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

Question

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, $4 \times 10^7 \mathrm{Nm}^{-2}$, $4 \times 10^8 \mathrm{Nm}^{-2}$, $4 \times 10^9 \mathrm{Nm}^{-2}$, $4 \times 10^{10} \mathrm{Nm}^{-2}$

MARKS App · Accepted Answer

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, $\begin{aligned} B & =\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}=\frac{8 \times 10^8 \times V_1}{V_1 / 5} \ & =4 \times 10^9 \mathrm{Nm}^{-2} \end{aligned}$

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