Question

A cube of metal is subjected to a hydrostatic pressure of $4GPa$. The percentage change in the length of the side of the cube is close to (Given bulk modulus of metal, $B=8\times 10^{10}Pa)$

• A. 0.6
• B. 1.67
• C. 5
• D. 20

Answer 1

Answer: (B)

$B=-\frac{\Delta P}{\frac{\Delta V}{V}}$
$\left|\frac{\Delta V}{V}\right|=\frac{\Delta P}{B}$
$=\frac{4\times 10^9}{8\times 10^{10}}=\frac{1}{20}$
$\frac{\Delta \ell }{\ell }=\frac{1}{3}\times \frac{\Delta V}{V}=\frac{1}{60}$
Percentage change $=\frac{\Delta \ell }{\ell }\times 100\%$
$=\frac{100}{60}\%=1.67\%$