Question

When a force is applied on a wire of uniform crosssectional area $3 \times 10^{-6} m ^{2}$ and length $4\, m$, the increase in length is $1\, mm$. Energy stored in it will be $\left(Y=2 \times 10^{11} N / m ^{2}\right)$

• A. 6250 J
• B. 0.177 J
• C. 0.075 J
• D. 0.150 J

Answer 1

Answer: (C)

$U =\frac{1}{2} \times \frac{Y A l^{2}}{L}$
$=\frac{1}{2} \times \frac{2 \times 10^{11} \times 3 \times 10^{-6} \times\left(1 \times 10^{-3}\right)^{2}}{4}$
$=0.075\, J$