Question

A circular disc has radius \( R_1 \) and thickness \( T_1 \). Another circular disc made of the same material has radius \( R_2 \) and thickness \( T_2 \). If the moments of inertia of both the discs are same and \[ \frac{R_1}{R_2} = 2, \quad \text{then} \quad \frac{T_1}{T_2} = \frac{1}{\alpha}. \] The value of \( \alpha \) is __________.

Hint:

For bodies made of the same material, mass can be replaced by density times volume to compare moments of inertia.

Answer 1

Correct Answer: 8

Step 1: Write expression for moment of inertia.
For a uniform circular disc about its central axis,
\[ I = \frac{1}{2}MR^2. \] Step 2: Express mass in terms of dimensions.
Since both discs are made of the same material, density \( \rho \) is same.
\[ M = \rho \times \text{Volume} = \rho \pi R^2 T. \] Thus,
\[ I = \frac{1}{2} \rho \pi R^4 T. \] Step 3: Equate moments of inertia.
\[ \frac{1}{2} \rho \pi R_1^4 T_1 = \frac{1}{2} \rho \pi R_2^4 T_2. \] Cancelling common terms,
\[ R_1^4 T_1 = R_2^4 T_2. \] Step 4: Substitute given ratio.
\[ \left(\frac{R_1}{R_2}\right)^4 = 2^4 = 16. \] So,
\[ \frac{T_1}{T_2} = \frac{1}{16}. \] Comparing with
\[ \frac{T_1}{T_2} = \frac{1}{\alpha}, \] we get
\[ \alpha = 8. \] Final Answer:
\[ \boxed{8} \]