The ratio of radii of two solid spheres of same material is `1:2`. The ratio of moments of inertia of smaller and larger spheres about axes passing through their centres is

Options

A. `1:4`
B. `1:8`
C. `1:16`
D. `1:32`

Explanation

To find the ratio of the moments of inertia of two solid spheres of the same material with a radius ratio of 1:2, we can follow these steps:

### Step 1: Understand the formula for the moment of inertia of a solid sphere
The moment of inertia \( I \) of a solid sphere about an axis passing through its center is given by the formula:
\[ I = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the sphere and \( r \) is its radius.

### Step 2: Express the mass of the spheres in terms of their radii
Since both spheres are made of the same material, their masses can be expressed in terms of their volumes and densities. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
The mass \( m \) can then be expressed as:
\[ m = \rho V = \rho \left(\frac{4}{3} \pi r^3\right) \]
where \( \rho \) is the density of the material.

### Step 3: Calculate the mass of both spheres
Let the radius of the smaller sphere be \( r_1 \) and the radius of the larger sphere be \( r_2 = 2r_1 \).

- For the smaller sphere:
\[ m_1 = \rho \left(\frac{4}{3} \pi r_1^3\right) \]

- For the larger sphere:
\[ m_2 = \rho \left(\frac{4}{3} \pi (2r_1)^3\right) = \rho \left(\frac{4}{3} \pi (8r_1^3)\right) = 8 \rho \left(\frac{4}{3} \pi r_1^3\right) = 8m_1 \]

### Step 4: Calculate the moments of inertia for both spheres
Using the formula for the moment of inertia:

- For the smaller sphere:
\[ I_1 = \frac{2}{5} m_1 r_1^2 = \frac{2}{5} \left(\rho \frac{4}{3} \pi r_1^3\right) r_1^2 = \frac{2}{5} \cdot \rho \cdot \frac{4}{3} \pi r_1^5 \]

- For the larger sphere:
\[ I_2 = \frac{2}{5} m_2 r_2^2 = \frac{2}{5} (8m_1) (2r_1)^2 = \frac{2}{5} (8m_1) (4r_1^2) = \frac{64}{5} m_1 r_1^2 \]

### Step 5: Find the ratio of the moments of inertia
Now we can find the ratio of the moments of inertia \( \frac{I_1}{I_2} \):
\[ \frac{I_1}{I_2} = \frac{\frac{2}{5} \cdot \rho \cdot \frac{4}{3} \pi r_1^5}{\frac{64}{5} m_1 r_1^2} \]
Substituting \( m_1 = \rho \frac{4}{3} \pi r_1^3 \):
\[ \frac{I_1}{I_2} = \frac{\frac{2}{5} \cdot \rho \cdot \frac{4}{3} \pi r_1^5}{\frac{64}{5} \cdot \rho \cdot \frac{4}{3} \pi r_1^3 \cdot r_1^2} = \frac{2}{64} = \frac{1}{32} \]

### Final Result
Thus, the ratio of the moments of inertia of the smaller sphere to the larger sphere is:
\[ \frac{I_1}{I_2} = \frac{1}{32} \]