Question

A hydraulic press contains $0.25{\text{m}}^3$ of oil. Find the percentage decrease in volume of the oil, when it is subjected to a pressure increase $\Delta P=1.6\times {10}^7\text{Pa}$. The bulk modulus of the oil is $B=5.0\times {10}^9\text{Pa}$.

• A. $0.32$%
• B. $0.64$%
• C. $0.82$%
• D. $0.96$%

Answer 1

The correct option is A $0.32$%

Given,
$B=-\frac{\Delta P}{\left(\frac{\Delta V}{V_0}\right)}$
$\Rightarrow \left(\frac{\Delta V}{V_0}\right)=-\frac{\Delta P}{B}=-\frac{1.6\times {10}^7}{5\times {10}^9}=-0.32\times {10}^{-2}$

$\therefore$ Percentage decrease in volume of the oil is given as
$\left(\frac{\Delta V}{V_0}\right)\times 100=(0.32\times {10}^{-2})\times 100=0.32\text{\%}$

Hence, $\left(A\right)$ is the correct answer.

Why this Q? Bulk Modulus: $B=\frac{\text{Volume Stress}}{\text{Volume strain}}=\frac{\Delta P}{-\frac{\Delta V}{V}}$ $\Rightarrow B=-V\times \frac{\Delta P}{\Delta V}$