Question

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

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

Answer 1

Answer: (A)

$16:1$

Elastic energy per unit volume: $E=\frac{1}{2}\times Stress\times Strain$

Using Hooke's law: $Strain=\frac{Stress}{Y}$

$⟹$ $E=\frac{(Stress{)}^2}{2Y}=\frac{F^2}{2Y{A}^2}$ where $A=\frac{\pi D^2}{4}$

$⟹$ $E=\frac{8F^2}{{\pi }^2{D}^{4}Y}$

$\therefore$ $\frac{E_1}{E_2}=(\frac{D_2}{D_1}{)}^4$ $(\because F$ and $Y$ are same for both$)$

GIven : $\frac{D_1}{D_2}=\frac{1}{2}$

$\frac{E_1}{E_2}=(\frac{2}{1}{)}^4=\frac{16}{1}$