The moment of inertia of a solid cylinder about its natural axis is `I`. If its moment of inertia about an axis `bot^r` to natural axis of cylinder and passing through one end of cylinder is `191//6` then the ratio of radius of cylinder and its length is

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `2:3`

Explanation

To solve the problem, we need to find the ratio of the radius (R) of a solid cylinder to its length (L) given the moment of inertia about different axes.

### Step-by-Step Solution:

1. **Identify the moment of inertia about the natural axis**:
The moment of inertia \( I \) of a solid cylinder about its natural axis (which is the axis through the center and along the length of the cylinder) is given by the formula:
\[ I = \frac{1}{2} m R^2 \]

2. **Moment of inertia about the perpendicular axis through one end**:
The moment of inertia \( I' \) about an axis perpendicular to the natural axis and passing through one end of the cylinder is given as:
\[ I' = \frac{1}{3} m L^2 + \frac{1}{4} m R^2 \]
According to the problem, \( I' = \frac{19}{6} I \).

3. **Substituting the value of \( I \)**:
Substitute \( I = \frac{1}{2} m R^2 \) into the equation for \( I' \):
\[ I' = \frac{19}{6} \left(\frac{1}{2} m R^2\right) = \frac{19}{12} m R^2 \]

4. **Setting the two expressions for \( I' \) equal**:
Now, we can set the two expressions for \( I' \) equal to each other:
\[ \frac{1}{3} m L^2 + \frac{1}{4} m R^2 = \frac{19}{12} m R^2 \]

5. **Canceling \( m \) from the equation**:
Since \( m \) appears in all terms, we can cancel it out:
\[ \frac{1}{3} L^2 + \frac{1}{4} R^2 = \frac{19}{12} R^2 \]

6. **Rearranging the equation**:
Rearranging gives:
\[ \frac{1}{3} L^2 = \frac{19}{12} R^2 - \frac{1}{4} R^2 \]
To combine the right-hand side, we need a common denominator:
\[ \frac{1}{4} R^2 = \frac{3}{12} R^2 \]
Thus:
\[ \frac{1}{3} L^2 = \frac{19}{12} R^2 - \frac{3}{12} R^2 = \frac{16}{12} R^2 = \frac{4}{3} R^2 \]

7. **Solving for \( L^2 \)**:
Multiply both sides by 3:
\[ L^2 = 4 R^2 \]

8. **Taking the square root**:
Taking the square root gives:
\[ L = 2R \]

9. **Finding the ratio \( \frac{R}{L} \)**:
The ratio of the radius to the length is:
\[ \frac{R}{L} = \frac{R}{2R} = \frac{1}{2} \]

### Final Answer:
The ratio of the radius of the cylinder to its length is \( 1:2 \).