The moment of inertia of a circular disc of mass `M` and radius `R` about an axis passing thrugh the center of mass is `I_(0)`. The moment of inertia of another circular disc of same mass and thickness but half the density about the same axis is

Options

A. `(I_(0))/(4)`
B. `(I_(0))/(2)`
C. `2I_(0)`
D. `4I_(0)`

Answer

C

Explanation

For the first disc, moment of inertia `I_(0)=(1)/(2)MR^(2)`
But M = Volume `xx` Density
`therefore M =piR_(1)^(2)xxt xx rho`
For the second disc, `M=piR_(2)^(2)txx(rho)/(2)`
As the masses are equal
`therefore piR_(2)^(2)t(rho)/(2)=piR_(1)^(2)t rho`
`therefore" "R_(2)^(2)=2R_(1)^(2)`
`therefore" "(I')/(I_(90))=((1)/(2)MR_(2)^(2))/((1)/(2)MR_(1)^(2))=(2R_(1)^(2))/(R_(1)^(2))=2`
`therefore I'=I_(0)`