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

Question

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

Dr. Abubacker · Accepted Answer

(c) Work done is stretching a wire,
$U = \frac{1}{2} 	imes \frac{{YAl}^{2}}{L}$
$= \frac{1}{2} 	imes \frac{2 	imes 10^{11} 	imes 3 	imes 10^{- 6} 	imes (1 	imes 10^{- 3})^{2}}{4}$
$= 0.075\ J$

Q. 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 1mm. Energy stored in it will be $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $\ (Y = 2{\times 10}^{11}{N/m}^{2}).$

Explanation

Reviewed by Dr. Abubacker

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