The moment of inertia of a circular disc of radius 2m and mass 1 kg about an axis passing through the centre of mass but perpendicular to the plane of the disc is `2" kg m"^(2)`. Its moment of inertia about an axis parallel to this axis but passing through the edge of the disc is . . . . . . . . . . (See the given figure)

Options

A. `8" kg m"^(2)`
B. `4" kg m"^(2)`
C. `10" kg m"^(2)`
D. `6" kg m"^(2)`

Answer

D

Explanation

Here ,
Mass of the disc, `M=1` Kg
Radius of the disc, `R=2` m
Moment of inertia of the circular disc about XY is

`I_(XY)=(MR^(2))/(2)=2" kg m"^(2)` (Given)
According to theorem of parallel axes the moment of inertia of the circular disc about `X_(1)Y_(1)` is
`I_(X_(1)Y_(1))=I_(XY)+MR^(2)=(MR^(2))/(2)+MR^(2)`
`=(3)/(2)MR^(2)=(3)/(2)xx(1" kg")xx("2 m")^(2)=6" kg m"^(2)`