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The ratio of angular speeds of minute hand and hour hand of a watch is
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Verified Answer
The correct answer is:
$12: 1$
$\omega_{\min }=\frac{2 \pi}{60} \frac{\mathrm{rad}}{\min }$
and $\omega_{\mathrm{hr}}=\frac{2 \pi}{12 \times 60} \frac{\mathrm{rad}}{\mathrm{min}}$
$$
\begin{aligned}
\therefore \frac{\omega_{\min }}{\omega_{\mathrm{hr}}} &=\frac{2 \pi / 60}{2 \pi / 12 \times 60} \\
&=\frac{12}{1}
\end{aligned}
$$
and $\omega_{\mathrm{hr}}=\frac{2 \pi}{12 \times 60} \frac{\mathrm{rad}}{\mathrm{min}}$
$$
\begin{aligned}
\therefore \frac{\omega_{\min }}{\omega_{\mathrm{hr}}} &=\frac{2 \pi / 60}{2 \pi / 12 \times 60} \\
&=\frac{12}{1}
\end{aligned}
$$
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