AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Consider the figure,
Let the mass of the entire disc is M.
The mass per unit area is, =

M

π(2R)2

=

M

4πR2

The mass of the removed disc of radius R is,
M1=

M

4πR2

×(πR2)
M1=

M

4

The remaining mass of the disc is,
M2=M−M1
M2=M−

M

4

M2=

3M

4

The centre of mass of combination of M1 and M2 will be C(0,0). So,

M1x1−M2x2

M1−M2

=0
M1x1=M2x2

M

4

R=

3M

4

(αR)
α=

1

3