Q. The length of a wire is $1.0\ m$ and the area of cross-section is $1.0 \times 10^{- 2}cm^{2}$. If the work done for increase in length by $0.2\ cm$ is $0.4\ joule$, then Young’s modulus of the material of the wire is
A) $2.0 \times 10^{10}N/m^{2}$
B) $4.0 \times 10^{10}N/m^{2}$
C) $2.0 \times 10^{11}N/m^{2}$
D) $4.0 \times 10^{11}N/m^{2}$
Explanation
(c) $W = \frac{1}{2}\frac{YAl^{2}}{L} \Rightarrow 0.4 = \frac{1}{2} \times \frac{Y \times 1^{- 6} \times \left( 0.2 \times 10^{- 2} \right)^{2}}{1}$
$\therefore Y = 2 \times 10^{11}N/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.