The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02 \%$ is______ $\mathrm{m}$.
(Take density of sea water $=10^3 \mathrm{kgm}^{-3}$, Bulk modulus of rubber $=9 \times 10^8 \mathrm{Nm}^{-2}$, and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
JEE MAIN 2024, Diffcult
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$\beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}$
$\Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}$
$\rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}}$
$10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right)$
$\Rightarrow \mathrm{h}=18 \mathrm{~m}$
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