JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions

Let σ be the mass per unit area of the disc.
Then the mass of the complete disc =σ(π(2R)2)
The mass of the removed disc =σ(πR2)=πσR2
Let us consider the above situation to be a complete disc of radius 2R on which a disc of radius R of negative mass is superimposed. Let O be the origin. Then the above figure can be redrawn keeping in mind the concept of centre of mass as :xc.m=

(6π(2R)2)×0+(−6(πR2))R

4πσR2−πσR2

∴xc.m=

−πσR2×R

3πσR2

∴xc.m=−

R

3

=αR ⇒α=

1

3