Question

Two discs having the same moment of inertia about their axis. Thickness is \( t_1 \) and \( t_2 \), and they have the same density. If \( \frac{R_1}{R_2} = \frac{1}{2} \), then find \( \frac{t_1}{t_2} \).

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

Hint:

For two discs with the same moment of inertia and density, the ratio of thicknesses is proportional to the square of the ratio of radii.

Answer 1

The Correct Option is B

Given:
Two discs have the same moment of inertia about their axes.
Thicknesses are \(t_1\) and \(t_2\).
Density of both discs is the same.
\(\dfrac{R_1}{R_2} = \dfrac{1}{2}\)

Step 1: Moment of inertia of a disc about its axis

\[ I = \frac{1}{2}MR^2 \]

Since the moments of inertia are equal:

\[ \frac{1}{2}M_1R_1^2 = \frac{1}{2}M_2R_2^2 \]

\[ M_1R_1^2 = M_2R_2^2 \]

Step 2: Express mass in terms of density

Mass of a disc:

\[ M = \rho \times \text{Volume} = \rho \pi R^2 t \]

Hence,

\[ M_1 = \rho \pi R_1^2 t_1,\quad M_2 = \rho \pi R_2^2 t_2 \]

Step 3: Substitute masses

\[ (\rho \pi R_1^2 t_1)R_1^2 = (\rho \pi R_2^2 t_2)R_2^2 \]

\[ \rho \pi R_1^4 t_1 = \rho \pi R_2^4 t_2 \]

Step 4: Simplify

\[ \frac{t_1}{t_2} = \frac{R_2^4}{R_1^4} \]

Step 5: Substitute the given ratio

\[ \frac{R_1}{R_2} = \frac{1}{2} \Rightarrow \frac{R_2}{R_1} = 2 \]

\[ \frac{t_1}{t_2} = (2)^4 = 16 \]

Final Answer:

\[ \boxed{\dfrac{t_1}{t_2} = 16} \]