Question

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

We know that,
Elastic potential energy in a stretched wire
$U=\frac{1}{2}$ Young's modulus $\times (\text{ strain }{)}^2$
$U=\frac{1}{2}\times \frac{F}{A}\times \frac{1}{l}\times \frac{Fl}{Y\cdot A}$
$U=\frac{1}{2}\frac{F^2}{A^2}\cdot \frac{1}{Y}$
$\Rightarrow U\propto \frac{1}{r^4}$
So, $\frac{U_1}{U_2}=\frac{1/r_1^4}{1/r_2^4}={\left(\frac{r_2}{r_1}\right)}^4\left[\because \frac{d_1}{d_2}=\frac{r_1}{r_2}=\frac{1}{2}\right]$
$\frac{U_1}{U_2}={\left(\frac{2}{1}\right)}^2=\frac{16}{1}$
$U_1:U_2=16:1$