Two copper wires A and B of lengths in the ratio $1: 2$ and diameters in the ratio $3: 2$ are stretched by…
Two copper wires A and B of lengths in the ratio $1: 2$ and diameters in the ratio $3: 2$ are stretched by forces in the ratio $3: 1$. The ratio of the elastic potential energies stored per unit volume in the wires A and B is
$2: 1$
$4: 9$
$16: 9$
$4: 3$
Solution
$l_1: l_2=1: 2, d_1: d_2=3: 2, \mathrm{~F}_1: \mathrm{F}_2=3: 1$
The elastic potential energy per unit volume is
$\begin{aligned} & U=\frac{\sigma^2}{2 y} \Rightarrow U \propto \sigma^2 \\ & \therefore \frac{U_1}{U_2}=\left(\frac{F_1}{d_1^2}\right)^2\left(\frac{\mathrm{~d}_2^2}{F_2}\right)^2=\left(\frac{3}{1}\right)^2 \times\left(\frac{2}{3}\right)^4=\frac{16}{9}=16: 9\end{aligned}$