For spheres each of mass M and radius R are placed with their centers on the four corners A,B,C and D of a square of side b. The spheres A and B are hollow and C and D are solids. The moment of inertia of the system about side AD of square is

Options

A. `8/3MR^2+2Mb^2`
B. `8/5MR^2+2Mb^2`
C. `32/15MR^2+2Mb^2`
D. `32Mr^2+4Mb^2`

Answer

To find the moment of inertia of the system about side AD of the square, we will follow these steps:

### Step 1: Identify the types of spheres and their moment of inertia formulas
- The spheres A and B are hollow spheres, and their moment of inertia about their center is given by:
\[ I_{\text{hollow}} = \frac{2}{3} M R^2 \]
- The spheres C and D are solid spheres, and their moment of inertia about their center is given by:
\[ I_{\text{solid}} = \frac{2}{5} M R^2 \]

### Step 2: Determine the distance of each sphere's center from the axis AD
- The distance from the axis AD to the centers of spheres A and B (which are hollow) is 0 (since they are on the axis).
- The distance from the axis AD to the centers of spheres C and D (which are solid) is equal to the side length of the square, \( b \).

### Step 3: Apply the parallel axis theorem
- For spheres A and B (hollow):
\[ I_{A, AD} = I_{\text{hollow}} + M \cdot d^2 = \frac{2}{3} M R^2 + M \cdot 0^2 = \frac{2}{3} M R^2 \]
\[ I_{B, AD} = I_{\text{hollow}} + M \cdot d^2 = \frac{2}{3} M R^2 + M \cdot 0^2 = \frac{2}{3} M R^2 \]

- For spheres C and D (solid):
\[ I_{C, AD} = I_{\text{solid}} + M \cdot d^2 = \frac{2}{5} M R^2 + M \cdot b^2 \]
\[ I_{D, AD} = I_{\text{solid}} + M \cdot d^2 = \frac{2}{5} M R^2 + M \cdot b^2 \]

### Step 4: Calculate the total moment of inertia about axis AD
- Total moment of inertia \( I_{AD} \) is the sum of the moments of inertia of all four spheres:
\[ I_{AD} = I_{A, AD} + I_{B, AD} + I_{C, AD} + I_{D, AD} \]
\[ I_{AD} = \left(\frac{2}{3} M R^2 + \frac{2}{3} M R^2\right) + \left(\frac{2}{5} M R^2 + M b^2 + \frac{2}{5} M R^2 + M b^2\right) \]
\[ I_{AD} = \frac{4}{3} M R^2 + \frac{4}{5} M R^2 + 2 M b^2 \]

### Step 5: Combine the fractions
- To combine \( \frac{4}{3} M R^2 \) and \( \frac{4}{5} M R^2 \), we find a common denominator:
\[ \text{LCM of 3 and 5 is 15.} \]
\[ \frac{4}{3} = \frac{20}{15}, \quad \frac{4}{5} = \frac{12}{15} \]
\[ I_{AD} = \left(\frac{20}{15} + \frac{12}{15}\right) M R^2 + 2 M b^2 \]
\[ I_{AD} = \frac{32}{15} M R^2 + 2 M b^2 \]

### Final Answer
The moment of inertia of the system about side AD of the square is:
\[ I_{AD} = \frac{32}{15} M R^2 + 2 M b^2 \]