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 establish the relationship between the length \( L \) and radius \( R \) of a solid cylinder based on the equality of its moments of inertia about two different axes.

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

### Step 2: Moment of inertia about an axis through the center of gravity and perpendicular to its length
For a solid cylinder, the moment of inertia about an axis passing through its center of gravity and perpendicular to its length (let's call this axis \( I_{\text{perpendicular}} \)) can be calculated using the parallel axis theorem. The formula is:
\[ I_{\text{perpendicular}} = \frac{1}{12} m (3R^2 + L^2) \]

### Step 3: Set the two moments of inertia equal
According to the problem, these two moments of inertia are equal:
\[ \frac{1}{2} m R^2 = \frac{1}{12} m (3R^2 + L^2) \]

### Step 4: Simplify the equation
Since \( m \) appears on both sides, we can cancel it out:
\[ \frac{1}{2} R^2 = \frac{1}{12} (3R^2 + L^2) \]

Now, multiply both sides by 12 to eliminate the fraction:
\[ 6 R^2 = 3 R^2 + L^2 \]

### Step 5: Rearrange the equation
Now, rearranging gives:
\[ 6 R^2 - 3 R^2 = L^2 \]
\[ 3 R^2 = L^2 \]

### Step 6: Solve for the relationship between \( L \) and \( R \)
Taking the square root of both sides, we find:
\[ L = \sqrt{3} R \]

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