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Q.

A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is I metre above the floor. The wire was elongated by I mm. The energy stored in the wire due to stretching is

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a

zero

b

0.05 joule

c

100 joule

d

500 joule

answer is B.

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Detailed Solution

W = 12×F×l = 12mgl     = 12×10×10×1×10-1 = 0.05 J
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