If the one end of a wire is fixed with a rigid support and the other end is stretched by a force of 10N, then the increase in length is 0.5mm. The ratio of the energy of the wire and the work done in displacing it through 0.5 mm by the same force
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answer is 3.
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Detailed Solution
work done in stretching the wire,W=12Fl=12×10×0.5×10−3=2.5×10−3 Work done to displace it through 0.5 mmW=F×l=5×10−3 So the required ratio is 1:2