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A wire, suspended vertically from one end, is stretched by attaching a weight of 20 \mathrm{~N} to the lower end. The weight stretches the wire by 1 \mathrm{~mm}. How much energy is gained by the wire?

Option: 1

0.01 J


Option: 2

0.02 J


Option: 3

0.04 J


Option: 4

1.0 J


Answers (1)

best_answer

We have seen above that work done by a force $F$ to produce an extension l in the wire \mathrm{=\frac{1}{2} F \times l}. Thus, the energy gained by the wire

\mathrm{=\frac{1}{2} F \times l=\frac{1}{2} \times 20 \times \left(1 \times 10^{-3}\right)=10^{-2} \mathrm{~J}=0.01 \mathrm{~J}}.

Hence the correct choice is (a).

Posted by

Suraj Bhandari

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