A circular disc of radius `R` and thickness `R//6` has moment of inertia `I` about an axis passing through its centre and perpendicular to its plane. It is melted and recast into a solid sphere. The `M.I` of the sphere about its diameter as axis of rotation is

Options

A. `(l)/(5)`
B. `(l)/(6)`
C. `(l)/(32)`
D. `(l)/(64)`

Answer

A

Explanation

Volume of disc `V=A.d=piR^(2).(R)/(6)=(pi)/(6)R^(3)`
Moment of interia of disc about an axis passing through its centre and perpendicular to its plane,
`l=(1)/(2)MR^(2)`
`because" Mass of disc = Mass of solid sphere"`
`therefore" "(pi)/(6)R^(3)rho=(4)/(3)pir^(3).rho" "(rho="density of the material")`
`rArr" "(R^(3))/(8)=r^(3)" "[therefore r=(R)/(2)]`
Thus moment of inertia of the sphere about its diameter
`l_(1)=(2)/(5)Mr^(2)=(2)/(5)M.(R^(2))/(4)`
`rArr" "=(MR^(2))/(10)`
`=(2l)/(10)" "(because l=(1)/(2)MR^(2))`
`l_(1)=l//5`