Question

Find the change in the volume of $1.0\mathrm{litre}$ kerosene when it is subjected to an extra pressure of$2.0\times {10}^5{\mathrm{Nm}}^{-2}$ from the following data. Density of kerosene = $800{\mathrm{kgm}}^{-3}$ and speed of sound in kerosene $1300{\mathrm{ms}}^{-1}$.

Answer 1

$$\begin{gathered} \mathbf{Step}\mathbf{}\mathbf{1}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{K} \\ \mathrm{v}=\surd (\mathrm{k}/\mathrm{p}) \\ \mathrm{where},\mathrm{k}={\mathrm{v}}^2\mathrm{\rho } \\ \Rightarrow \mathrm{k}={\left(1330\right)}^2\mathrm{x}800\mathrm{N}/{\mathrm{m}}^2 \\ \mathbf{Step}\mathbf{}\mathbf{2}\mathbf{:}\mathbf{}\mathbf{The}\mathbf{}\mathbf{velocity}\mathbf{}\mathbf{in}\mathbf{}\mathbf{terms}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{bulk}\mathbf{}\mathbf{modulus}\mathbf{}\mathbf{and}\mathbf{}\mathbf{density}\mathbf{}\mathbf{:} \\ \mathrm{K}=\frac{(\mathrm{F}/\mathrm{A})}{(\mathrm{\Delta v}/\mathrm{\Delta v})} \\ \mathrm{Therefore},\mathrm{\Delta V}=\frac{[\mathrm{pressure}\mathrm{x}\mathrm{v}]}{\mathrm{k\Delta v}} \\ =\frac{[2\times 105\mathrm{x}1\mathrm{x}{10}^{-3}]}{\left[1330\mathrm{x}1330\mathrm{x}800\right]} \\ =\mathbf{}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{}{\mathbf{cm}}^{\mathbf{3}} \\ \mathrm{The}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{kerosene}\mathrm{is}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{}{\mathbf{cm}}^{\mathbf{3}}\mathbf{.} \end{gathered}$$