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(B) When a load $W$ is suspended from a wire,it moves down by a distance $x$. The loss in potential energy of the load is $W 	imes x$. However,the en

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The energy stored in the wire can be calculated from the area between the force-extension graph and the extension axis.

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(B) When a load $W$ is suspended from a wire,it moves down by a distance $x$. The loss in potential energy of the load is $W 	imes x$. However,the energy stored in the stretched wire (elastic potential energy) is given by $U = \frac{1}{2} 	imes W 	imes x$. The remaining energy,which is $\frac{1}{2}Wx$,is dissipated as heat. Therefore,option $A$ is incorrect. Option $B$ is a correct statement because the work done in stretching the wire is the area under the force-extension graph. Option $C$ is incorrect because the energy per unit volume is $\frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} = \frac{1}{2} 	imes \frac{W}{A} 	imes \frac{x}{L} = \frac{1}{2} \frac{Wx}{AL}$,not $\frac{1}{2}Wx$.

$A$ suspended long metal wire is stretched a small distance $x$ by a load $W$ in newton suspended at the other end. Select the best answer out of the following.

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