Two circular discs `A` and `B` of equal masses and thicknesses. But are made of metals with densities `d_A and d_B (d_A gt d_B)`. If their moments of inertia about an axis passing through the centre and normal to the circular faces be `I_A and I_B`, then.

Explanation

Let `r_(A) and r_(B)` be the radii of disc `A & B`. As their mass `(m)` and thickness `(t)` are same, therefore,
`m = (pi r_(A)^(2)) t xx d_(A) = (pir_(B)^(2)) xx t xx d_(B)`
`:. (r_(A)^(2))/(r_(B)^(2)) = (d_(B))/(d_(A))`
Now `(I_(A))/(I_(B)) = ((1)/(2) m r_(A)^(2))/((1)/(2)m r_(B)^(2)) = (r_(A)^(2))/(r_(B)^(2)) = (d_(B))/(d_(A))`
As `d_(A) gt d_(B), :. I_(B) gt I_(A)`