JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 24)
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $$0.02 \%$$ is _______ $$m$$.
(Take density of sea water $$=10^3 \mathrm{kgm}^{-3}$$, Bulk modulus of rubber $$=9 \times 10^8 \mathrm{~Nm}^{-2}$$, and $$g=10 \mathrm{~ms}^{-2}$$)
Answer
18
Explanation
$$\begin{aligned}
& \beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} \\
& \Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\
& \rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\
& 10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right) \\
& \Rightarrow \mathrm{h}=18 \mathrm{~m}
\end{aligned}$$
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