An object is taken $1.0\,km$ deep in sea. Density of sea $= 1.025 \times 10^3\,kg/m^3$ , Bulk modulus of object $= 1.6 \times 10^6\, KPa$ Find out percentage change in density of object $....... \%$
Medium
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$\mathrm{B}=\frac{-\mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} {\rho \mathrm{V}=\mathrm{const}}$
${\frac{\Delta \rho}{\rho}=-\frac{\Delta \mathrm{V}}{\mathrm{V}}}$
$\frac{\Delta \rho}{\rho}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{dgh}}{\mathrm{B}}$
$=\frac{1.025 \times 10^{3} \times 10 \times 10^{3}}{1.6 \times 10^{6} \times 10^{3}}$
$=0.64 \times 10^{-2}$
$\%$ change $=\frac{\Delta \rho}{\rho} \times 100=0.64 \%$
${\frac{\Delta \rho}{\rho}=-\frac{\Delta \mathrm{V}}{\mathrm{V}}}$
$\frac{\Delta \rho}{\rho}=\frac{\mathrm{P}}{\mathrm{B}}=\frac{\mathrm{dgh}}{\mathrm{B}}$
$=\frac{1.025 \times 10^{3} \times 10 \times 10^{3}}{1.6 \times 10^{6} \times 10^{3}}$
$=0.64 \times 10^{-2}$
$\%$ change $=\frac{\Delta \rho}{\rho} \times 100=0.64 \%$
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