Question

When a force is applied on a wire of uniform cross-sectional area $3\times {10}^{-6}{m}^2$and length $4m$, the increase in length is $1mm$. Energy stored in it will be

$(Y=2\times {10}^{11}N{m}^{-2})$

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

Answer 1

The correct option is C

0.075 J

The energy stored can be written as,

$U=\frac{1}{2}\times \frac{YA{l}^2}{L}$

$=\frac{1}{2}\times \frac{2\times {10}^{11}\times 3\times {10}^{-6}\times (1\times {10}^{-3}{)}^2}{4}$

$=0.075J$