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If the bulk modulus of water is $2 \times 10^9 \mathrm{Nm}^{-2}$, then the required pressure to reduce the given volume of water by $2 \%$ is
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$4 \times 10^7 \mathrm{Nm}^{-2}$
Given that, $B=2 \times 10^9 \mathrm{~N} / \mathrm{m}^2$
$\frac{\Delta V}{V} \times 100 \%=-2 \% \Rightarrow \frac{\Delta V}{V}=\frac{-2}{100}=\frac{-1}{50}$
Now, $\begin{aligned} B & =\frac{\Delta p}{\left(\frac{-\Delta V}{V}\right)} \\ \Rightarrow \quad 2 \times 10^9 & =\frac{\Delta p}{\left[-\left(-\frac{1}{50}\right)\right]} \Rightarrow 2 \times 10^9=\frac{\Delta p}{\left[\frac{1}{50}\right]} \\ \Rightarrow \quad 2 \times 10^9 & =50(\Delta p) \Rightarrow \Delta p=\frac{2 \times 10^9}{50}=\frac{1}{25} \times 10^9 \\ & =0.04 \times 10^9 \mathrm{~N} / \mathrm{m}^2=4 \times 10^7 \mathrm{~N} / \mathrm{m}^2\end{aligned}$
$\frac{\Delta V}{V} \times 100 \%=-2 \% \Rightarrow \frac{\Delta V}{V}=\frac{-2}{100}=\frac{-1}{50}$
Now, $\begin{aligned} B & =\frac{\Delta p}{\left(\frac{-\Delta V}{V}\right)} \\ \Rightarrow \quad 2 \times 10^9 & =\frac{\Delta p}{\left[-\left(-\frac{1}{50}\right)\right]} \Rightarrow 2 \times 10^9=\frac{\Delta p}{\left[\frac{1}{50}\right]} \\ \Rightarrow \quad 2 \times 10^9 & =50(\Delta p) \Rightarrow \Delta p=\frac{2 \times 10^9}{50}=\frac{1}{25} \times 10^9 \\ & =0.04 \times 10^9 \mathrm{~N} / \mathrm{m}^2=4 \times 10^7 \mathrm{~N} / \mathrm{m}^2\end{aligned}$
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