The moment of inerta of a ring of mass `1kg` about an axis passing through its centre perpendicular to its surface is `4kgm^(2)`. Calculate the radius of the ring.

Explanation

To solve the problem, we will use the formula for the moment of inertia of a ring about an axis passing through its center and perpendicular to its surface. The formula is given by:

\[ I = m r^2 \]

Where:
- \( I \) is the moment of inertia,
- \( m \) is the mass of the ring,
- \( r \) is the radius of the ring.
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