Question

A wire suspended vertically from one of its ends is stretched by attaching a weight of $200 \, N$ to the lower end. The weight stretches the wire by $1 \, mm$ . Then the elastic energy stored in the wire is

• A. $0.2 \, J$
• B. $10 \, J$
• C. $20 \, J$
• D. $0.1 \, J$

Answer 1

Answer: (D)

Elastic energy per unit volume = $\frac{1}{2} \times \text{stress} \times \text{strain}$
$\therefore $ Elastic energy = $\frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$
= $\frac{1}{2}\times \frac{\text{F}}{\text{A}}\times \frac{\Delta \text{L}}{\text{L}}\times \left(\text{AL}\right)$
$=\frac{1}{2}\text{F}\Delta \text{L}=\frac{1}{2}\times \text{200}\times \text{10}^{- 3}\text{=0.1 J}$