When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L+l). The elastic potential energy stored in the extended wire is

Options

A. `(1)/(2) MgL`
B. Mgl
C. MgL
D. `(1)/(2)` Mgl

Answer

D

Explanation

Stress `= (F)/(A) = (Mg)/(A)`, Strain `=(Delta L)/(L) = (L + l-L)/(L) = (l)/(L)`
Energy stored in the wire is, `U =(1)/(2) xx` Stress `xx` Strain `xx` Volume
`=(1)/(2) xx (Mg)/(A) xx (l)/(L) xx A xx L = (1)/(2) Mgl`