The moment of inertia of a hollow sphere of mass `M` having internal and external radii `R` and `2R` about an axis passing through its centre and perpendicular to its plane is

Options

A. `(13)/(32)MR^(2)`
B. `(31)/(35)MR^(2)`
C. `(62)/(35)MR^(2)`
D. `(3)/(2)MR^(2)`

Answer

C

Explanation

The given hollow sphere can be obtained by removing a solid sphere of radius R and mass `M_(1)` from a solid sphere of radius 2R and mass `M_(2)`. If `rho` is the density of the material of the sphere then the mass of the hollow sphere
`M=M_(2)-M_(1)`
`=(4)/(3)pi(2R)^(3)rho-(4)/(3)pi(R)^(3)rho`
`M=(4)/(3)pi rho(8R^(3)-R^(3))=(28)/(3)piR^(3)rho" ...(1)"`
The M.I. of the hollow sphere about its diameter is
`I=(2)/(5)M_(2)(2R)^(2)-(2)/(5)M_(1)(R)^(2)`
`=(2)/(5)[(4)/(3)pi(2R)^(3)rho.4R^(2)-(4)/(3)pi rho(R^(3))R^(2)]`
`=(2)/(5).(4)/(3)pi rho[32R^(5)-R^(5)]=(8)/(15)pi rho.31R^(5)" ...(2)"`
`therefore" "(I)/(M)=((8)/(15)xx31xxpi rhoR^(5))/((28)/(3)xxpiR^(3) rho)`
`=(8xx3xx31)/(28xx15).R^(2)=(62)/(35)R^(2)`
`therefore I=(62)/(35)MR^(2)`.