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(C) The moment of inertia of a circular disc about an axis passing through its centre and normal to its circular face is given by $I = \frac{1}{2}MR^2

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${I_A} < {I_B}$

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(C) The moment of inertia of a circular disc about an axis passing through its centre and normal to its circular face is given by $I = \frac{1}{2}MR^2$.Since the mass $M$ of the disc is given by $M = V\rho = \pi R^2 t \rho$,where $t$ is the thickness and $\rho$ is the density,we can write $R^2 = \frac{M}{\pi t \rho}$.Substituting this into the expression for $I$,we get $I = \frac{1}{2} M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.Given that the mass $M$ and thickness $t$ are constant for both discs,we have $I \propto \frac{1}{\rho}$.Therefore,$\frac{I_A}{I_B} = \frac{d_B}{d_A}$.Since it is given that $d_A > d_B$,it follows that $I_A < I_B$.

Two circular discs $A$ and $B$ are of equal masses and thickness but made of metals with densities ${d_A}$ and ${d_B}$ $({d_A} > {d_B})$. If their moments of inertia about an axis passing through their centres and normal to the circular faces are ${I_A}$ and ${I_B}$ respectively,then:

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