When a force is applied on a wire of uniform cross-sectional are `3 xx 10^(-6) m^(2)` and length 4 m, the increase in length is 1 mm. Energy stored in it will be (Given, `Y = 2 xx 10^(11) Nm^(-2)`)

Options

A. 6250 J
B. 0.177 J
C. 0.075 J
D. 0.150 J

Answer

c

Explanation

Energy, `U = (1)/(2) xx (4AI^(2))/(L)`
`= (1)/(2) xx (2 xx 10^(11) xx 3 xx 10^(-6) xx (1 xx 1-^(-3))^(2))/(4) = 0.075 J`