A massless wire 2m in length suspended vertically stretches by 10mm when mass of 10kg is attached to the lower end. The elastic potential energy gained by the wire is (Take g=10m/s2)
A
0.5J
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B
5J
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C
50J
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D
500J
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Solution
The correct option is A0.5J Change in elastic potential energy is given by: ΔU=12×F×ΔL...(i)
Change in length of wire is, ΔL=10mm=10×10−3m
Force/Load acting on wire is its weight: F=mg=10×10=100N
From Eq. (i) ΔU=12×100×10×10−3 ∴ΔU=0.5J