Question

Two discs of same mass and same thickness (t) are made from two different materials of densities $d_{1}$ and $d_{2}$ respectively. The ratio of the moment of inertia $I_{1}$ to $I_{2}$ of two discs about an axis passing through the centre and perpendicular to the plane of disc is

• A. $d_{1}:d_{2}$
• B. $d_{2}:d_{1}$
• C. $1:d_{1}d_{2}$
• D. $1:d_{1}^2d_{2}$

Hint:

Logic Tip: A denser material will have a smaller radius for the same mass, leading to a smaller moment of inertia.

Answer 1

The Correct Option is B

Concept: The moment of inertia of a solid disc about its central axis is: \[ I = \frac{1}{2}MR^2 \] For discs of equal mass, $I \propto R^2$.
Step 1: Relate mass, radius, and density Mass of a disc: \[ M = \text{Volume} \times \text{Density} \] \[ M = (\pi R^2 t)\, d \] where $t$ is thickness and $d$ is density. Since $M$ and $t$ are constant: \[ \pi R^2 t \cdot d = \text{constant} \] \[ R^2 \propto \frac{1}{d} \]
Step 2: Compare two discs For two discs: \[ \frac{R_1^2}{R_2^2} = \frac{d_2}{d_1} \]
Step 3: Ratio of moments of inertia Since $I \propto R^2$: \[ \frac{I_1}{I_2} = \frac{R_1^2}{R_2^2} \] \[ \Rightarrow \frac{I_1}{I_2} = \frac{d_2}{d_1} \] Final Answer: \[ \boxed{\frac{I_1}{I_2} = \frac{d_2}{d_1 \]