A solid uniform sphere rotating about its axis with kinetic energy `E_(1)` is gently placed on a rough horizontal plane at time t=0, Assume that, at time `t=t_(1)`, it starts pure rolling and at that instant total KE of the sphere is `E_(2).` After sometime, at time `t=t_(2)`. KE of the sphere is `E_(3)`. Then

Options

A. `E_(1)=E_(2)=E_(3)`
B. `E_(1)gtE_(2)=E_(3)`
C. `E_(1)gtE_(2)gtE_(3)`
D. `E_(1)ltE_(2)=E_(3)`

Answer

To solve the problem step by step, we will analyze the kinetic energy of the solid uniform sphere at different times as it transitions from pure rotation to pure rolling.

### Step 1: Initial Kinetic Energy (E1)
At time \( t = 0 \), the solid uniform sphere is rotating about its axis. The total kinetic energy \( E_1 \) consists of two components:
- Rotational kinetic energy due to its rotation about its axis.
- Translational kinetic energy is zero since it is not moving linearly.

The formula for the rotational kinetic energy \( E_{rot} \) of a solid sphere is given by:
\[ E_{rot} = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid sphere, \( I = \frac{2}{5} m r^2 \).

Thus, we can express \( E_1 \) as:
\[ E_1 = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 = \frac{1}{5} m r^2 \omega^2 \]

### Step 2: Transition to Pure Rolling (E2)
At time \( t = t_1 \), the sphere is placed on a rough horizontal plane. Due to friction, the sphere will start to roll without slipping. As it transitions to pure rolling, some of the rotational kinetic energy is converted into translational kinetic energy due to the work done against friction.

At this point, the total kinetic energy \( E_2 \) will be less than \( E_1 \) because energy is lost to friction:
\[ E_2 < E_1 \]

### Step 3: Pure Rolling Condition
When the sphere starts pure rolling, the condition for pure rolling is:
\[ v = r \omega \]
where \( v \) is the linear velocity of the center of mass. At this moment, the kinetic energy \( E_2 \) can be expressed as:
\[ E_2 = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]
Substituting \( v = r \omega \) into the equation gives:
\[ E_2 = \frac{1}{2} m (r \omega)^2 + \frac{1}{2} \left(\frac{2}{5} m r^2\right) \omega^2 \]
This simplifies to:
\[ E_2 = \frac{1}{2} m r^2 \omega^2 \left(1 + \frac{2}{5}\right) = \frac{7}{10} m r^2 \omega^2 \]

### Step 4: Kinetic Energy After Some Time (E3)
At time \( t = t_2 \), the sphere continues to roll without slipping. Since there is no friction acting on it anymore (as it is already rolling), the total kinetic energy \( E_3 \) will remain constant:
\[ E_3 = E_2 \]

### Conclusion
From the analysis, we can conclude:
- \( E_1 > E_2 \) (because energy is lost due to friction)
- \( E_2 = E_3 \) (because there is no further energy loss once pure rolling starts)

Thus, the correct relationship is:
\[ E_1 > E_2 = E_3 \]

### Final Answer
The correct option is:
\[ E_1 > E_2 = E_3 \]