Q. A wire $(Y = 2 \times 10^{11}\ Nm^{- 2})$ has length 1 m and cross-sectional area 1 $mm^{- 2}.$ The work required to increase the length by 2 mm is
A) 0.4 J
B) 4 J
C) 40 J
D) 400 J
Explanation
(a) Work done = $\frac{1}{2}\ F\mathrm{\Delta}l$
$\binom{= \ \frac{1}{2}\frac{YA\mathrm{\Delta}l^{2}}{l}}{\frac{2 \times 10^{11} \times 10^{- 6}\left( 2 \times 10^{- 3} \right)^{2}}{2 \times 1}}$ $\left| \begin{array}{r}
Y = \frac{Fl}{A\mathrm{\Delta}l} \\
or\ \ F = \frac{YA\mathrm{\Delta}l}{l}
\end{array} \right.\ $
= 4$\times 10^{- 1}\ J = 0.4\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.