A clock a minute-hand `10 cm` long. Find the average velocity between `6.00 AM` to `6.30 AM` for the tip of minute-hand.

Options

A. `22/21 cm "min"^(-1)`
B. `2/21 cm "min"^(-1)`
C. `12/21 cm "min"^(-1)`
D. `2/3 cm "min"^(-1)`

Explanation

To find the average velocity of the tip of the minute hand of a clock between 6:00 AM and 6:30 AM, we can follow these steps:

### Step 1: Understand the Movement of the Minute Hand
The minute hand of a clock moves in a circular path. Given that the length of the minute hand is 10 cm, it describes a circle with a radius of 10 cm.

### Step 2: Determine the Initial and Final Positions
At 6:00 AM, the minute hand points at the 12 o'clock position (0 degrees). By 6:30 AM, the minute hand points at the 6 o'clock position (180 degrees).

### Step 3: Calculate the Displacement
Displacement is the shortest distance between the initial and final positions. The initial position is at the top of the circle (12 o'clock), and the final position is at the bottom (6 o'clock). The straight-line distance between these two points can be visualized as the diameter of the circle.

- Diameter of the circle = 2 × radius = 2 × 10 cm = 20 cm

Thus, the displacement of the tip of the minute hand is 20 cm.

### Step 4: Calculate the Time Interval
The time interval from 6:00 AM to 6:30 AM is 30 minutes.

### Step 5: Calculate the Average Velocity
Average velocity is defined as the total displacement divided by the total time taken.

\[ \text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{20 \text{ cm}}{30 \text{ minutes}} = \frac{2}{3} \text{ cm/min} \]

### Conclusion
The average velocity of the tip of the minute hand between 6:00 AM and 6:30 AM is \(\frac{2}{3} \text{ cm/min}\).

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