Question

Let ω1, ω2 and ω3 be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If x1, x2 and x3 are their respective angular distances in 1 minute then the factor which remains constant (k) is

• A. $\frac{\omega_1}{x_1}$ = $\frac{\omega_2}{x_2} = \frac{\omega_3}{x_3} = k$
• B. $\omega_1 x_1 = \omega_2 x_2 = \omega_3 x_3 = k$
• C. $\omega_1 x_1^2 = \omega_2 x_2^2 = \omega_3 x_3^2 = k$
• D. $\omega_1 x_1^2 = \omega_2 x_2^2 = \omega_3 x_3^2 = k$

Answer 1

The Correct Option is A

Angular speed ($\omega$) = $\frac{\text{Angular displacement}}{\text{time}}$

For the second hand: $\omega_1 = \frac{2\pi}{60}$ rad/s x₁ = ω₁ × 60 = 2π rad

For the minute hand: $\omega_2 = \frac{2\pi}{3600}$ rad/s x₂ = ω₂ × 60 = $\frac{2\pi}{60}$ rad

For the hour hand:

$\omega_3 = \frac{2\pi}{3600 \times 12}$ rad/s

x₃ = ω₃ × 60 = $\frac{2\pi}{720}$ rad

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