Two wires $\mathrm{A}$ and $\mathrm{B}$ having same length and material are stretched by the same force.…
Two wires $\mathrm{A}$ and $\mathrm{B}$ having same length and material are stretched by the same force.
Their diameters are in the ratio $1: 3$. The ratio of energy density of wire $A$ to that of wire B when stretched, is
$27: 1$
$9: 1$
$81: 1$
$3: 1$
Solution
Strain energy per unit volume $U=\frac{1}{2} \times$ Stress $\times$ Strain From Hooke's law, Stress $=\mathrm{Y} \times$ Strain
$\Longrightarrow \mathrm{U}=\frac{1}{2 \mathrm{Y}} \times(\text { Stress })^{2}$
where Stress $=\frac{\mathrm{F}}{\mathrm{A}}=\frac{\mathrm{F}}{\pi \mathrm{d}^{2} / 4}$
$\Longrightarrow \mathrm{U} \propto \frac{1}{\mathrm{~d}^{4}}$
Given $: \frac{\mathrm{d}_{\mathrm{s}}}{\mathrm{d}_{\mathrm{l}}}=\frac{1}{3}$
Thus ratio of strain energy per unit volume $\frac{\mathrm{U}_{\mathrm{s}}}{\mathrm{U}_{1}}=\left(\frac{\mathrm{d}_{\mathrm{l}}}{\mathrm{d}_{\mathrm{s}}}\right)^{4}$ $\Longrightarrow \frac{\mathrm{U}_{\mathrm{s}}}{\mathrm{U}_{1}}=\left(\frac{3}{1}\right)^{4}=81$