Question

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

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

Hint:

For objects of the same material:
Always express mass in terms of density and volume
Look for proportional relationships to simplify calculations

Answer 1

The Correct Option is B

Concept:  
The moment of inertia \( I \) of a uniform solid disc about its central axis is given by: \[ I = \frac{1}{2} M R^2 \] For a disc with density \( \rho \), radius \( R \), and thickness \( t \): \[ M = \rho \times \text{Volume} = \rho \pi R^2 t \] Substituting this into the formula for the moment of inertia: \[ I = \frac{1}{2} \rho \pi R^4 t \] Thus, for discs of the same material: \[ I \propto R^4 t \] 
Step 1: Use the condition of equal moments of inertia.
Given that the moments of inertia of both discs are equal: \[ I_1 = I_2 \quad \Rightarrow \quad R_1^4 t_1 = R_2^4 t_2 \] 
Step 2: Substitute the given radius ratio.
We are given the radius ratio as \( \frac{R_1}{R_2} = \frac{1}{2} \). Therefore: \[ \left(\frac{R_1}{R_2}\right)^4 = \frac{1}{16} \] Substitute this into the equation: \[ \frac{1}{16} \cdot t_1 = t_2 \] 
Step 3: Find the required ratio.
Rearranging the equation to find the ratio of thicknesses: \[ \frac{t_1}{t_2} = 16 \] 
Final Answer: \[ \boxed{\frac{t_1}{t_2} = 16}\]