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If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \mathrm{~N}$, then the increase in length is $0.5 \mathrm{~mm}$. The ratio of the energy of the wire and the work done in displacing it through $1.5 \mathrm{~mm}$ by the weight is
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The correct answer is:
$\frac{1}{2}$
Work done in stretching a wire
$W=\frac{1}{2} F l=\frac{1}{2} \times 10 \times 0.5 \times 10^{-3}=2.5 \times 10^{-3} \mathrm{~J}$
Work done to displace it through $1.5 \mathrm{~mm}$
$W=F \times l=5 \times 10^{-3} \mathrm{~J}$
The ratio of above two work $=1: 2$
$W=\frac{1}{2} F l=\frac{1}{2} \times 10 \times 0.5 \times 10^{-3}=2.5 \times 10^{-3} \mathrm{~J}$
Work done to displace it through $1.5 \mathrm{~mm}$
$W=F \times l=5 \times 10^{-3} \mathrm{~J}$
The ratio of above two work $=1: 2$
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