MHT CET 2020 Physics Solved Paper

Let the mass per unit area be Then the mass of the complete disc =σ[π(2R)2]=4πσR2
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 discof 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 :
4πσ

R

O

2

R ↔

πσR2

Xc⋅m=‌

(4πσR2)×0+(−πσR2)R

4πσR2−πσR2

∴xc.m=‌

−πσR2×R

3πσR2

∴xc.m=−‌

R

3

⇒α=‌

1

3