A solid sphere of mass `m &` rdius `R` is divided in two parts of `m` mass `(7m)/(8) & (m)/(8)`, and converted to a disc of radius `2R &` solid sphere of radius `'r'` respectively. Find `(I_(1))/(I_(2))` , If `I_(1) & I_(2)` are moment of inertia of disc `&` solid sphere respectively

Options

A. 200
B. 140
C. 120
D. 180

Answer

To solve the problem, we need to find the ratio of the moments of inertia \( \frac{I_1}{I_2} \) for a disc and a solid sphere, respectively.

### Step-by-Step Solution:

1. **Identify the Masses and Shapes**:
- We have a solid sphere of mass \( m \) and radius \( R \).
- The sphere is divided into two parts:
- Part 1: Mass \( \frac{7m}{8} \) converted into a disc of radius \( 2R \).
- Part 2: Mass \( \frac{m}{8} \) converted into a solid sphere of radius \( r \).

2. **Calculate the Moment of Inertia of the Disc \( I_1 \)**:
- The formula for the moment of inertia of a disc about its central axis is given by:
\[ I_1 = \frac{1}{2} M R^2 \]
- Here, \( M = \frac{7m}{8} \) and \( R = 2R \).
- Substituting these values:
\[ I_1 = \frac{1}{2} \left(\frac{7m}{8}\right) (2R)^2 = \frac{1}{2} \left(\frac{7m}{8}\right) (4R^2) = \frac{7m}{4} \]

3. **Calculate the Moment of Inertia of the Solid Sphere \( I_2 \)**:
- The formula for the moment of inertia of a solid sphere about its central axis is:
\[ I_2 = \frac{2}{5} m R^2 \]
- Here, the mass is \( \frac{m}{8} \) and we need to find the radius \( r \).
- To find \( r \), we use the fact that the densities of both spheres are equal:
\[ \frac{m}{V} = \frac{\frac{m}{8}}{\frac{4}{3} \pi r^3} \]
- Setting the densities equal:
\[ \frac{m}{\frac{4}{3} \pi R^3} = \frac{\frac{m}{8}}{\frac{4}{3} \pi r^3} \]
- This simplifies to:
\[ R^3 = \frac{r^3}{8} \implies r^3 = 8R^3 \implies r = \frac{R}{2} \]
- Now substituting \( r \) back into the moment of inertia formula:
\[ I_2 = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R}{2}\right)^2 = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R^2}{4}\right) = \frac{mR^2}{80} \]

4. **Calculate the Ratio \( \frac{I_1}{I_2} \)**:
- Now we can find the ratio:
\[ \frac{I_1}{I_2} = \frac{\frac{7m}{4}}{\frac{mR^2}{80}} = \frac{7m}{4} \cdot \frac{80}{mR^2} = \frac{7 \cdot 80}{4R^2} = \frac{560}{4R^2} = \frac{140}{R^2} \]

5. **Final Result**:
- The final answer is:
\[ \frac{I_1}{I_2} = 140 \]