The relation between length L and radius R of the cylinder , if its moment of inertia about its axis is equal to that about the equatorial axis, will be

Options

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

Answer

To solve the problem, we need to find the relationship between the length \( L \) and the radius \( R \) of a cylinder when its moment of inertia about its axis is equal to that about its equatorial axis.

### Step-by-Step Solution:

1. **Identify the Moments of Inertia**:
- The moment of inertia of a solid cylinder about its own axis (longitudinal axis) is given by:
\[ I_1 = \frac{1}{2} m R^2 \]
- The moment of inertia about the equatorial axis (perpendicular to the length of the cylinder) can be calculated using the parallel axis theorem. The moment of inertia about the equatorial axis is:
\[ I_2 = \frac{1}{12} m L^2 + m \left(\frac{L}{2}\right)^2 \]
Simplifying this gives:
\[ I_2 = \frac{1}{12} m L^2 + \frac{1}{4} m L^2 = \left(\frac{1}{12} + \frac{3}{12}\right) m L^2 = \frac{1}{3} m L^2 \]

2. **Set the Moments of Inertia Equal**:
- According to the problem, we have:
\[ I_1 = I_2 \]
- Substituting the expressions we derived:
\[ \frac{1}{2} m R^2 = \frac{1}{3} m L^2 \]

3. **Cancel the Mass**:
- Since mass \( m \) appears on both sides, we can cancel it out:
\[ \frac{1}{2} R^2 = \frac{1}{3} L^2 \]

4. **Cross-Multiply to Solve for L and R**:
- Cross-multiplying gives:
\[ 3R^2 = 2L^2 \]

5. **Rearranging the Equation**:
- Rearranging the equation to express \( L \) in terms of \( R \):
\[ L^2 = \frac{3}{2} R^2 \]
- Taking the square root of both sides:
\[ L = R \sqrt{\frac{3}{2}} = R \frac{\sqrt{6}}{2} \]

6. **Final Relationship**:
- Therefore, the relationship between the length \( L \) and the radius \( R \) of the cylinder is:
\[ L = \frac{\sqrt{6}}{2} R \]