Q. A wire of length $50\ cm$ and cross sectional area of $1\ sq.\ mm$ is extended by $1\ mm$. The required work will be $(Y = 2 \times 10^{10}Nm^{- 2})$
A) $6 \times 10^{- 2}\ J$
B) $4 \times 10^{- 2}\ J$
C) $2 \times 10^{- 2}\ J$
D) $1 \times 10^{- 2}\ J$
Explanation
(c) $W = \frac{YAl^{2}}{2L} = \frac{2 \times 10^{10} \times 10^{- 6} \times \left( 10^{- 3} \right)^{2}}{2 \times 50 \times 10^{- 2}} = 2 \times 10^{- 2}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.