The moment of inertia of a uniform cicrular disc about its diameter is `200g cm^(2)`. Its moment of inertia about an axis passing through its centre and perpendicular to its circular face is

Options

A. `100 gm cm^2`
B. `200 gm cm^2`
C. `400 gm cm^2`
D. `1000 gm cm^2`

Answer

To find the moment of inertia of a uniform circular disc about an axis passing through its center and perpendicular to its circular face, we can use the relationship between the moment of inertia about the diameter and the moment of inertia about the perpendicular axis.

### Step-by-step Solution:

1. **Understanding the Moment of Inertia**:
The moment of inertia \( I \) of a body depends on the axis about which it is calculated. For a uniform circular disc, we have two important axes:
- About its diameter (let's call this \( I_d \))
- About an axis passing through its center and perpendicular to its face (let's call this \( I_c \))

2. **Given Data**:
- The moment of inertia about the diameter \( I_d = 200 \, g \, cm^2 \).

3. **Using the Perpendicular Axis Theorem**:
The perpendicular axis theorem states that for a planar body, the moment of inertia about an axis perpendicular to the plane (through the center) is equal to the sum of the moments of inertia about two perpendicular axes in the plane of the body (both passing through the center).
\[ I_c = I_d + I_d \]
This means:
\[ I_c = 2I_d \]

4. **Calculating the Moment of Inertia about the Center**:
Now substituting the given value of \( I_d \):
\[ I_c = 2 \times 200 \, g \, cm^2 = 400 \, g \, cm^2 \]

5. **Final Result**:
The moment of inertia of the uniform circular disc about an axis passing through its center and perpendicular to its circular face is:
\[ I_c = 400 \, g \, cm^2 \]