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A uniform disc of radius R has a hole cu...

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 :

A

`(R-r)/(R )`

B

`(Rr^(2))/(2(R^(2)-r^(2)))`

C

`(Rr^(2))/(2(R^(2)+r^(2)))`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
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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Knowledge Check

  • A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the disc coincide. The centre of mass of the new disc is alphaR from the centre of the bigger disc. The value of alpha is :

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