A uniform disc of radius R has a hole cut out which has a radius r. The centre of hole is at a distance `(R )/(2)` from the centre of disc. The position of centre of mass is :

Options

A. `(R-r)/(R )`
B. `(Rr^(2))/(2(R^(2)-r^(2)))`
C. `(Rr^(2))/(2(R^(2)+r^(2)))`
D. None of these

Answer

B

Explanation

Mass of complete disc =`piR^(2)t rho`
(t = thickness)
Mass of scooped out disc = `pir^(2)trho`
Mass of remaining disc = `pi(R^(2)-r^(2))trho`
Now `pi(R^(2)-r^(2))trhoxx x=pir^(2)trho(R)/(2)x=(r^(2)R)/(2(R^(2)-r^(2)))`