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The pressure required to decrease the volume of $4000 \mathrm{cc}$ water by $0.05 \%$ is
(Bulk modulus of water $=2.2 \times 10^9 \mathrm{Nm}^{-2}$ )
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(Bulk modulus of water $=2.2 \times 10^9 \mathrm{Nm}^{-2}$ )
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Verified Answer
The correct answer is:
$1.1 \times 10^6 \mathrm{Nm}^{-2}$
Volume of water, $\mathrm{V}=4000 \mathrm{cc}$ decrease of volume, $\frac{\Delta V}{V}=0.05 \%=0.0005$ Bulk modulus, $B=2.2 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ Bulk modulus is given as:
$$
\begin{aligned}
& B=-\frac{P}{\frac{\Delta V}{V}} \Rightarrow P=B \times \frac{\Delta V}{V} \\
& P=2.2 \times 10^9 \times 0.0005=0.0011 \times 10^9 \\
& P=1.1 \times 10^6 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
$$
\begin{aligned}
& B=-\frac{P}{\frac{\Delta V}{V}} \Rightarrow P=B \times \frac{\Delta V}{V} \\
& P=2.2 \times 10^9 \times 0.0005=0.0011 \times 10^9 \\
& P=1.1 \times 10^6 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
$$
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