When a block of mass 2kg is suspended by a long wire of length 1m, the length of wire becomes 1.02m. The elastic potential energy stored in wire is
(Take g=10m/s2)
A
0.1J
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B
0.2J
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C
0.4J
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D
0.8J
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Solution
The correct option is B0.2J Elastic potential energy of stretched wire, ΔU=12×Stretching force×extension in wire
Here body is stretched due to its weight mg ⇒ΔU=12×F×ΔL ΔU=12×mg×(1.02−1) ΔU=12×2×10×0.02 ∴ΔU=0.2J