Question

Find the change in the volume of 1.0 litre kerosene when it is subjected to an extra pressure of 2.0 × 10⁵ N m⁻² from the following data. Density of kerosene = 800 kg m⁻³ and speed of sound in kerosene = 1330 ms⁻¹.

Answer 1

Given:
Volume of kerosene V = 1 litre = $1\times {10}^{-3}{\mathrm{m}}^3$
Pressure applied P = 2.0 × 10⁵ Nm^$-2$
Density of kerosene ρ = 800 kgm⁻³
Speed of sound in kerosene v = 1330 ms⁻¹
Change in volume of kerosene $∆V$ = ?
The velocity in terms of the bulk modulus $\left(K\right)$ and density $\left(\rho \right)$ is given by:
$v=\sqrt{\left(\frac{\mathrm{K}}{p}\right)}$,
$$\begin{gathered} w\mathrm{here} \\ K=v^2\rho . \\ \Rightarrow K={\left(1330\right)}^2\times 800\mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{As}\mathrm{we}\mathrm{know}, \\ \mathit{}K=\frac{\left({\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}\right)}{\left({\displaystyle \raisebox{1ex}{$∆V$}\!\left/ \!\raisebox{-1ex}{$V$}\right.}\right)}. \\ \therefore ∆V=\frac{\mathrm{Pressure}\times V}{K}\left(\because P=\frac{F}{A}\right) \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{respective}\mathrm{values},\mathrm{we}\mathrm{get}: \\ ∆V=\frac{2\times {10}^5\times 1\times {10}^{-3}}{1330\times 1330\times 800}=0.14{\mathrm{cm}}^3 \end{gathered}$$

Therefore, the change in the volume of kerosene ∆V = 0.14 cm³.