The mass and radius of a solid cylinder be M and R respectively. Its M.I. about a generator line will be:

Options

A. `M((L^(2))/(12)+(R^(2))/(4))`
B. `M((L^(2))/(3)+(R^(2))/(4))`
C. `(MR^(2))/(2)`
D. `(3MR^(2))/(2)`

Answer

To find the moment of inertia (M.I.) of a solid cylinder about its generator line, we will follow these steps:

### Step 1: Understand the Geometry of the Cylinder
A solid cylinder has a mass \( M \) and a radius \( R \). The generator line is an axis that runs along the length of the cylinder.

### Step 2: Use the Parallel Axis Theorem
The moment of inertia about the generator line can be calculated using the parallel axis theorem. The parallel axis theorem states that:
\[ I = I_{cm} + Md^2 \]
where:
- \( I \) is the moment of inertia about the new axis (generator line).
- \( I_{cm} \) is the moment of inertia about the center of mass axis.
- \( M \) is the mass of the object.
- \( d \) is the distance between the center of mass axis and the new axis.

### Step 3: Calculate \( I_{cm} \) for the Cylinder
For a solid cylinder, the moment of inertia about its central axis (which is perpendicular to the height) is given by:
\[ I_{cm} = \frac{1}{2} M R^2 \]

### Step 4: Determine the Distance \( d \)
In this case, the distance \( d \) from the center of mass axis to the generator line is \( \frac{R}{2} \) since the generator line runs along the length of the cylinder at its surface.

### Step 5: Substitute Values into the Parallel Axis Theorem
Now, substituting \( I_{cm} \) and \( d \) into the parallel axis theorem:
\[ I = I_{cm} + Md^2 \]
\[ I = \frac{1}{2} M R^2 + M \left(\frac{R}{2}\right)^2 \]
\[ I = \frac{1}{2} M R^2 + M \frac{R^2}{4} \]

### Step 6: Simplify the Expression
Now, combine the terms:
\[ I = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 \]
To combine these, find a common denominator:
\[ I = \frac{2}{4} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2 \]

### Final Answer
Thus, the moment of inertia of the solid cylinder about its generator line is:
\[ I = \frac{3}{4} M R^2 \]

---