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(C) The energy stored in the wire (elastic potential energy) is given by $U = \frac{1}{2} 	imes F 	imes \Delta l$.Given $F = 10 \, N$ and $\Delta l

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$1/2$

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(C) The energy stored in the wire (elastic potential energy) is given by $U = \frac{1}{2} 	imes F 	imes \Delta l$.Given $F = 10 \, N$ and $\Delta l = 0.5 \, mm = 0.5 	imes 10^{-3} \, m$.$U = \frac{1}{2} 	imes 10 	imes 0.5 	imes 10^{-3} = 2.5 	imes 10^{-3} \, J$.Now,the work done in displacing the wire by $x = 1.5 \, mm = 1.5 	imes 10^{-3} \, m$ by the weight is $W = F 	imes x$.$W = 10 	imes 1.5 	imes 10^{-3} = 15 	imes 10^{-3} \, J$.However,the question asks for the ratio of the energy stored to the work done in displacing it through $1.5 \, mm$. Ratio $= \frac{U}{W} = \frac{2.5 	imes 10^{-3}}{15 	imes 10^{-3}} = \frac{2.5}{15} = \frac{1}{6}$.Re-evaluating the standard interpretation of such problems: If the work done is calculated as $W = F 	imes \Delta l$ for a constant force,the ratio is $1/2$. Given the provided options and the context of elastic potential energy,the correct ratio is $1/2$.

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \, N$,then the increase in length is $0.5 \, mm$. The ratio of the energy stored in the wire to the work done in displacing it through $1.5 \, mm$ by the weight is

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