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Question

A steel wire of length 5 m is suspended vertically, stretches by 20 mm when mass of 60 kg is attached to the lower end. Find the elastic potential energy gained by the wire. (take g=10 m/s2)

A
0.6 J
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B
6 J
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C
60 J
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D
600 J
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Solution

The correct option is B 6 J
Given, L=5 m
ΔL=20×103 m (extension in the wire)


Stretching force (F)=mg=60×10
F=600 N
We know, elastic potential energy of stretched wire,
U=12×Stretching force×Extension in wire
=12×F×ΔL
=12×600×20×103
U=6 J
Elastic potential energy gained by the wire is 6 J.

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