AIEEE 2007 Solved Paper

Let the mass per unit area be σ.
Then the mass of the complete disc
=σ[π(2R)2]=4πσR2
The mass of the removed disc =σ(πR2)=πσR2
So mass of the remaining disc =4πσR2−πσR2=3πσR2
Let center of mass of 3πσR2 mass is at x distance from origin 0
∴‌

3πR2σ.x+πR2σ.R

4πR2σ

=0
As center of mass of full disc is at Origin.
∴x=−‌

R

3

According to the question, x=αR
∴α=−‌

1

3

⇒|α|=‌

1

3