The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and a circular ring of the same radius about a tengential axis in the plane of the ring is

Options

A. `2 : 1`
B. `sqrt5:sqrt6`
C. `2 : 3`
D. `1:sqrt2`

Explanation

To solve the problem, we need to find the ratio of the radii of gyration of a circular disc and a circular ring about a tangential axis in their respective planes. Let's break this down step by step.

### Step 1: Understand the Moment of Inertia for the Disc
1. The moment of inertia \( I \) of a circular disc about its center is given by:
\[ I_{\text{disc}} = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the disc and \( R \) is its radius.

2. To find the moment of inertia about a tangential axis (using the parallel axis theorem):
\[ I_1 = I_{\text{disc}} + M \cdot d^2 \]
Here, \( d = R \) (the distance from the center to the tangential axis).
\[ I_1 = \frac{1}{2} M R^2 + M R^2 = \frac{1}{2} M R^2 + 1 M R^2 = \frac{3}{2} M R^2 \]

### Step 2: Find the Radius of Gyration for the Disc
3. The radius of gyration \( k_1 \) for the disc is defined as:
\[ I_1 = M k_1^2 \]
Substituting \( I_1 \):
\[ \frac{3}{2} M R^2 = M k_1^2 \]
Dividing both sides by \( M \):
\[ k_1^2 = \frac{3}{2} R^2 \]
Taking the square root:
\[ k_1 = R \sqrt{\frac{3}{2}} \]

### Step 3: Understand the Moment of Inertia for the Ring
4. The moment of inertia \( I \) of a circular ring about its center is given by:
\[ I_{\text{ring}} = M R^2 \]

5. Again, using the parallel axis theorem for the tangential axis:
\[ I_2 = I_{\text{ring}} + M \cdot d^2 \]
Here, \( d = R \):
\[ I_2 = M R^2 + M R^2 = 2 M R^2 \]

### Step 4: Find the Radius of Gyration for the Ring
6. The radius of gyration \( k_2 \) for the ring is defined as:
\[ I_2 = M k_2^2 \]
Substituting \( I_2 \):
\[ 2 M R^2 = M k_2^2 \]
Dividing both sides by \( M \):
\[ k_2^2 = 2 R^2 \]
Taking the square root:
\[ k_2 = R \sqrt{2} \]

### Step 5: Find the Ratio of the Radii of Gyration
7. Now we find the ratio of the radii of gyration \( k_1 \) and \( k_2 \):
\[ \frac{k_1}{k_2} = \frac{R \sqrt{\frac{3}{2}}}{R \sqrt{2}} = \frac{\sqrt{\frac{3}{2}}}{\sqrt{2}} = \frac{\sqrt{3}}{2} \]

### Final Answer
The ratio of the radii of gyration of the circular disc to the circular ring about their respective tangential axes is:
\[ \frac{\sqrt{3}}{2} \]