Question

If a pressure of $8 \times 10^{8} N \cdot m ^{-2}$ is applied to a lead block so that its volume reduces by $20 \%$. The Bulk modulus of lead block is______

• A. $4 \times 10^{7} N \cdot m^{-2}$
• B. $4 \times 10^{8} N \cdot m ^{-2}$
• C. $4 \times 10^{9} N \cdot m ^{-2}$
• D. $4 \times 10^{10} N \cdot m^{-2}$

Answer 1

Answer: (C)

Pressure applied on the block,
$p=8 \times 10^{8} Nm ^{-2}$
If $V_{1}$ be initial volume, then final volume of block,
$V_{2}=V_{1}-20 \%$ of $V_{1}=V_{1}-\frac{20}{100} V_{1}$
$V_{2}=\frac{4}{5} V_{1}$
$\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,
$B =\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}$
$=\frac{8 \times 10^{8} \times V_{1}}{V_{1} / 5}$
$=4 \times 10^{9} Nm ^{-2}$