Moment of inertia of a uniform circular disc about a diameter is `I`. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be.

Options

A. 5 I
B. 3 I
C. 6 I
D. 4 I

Explanation

To find the moment of inertia of a uniform circular disc about an axis perpendicular to its plane and passing through a point on its rim, we can use the parallel axis theorem. Here’s a step-by-step solution:

### Step 1: Understand the given moment of inertia
The moment of inertia of the disc about a diameter is given as \( I \). For a uniform circular disc, the moment of inertia about a diameter is calculated using the formula:
\[ I = \frac{1}{4} m r^2 \]
where \( m \) is the mass of the disc and \( r \) is the radius.

### Step 2: Use the parallel axis theorem
The parallel axis theorem states that if you know the moment of inertia about an axis through the center of mass, you can find the moment of inertia about any parallel axis by:
\[ I' = I + md^2 \]
where \( I' \) is the moment of inertia about the new axis, \( I \) is the moment of inertia about the center of mass axis, \( m \) is the mass, and \( d \) is the distance between the two axes.

### Step 3: Identify the axes
In this case:
- The moment of inertia about the diameter (center of mass axis) is \( I \).
- The new axis is perpendicular to the plane of the disc and passes through a point on its rim. The distance \( d \) from the center of the disc to the rim is equal to the radius \( r \).

### Step 4: Substitute the values into the parallel axis theorem
Using the parallel axis theorem:
\[ I' = I + m r^2 \]
We already know that \( I = \frac{1}{4} m r^2 \).

### Step 5: Substitute \( I \) into the equation
Substituting \( I \) into the equation:
\[ I' = \frac{1}{4} m r^2 + m r^2 \]
\[ I' = \frac{1}{4} m r^2 + \frac{4}{4} m r^2 = \frac{5}{4} m r^2 \]

### Step 6: Relate \( m r^2 \) to \( I \)
Since we know \( I = \frac{1}{4} m r^2 \), we can express \( m r^2 \) in terms of \( I \):
\[ m r^2 = 4I \]

### Step 7: Substitute back into the equation for \( I' \)
Now substituting \( m r^2 \) back into the equation for \( I' \):
\[ I' = \frac{5}{4} (4I) = 5I \]

### Step 8: Final calculation
Now we need to add the moment of inertia about the center of mass axis (which is \( \frac{1}{4} m r^2 \)) to the moment of inertia about the new axis:
\[ I' = \frac{1}{4} m r^2 + m r^2 = \frac{1}{4} m r^2 + \frac{4}{4} m r^2 = \frac{5}{4} m r^2 \]

### Step 9: Conclusion
Thus, the moment of inertia of the disc about the axis perpendicular to its plane and passing through a point on its rim is:
\[ I' = 5I \]