Question

If force $F$ is applied for stretching a wire of length $L$ by $\Delta L$, then elastic potential energy stored per unit volume is
($A$ is cross sectional area of wire)

• A. $\frac{F\Delta L}{2AL}$
• B. $\frac{FA}{2L}$
• C. $\frac{FL}{2A}$
• D. $\frac{FL}{2}$

Answer 1

The correct option is A $\frac{F\Delta L}{2AL}$

Elastic potential energy stored per unit volume is given by,
$\Delta U=\frac{1}{2}\times \text{stress}\times \text{strain}...\left(i\right)$
where,
$\text{Stress}=\frac{F}{A},\text{Strain}=\frac{\Delta L}{L}$
$F=$ Force applied $\perp$ to the cross-sectional area $A$ of the wire
$L=$Original Length of the wire
$\Delta L=$Elongation in length of wire due to force $F$
Hence from Eq.$\left(i\right)$
$\Delta U=\frac{1}{2}\times \left(\frac{F}{A}\right)\times \left(\frac{\Delta L}{L}\right)$
$\therefore \Delta U=\frac{F\Delta L}{2AL}$