A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is __________ .

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KollegeApply · Accepted Answer

IA < IB Moment of inertia of circular disc of radius r: I = $\frac{1}{2}$m r^2 Mass = Volume $\times$ Density Volume of disc = $\pi$ r^2 t Here, t is the thickness of the disc. As density is same for both the rods, we have Moment of inertia, I \propto\text{ thickness } $\times$ \left(\text{radius} \right)^4 I_A/I_B = \frac{t . \left( r \right)^4}{t/4 \left( 4r \right)^4} < 1 \Rightarrow I_A/I_B < 1 \Rightarrow I_A < I_B

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