The elastic potential energy stored in a copp | vedclass

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Question

(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula: $U = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain}

Vedclass · Accepted Answer

$6 	imes 10^{-2} \,J$

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(A) The elastic potential energy $U$ stored in a stretched wire is given by the formula: $U = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} 	imes 	ext{Volume}$.Alternatively,$U = \frac{1}{2} 	imes Y 	imes A 	imes \frac{(\Delta L)^2}{L}$,where $Y$ is Young's modulus,$A$ is the area of cross-section,$\Delta L$ is the extension,and $L$ is the original length.Given:$Y = 1.2 	imes 10^{11} \,N/m^2$$A = 1 \,mm^2 = 1 	imes 10^{-6} \,m^2$$L = 1 \,m$$\Delta L = 1 \,mm = 10^{-3} \,m$Substituting the values:$U = \frac{1}{2} 	imes (1.2 	imes 10^{11}) 	imes (10^{-6}) 	imes \frac{(10^{-3})^2}{1}$$U = 0.6 	imes 10^5 	imes 10^{-6} 	imes 10^{-6}$$U = 0.6 	imes 10^{-7} 	imes 10^5 = 0.6 	imes 10^{-2} \,J = 6 	imes 10^{-3} \,J$.Wait,re-calculating: $U = 0.5 	imes 1.2 	imes 10^{11} 	imes 10^{-6} 	imes 10^{-6} = 0.6 	imes 10^{-1} = 0.06 \,J = 6 	imes 10^{-2} \,J$.Thus,the correct option is $A$.

The elastic potential energy stored in a copper rod of length $1 \,m$ and area of cross-section $1 \,mm^2$ when stretched by $1 \,mm$ is (Young's modulus of copper $= 1.2 \times 10^{11} \,N/m^2$).

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