Determine the change in volume of 1.0 litre of kerosene upon being...

The speed of sound in a liquid is given by $v = \sqrt{\dfrac{B}{\rho}}$, where $B$ is the bulk modulus and $\rho$ is the density.

Squaring both sides, the bulk modulus is $B = \rho v^{2}$.

The bulk modulus is also defined as $B = \dfrac{\Delta P}{\Delta V / V}$, where $\Delta P$ is the change in pressure, $V$ is the initial volume, and $\Delta V$ is the change in volume.

Equating the two expressions for $B$:

$\dfrac{\Delta P \cdot V}{\Delta V} = \rho v^{2}$

Rearranging for $\Delta V$:

$\Delta V = \dfrac{\Delta P \cdot V}{\rho v^{2}}$

Substituting the given values $\Delta P = 2.0 \times 10^{5} \text{ N m}^{-2}$, $V = 1.0 \text{ litre} = 10^{-3} \text{ m}^{3}$, $\rho = 800 \text{ kg m}^{-3}$, and $v = 1330 \text{ m s}^{-1}$:

$\Delta V = \dfrac{2.0 \times 10^{5} \times 10^{-3}}{800 \times (1330)^{2}}$

$\Delta V = \dfrac{200}{800 \times 1768900}$

$\Delta V = \dfrac{1}{4 \times 1768900} = \dfrac{1}{7075600} \text{ m}^{3}$

$\Delta V \approx 1.413 \times 10^{-7} \text{ m}^{3}$

Converting to $\text{cm}^{3}$:

$\Delta V = 1.413 \times 10^{-7} \times 10^{6} \text{ cm}^{3} \approx 0.14 \text{ cm}^{3}$