When a force is applied on a wire of uniform crosssectional area 3 x 10-6m2 and length 4 m, the increase in length is I mm. Energy stored in it will be (Y = 2 x 1011 N/m2)
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a
6250 J
b
0.177 J
c
0.075 J
d
0.150 J
answer is C.
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Detailed Solution
U = 12×YAl2L = 12×2×1011×3×10-6×(1×10-3)24 = 0.075 J