A disc having radius `R` is rolling without slipping on a horizontal (`x-z`) plane. Centre of the disc has a velocity ` v` and acceleration `a` as shown.
If `v=sqrt(2aR)` the angle `theta` between acceleration of the top most point and the horizontal is

Options

A. `0`
B. `45^@`
C. `tan^(-1)2`
D. `tan^(-1)(1/2)`

Explanation

`a=alphaR`

`a_(C)=omega^(2)R=(v/R)^(2)=(v^(2))/R=(2aR)/R=2a`
`tantheta=(2a)/(2a)=1impliestheta=45^@`