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