The relative angular speed of hour hand and second hand of a clock is (in rad/s)
The relative angular speed of hour hand and second hand of a clock is (in rad/s)
$\frac{311 \pi}{578}$
$\frac{421 \pi}{11600}$
$\frac{719 \pi}{21600}$
$\frac{919 \pi}{15600}$
Solution
The correct option is (C).
Concept: The angular speed is given by $\omega=\frac{\Delta \theta}{\Delta t}$
Relative angular speed is given by, $\omega=\omega_{\mathrm{s}}-\omega_{\mathrm{h}}$
The angular speed of the second dial is $\omega_{\mathrm{s}}=\frac{2 \pi}{60}=\frac{\pi}{30} \mathrm{rad} / \mathrm{s}$
The angular speed of the hour dial is
$\omega_{\mathrm{h}}=\frac{2 \pi}{3600 \times 12}=\frac{\pi}{21600} \mathrm{rad} / \mathrm{s}$
Therefore, relative angular speed is given by,
$\omega=\omega_{\mathrm{s}}-\omega_{\mathrm{h}}=\frac{\pi}{30}-\frac{\pi}{21600}=\frac{719 \pi}{21600} \mathrm{rad} / \mathrm{s}$