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Bulk modulus of water is $2 \times 10^9 \mathrm{~N} / \mathrm{m}^2$. The pressure required to increase the volume of water by $0.1 \%$ in $\mathrm{N} / \mathrm{m}^2$ is :
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The correct answer is:
$2 \times 10^6$
Bulk modulus, $K=2 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
Change in volume, $\Delta V=0.1 \%$ of initial volume
$=0.1 \%$ of $V=\frac{V \times 0.1}{100}=V \times 10^{-3}$
$\therefore \quad K=\frac{p V}{\Delta V}$
$2 \times 10^9=\frac{p \times V}{V \times 10^{-3}}$
$p=2 \times 10^9 \times 10^{-3}$
$=2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
Change in volume, $\Delta V=0.1 \%$ of initial volume
$=0.1 \%$ of $V=\frac{V \times 0.1}{100}=V \times 10^{-3}$
$\therefore \quad K=\frac{p V}{\Delta V}$
$2 \times 10^9=\frac{p \times V}{V \times 10^{-3}}$
$p=2 \times 10^9 \times 10^{-3}$
$=2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
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