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(D) The energy stored in the stretched wire (elastic potential energy) is given by $U = \frac{1}{2} F \Delta l$,where $F = 20\, N$ is the applied forc

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$1/2$

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(D) The energy stored in the stretched wire (elastic potential energy) is given by $U = \frac{1}{2} F \Delta l$,where $F = 20\, N$ is the applied force and $\Delta l = 1.0\, mm$ is the extension.The decrease in gravitational potential energy of the load as it moves downwards by $\Delta l = 1.0\, mm$ is given by $W_g = F \Delta l$.The ratio of the increase in energy of the wire to the decrease in gravitational potential energy is:Ratio $= \frac{U}{W_g} = \frac{\frac{1}{2} F \Delta l}{F \Delta l} = \frac{1}{2}$.

$A$ wire is suspended by one end. At the other end,a weight equivalent to $20\, N$ force is applied. If the increase in length is $1.0\, mm$,the ratio of the increase in energy of the wire to the decrease in gravitational potential energy when the load moves downwards by $1\, mm$ will be:

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