The moment of inertia of a solid cylinder about its own axis is the same as its moment of inertia about an axis passing through its centre of gravity and perpendicular to its length. The relation between its length L and radius R is :

Options

A. `L=sqrt(2)R`
B. `L=sqrt(3)R`
C. `L=3R`
D. `L=R`

Answer

To solve the problem, we need to find the relationship between the length \( L \) and the radius \( R \) of a solid cylinder, given that its moment of inertia about its own axis is the same as its moment of inertia about an axis passing through its center of gravity and perpendicular to its length.

### Step-by-Step Solution:

1. **Identify the Moments of Inertia**:
- The moment of inertia of a solid cylinder about its own axis (let's call this \( I_1 \)) is given by:
\[ I_1 = \frac{1}{2} MR^2 \]
- The moment of inertia about an axis passing through its center of gravity and perpendicular to its length (let's call this \( I_2 \)) is given by:
\[ I_2 = \frac{1}{12} ML^2 + \frac{1}{4} MR^2 \]

2. **Set the Moments of Inertia Equal**:
- According to the problem, \( I_1 = I_2 \). Therefore, we can set up the equation:
\[ \frac{1}{2} MR^2 = \frac{1}{12} ML^2 + \frac{1}{4} MR^2 \]

3. **Simplify the Equation**:
- First, we can cancel \( M \) from both sides since \( M \neq 0 \):
\[ \frac{1}{2} R^2 = \frac{1}{12} L^2 + \frac{1}{4} R^2 \]
- Rearranging gives:
\[ \frac{1}{2} R^2 - \frac{1}{4} R^2 = \frac{1}{12} L^2 \]
- Simplifying the left side:
\[ \frac{2}{4} R^2 - \frac{1}{4} R^2 = \frac{1}{12} L^2 \]
\[ \frac{1}{4} R^2 = \frac{1}{12} L^2 \]

4. **Cross Multiply to Solve for \( L \)**:
- Cross multiplying gives:
\[ 12 \cdot \frac{1}{4} R^2 = L^2 \]
\[ 3R^2 = L^2 \]

5. **Take the Square Root**:
- Taking the square root of both sides results in:
\[ L = R \sqrt{3} \]

### Final Answer:
The relation between the length \( L \) and the radius \( R \) of the solid cylinder is:
\[ L = R \sqrt{3} \]