In the above problem, if coefficient of friction for both the spheres is same and let `t_(1)` and `t_(2)` be the times when pure rolling of solid sphere and of hollow sphere is started. Then

Options

A. `t_(1)=t_(2)`
B. `t_(1)ltt_(2)`
C. `t_(1)gtt_(2)`
D. None of these

Answer

To solve the problem, we need to analyze the motion of both the solid sphere and the hollow sphere under the influence of friction until they start pure rolling. We will derive the expressions for the times \( T_1 \) and \( T_2 \) for the solid and hollow spheres respectively.

### Step-by-Step Solution:

1. **Understanding the Forces Acting on the Spheres:**
- Both spheres experience a frictional force due to the coefficient of friction \( \mu \). This force acts in the forward direction and is given by \( F_f = \mu mg \), where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity.

2. **Linear Acceleration:**
- The linear acceleration \( a \) of the spheres can be expressed as:
\[ a = \frac{F_f}{m} = \frac{\mu mg}{m} = \mu g \]

3. **Angular Acceleration:**
- The angular acceleration \( \alpha \) for both spheres is related to the linear acceleration by the equation:
\[ \alpha = \frac{a}{r} \]
- For the solid sphere, the moment of inertia \( I \) is \( \frac{2}{5}mr^2 \), and for the hollow sphere, it is \( \frac{2}{3}mr^2 \).

4. **Relating Linear and Angular Motion:**
- The relationship between linear and angular motion leads to:
\[ \tau = I \alpha \]
- The torque \( \tau \) due to friction is \( \tau = F_f \cdot r = \mu mg \cdot r \).

5. **Finding Time for Pure Rolling:**
- For the solid sphere:
\[ \mu mg \cdot r = \frac{2}{5}mr^2 \cdot \alpha \implies \alpha = \frac{5\mu g}{2r} \]
- For the hollow sphere:
\[ \mu mg \cdot r = \frac{2}{3}mr^2 \cdot \alpha \implies \alpha = \frac{3\mu g}{2r} \]

6. **Calculating Time \( T_1 \) and \( T_2 \):**
- The time taken for the solid sphere to start pure rolling \( T_1 \):
\[ T_1 = \frac{v}{a} = \frac{r\omega_0}{\mu g} = \frac{r \cdot \frac{5\mu g}{2r}}{\mu g} = \frac{5}{2\mu g} \]
- The time taken for the hollow sphere to start pure rolling \( T_2 \):
\[ T_2 = \frac{v}{a} = \frac{r\omega_0}{\mu g} = \frac{r \cdot \frac{3\mu g}{2r}}{\mu g} = \frac{3}{2\mu g} \]

7. **Comparing \( T_1 \) and \( T_2 \):**
- Since \( T_1 = \frac{5}{2\mu g} \) and \( T_2 = \frac{3}{2\mu g} \), we can conclude:
\[ T_1 < T_2 \]

### Conclusion:
Thus, the time when pure rolling starts for the solid sphere \( T_1 \) is less than the time for the hollow sphere \( T_2 \). Therefore, \( T_1 < T_2 \).