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. 185
B. 140
C. 285
D. 65

Answer

To solve the problem, we need to find the ratio of the moments of inertia \( I_1 \) and \( I_2 \) for a disc and a solid sphere, respectively. Let's break down the solution step by step.

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

The moment of inertia \( I \) of a disc about its central axis is given by the formula:

\[ I = \frac{1}{2} m r^2 \]

In this case, the mass of the disc is \( \frac{7m}{8} \) and the radius is \( 2R \). Therefore, we substitute these values into the formula:

\[ I_1 = \frac{1}{2} \left(\frac{7m}{8}\right) (2R)^2 \]

Calculating \( (2R)^2 \):

\[ (2R)^2 = 4R^2 \]

Now substituting back into the equation for \( I_1 \):

\[ I_1 = \frac{1}{2} \left(\frac{7m}{8}\right) (4R^2) = \frac{7m \cdot 4R^2}{16} = \frac{28mR^2}{16} = \frac{7mR^2}{4} \]

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

The moment of inertia \( I \) of 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 \( r \). We can find \( r \) by equating the densities of the original sphere and the new sphere.

The density \( \rho \) is given by:

\[ \rho = \frac{m}{\text{Volume}} = \frac{m}{\frac{4}{3} \pi R^3} \]

For the solid sphere of mass \( \frac{m}{8} \) and radius \( r \):

\[ \rho = \frac{\frac{m}{8}}{\frac{4}{3} \pi r^3} \]

Setting the two densities equal:

\[ \frac{m}{\frac{4}{3} \pi R^3} = \frac{\frac{m}{8}}{\frac{4}{3} \pi r^3} \]

Cancelling \( \frac{4}{3} \pi \) and \( m \) from both sides gives:

\[ \frac{1}{R^3} = \frac{1/8}{r^3} \]

Cross-multiplying yields:

\[ r^3 = \frac{R^3}{8} \implies r = \frac{R}{2} \]

Now substituting \( r = \frac{R}{2} \) into the formula for \( I_2 \):

\[ I_2 = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R}{2}\right)^2 \]

Calculating \( \left(\frac{R}{2}\right)^2 \):

\[ \left(\frac{R}{2}\right)^2 = \frac{R^2}{4} \]

Substituting back into the equation for \( I_2 \):

\[ I_2 = \frac{2}{5} \left(\frac{m}{8}\right) \left(\frac{R^2}{4}\right) = \frac{2mR^2}{160} = \frac{mR^2}{80} \]

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

Now we can find the ratio of the moments of inertia:

\[ \frac{I_1}{I_2} = \frac{\frac{7mR^2}{4}}{\frac{mR^2}{80}} \]

Cancelling \( mR^2 \) from the numerator and denominator gives:

\[ \frac{I_1}{I_2} = \frac{7}{4} \cdot 80 = \frac{560}{4} = 140 \]

### Final Answer

Thus, the ratio of the moments of inertia \( \frac{I_1}{I_2} \) is:

\[ \frac{I_1}{I_2} = 140 \]