Q. The Young’s modulus of the material of a wire is $2 \times 10^{10}{Nm}^{- 2}.$ If the elongation strain is 1%, then the energy stored in the wire per unit volume in ${Jm\ }^{- 3}$ is
A) $10^{6}$
B) $10^{8}$
C) ${2 \times 10}^{6}$
D) ${2\ \times 10}^{8}$
Explanation
(a) Per unit volume energy stored
$= \frac{1}{2} \times Y \times (strain)^{2} = \frac{1}{2} \times Y \times \left( \frac{l}{L} \right)^{2}$
given $l = L \times 1\%$
or $l = \frac{L}{100} \times stored\ energy$
$Y = \frac{1}{2} \times 2 \times 10^{10} \times \left( \frac{L}{100L} \right)^{2}
$ $= 10^{6}{Jm}^{- 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.