Q128 / 150 · MHT CET 2026 (13 April Shift 1) · Physics
The relative angular speed of hour hand and minute hand of clock is
- A$\dfrac{2\pi}{60}$
- B$\dfrac{11\pi}{360}$
- Correct answerC$\dfrac{11\pi}{21600}$
- D$\dfrac{2\pi}{3600}$
Solution
The angular speed of the minute hand is the angle covered in one revolution divided by the time taken (1 hour = 3600 s):
$\omega_m = \dfrac{2\pi}{3600} = \dfrac{\pi}{1800}$ rad/s
The angular speed of the hour hand is the angle covered in one revolution divided by the time taken (12 hours = $12 \times 3600$ s = 43200 s):
$\omega_h = \dfrac{2\pi}{43200} = \dfrac{\pi}{21600}$ rad/s
The relative angular speed of the minute hand with respect to the hour hand is the difference between their angular speeds:
$\omega_{rel} = \omega_m - \omega_h$
$\omega_{rel} = \dfrac{\pi}{1800} - \dfrac{\pi}{21600}$
$\omega_{rel} = \dfrac{12\pi - \pi}{21600} = \dfrac{11\pi}{21600}$ rad/s
Answer: $\dfrac{11\pi}{21600}$
$\omega_m = \dfrac{2\pi}{3600} = \dfrac{\pi}{1800}$ rad/s
The angular speed of the hour hand is the angle covered in one revolution divided by the time taken (12 hours = $12 \times 3600$ s = 43200 s):
$\omega_h = \dfrac{2\pi}{43200} = \dfrac{\pi}{21600}$ rad/s
The relative angular speed of the minute hand with respect to the hour hand is the difference between their angular speeds:
$\omega_{rel} = \omega_m - \omega_h$
$\omega_{rel} = \dfrac{\pi}{1800} - \dfrac{\pi}{21600}$
$\omega_{rel} = \dfrac{12\pi - \pi}{21600} = \dfrac{11\pi}{21600}$ rad/s
Answer: $\dfrac{11\pi}{21600}$
Hence, the correct answer is option: 'C'.