Question:
When a rubber ball is taken to a depth of $m$ in deep sea, its volume decreases by $0.5 \%$
(The bulk modulus of rubber $=9.8 \times 10^{8} \mathrm{Nm}^{-2}$ Density of sea water $=10^{3} \mathrm{kgm}^{-3}$
$\left.\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$
Solution:
$\mathrm{B}=-\frac{\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)}=-\frac{\rho \mathrm{gh}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)}$
$-\frac{\mathrm{B} \frac{\Delta \mathrm{V}}{\mathrm{V}}}{\rho \mathrm{g}}=\mathrm{h}$
$\frac{9.8 \times 10^{8} \times 0.5}{100 \times 10^{3} \times 9.8}=\mathrm{h}$
$h=500$
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