1.

Moment of inertia of a disc of radius R about a diametric axis is 35 "kg m"^(-2). The MI of the disc about a parallel axis at a distance R//2 from the centre is

Answer»

`31.25 "kg-m"^(2)`
`37.5 "kg-m"^(2)`
`50 "kg-m"^(2)`
`62.5 "kg-m"^(2)`

Solution :MOMENT of inertia of a disc about a parallel axis
`I=I_(V)+md^(2)=(ma^(2))/(2)+md^(2)""(because a=R,d=R)`
`rArr` Moment of inertia `= (3)/(2) ma^(2)=(3)/(2)(25)`
`:.` Moment of inertia `=(75)/(2)=37.5 "kg m"^(-2)`.


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