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When a load of 10 kg is hung from the wire, then extension of 2m is produced. Then work done by restoring force is
A. 200 J
B. 100 J
C. 50 J
D. 25 J

1 Answer

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Best answer
Correct Answer - B
(b) `W=(1)/(2)` (Maximum stretching force)`xx`(Elongation produced)
`=(1)/(2)F Delta l`
Here, `F=mg=10xx10`
`Deltal=2m`
`therefore W=(1)/(2)(10xx10)xx2=100J`

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