Question

When a force is applied on a wire of uniform cross-sectional area $3\times {10}^{-6}{m}^2$ and length 4 m, the increase in length is 1 mm. 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. 0150 J

Answer 1

Answer: (C)

0.075 J

cross sectional Area $=3\times {10}^{-6}{m}^2$ length =$=4m$

increased in length =$1mm$

$\text{Young modulus}\left(y\right)=2\times {10}^{11}N/m^2$ We Know

Energy stored (U) $=\frac{1}{2}$ (Stress) (Strain) (volume)

$\begin{array}{c}\therefore y=\frac{\text{stress}}{\text{strain}}\\ \to \text{stress}=y\left(\text{strain}\right)\\ u=\frac{1}{2}y\left(\text{strain}{)}^2\right(\text{volume})\\ =\frac{1}{2}2\times {10}^{11}\left(\frac{{10}^{-3}}{4}\right)(3\times {10}^{-6}\times 4^{-})\end{array}$

$U=0.075J$