JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Elasticity & SHM

The elastic energy stored per unit volume is $u=\frac{1}{2}\times \text{stress}\times \text{strain}=\frac{F^2}{2A^{2}Y}$. Since both wires carry the same load $F$ and share the same material $Y$, $u$ depends only on the cross-sectional area, and with $A=\frac{\pi d^2}{4}$ this means $u\propto \frac{1}{d^4}$. A stored-energy ratio of $1:4$ gives $\frac{d_2^4}{d_1^4}=\frac{1}{4}$, so $\frac{d_1}{d_2}=4^{1/4}=\sqrt{2}$, and the diameters are in the ratio $\sqrt{2}:1$.