Question

Two discs have the same moment of inertia about their axes. Their thicknesses are \(t_1\) and \(t_2\) and they have the same density. If \( \dfrac{R_1}{R_2} = \dfrac{1}{2} \), then find \( \dfrac{t_1}{t_2} \).

• A. \( \dfrac{1}{16} \)
• B. \( 16 \)
• C. \( \dfrac{1}{4} \)
• D. \( 4 \)

Hint:

For objects of same density, always express mass in terms of geometry first. In disc problems, remember: \( M \propto R^2 t \) and \( I \propto R^4 t \).

Answer 1

The Correct Option is B

Concept: The moment of inertia of a uniform solid disc about its central axis is given by: \[ I = \frac{1}{2}MR^2 \] Mass of a disc depends on its density (\(\rho\)), radius, and thickness: \[ M = \rho \times \text{Volume} = \rho \pi R^2 t \]
Step 1: Write expressions for moment of inertia of both discs. \[ I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2} (\rho \pi R_1^2 t_1) R_1^2 = \frac{1}{2}\rho \pi R_1^4 t_1 \] \[ I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2} (\rho \pi R_2^2 t_2) R_2^2 = \frac{1}{2}\rho \pi R_2^4 t_2 \]
Step 2: Use the condition that the moments of inertia are equal. \[ I_1 = I_2 \] \[ \frac{1}{2}\rho \pi R_1^4 t_1 = \frac{1}{2}\rho \pi R_2^4 t_2 \] Cancel common terms: \[ R_1^4 t_1 = R_2^4 t_2 \]
Step 3: Express the ratio \( \dfrac{t_1}{t_2} \). \[ \frac{t_1}{t_2} = \left(\frac{R_2}{R_1}\right)^4 \] Given: \[ \frac{R_1}{R_2} = \frac{1}{2} \Rightarrow \frac{R_2}{R_1} = 2 \] \[ \frac{t_1}{t_2} = 2^4 = 16 \]