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Q.

When a force is applied on a wire of uniform cross-sectional area 3×10−6m2 and length 4m, the increase in length is 1 mm. Energy stored in it will be Y=2×1011N/m2

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An Intiative by Sri Chaitanya

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=12×2×1011×3×10−6×1×10−324=0.075J.

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