A spring when loaded has a potential energy 'E'. Then 'm' turns out of 'n' turns are removed form the spring. If the same load is suspended, then the energy stored in the spring is

Options

A. `(n)/((n-m))E`
B. `(mE)/(n)`
C. `((n-m))/(m)E`
D. `((n-m))/(n)E`

Answer

D

Explanation

`PE = (1)/(2)Kx^(2) = (F^(2))/(2K), K alpha(1)/(l)`