Q. The work done in increasing the length of a one metre long wire of cross-sectional area 1$mm^{2}$ through 1 mm will be $(Y = 2 \times 10^{11}\ Nm^{- 2})$
A) 0.1 J
B) 5 J
C) 10 J
D) 250 J
Explanation
(a) Work done, W $= \frac{1}{2}F \times l = \frac{1}{2} \times stress\ \times strain\ \times volume\ $
Or $W = \frac{1}{2}Y\ \times (stress)^{2} \times volume\ $
= $\frac{1}{2}Y\left( \frac{\mathrm{\Delta}l}{l} \right)^{2} \times Al = \frac{1}{2}Y\frac{\mathrm{\Delta}l^{2}A}{l}$
= $\frac{2 \times 10^{11} \times 10^{- 6} \times 10^{- 6}}{2 \times 1\ } = 0.1\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.