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 load moves downwards by $1\, mm,$ will be
A$1$
B$\frac{1}{4}$
C$\frac{1}{3}$
D$\frac{1}{2}$
Medium
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D$\frac{1}{2}$
d (d) Ratio of work done $=$ $\frac{{1/2Fl}}{{Fl}} = \frac{1}{2}$
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