A spherical shell of mass `M` and radius `R` filed completely with a liquid of same mass and set to rotate about a vertical axis through its centre has a moment of inertia `I_(1)` about the axis. The bottom. If moment of inertia of the system is `I_(2)` when the shell is half filled and `I_(3)` is the moment of inertia when entire water drained off then :

Options

A. `(I_(1))/(I_(2)) ~~ 1.5`
B. `(I_(1))/(I_(2)) ~~0.67`
C. `(I_(1))/(I_(3)) ~~ 1.6`
D. `(I_(2))/(I_(3)) ~~ 1.4`

Explanation

To solve the problem, we need to calculate the moment of inertia for three different scenarios involving a spherical shell filled with liquid. We will denote the mass of the shell as \( M \) and its radius as \( R \).

### Step 1: Calculate \( I_1 \) (Moment of Inertia when the shell is fully filled)
The moment of inertia of a hollow sphere about an axis through its center is given by:

\[ I_{\text{hollow}} = \frac{2}{3} M R^2 \]

When the shell is filled with liquid of the same mass \( M \), it behaves like a solid sphere. The moment of inertia of a solid sphere is given by:

\[ I_{\text{solid}} = \frac{2}{5} M R^2 \]

Thus, the total moment of inertia \( I_1 \) when the shell is fully filled is:

\[ I_1 = I_{\text{hollow}} + I_{\text{solid}} = \frac{2}{3} M R^2 + \frac{2}{5} M R^2 \]

To combine these fractions, we find a common denominator (15):

\[ I_1 = \left(\frac{10}{15} M R^2 + \frac{6}{15} M R^2\right) = \frac{16}{15} M R^2 \]

### Step 2: Calculate \( I_2 \) (Moment of Inertia when the shell is half-filled)
When the shell is half-filled, we treat the system as a hollow hemisphere plus a solid hemisphere of mass \( \frac{M}{2} \).

The moment of inertia of a hollow hemisphere is:

\[ I_{\text{hollow hemisphere}} = \frac{1}{2} I_{\text{hollow}} = \frac{1}{2} \cdot \frac{2}{3} M R^2 = \frac{1}{3} M R^2 \]

The moment of inertia of the solid hemisphere is:

\[ I_{\text{solid hemisphere}} = \frac{2}{5} \cdot \frac{M}{2} R^2 = \frac{1}{5} M R^2 \]

Thus, the total moment of inertia \( I_2 \) when the shell is half-filled is:

\[ I_2 = I_{\text{hollow hemisphere}} + I_{\text{solid hemisphere}} = \frac{1}{3} M R^2 + \frac{1}{5} M R^2 \]

Finding a common denominator (15):

\[ I_2 = \left(\frac{5}{15} M R^2 + \frac{3}{15} M R^2\right) = \frac{8}{15} M R^2 \]

### Step 3: Calculate \( I_3 \) (Moment of Inertia when the shell is empty)
When the shell is empty, the moment of inertia is just that of the hollow shell:

\[ I_3 = \frac{2}{3} M R^2 \]

### Step 4: Find the ratios \( \frac{I_1}{I_2} \) and \( \frac{I_1}{I_3} \)
Now we can find the ratios:

1. **For \( \frac{I_1}{I_2} \)**:

\[ \frac{I_1}{I_2} = \frac{\frac{16}{15} M R^2}{\frac{8}{15} M R^2} = \frac{16}{8} = 2 \]

2. **For \( \frac{I_1}{I_3} \)**:

\[ \frac{I_1}{I_3} = \frac{\frac{16}{15} M R^2}{\frac{2}{3} M R^2} = \frac{16}{15} \cdot \frac{3}{2} = \frac{16 \cdot 3}{15 \cdot 2} = \frac{48}{30} = \frac{8}{5} = 1.6 \]

### Final Results
- \( I_1 = \frac{16}{15} M R^2 \)
- \( I_2 = \frac{8}{15} M R^2 \)
- \( I_3 = \frac{2}{3} M R^2 \)
- \( \frac{I_1}{I_2} = 2 \)
- \( \frac{I_1}{I_3} = 1.6 \)