Question

A brass rod of cross-sectional area $1\, cm ^{2}$ and length $0.2\, m$ 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 $g=10\, m / s$, 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. $2.5 \times 10^{-4} J$

Answer 1

Answer: (B)

$U=\frac{1}{2} \times \frac{(\text { Stress })^{2}}{Y} \times \text { Volume }$
$=\frac{1}{2} \times \frac{F^{2} \times A \times L}{A^{2} \times Y}$
$=\frac{1}{2} \times \frac{F^{2} L}{A Y}=\frac{1}{2} \times \frac{(50)^{2} \times 0.2}{1 \times 10^{-4} \times 1 \times 10^{11}}$
$=2.5 \times 10^{-5} J$