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Compute the bulk modulus of water from the following data : Initial volume $=100$ litre, pressure increase $=100$ atmosphere, final volume $=100.5$ litre $(1$ atmosphere $=1.013$ $\left.\times 10^5 \mathrm{~Pa}\right)$
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Here,
$$
\begin{aligned}
&\Delta V=100.5-100=0.5 \text { litre }=0.5 \times 10^{-3} \mathrm{~m}^3 \\
&\begin{array}{l}
P=100 \mathrm{~atm}=100 \times 1.013 \times 10^5 \mathrm{~Pa} \\
V=100 \text { litre }=100 \times 10^{-3} \mathrm{~m}^3
\end{array} \\
&\text { Bulk modulus }=B=\frac{-P}{\Delta V / V}=\frac{\mathrm{PV}}{\Delta \mathrm{V}} \\
&=\frac{100 \times 1.013 \times 10^5 \times 100 \times 10^{-3}}{0.5 \times 10^{-3}} \\
&\Rightarrow \quad B=2.026 \times 10^9 \mathrm{~Pa}
\end{aligned}
$$
$$
\begin{aligned}
&\Delta V=100.5-100=0.5 \text { litre }=0.5 \times 10^{-3} \mathrm{~m}^3 \\
&\begin{array}{l}
P=100 \mathrm{~atm}=100 \times 1.013 \times 10^5 \mathrm{~Pa} \\
V=100 \text { litre }=100 \times 10^{-3} \mathrm{~m}^3
\end{array} \\
&\text { Bulk modulus }=B=\frac{-P}{\Delta V / V}=\frac{\mathrm{PV}}{\Delta \mathrm{V}} \\
&=\frac{100 \times 1.013 \times 10^5 \times 100 \times 10^{-3}}{0.5 \times 10^{-3}} \\
&\Rightarrow \quad B=2.026 \times 10^9 \mathrm{~Pa}
\end{aligned}
$$
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