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

e=ΔYl=1×10−34 u=e2Y2=(10−34)2×2×10112= 0.075 J

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

e=ΔYl=1×10−34 u=e2Y2=(10−34)2×2×10112= 0.075 J

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When a force is applied on a wire of uniform cross-section of area $3 \times 10^{- 6} m^{2}$ and length 4m, the increase in length is 1 mm. Energy stored in it will be $(Y = 2 \times 10^{11} N / m^{2})$

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0.075 J

6250 J

0.177 J

0.150 J

answer is C.

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Detailed Solution

$e = \frac{Δ Y}{l} = \frac{1 \times 10^{- 3}}{4}$

$u = \frac{e^{2} Y}{2} = {(\frac{10^{- 3}}{4})}^{2} \times \frac{2 \times 10^{11}}{2}$

$= 0.075 J$

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

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When a force is applied on a wire of uniform cross-section of area $3 \times 10^{- 6} m^{2}$ and length 4m, the increase in length is 1 mm. Energy stored in it will be $(Y = 2 \times 10^{11} N / m^{2})$