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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