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{F l}{Y \cdot A} $
$U =\frac{1}{2} \frac{F^{2}}{A^{2}} \cdot \frac{1}{Y} $
$ \Rightarrow U \propto \frac{1}{r^{4}} $
So, $\frac{U_{1}}{U_{2}}=\frac{1 / r_{1}^{4}}{1 / r_{2}^{4}}=\left(\frac{r_{2}}{r_{1}}\right)^{4}\,\,\,\left[\because \frac{d_{1}}{d_{2}}=\frac{r_{1}}{r_{2}}=\frac{1}{2}\right]$
$\frac{U_{1}}{U_{2}}=\left(\frac{2}{1}\right)^{2}=\frac{16}{1}$
$U_{1}: U_{2}=16: 1$