Question

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

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

Answer 1

Answer: (D)

16 : 1

Answer 2

: Elastic potential energy per unit volume is

$u = \frac{1}{2}\frac{(stress)^{2}}{Y}$

As both the wires are made up of same material, so their Young’s modulus is same for both the wires.

$\therefore u \propto (stress)^{2}$

$\therefore\frac{u_{1}}{u_{2}} = \frac{(stress)_{1}^{2}}{(stress)_{2}^{2}} = \frac{(F_{1}/A_{1})^{2}}{(F_{2}/A_{2})^{2}}$

As both the wires are stretched by the same load, therefore

$F_{1} = F_{2}$

$\therefore\frac{u_{1}}{u_{2}} = \left( \frac{A_{2}}{A_{1}} \right)^{2} = \left( \frac{D_{2}^{2}}{D_{1}^{2}} \right)^{2} = \left( \frac{D_{2}}{D_{1}} \right)^{4} = \left( \frac{2}{1} \right)^{4} = \frac{16}{1}$