A thin disc of mass M and radius R has mass per unit area `sigma( r)=kr^2` where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is : a) `(MR^3)/3` b) `(2MR^2)/3` c) `(MR^2)/6` d) `(MR^2)/2`

Options

A. `(MR^(2))/3`
B. `(2MR^(2))/3`
C. `(MR^(2))/6`
D. `(MR^(2))/2`

Answer

B

Explanation

Disc can be underlined as the combination of co-axial rings.
M.I. of element ring of radius r and infinitesimally small thickness dr about the axis, is
`dI=(dm)x^(2) = (kr^(2)) 2pi rdrr^(2) therefore` M.I. of disc `I=int dI = int_(r=0)^(r ) kr^(2).(2pi rdr)x^(2)`
`rArr I=2pi kR^(6)//6` .......(i)
Also, mass of ring, `M = int dm = int_(r=0)^(R) (kr^(2)) (2pi r dr)`
Or, `M=2pi kR^(4)/4`.......(ii)
From (i) & (ii) `I=4/6 MR^(2) = 2/3 MR^(2)`