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 of the wire and the work done in displacing it through $1.5\, mm$ by the weight is
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(c) Work done in stretching a wire
$W = \frac{1}{2}Fl = \frac{1}{2} \times 10 \times 0.5 \times {10^{ - 3}}$= $2.5 \times {10^{ - 3}}J$
Work done to displace it through $1.5 \,mm$
$W = F \times l = 5 \times {10^{ - 3}}J$
The ratio of above two work $= 1 : 2$
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