The second's hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be

Options

A. `2 pi` & `0 mm//s`
B. `2sqrt(2)pi` & `4.44 mm//s`
C. `2sqrt(2) pi` & `2pi mm//s`
D. `2pi` & `2sqrt(2) mm//s`

Explanation

To solve the problem, we need to find two things: the speed of the endpoint of the second's hand of a watch and the magnitude of the difference of velocities at two perpendicular positions.

### Step 1: Calculate the speed of the endpoint of the second's hand

1. **Identify the length of the second's hand**: The length \( r \) of the second's hand is given as 6 cm.
2. **Determine the angular velocity \( \omega \)**: The second's hand completes one full revolution (which is \( 2\pi \) radians) in 60 seconds. Therefore, the angular velocity \( \omega \) can be calculated as:
\[ \omega = \frac{2\pi \text{ radians}}{60 \text{ seconds}} = \frac{\pi}{30} \text{ radians/second} \]
3. **Calculate the linear speed \( v \)**: The linear speed \( v \) at the endpoint of the second's hand can be calculated using the formula:
\[ v = r \cdot \omega \]
Substituting the values:
\[ v = 6 \text{ cm} \cdot \frac{\pi}{30} \text{ radians/second} = \frac{6\pi}{30} \text{ cm/second} = \frac{\pi}{5} \text{ cm/second} \]

### Step 2: Calculate the magnitude of the difference of velocities at two perpendicular positions

1. **Understanding the velocities at perpendicular positions**: At two perpendicular positions, the velocities can be represented as vectors \( \vec{v}_1 \) and \( \vec{v}_2 \). Since the magnitudes of both velocities are the same, we have:
\[ |\vec{v}_1| = |\vec{v}_2| = \frac{\pi}{5} \text{ cm/second} \]
2. **Using vector addition to find the difference**: The magnitude of the difference of the velocities at two perpendicular positions can be calculated using the Pythagorean theorem:
\[ |\vec{v}_1 - \vec{v}_2| = \sqrt{|\vec{v}_1|^2 + |\vec{v}_2|^2} \]
Substituting the values:
\[ |\vec{v}_1 - \vec{v}_2| = \sqrt{\left(\frac{\pi}{5}\right)^2 + \left(\frac{\pi}{5}\right)^2} = \sqrt{2 \left(\frac{\pi}{5}\right)^2} = \frac{\pi}{5} \sqrt{2} \]

### Final Answers
- The speed of the endpoint of the second's hand is \( \frac{\pi}{5} \) cm/second.
- The magnitude of the difference of velocities at two perpendicular positions is \( \frac{\pi \sqrt{2}}{5} \) cm/second.