The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as `rho(r)=A+Br.` Find its moment of inertia about the line perpendicular to the plane of the disc through its centre.

Answer

A, B, D

Explanation

The surface density of a circular disc of radius a depends upon the distance from the centre as P(r)=A+Br.
Therefore the mass of the ring of radius r will be
`theta=(a+Br)xx2pirxxdr`

Therefore moment of inertila about the centre
`=int_a^0(A+Br)2pir.drxxr^2`
`=aint_a^02piAr^3 dr+int_a^0 2pi Br^4dr`
`=[2piA(r^4/4)+2piB(r^5/5)]_a^0`
`=2pi((Aa^4)/4+(Ba^5)/5)`