Q. 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 1mm. Then, the elastic energy stored in the wire is
A) 0.2 J
B) 10 J
C) 20 J
D) 0.1 J
Explanation
(d) Elastic energy stored in the wire is
$U = \frac{1}{2} \times stress\ \times strain\ \times volume$
= $\frac{1}{2} \times \frac{F}{A} \times \frac{\mathrm{\Delta}l}{l}\ \times Al\ $
= $\frac{1}{2}\ F\mathrm{\Delta}l$
= $\frac{1}{2}\ \times 200\ \times 1 \times 10^{- 3} = 0.1\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.