Q. A brass rod of cross-sectional area $1cm^{2}$ and length $0.2m$ is compressed lengthwise by a weight of $5\ kg$. If Young’s modulus of elasticity of brass is $1 \times 10^{11}N/m^{2}$ and $g = 10m/\sec^{2}\ $, then increase in the energy of the rod will be
A) $10^{- 5}\ J$
B) $2.5 \times 10^{- 5}\ J$
C) $5 \times 10^{- 5}\ J$
D) $2.5 \times 10^{- 4}\ J$
Explanation
(b) $U = \frac{1}{2} \times \frac{(stress)^{2}}{Y} \times volume = \frac{1}{2} \times \frac{F^{2} \times A \times L}{A^{2} \times Y}$
$= \frac{1}{2} \times \frac{F^{2}L}{AY} = \frac{1}{2} \times \frac{(50)^{2} \times 0.2}{1 \times 10^{- 4} \times 1 \times 10^{11}} = 2.5 \times 10^{- 5}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.