The ratio of angular speeds of the minute han | vedclass

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(C) The angular speed $\omega$ is defined as the rate of change of angular displacement,given by $\omega = \frac{2\pi}{T}$,where $T$ is the time perio

Vedclass · Accepted Answer

$12 : 1$

Vedclass · Answer

(C) The angular speed $\omega$ is defined as the rate of change of angular displacement,given by $\omega = \frac{2\pi}{T}$,where $T$ is the time period.For the minute hand,the time period $T_m = 1 	ext{ hour}$. Thus,$\omega_m = \frac{2\pi}{1} = 2\pi 	ext{ rad/hr}$.For the hour hand,the time period $T_h = 12 	ext{ hours}$. Thus,$\omega_h = \frac{2\pi}{12} = \frac{\pi}{6} 	ext{ rad/hr}$.The ratio of the angular speed of the minute hand to the hour hand is $\frac{\omega_m}{\omega_h} = \frac{2\pi}{\frac{2\pi}{12}} = \frac{12}{1}$.Therefore,the ratio is $12 : 1$.

The ratio of angular speeds of the minute hand and the hour hand of a watch is

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