The moment of inertia of a solid cylinder of mass M, length 2 R and radius about R an axis passing through the centre of mass and perpendicular to the axis of the cylinder is I and about an axis passing through one end of the cylinder and perpendicular to the axis of cylinder is `I_(2)`,then

Options

A. `I_(2)ltI_(1)`
B. `I_(2)-I_(1)=MR^(2)`
C. `(I_(2))/(I_(1))=(19)/(12)`
D. `(I_(2))/(I_(1))=(7)/(6)`

Answer

(b)

Explanation

`I_(1)=(M)/(12)[4R^(2)+R^(2)]=(5MR^(2))/(12)`
`I_(2)=I_(1)+MR^(2)`
`=(5MR^(2))/(12)+MR^(2)`
`I_(2)=(17MR^(2))/(12)`
`I_(2)-I_(1)=(17MR^(2))/(12)-(5)/(12)MR^(2)`
`=(12MR^(2))/(12)=MR^(2)`
because `I_(2)-I_(1)=MR^(2)`