A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end.The weight stretches the wire by 1 mm. Then the energy stored in the wire is

Options

A. 0.2 J
B. 10 J
C. 20 J
D. 0.1 J

Answer

D

Explanation

Elastic energy stored in the wire is `U=(1)/(2)xx"Stress"xx"Strain"xx"Volume"=(1)/(2)xx(F)/(A)xx(Al)/(L)xxAL=(1)/(2)FDeltal=(1)/(2)xx200xx1xx10^(-3)=0.1J`