Question

A copper rod of cross-sectional area $2c{m}^2$ and length $2m$ is compressed length wise by a weight of $10kg$. If Young's modulus of elasticity of brass is $10^{\prime \prime }N/m^2$ and $g=9.8m/s^2$ then increase in energy of the rod will be

• A. $480.2\mu J$
• B. $490.1\mu J$
• C. $1500\mu$ J
• D. None

Answer 1

Answer: (A)

Work done
$=\frac{1}{2}\times \frac{(\text{ Stress }{)}^2}{Y}\times$ Volume
$=\frac{1}{2}\times \frac{F^2\times A\times L}{A^2\times Y}$
$\therefore W=\frac{1}{2}\times \frac{F^{2}L}{AY}=\frac{1}{2}\times \frac{(10\times 9.8{)}^2\times 2}{2\times 10^{-4}\times 10^{11}}$
$=\frac{9604}{2\times 10^7}$
$W=4802\times 10^{-7}$
$W=480.2\mu J$