Circular Motion
1 Board Physics previous year questions on Circular Motion — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.
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 :
\(\frac{{\omega }_{1}}{{x}_{1}}=\frac{{\omega }_{2}}{{x}_{2}}=\frac{{\omega }_{3}}{{x}_{3}}=\mathrm{k}\)
\({\omega }_{1}=\frac{2\pi }{60};{x}_{1}=\frac{2\pi }{60}\times 60=2\pi\)
\({\omega }_{2}=\frac{2\pi }{3600};{x}_{2}=\frac{2\pi }{3600}\times 60=\frac{2\pi }{60}\)
\({\omega }_{3}=\frac{2\pi }{3600\times 12};{x}_{3}=\frac{2\pi }{3600\times 12}\times 60=\frac{7\pi }{720}\)
\(\frac{{\omega }_{1}}{{x}_{1}}=\frac{{\omega }_{2}}{{x}_{2}}=\frac{{\omega }_{3}}{{x}_{3}}=\frac{1}{60}=k\)
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