Question

A circular disc \( X \) of radius \( R \) made from iron plate of thickness \( t \) has moment of inertia \( I_x \) about an axis passing through the centre of the disc and perpendicular to its plane. Another disc \( Y \) of radius \( 3R \) made from an iron plate of thickness \( \frac{t}{3} \) has moment of inertia \( I_y \) about the axis same as that of disc X. The relation between \( I_x \) and \( I_y \) is:

• A. \( I_y = 9 I_x \)
• B. \( I_y = I_x \)
• C. \( I_y = 27 I_x \)
• D. \( I_y = 31 I_x \)

Hint:

The moment of inertia of a disc is directly proportional to the square of its radius and its thickness. If both radius and thickness change, the effect on moment of inertia can be calculated using their proportional relationships.

Answer 1

The Correct Option is C

Step 1: Moment of Inertia of a Disc.
The moment of inertia of a disc about an axis through its centre and perpendicular to its plane is given by: \[ I = \frac{1}{2} M R^2 \] where \( M \) is the mass of the disc and \( R \) is its radius. The mass of the disc is proportional to its volume, and since the thickness \( t \) of the discs is involved, we have: \[ I_x \propto R^2 t \] For disc \( X \), the moment of inertia is proportional to \( R^2 t \). For disc \( Y \), with radius \( 3R \) and thickness \( \frac{t}{3} \), the moment of inertia becomes: \[ I_y \propto (3R)^2 \cdot \frac{t}{3} = 9R^2 \cdot \frac{t}{3} = 27 R^2 t \] Thus, the ratio of the moments of inertia is: \[ I_y = 27 I_x \] Step 2: Final Answer.
Thus, the relation between \( I_x \) and \( I_y \) is \( I_y = 27 I_x \).