If a pressure of \(8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}\) is applied to a lead block, so that its…
If a pressure of \(8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}\) is applied to a lead block, so that its volume reduces by \(20 \%\). The Bulk modulus of lead block is
\(4 \times 10^7 \mathrm{Nm}^{-2}\)
\(4 \times 10^8 \mathrm{Nm}^{-2}\)
\(4 \times 10^9 \mathrm{Nm}^{-2}\)
\(4 \times 10^{10} \mathrm{Nm}^{-2}\)
Solution
Pressure applied on the block,
\(p=8 \times 10^8 \mathrm{Nm}^{-2}\)
If \(V_1\) be initial volume, then final volume of block,
\(\begin{aligned}
& V_2=V_1-20 \% \text { of } V_1=V_1-\frac{20}{100} V_1 \\
& V_2=\frac{4}{5} V_1
\end{aligned}\)
\(\therefore\) Change in volume,
\(\Delta V=V_1-V_2=V_1-\frac{4}{5} V_1 \Rightarrow=\frac{V_1}{5}\)
\(\therefore\) Bulk modulus,
\(\begin{aligned}
B & =\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}=\frac{8 \times 10^8 \times V_1}{V_1 / 5} \\
& =4 \times 10^9 \mathrm{Nm}^{-2}
\end{aligned}\)