Question

Two wires of same material and length but diameters in the ratio $1:2$, are stretched by the same force. The potential energy per unit volume for the two wires will be in the ratio,

• A. $16:1$
• B. $4:1$
• C. $2:1$
• D. $1:1$

Answer 1

The correct option is A $16:1$

For an elastic wire,
$\Delta l=\frac{Fl}{AY}$

$\frac{\Delta l}{l}=\left(\frac{F}{AY}\right)$

Potential energy per unit volume $=\frac{1}{2}\times \text{stress}\times \text{strain}$

$U=\frac{1}{2}\times \frac{F}{A}\times \frac{F}{AY}$

From the above equation, we can write, $U\propto \frac{1}{A^2}$ or $U\propto \frac{1}{d^4}$

$\therefore \frac{U_1}{U_2}={\left(\frac{d_2}{d_1}\right)}^4=\frac{16}{1}$

Hence, $\left(A\right)$ is the correct answer.

Why this Q? Strain energy density $U=\frac{\text{Strain~energy}}{\text{Volume}}=\frac{1}{2}\frac{{\text{(Stress)}}^2}{Y}$ $U=\frac{1}{2}Y{\text{(Strain)}}^2=\frac{1}{2}\text{Stress}\times \text{Strain}$