Solution:
The bulk modulus of elasticity κ of the liquid is given by
[tex]\kappa=\dfrac{\Delta P}{\left(-\dfrac{\Delta V}{V}\right) } =-V\left(\dfrac{\Delta P}{\Delta V}\right) =-\dfrac{V_1(P_2-P_1)}{V_2-V_1}[/tex]
Given:
[tex]V_o=1\text{ ft}^3\\\kappa=325,000\text{ psi}\\\Delta P=300\text{ psi}\\\Delta V=\;?[/tex]
Then,
[tex]\Delta V=-\dfrac{V\Delta P}{\kappa}\\\Delta V=-\dfrac{\left[1\text{ ft}^3\times\left(\dfrac{1\text{ m}}{3.281\text{ ft}}\right)^3\right](300\text{ psi})}{325,000\text{ psi}} \\\\\Delta V=-26.1\times10^{-6}\text{ m}^3\quad\textsf{(ANSWER)}[/tex]
