Question

A copper rod of cross-sectional area $2 cm ^{2}$ and length $2 m$ is compressed length wise by a weight of $10 kg$. If Young's modulus of elasticity of brass is $10^{\prime \prime} N / m ^{2}$ and $g =9.8 m / s ^{2}$ 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
$=\frac{1}{2} \times \frac{F^{2} \times A \times L}{A^{2} \times Y}$
$\therefore W=\frac{1}{2} \times \frac{F^{2} L}{A Y} =\frac{1}{2} \times \frac{(10 \times 9.8)^{2} \times 2}{2 \times 10^{-4} \times 10^{11}}$
$=\frac{9604}{2 \times 10^{7}}$
$W=4802 \times 10^{-7}$
$W=480.2 \,\mu J$