A long spring when stretched by `x cm`, has a potential energy `U`. On increasing the stretching to `nx cm,` the potential energy stored in spring will be

Options

A. `(U)/(N)`
B. `NU`
C. `N^(2)U`
D. `(U)/(N^(3))`

Answer

C

Explanation

`U = (1)/(2) kx^(2)` & `U' = (1)/(2) K(Nx)^(2)`.