A solid cylinder of mass 20kg has length 1 m and radius 0.2 m. Then its moment of inertia (in kg-`m^(2)`) about its geometrical axis is :

Options

A. 0.8kg-`m^(2)`
B. 0.4kg-`m^(2)`
C. 0.2 kg-`m^(2)`
D. 20.2 kg-`m^(2)`

Answer

To find the moment of inertia of a solid cylinder about its geometrical axis, we can use the formula:

\[ I = \frac{1}{2} m r^2 \]

where:
- \( I \) is the moment of inertia,
- \( m \) is the mass of the cylinder,
- \( r \) is the radius of the cylinder.

### Step 1: Identify the given values
From the problem, we have:
- Mass \( m = 20 \, \text{kg} \)
- Radius \( r = 0.2 \, \text{m} \)

### Step 2: Substitute the values into the formula
Now, we substitute the values into the moment of inertia formula:

\[ I = \frac{1}{2} \times 20 \, \text{kg} \times (0.2 \, \text{m})^2 \]

### Step 3: Calculate \( r^2 \)
First, calculate \( r^2 \):

\[ (0.2 \, \text{m})^2 = 0.04 \, \text{m}^2 \]

### Step 4: Substitute \( r^2 \) back into the equation
Now substitute \( r^2 \) back into the equation:

\[ I = \frac{1}{2} \times 20 \, \text{kg} \times 0.04 \, \text{m}^2 \]

### Step 5: Calculate the moment of inertia
Now, perform the multiplication:

\[ I = \frac{1}{2} \times 20 \times 0.04 = 10 \times 0.04 = 0.4 \, \text{kg m}^2 \]

### Final Answer
Thus, the moment of inertia of the solid cylinder about its geometrical axis is:

\[ I = 0.4 \, \text{kg m}^2 \]