Q 11-08-091JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
A cube of metal is subjected to a hydrostatic pressure of $4\ \text{GPa}$. The percentage change in the length of the side of the cube is close to: (Given bulk modulus of metal, $B = 8\times10^{10}\ \text{Pa}$)
Answer: (D) $1.67$
$\dfrac{\Delta V}{V} = \dfrac PB = \dfrac{4\times10^{9}}{8\times10^{10}} = 0.05 = 5\%$.
For a cube $\dfrac{\Delta V}{V} = 3\dfrac{\Delta l}{l}$, so $\dfrac{\Delta l}{l} \approx \dfrac53\% \approx 1.67\%$.
Solution by Sreeraj P, M.Sc Physics