A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end at is 1 metre above the floor. The wire was alongated by 1 mm. The energy stored in the wire due to stretching is

Options

A. 0.01 J
B. 0.05 J
C. 0,02 J
D. 0,04 J

Answer

b

Explanation

`W = (1)/(2) xx F xx I = (1)/(2) mg I`
`= (1)/(2) xx 10 xx 10 xx 1 xx 10^(-3) = 0.05 J`