For the same total mass, which of the following will have the largest moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the body

Options

A. a disc of radius R
B. a ring of radius R
C. a square lamina of side 2 R
D. four rods forming a square of side 2R

Explanation

To determine which of the given shapes has the largest moment of inertia about an axis passing through the center of mass and perpendicular to the plane of the body, we will calculate the moment of inertia for each shape. Let's denote the total mass of each shape as \( m \).

### Step 1: Moment of Inertia of the Disc
The moment of inertia \( I \) of a disc about an axis through its center and perpendicular to its plane is given by:
\[ I_{\text{disc}} = \frac{1}{2} m r^2 \]

### Step 2: Moment of Inertia of the Ring
The moment of inertia \( I \) of a ring about an axis through its center and perpendicular to its plane is given by:
\[ I_{\text{ring}} = m r^2 \]

### Step 3: Moment of Inertia of the Square Lamina
For a square lamina with side length \( 2r \), the moment of inertia about an axis through its center and perpendicular to its plane is given by:
\[ I_{\text{square}} = \frac{1}{6} m (2r)^2 = \frac{2}{3} m r^2 \]

### Step 4: Moment of Inertia of the Square Formed by Four Rods
For a square formed by four rods, we can calculate the moment of inertia using the parallel axis theorem. Each rod has a mass of \( \frac{m}{4} \) and length \( 2r \). The moment of inertia of one rod about its center is:
\[ I_{\text{rod}} = \frac{1}{12} \left(\frac{m}{4}\right) (2r)^2 = \frac{mr^2}{12} \]
Using the parallel axis theorem, the distance from the center of the square to the center of each rod is \( r \). Thus, the total moment of inertia for one rod is:
\[ I_{\text{total, rod}} = I_{\text{rod}} + \left(\frac{m}{4}\right) r^2 = \frac{mr^2}{12} + \frac{mr^2}{4} = \frac{mr^2}{12} + \frac{3mr^2}{12} = \frac{4mr^2}{12} = \frac{mr^2}{3} \]
Since there are four rods, the total moment of inertia for the square is:
\[ I_{\text{square of rods}} = 4 \cdot \frac{mr^2}{3} = \frac{4mr^2}{3} \]

### Step 5: Comparing the Moments of Inertia
Now we can compare the moments of inertia calculated:
1. Disc: \( \frac{1}{2} m r^2 \)
2. Ring: \( m r^2 \)
3. Square Lamina: \( \frac{2}{3} m r^2 \)
4. Square of Rods: \( \frac{4}{3} m r^2 \)

### Conclusion
From the calculations:
- Moment of inertia of the disc: \( \frac{1}{2} m r^2 \)
- Moment of inertia of the ring: \( m r^2 \)
- Moment of inertia of the square lamina: \( \frac{2}{3} m r^2 \)
- Moment of inertia of the square formed by four rods: \( \frac{4}{3} m r^2 \)

The largest moment of inertia is from the square formed by four rods, which is \( \frac{4}{3} m r^2 \).

Thus, the answer is **the square formed by four rods**.