Question

If the radius of gyration of a thin circular ring about an axis passing through its centre and perpendicular to its plane is $10\sqrt{2}$ cm, then its radius about its diameter is
Identify the correct option from the following:

• A. 10 cm
• B. 20 cm
• C. $10\sqrt{2}$ cm
• D. $20\sqrt{2}$ cm

Hint:

Use the perpendicular axis theorem for a ring: the moment of inertia about a diameter is half the moment of inertia about the perpendicular axis through the center.

Answer 1

The Correct Option is A

Step 1: Relate radius of gyration to moment of inertia
The radius of gyration $k$ about an axis is given by $I = m k^2$, where $I$ is the moment of inertia and $m$ is the mass. For a thin circular ring of radius $r$ about an axis through its center and perpendicular to its plane, $I = m r^2$, so $k = \sqrt{\frac{I}{m}} = r$. Given $k = 10\sqrt{2}$ cm, the radius $r = 10\sqrt{2}$ cm. Step 2: Find the radius of gyration about the diameter
For a thin circular ring, the moment of inertia about a diameter (using the perpendicular axis theorem) is $I_{\text{diameter}} = \frac{m r^2}{2}$. The radius of gyration about the diameter is $k_{\text{diameter}} = \sqrt{\frac{I_{\text{diameter}}}{m}}} = \sqrt{\frac{m r^2 / 2}{m}}} = \frac{r}{\sqrt{2}}$. Substitute $r = 10\sqrt{2}$ cm: $k_{\text{diameter}} = \frac{10\sqrt{2}}{\sqrt{2}}} = 10$ cm. Step 3: Match with options
The radius of gyration about the diameter is 10 cm, matching option (1).