Question
The increase in pressure required to decrease the volume of a water sample by $0.2 \%$ is $P \times 10^{\circ} Nm ^{-2}$. Bulk modulus of water is $2.15 \times 10^9 Nm ^{-2}$. The value of $P$ is _____________
✓
Answer
(43)
Sol. Since bulk modulus is given as
$
\begin{array}{l}
B=\frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)} \\
2.15 \times 10^9=\frac{-\Delta P}{-\left(\frac{0.2}{100}\right)} \\
\Delta P=2.15 \times 10^9 \times 2 \times 10^{-3} \\
=4.3 \times 10^6=43 \times 10^5 N / m^2
\end{array}
$
Sol. Since bulk modulus is given as
$
\begin{array}{l}
B=\frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)} \\
2.15 \times 10^9=\frac{-\Delta P}{-\left(\frac{0.2}{100}\right)} \\
\Delta P=2.15 \times 10^9 \times 2 \times 10^{-3} \\
=4.3 \times 10^6=43 \times 10^5 N / m^2
\end{array}
$
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