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. `l_(1)ltl_(2)`
B. `l_(2)-l_(1)=MR^(2)`
C. `(l_(2))/(l_(1))=(19)/(12)`
D. `(l_(2))/(l_(1))=(7)/(6)`

Explanation

Moment of inertia of a solid cylinder of mass M, length 2R and radius R about an axis passing through the centre of mass and perpendicular to the axis is
`I_(1)=M(L^(2)/(12)+(R^(2))/(4))`
`I_(2)=M[(4R^(2))/(12)+(R^(2))/(4)]=M[(7R^(2))/(12)]`
Moment of inertia passing through one end of the cylinder and perpendicular to the axis of the cylinder
`=M[(19R^(2))/(12)]`
`I_(2)-I_(1)=MR^(2)`