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. 160
B. 140
C. 240
D. 120

Answer

To find the ratio of the moments of inertia \( \frac{I_1}{I_2} \) for the given solid sphere and the disk, we will follow these steps:

### Step 1: Calculate the Moment of Inertia of the Disk \( I_1 \)

The moment of inertia \( I \) for a disk about its central axis is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
In this case, the mass of the disk is \( \frac{7m}{8} \) and the radius of the disk is \( 2R \). Therefore, we can calculate \( I_1 \):
\[ I_1 = \frac{1}{2} \left(\frac{7m}{8}\right) (2R)^2 \]
Calculating \( (2R)^2 \):
\[ (2R)^2 = 4R^2 \]
Now substituting this back into the equation for \( I_1 \):
\[ I_1 = \frac{1}{2} \left(\frac{7m}{8}\right) (4R^2) = \frac{7m}{8} \cdot 2R^2 = \frac{14mR^2}{8} = \frac{7mR^2}{4} \]

### Step 2: Calculate the Moment of Inertia of the Solid Sphere \( I_2 \)

The moment of inertia \( I \) for a solid sphere about its central axis is given by the formula:
\[ I = \frac{2}{5} m r^2 \]
Here, the mass of the solid sphere is \( \frac{m}{8} \) and we need to find the radius \( r \) of this smaller sphere. To find \( r \), we will use the fact that the densities of the two spheres are equal.

The density \( \rho \) of the larger sphere is:
\[ \rho = \frac{m}{\frac{4}{3} \pi R^3} \]
The density \( \rho \) of the smaller sphere is:
\[ \rho = \frac{\frac{m}{8}}{\frac{4}{3} \pi r^3} \]
Setting these equal gives:
\[ \frac{m}{\frac{4}{3} \pi R^3} = \frac{\frac{m}{8}}{\frac{4}{3} \pi r^3} \]
Cancelling \( \frac{4}{3} \pi \) and \( m \) from both sides, we have:
\[ \frac{1}{R^3} = \frac{1/8}{r^3} \]
Cross-multiplying gives:
\[ r^3 = \frac{R^3}{8} \implies r = \frac{R}{2} \]

Now substituting \( r \) back into the equation for \( I_2 \):
\[ I_2 = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R}{2}\right)^2 \]
Calculating \( \left(\frac{R}{2}\right)^2 \):
\[ \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]
Now substituting this back into the equation for \( I_2 \):
\[ I_2 = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R^2}{4}\right) = \frac{2mR^2}{5 \cdot 32} = \frac{mR^2}{80} \]

### Step 3: Calculate the Ratio \( \frac{I_1}{I_2} \)

Now we can find the ratio:
\[ \frac{I_1}{I_2} = \frac{\frac{7mR^2}{4}}{\frac{mR^2}{80}} \]
Cancelling \( mR^2 \) from the numerator and denominator:
\[ \frac{I_1}{I_2} = \frac{7}{4} \cdot 80 = \frac{560}{4} = 140 \]

### Final Answer

Thus, the ratio of the moments of inertia is:
\[ \frac{I_1}{I_2} = 140 \]