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 tangential axis in the plane of the ring is

Options

A. `sqrt(3) : sqrt(4)`
B. `sqrt(5) : sqrt(6)`
C. `sqrt(6) : sqrt(5)`
D. `sqrt(4) : sqrt(3)`

Explanation

To solve the problem of finding the ratio of the radii of gyration of a circular disc and a circular ring about a tangential axis in their respective planes, we can follow these steps:

### Step 1: Define the Variables
Let:
- \( R \) = radius of both the circular disc and the circular ring
- \( M_D \) = mass of the disc
- \( M_R \) = mass of the ring

### Step 2: Moment of Inertia about the Center of Mass
The moment of inertia about the center of mass for each shape is given by:
- For the circular disc:
\[ I_{D, O} = \frac{1}{4} M_D R^2 \]
- For the circular ring:
\[ I_{R, O} = \frac{1}{2} M_R R^2 \]

### Step 3: Apply the Parallel Axis Theorem
Since we need the moment of inertia about a tangential axis (let's call it axis AB), we can use the parallel axis theorem which states:
\[ I_{AB} = I_{O} + M d^2 \]
where \( d \) is the distance from the center of mass to the new axis. For both the disc and the ring, \( d = R \).

- For the circular disc:
\[ I_{D, AB} = I_{D, O} + M_D R^2 = \frac{1}{4} M_D R^2 + M_D R^2 = \frac{5}{4} M_D R^2 \]

- For the circular ring:
\[ I_{R, AB} = I_{R, O} + M_R R^2 = \frac{1}{2} M_R R^2 + M_R R^2 = \frac{3}{2} M_R R^2 \]

### Step 4: Calculate the Radii of Gyration
The radius of gyration \( K \) is defined as:
\[ K = \sqrt{\frac{I}{M}} \]

- For the circular disc:
\[ K_D = \sqrt{\frac{I_{D, AB}}{M_D}} = \sqrt{\frac{\frac{5}{4} M_D R^2}{M_D}} = \sqrt{\frac{5}{4}} R = R \frac{\sqrt{5}}{2} \]

- For the circular ring:
\[ K_R = \sqrt{\frac{I_{R, AB}}{M_R}} = \sqrt{\frac{\frac{3}{2} M_R R^2}{M_R}} = \sqrt{\frac{3}{2}} R = R \frac{\sqrt{3}}{\sqrt{2}} \]

### Step 5: Find the Ratio of the Radii of Gyration
Now, we can find the ratio of the radii of gyration \( K_D \) to \( K_R \):
\[ \frac{K_D}{K_R} = \frac{R \frac{\sqrt{5}}{2}}{R \frac{\sqrt{3}}{\sqrt{2}}} = \frac{\frac{\sqrt{5}}{2}}{\frac{\sqrt{3}}{\sqrt{2}}} = \frac{\sqrt{5} \cdot \sqrt{2}}{2 \cdot \sqrt{3}} = \frac{\sqrt{10}}{2\sqrt{3}} \]

### Step 6: Simplify the Ratio
To express it in a more standard form:
\[ \frac{K_D}{K_R} = \frac{\sqrt{10}}{2\sqrt{3}} = \frac{\sqrt{10}}{\sqrt{12}} = \frac{\sqrt{10}}{\sqrt{4 \cdot 3}} = \frac{\sqrt{10}}{2\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{6}} \]

### Final Answer
Thus, the ratio of the radii of gyration of the circular disc to that of the circular ring is:
\[ \frac{K_D}{K_R} = \frac{\sqrt{5}}{\sqrt{6}} \]