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)

0.1 J

Answer 2

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