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 find the relation between the length \( L \) and radius \( R \) of a cylinder when its moment of inertia about its axis is equal to that about the equatorial axis, we can follow these steps:

### Step 1: Write the moment of inertia about the axis of the cylinder
The moment of inertia \( I_1 \) of a solid cylinder about its own axis is given by the formula:
\[ I_1 = \frac{1}{2} m R^2 \]
where \( m \) is the mass of the cylinder and \( R \) is its radius.

### Step 2: Write the moment of inertia about the equatorial axis
The moment of inertia \( I_2 \) of a solid cylinder about an axis through its center and perpendicular to its length (equatorial axis) can be calculated using the parallel axis theorem:
\[ I_2 = \frac{1}{12} m L^2 + m R^2 \]
Here, \( L \) is the length of the cylinder.

### Step 3: Set the two moments of inertia equal
According to the problem, we have:
\[ I_1 = I_2 \]
Substituting the expressions we found:
\[ \frac{1}{2} m R^2 = \frac{1}{12} m L^2 + m R^2 \]

### Step 4: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} R^2 = \frac{1}{12} L^2 + R^2 \]
Now, rearranging gives:
\[ \frac{1}{2} R^2 - R^2 = \frac{1}{12} L^2 \]
This simplifies to:
\[ -\frac{1}{2} R^2 = \frac{1}{12} L^2 \]

### Step 5: Multiply through by -1
To make the equation easier to work with, multiply both sides by -1:
\[ \frac{1}{2} R^2 = -\frac{1}{12} L^2 \]

### Step 6: Clear fractions by multiplying by 12
Multiply through by 12 to eliminate the fractions:
\[ 6 R^2 = -L^2 \]

### Step 7: Rearranging and solving for \( L \)
Rearranging gives:
\[ L^2 = 6 R^2 \]
Taking the square root of both sides results in:
\[ L = \sqrt{6} R \]

### Conclusion
Thus, the relation between the length \( L \) and radius \( R \) of the cylinder is:
\[ L = \sqrt{6} R \]