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.

Options

A. `I_(A)=I_(B)`
B. `I_(A)gtI_(B)`
C. `I_(A)ltI_(B)`
D. `I_(A)geI_(B)`

Answer

C

Explanation

Let `M` be the mass of each disc. Let `R_(A)` and `R_(B)` be the radii of discs `A` and `B`, respectively. Then `M=piR_(A)^(2)td_(A)=piR_(s)^(2) td_(B)`
As `d_(A)=d_(B)`, so `R_(A)^(2)ltR_(s)^(2)`. Now
`(I_(A))/(I_(B))=(R_(A)^(2))/(R_(B)^(2))lt1,` i.e. `I_(A)ltI_(B)`