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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 isA. 200 JB. 100 JC. 50 JD. 25 J |
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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