Q. A wire fixed at the upper end stretches by length $l$ by applying a force F. The work done in stretching is
A) $\frac{F}{2l}$
B) $Fl$
C) $2Fl$
D) $\frac{Fl}{2}$
Explanation
(d) Work done in stretching the wire
$= potential\ energy\ stored$
$= \frac{1}{2} \times stress \times strain \times volume$
$= \frac{1}{2} \times \frac{F}{A} \times \frac{l}{L} \times AL = \frac{1}{2}Fl$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.