A uniform circular disc of radius `R` lies in the `XY` plane with its centre coinciding with the origin.The moment of inertia about an axis passing through a point on the X-axis at a distance `x=2R` and perpendicular to the X-Y plane is equal to its moment of inertia about an axis passing through a pint on the Y-axis at a distance `y=d` and parallel to the X-axis in the X-Y plane. The value of `d` is

Options

A. `(4R)/3`
B. `sqrt(17)(R/2)`
C. `sqrt(15)(R/2)`
D. `sqrt(13)(R/2)`

Answer

B

Explanation

On axis at a distance `2R`,
`I=(mR^(2))/2+m(4R^(2))=9/2 mR^(2)`
On y-axis at a distance 'd' `I=(mR^(2))/4+md^(2)` equating both, `d=sqrt(17)R/2`