Let ${\omega }_{1},{\omega }_{2}$ and ${\omega }_{3}$ be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If ${x}_{1},{x}_{2}$ and ${x}_{3}$ are their respective angular distances in 1 minute then the factor which remains constant$(k)$ is :

Question

YESPYQ · Accepted Answer

Correct answer: a) $\frac{{\omega }_{1}}{{x}_{1}}=\frac{{\omega }_{2}}{{x}_{2}}=\frac{{\omega }_{3}}{{x}_{3}}=\mathrm{k}$. ${\omega }_{1}=\frac{2\pi }{60};{x}_{1}=\frac{2\pi }{60}\times 60=2\pi$ ${\omega }_{2}=\frac{2\pi }{3600};{x}_{2}=\frac{2\pi }{3600}\times 60=\frac{2\pi }{60}$ ${\omega }_{3}=\frac{2\pi }{3600\times 12};{x}_{3}=\frac{2\pi }{3600\times 12}\times 60=\frac{7\pi }{720}$ $\frac{{\omega }_{1}}{{x}_{1}}=\frac{{\omega }_{2}}{{x}_{2}}=\frac{{\omega }_{3}}{{x}_{3}}=\frac{1}{60}=k$

Let \({\omega }_{1},{\omega }_{2}\) and \({\omega }_{3}\) be the angular speed of the second hand, minute hand and hour …

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