Question:
A cube of metal is subjected to a hydrostatic pressure of $4 \mathrm{GPa}$. The percentage change in the length of the side of the cube is close to :
(Given bulk modulus of metal, $B=8 \times 10^{10} \mathrm{~Pa}$ )
Correct Option: , 4
Solution:
(4) Bulk modulus, $B=\frac{P}{\frac{\Delta V}{V}}$
$\Rightarrow \frac{\Delta V}{V}=\frac{P}{B}$ ...(i)
If the side of cube is $L$ then $V=L^{3}$
$\frac{\Delta V}{V}=\frac{3 \Delta L}{L}=\frac{P}{B}$
$\Rightarrow \frac{\Delta L}{L}=\frac{1}{3} \times \frac{P}{B}=\frac{4 \times 10^{9} \mathrm{~Pa}}{3 \times 8 \times 10^{10} \mathrm{~Pa}}=\frac{1}{60}$
$\Rightarrow \frac{\Delta L}{L} \times 100=\frac{1}{60} \times 100=1.67 \%$