Question

Two disc of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio \(1: 3\). The moment of inertia of these disc about the respective axes passing through their centres and perpendicular to their planes will be in the ratio. \(\\{\mathrm{A}\\} 1: 3\) \\{B\\} \(3: 1\) \\{C\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)

Short answer

The ratio of the moments of inertia of the two discs is \({B}\\ \, 3:1 \).

Step-by-step solution

Determine the formula for the moment of inertia of a disc

The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is given by the formula: \[I = \frac{1}{2}MR^2\] Where: \(I\) is the moment of inertia \(M\) is the mass of the disc \(R\) is the radius of the disc

Set up the relationship between the given densities and radii

The density \(\rho\) of a material is defined as the mass per unit volume: \(\rho = \frac{M}{V}\). For a circular disc, the volume can be expressed in terms of the radius and thickness: \(V = \pi R^2 h\), where \(h\) is the thickness. Since both discs have the same mass, we can write: \[\frac{\rho_1}{\rho_2} = \frac{3}{1} = \frac{M_1 R_1^2 h}{M_2 R_2^2 h}\] Since \(M_1 = M_2\) and the thicknesses are equal, we can cancel them out: \[\frac{3}{1} = \frac{R_1^2}{R_2^2}\]

Find the ratio of the moments of inertia of the two discs

We want to find the ratio \(\frac{I_1}{I_2}\), where \(I_1 = \frac{1}{2}M_1R_1^2\) and \(I_2 = \frac{1}{2}M_2R_2^2\). Using the known ratio of densities, we can calculate the ratio of moments of inertia: \[\frac{I_1}{I_2} = \frac{\frac{1}{2}M_1R_1^2}{\frac{1}{2}M_2R_2^2} = \frac{R_1^2}{R_2^2}\] Using the relation obtained in Step 2 for \(\frac{R_1^2}{R_2^2}\), we can write: \[\frac{I_1}{I_2} = \frac{3}{1} = 3:1\] So the answer to the problem is: \[{B}\\ \, 3:1 \]