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Two circular discs A and B of equal mass...

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.

A

`I_(A)=I_(B)`

B

`I_(A)gtI_(B)`

C

`I_(A)ltI_(B)`

D

`I_(A)gt=ltI_(B)`

Text Solution

Verified by Experts

The correct Answer is:
C
Moment of inertia of circular disc about an axis passing through centre and normal to the circular face
`I=(1)/(2)MR^(2)=(1)/(2)M((M)/(pitrho))["As"M=Vrho=piR^(2)trhotherefore R^(2)=(M)/(pitrho)]`
`impliesI=(M^(2))/(2pitrho)orIprop(1)/(rho)` If mass and thickness are constant.
So, in the problem `(I_(A))/(I_(B))=(d_(B))/(d_(A)) therefore I_(A)ltI_(B)["As "d_(A)gtd_(B)]`
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