1.

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 :

Answer»

`(R-r)/(R )`
`(Rr^(2))/(2(R^(2)-r^(2)))`
`(Rr^(2))/(2(R^(2)+r^(2)))`
None of these

Solution :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)))`


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