A disc of mass `m` and radius `r` placed on a routh horizontal surface. A cue of mass `m` hits the disc at a height `h` from the axis passing through centre and parallel to the surface. The disc will start pure rolling for.

Options

A. `hltr/3`
B. `h=r/2`
C. `hgtr/2`
D. `hger/2`

Explanation

To solve the problem, we need to analyze the situation where a cue of mass `m` hits a disc of mass `m` at a height `h` from the center. The goal is to determine the height `h` at which the cue must hit the disc for the disc to start pure rolling.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a disc of mass `m` and radius `r` resting on a rough horizontal surface.
- A cue of mass `m` strikes the disc at a height `h` from the center.

2. **Conservation of Angular Momentum**:
- Since there are no external torques acting on the system, the angular momentum of the system is conserved.
- The initial angular momentum of the cue about the center of the disc (just before the collision) is given by:
\[ L_{initial} = m \cdot h \cdot v \]
where `v` is the velocity of the cue just before the collision.

3. **Final Angular Momentum**:
- After the collision, the disc will start rotating with an angular velocity `ω`.
- The moment of inertia `I` of the disc about its center is:
\[ I = \frac{1}{2} m r^2 \]
- The final angular momentum of the system after the collision is:
\[ L_{final} = I \cdot \omega = \frac{1}{2} m r^2 \cdot \omega \]

4. **Setting Up the Equation**:
- By conservation of angular momentum, we have:
\[ m \cdot h \cdot v = \frac{1}{2} m r^2 \cdot \omega \]
- We can cancel `m` from both sides (assuming `m ≠ 0`):
\[ h \cdot v = \frac{1}{2} r^2 \cdot \omega \]

5. **Relating Linear and Angular Velocity**:
- For pure rolling to occur, the relationship between linear velocity `v` and angular velocity `ω` is given by:
\[ v = r \cdot \omega \]
- Substituting this into the previous equation gives:
\[ h \cdot (r \cdot \omega) = \frac{1}{2} r^2 \cdot \omega \]

6. **Solving for Height `h`**:
- We can cancel `ω` from both sides (assuming `ω ≠ 0`):
\[ h \cdot r = \frac{1}{2} r^2 \]
- Rearranging gives:
\[ h = \frac{1}{2} r \]

### Final Answer:
The height `h` at which the cue must hit the disc for it to start pure rolling is:
\[ h = \frac{r}{2} \]