A rod mass (M) hinged at (O) is kept in equilibrium with a spring of stiffness (k) as shown in figure. The potential energy stored in the spring is .

Options

A. `(mg)^(2)/(4k)`
B. `(mg)^(2)/(2k)`
C. `(mg)^(2)/(8k)`
D. `(mg)^(2)/(k)`.

Answer

C

Explanation

`sum ("Moment about" O)=0`
`:. (kx) l = mg (l/2)` or `x=(mg)/(2k)`
`U=(1)/(2)kx^(2) =(mg)^(2)/(8k)`