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 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
Explanation
(d) Elastic energy stored in the wire is
$U = \frac{1}{2}stess \times strain \times volume$
$= \frac{1}{2}\frac{F}{A} \times \frac{l}{L} \times AL = \frac{1}{2}Fl$
$= \frac{1}{2} \times 200 \times 1 \times 10^{- 3} = 0.1\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.