Let $\omega_1$, $\omega_2$ and $\omega_3$ be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If $x_1$, $x_2$ and $x_3$ are their respective angular distances in $1$ minute then the factor which remains constant $(k)$ is

Question

Accepted Answer

A. 1. **Angular speed** of a hand is the angle it sweeps per unit time, and it completes one full revolution ($2\pi$ radians) in the time it takes to go around the clock once.

For the **second hand**, it completes one revolution in $60$ seconds:

$$\omega_1=\frac{2\pi}{60}$$

Since the angular distance covered in $1$ minute ($60$ s) is $x_1=\omega_1	imes60$:

$$x_1=2\pi$$

2. For the **minute hand**, it completes one revolution in $3600$ seconds (one hour):

$$\omega_2=\frac{2\pi}{3600}$$

So the angular distance in $1$ minute is:

$$x_2=\omega_2	imes60=\frac{2\pi}{60}$$

3. For the **hour hand**, it completes one revolution in $3600	imes12$ seconds (12 hours):

$$\omega_3=\frac{2\pi}{3600	imes12}$$

So the angular distance in $1$ minute is:

$$x_3=\omega_3	imes60=\frac{2\pi}{720}$$

4. Now check which ratio stays the same for all three hands. Dividing $\omega$ by $x$ for each hand:

$$\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=\frac{1}{60}=k$$

Hence, the answer is **A**.