Moment of inertia of a thin circular plate of mass M, radius R about an axis passing through its diameter is I. The moment of inertia of a circular ring of mass M, radius R about an axis perpendicular to its plane and passing through its centre is:
1) 2I
2) I/2
3) 4I
4) I/4

Problem Analysis:
We are given the moment of inertia (II) of a thin circular plate of mass MM and radius RR about an axis passing through its diameter. The task is to find the moment of inertia of a circular ring of the same mass MM and radius RR about an axis perpendicular to its plane and passing through its center.
Key Concepts:
(1) Moment of Inertia of a Thin Circular Plate:
• The moment of inertia of a thin circular plate about its diameter is given by:
Iplate (diameter)=14MR2.I_{\text{plate (diameter)}} = \frac{1}{4} M R^2.
This is provided in the problem as II, so:
I=14MR2.(1)I = \frac{1}{4} M R^2. \tag{1}
(2) Moment of Inertia of a Circular Ring:
• The moment of inertia of a circular ring of mass MM and radius RR about an axis perpendicular to its plane and passing through its center is:
Iring (perpendicular)=MR2.(2)I_{\text{ring (perpendicular)}} = M R^2. \tag{2}
Step 1: Relating the Moments of Inertia
From Equation (1):
I=14MR2.I = \frac{1}{4} M R^2.
From Equation (2):
Iring (perpendicular)=MR2.I_{\text{ring (perpendicular)}} = M R^2.
Divide Iring (perpendicular)I_{\text{ring (perpendicular)}} by II:
Iring (perpendicular)I=MR214MR2.\frac{I_{\text{ring (perpendicular)}}}{I} = \frac{M R^2}{\frac{1}{4} M R^2}.
Simplify:
Iring (perpendicular)I=4.\frac{I_{\text{ring (perpendicular)}}}{I} = 4.
Thus:
Iring (perpendicular)=4I.I_{\text{ring (perpendicular)}} = 4I.
The moment of inertia of the circular ring about an axis perpendicular to its plane and passing through its center is:
4I.\boxed{4I}.