Ans.

For disc 1, mass and thickness are M and t respectively. Let, \( r_1 \) be the radius of the circular disc and \( {\rho}_1 \) be the uniform density of the disc. The volume of this disc is \( \left(\pi{r_1}^2t\right) \).

So the mass of disc 1 is \( M=\left(\pi{r_1}^2t\right){\rho}_1\tag{1} \).

For disc 2, mass and thickness are M and t respectively. Let \( r_2 \) be the radius of the circular disc and \( {\rho}_2 \) be the uniform density of the disc. The volume of this disc is \( \left(\pi{r_2}^2t\right) \).

So the mass of disc 2 is \( M= \left(\pi{r_2}^2t\right){\rho}_2\tag{2} \).

Since the two discs have the same mass M, so from equation (1) and equation (2) we can write,

\( \left(\pi{r_1}^2t\right){\rho}_1=\left(\pi{r_2}^2t\right){\rho}_2 \)

\( or,\ \displaystyle{\frac{{r_1}^2}{{r_2}^2}=\frac{{\rho}_2}{{\rho}_1}}\tag{3} \)

The moment of inertia of the disc 1 is

\( I_1=\frac{1}{2}M{r_1}^2 \)

[ To know the derivation of the moment of inertia of a solid circular disc about an axis passing through the centre of the disc and perpendicular to the plane of it, ( CLICK HERE ) ]

The moment of inertia of the disc 2 is

\( I_2=\frac{1}{2}M{r_2}^2 \)

Therefore, \( \displaystyle{\frac{I_1}{I_2}=\frac{{r_1}^2}{{r_2}^2}} \)

\( or,\ \displaystyle{\frac{I_1}{I_2}=\frac{{\rho}_2}{{\rho}_1}} \) [using equation (3)]

Since, \( {\rho}_2>{\rho}_1 \)

Therefore, \( I_1>I_2 \)

So. disc 1 has a larger moment of inertia than that of disc 2.