Where will the hand of a clock stop if it
starts at 12 and makes `(1)/(2)` of a revolution, clockwise ?

Answer

To solve the problem of where the hand of a clock will stop after making \( \frac{1}{2} \) of a revolution clockwise from the 12 o'clock position, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Clock Positions**:
A clock is divided into 12 hours. The positions are as follows:
- 12 o'clock is at the top.
- 3 o'clock is to the right.
- 6 o'clock is at the bottom.
- 9 o'clock is to the left.

2. **Identify the Starting Position**:
The hand of the clock starts at the 12 o'clock position.

3. **Determine the Meaning of \( \frac{1}{2} \) Revolution**:
A full revolution of the clock is 360 degrees. Therefore, \( \frac{1}{2} \) of a revolution is:
\[ \frac{1}{2} \times 360^\circ = 180^\circ \]

4. **Calculate the New Position**:
Moving clockwise from the 12 o'clock position by 180 degrees means the hand will move to the opposite side of the clock face.
- Starting at 12 and moving 180 degrees clockwise lands the hand at the 6 o'clock position.

5. **Conclusion**:
Therefore, after making \( \frac{1}{2} \) of a revolution clockwise from 12, the hand of the clock will stop at 6.

### Final Answer:
The hand of the clock will stop at 6.