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`

Answer

0.67

Explanation

Disc can be understood 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).r^(2) ={(kr^(2))2pi rdr}r^(2)`

`rArr I=2pi kR^(6) //6` ...........(1)
Also, mass of ring, `M = int dm = int_(r=0)^(R) (kr^(2))(2pi r dr)`
Or, `M=2pi k R^(4)/4`........(2)
From (2) and (2)
`I= 4/6 MR^(2) = 2/3 MR^(2)`