The length of second's hand in watch is `1 cm`. The change in Velocity of its tip in 15 seconds is

Options

A. Zero
B. `(pi)/(30sqrt(2))cm//sec`
C. `(pi)/(30)cm//sec`
D. `(pisqrt(2))/(30)cm//sec`

Explanation

To solve the problem of finding the change in velocity of the tip of the second's hand of a watch in 15 seconds, we can follow these steps:

### Step 1: Understand the motion of the second's hand
The second's hand of the watch moves in a circular path. The length of the second's hand (radius \( R \)) is given as 1 cm. The hand completes a full revolution (360 degrees) in 60 seconds.

### Step 2: Determine the angular velocity
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period for one complete revolution (60 seconds). Thus,
\[ \omega = \frac{2\pi}{60} = \frac{\pi}{30} \text{ radians per second} \]

### Step 3: Calculate the linear velocity
The linear velocity \( v \) of the tip of the second's hand is given by:
\[ v = R \cdot \omega \]
Substituting the values:
\[ v = 1 \cdot \frac{\pi}{30} = \frac{\pi}{30} \text{ cm/s} \]

### Step 4: Determine the position after 15 seconds
In 15 seconds, the second's hand will move:
\[ \text{Angle} = \omega \cdot t = \frac{\pi}{30} \cdot 15 = \frac{\pi}{2} \text{ radians} = 90 \text{ degrees} \]
This means the hand moves from the 12 o'clock position to the 3 o'clock position.

### Step 5: Find the change in velocity
At the 12 o'clock position, the velocity vector is directed horizontally to the right (positive x-direction). After 15 seconds, at the 3 o'clock position, the velocity vector is directed vertically downward (negative y-direction).

The change in velocity \( \Delta v \) can be calculated using vector subtraction:
\[ \Delta v = v_f - v_i \]
Where:
- \( v_i = \frac{\pi}{30} \hat{i} \) (initial velocity)
- \( v_f = -\frac{\pi}{30} \hat{j} \) (final velocity)

The magnitude of the change in velocity is given by:
\[ |\Delta v| = \sqrt{(v_f)^2 + (v_i)^2} \]
Substituting the values:
\[ |\Delta v| = \sqrt{\left(-\frac{\pi}{30}\right)^2 + \left(\frac{\pi}{30}\right)^2} = \sqrt{2 \left(\frac{\pi}{30}\right)^2} = \frac{\pi}{30} \sqrt{2} \]

### Step 6: Final answer
Thus, the change in velocity of the tip of the second's hand in 15 seconds is:
\[ |\Delta v| = \frac{\pi \sqrt{2}}{30} \text{ cm/s} \]