A solid sphere and a hollow sphere of equal mass and radius are placed over a rough horizontal surface after rotating it about its mass centre with same angular velocity `omega_(0)`. Once the pure rolling starts let `v_(1)` and `v_(2)` be the linear speeds of their centres of mass. Then

Options

A. `v_(1)=v_(2)`
B. `v_(1)gtv_(2)`
C. `v_(1)ltv_(2)`
D. data is insufficient

Answer

To solve the problem, we need to analyze the motion of both the solid sphere and the hollow sphere after they start rolling without slipping. We will derive the relationship between their linear speeds \(v_1\) and \(v_2\) based on their moments of inertia.

### Step-by-Step Solution:

1. **Identify the Moments of Inertia**:
- The moment of inertia \(I\) for a solid sphere about its center of mass is given by:
\[ I_{\text{solid}} = \frac{2}{5} m R^2 \]
- The moment of inertia \(I\) for a hollow sphere about its center of mass is given by:
\[ I_{\text{hollow}} = \frac{2}{3} m R^2 \]

2. **Initial Angular Momentum**:
- Both spheres are initially rotating with the same angular velocity \(\omega_0\). The initial angular momentum \(L_i\) for both can be expressed as:
\[ L_{i,\text{solid}} = I_{\text{solid}} \cdot \omega_0 = \left(\frac{2}{5} m R^2\right) \omega_0 \]
\[ L_{i,\text{hollow}} = I_{\text{hollow}} \cdot \omega_0 = \left(\frac{2}{3} m R^2\right) \omega_0 \]

3. **Final Angular Momentum**:
- Once pure rolling starts, the relationship between linear speed \(v\) and angular speed \(\omega\) is given by:
\[ v = \omega R \]
- The final angular momentum \(L_f\) for both spheres can be expressed as:
\[ L_{f,\text{solid}} = I_{\text{solid}} \cdot \omega + m R v = I_{\text{solid}} \cdot \omega + m R \cdot \left(\frac{\omega R}{R}\right) = I_{\text{solid}} \cdot \omega + m R^2 \cdot \omega \]
\[ L_{f,\text{hollow}} = I_{\text{hollow}} \cdot \omega + m R v = I_{\text{hollow}} \cdot \omega + m R^2 \cdot \omega \]

4. **Conservation of Angular Momentum**:
- According to the law of conservation of angular momentum, we have:
\[ L_{i,\text{solid}} = L_{f,\text{solid}} \]
\[ L_{i,\text{hollow}} = L_{f,\text{hollow}} \]

5. **Setting Up the Equations**:
- For the solid sphere:
\[ \frac{2}{5} m R^2 \omega_0 = \left(\frac{2}{5} m R^2\right) \omega + m R^2 \cdot \omega \]
- For the hollow sphere:
\[ \frac{2}{3} m R^2 \omega_0 = \left(\frac{2}{3} m R^2\right) \omega + m R^2 \cdot \omega \]

6. **Solving for Linear Speeds**:
- From the equations, we can derive expressions for \(v_1\) and \(v_2\):
\[ v_1 = \frac{\omega_0 R}{\frac{2}{5} + 1} = \frac{\omega_0 R}{\frac{7}{5}} = \frac{5 \omega_0 R}{7} \]
\[ v_2 = \frac{\omega_0 R}{\frac{2}{3} + 1} = \frac{\omega_0 R}{\frac{5}{3}} = \frac{3 \omega_0 R}{5} \]

7. **Comparing Linear Speeds**:
- Now we can compare \(v_1\) and \(v_2\):
\[ v_1 = \frac{5 \omega_0 R}{7}, \quad v_2 = \frac{3 \omega_0 R}{5} \]
- To compare, we can express both in terms of a common denominator:
\[ v_1 = \frac{25 \omega_0 R}{35}, \quad v_2 = \frac{21 \omega_0 R}{35} \]
- Thus, we find that:
\[ v_2 > v_1 \]

### Conclusion:
The linear speed of the hollow sphere \(v_2\) is greater than that of the solid sphere \(v_1\). Therefore, the correct relationship is:
\[ v_2 > v_1 \]