Two circular discs A and B are of equal masses and thickness but made of metals with densities dA and dB ( dA > dB ) . If their moments of inertia about an axis passing through centres and normal to the circular faces be IA and IB , then

The correct answer is I_A < I_B (Option 3 / "IA<IB").

Let us analyze the problem step-by-step:
Let $M$ be the mass and $t$ be the thickness of both circular discs $A$ and $B$.
Let $d_A$ and $d_B$ be the densities of the materials of discs $A$ and $B$ respectively, where it is given that:
$d_A>d_B$

The mass of a circular disc of radius $R$, thickness $t$, and density $d$ is given by:
$M=\text{Volume}\times \text{density}=\pi R^{2}td$

Since both discs have the same mass $M$ and thickness $t$, we can write:
$M_A=M_B=M$
$\pi R_A^{2}t{d}_A=\pi R_B^{2}t{d}_B$

Canceling common terms $\pi$ and $t$ from both sides, we get:
$R_A^2{d}_A=R_B^2{d}_B$
$\frac{R_A^2}{R_B^2}=\frac{d_B}{d_A}$

Since we are given that $d_A>d_B$, the ratio $\frac{d_B}{d_A}<1$.
Therefore:
$\frac{R_A^2}{R_B^2}<1\Rightarrow R_A^2<R_B^2$

Now, the moment of inertia $I$ of a circular disc of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its face is:
$I=\frac{1}{2}M{R}^2$

For discs $A$ and $B$, their moments of inertia are:
$I_A=\frac{1}{2}M{R}_A^2$
$I_B=\frac{1}{2}M{R}_B^2$

Dividing the two equations, we obtain:
$\frac{I_A}{I_B}=\frac{R_A^2}{R_B^2}$

Since we established that $R_A^2<R_B^2$, it follows that:
$I_A<I_B$