The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is 2 : 1. If their radii are in the ratio 2 : 1, then the ratio of their masses will be

Options

A. `2:1`
B. `1:8`
C. `1:7`
D. `2sqrt2:1`

Answer

To solve the problem, we need to find the ratio of the masses of two spheres given the ratio of their kinetic energies and the ratio of their radii.

### Step-by-Step Solution:

1. **Understanding the Given Ratios**:
- The ratio of the kinetic energies of the two spheres is given as \( KE_1 : KE_2 = 2 : 1 \).
- The ratio of their radii is given as \( R_1 : R_2 = 2 : 1 \).

2. **Expressing Kinetic Energy**:
- The total kinetic energy \( KE \) of a rolling sphere is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2 \]
- For a solid sphere, the moment of inertia \( I \) is \( \frac{2}{5} m r^2 \).
- The relationship between angular velocity \( \omega \) and linear velocity \( v \) for rolling without slipping is \( \omega = \frac{v}{r} \).

3. **Calculating Kinetic Energy for Each Sphere**:
- For sphere 1:
\[ KE_1 = \frac{1}{2} \left(\frac{2}{5} m_1 R_1^2\right) \left(\frac{v}{R_1}\right)^2 + \frac{1}{2} m_1 v^2 \]
Simplifying this gives:
\[ KE_1 = \frac{1}{5} m_1 v^2 + \frac{1}{2} m_1 v^2 = \frac{7}{10} m_1 v^2 \]

- For sphere 2:
\[ KE_2 = \frac{1}{2} \left(\frac{2}{5} m_2 R_2^2\right) \left(\frac{v}{R_2}\right)^2 + \frac{1}{2} m_2 v^2 \]
Simplifying this gives:
\[ KE_2 = \frac{1}{5} m_2 v^2 + \frac{1}{2} m_2 v^2 = \frac{7}{10} m_2 v^2 \]

4. **Setting Up the Ratio of Kinetic Energies**:
- From the problem, we know:
\[ \frac{KE_1}{KE_2} = \frac{2}{1} \]
- Substituting the expressions for kinetic energies:
\[ \frac{\frac{7}{10} m_1 v^2}{\frac{7}{10} m_2 v^2} = \frac{2}{1} \]
- The \( \frac{7}{10} v^2 \) cancels out:
\[ \frac{m_1}{m_2} = 2 \]

5. **Conclusion**:
- Therefore, the ratio of the masses \( m_1 : m_2 \) is \( 2 : 1 \).

### Final Answer:
The ratio of their masses is \( 2 : 1 \).