A solid sphere of mass `M`, radius `R` and having moment of inertia about as axis passing through the centre of mass as `I`, is recast into a disc of thickness `t`, whose moment of inertia about an axis passing through its edge and perpendicular to its plance remains `I`. Then, radius of the disc will be.

Options

A. `r=sqrt(2/15)R`
B. `r=sqrt(2/15)R`
C. `r=2/15R`
D. `r=2/(sqrt(5))R`

Answer

B

Explanation

`I_(AB)=2/5MR^(2)=I` (given)…….i
`I_(A'B)=I_(YY') =I_(YY')+Mr^(2) =I1/2Mr^(2)+Mr^(2)`
`=3/2Mr^(2)=1` (given)
From eqn i and ii `2/5MR^(2)=3/2Mr^(2)implies r=2/(sqrt(15))R`