If the force constant of a wire is K,the work | vedclass

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(C) The force required to stretch a wire by an extension $l$ is given by $F = Kl$,where $K$ is the force constant of the wire.The work done $W$ in str

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$Kl^2/2$

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(C) The force required to stretch a wire by an extension $l$ is given by $F = Kl$,where $K$ is the force constant of the wire.The work done $W$ in stretching the wire by a small extension $dl$ is $dW = F \cdot dl = (Kl) \cdot dl$.To find the total work done in increasing the length by $l$,we integrate the expression from $0$ to $l$:$W = \int_{0}^{l} Kl \, dl = K \left[ \frac{l^2}{2} ight]_{0}^{l} = \frac{1}{2}Kl^2$.Alternatively,since the force increases linearly from $0$ to $Kl$,the average force is $\frac{0 + Kl}{2} = \frac{Kl}{2}$.Thus,the work done is $W = 	ext{Average Force} 	imes 	ext{Extension} = \left( \frac{Kl}{2} ight) 	imes l = \frac{1}{2}Kl^2$.

If the force constant of a wire is $K$,the work done in increasing the length of the wire by $l$ is

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