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

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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 :

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a

\(\frac{{\omega }_{1}}{{x}_{1}}=\frac{{\omega }_{2}}{{x}_{2}}=\frac{{\omega }_{3}}{{x}_{3}}=\mathrm{k}\)

b

\({\omega }_{1}{x}_{1}={\omega }_{2}{x}_{2}={\omega }_{3}{x}_{3}=\mathrm{k}\)

c

\({\omega }_{1}{x}_{1}^{2}={\omega }_{2}{x}_{2}^{2}={\omega }_{3}{x}_{3}^{2}=\mathrm{k}\)

d

\({\omega }_{1}^{2}{x}_{1}={\omega }_{2}^{2}{x}_{2}={\omega }_{3}^{2}{x}_{3}=\mathrm{k}\)

✓ Correct answer: a)

\(\frac{{\omega }_{1}}{{x}_{1}}=\frac{{\omega }_{2}}{{x}_{2}}=\frac{{\omega }_{3}}{{x}_{3}}=\mathrm{k}\)

Explanation

\({\omega }_{1}=\frac{2\pi }{60};{x}_{1}=\frac{2\pi }{60}\times 60=2\pi\)

\({\omega }_{2}=\frac{2\pi }{3600};{x}_{2}=\frac{2\pi }{3600}\times 60=\frac{2\pi }{60}\)

\({\omega }_{3}=\frac{2\pi }{3600\times 12};{x}_{3}=\frac{2\pi }{3600\times 12}\times 60=\frac{7\pi }{720}\)

\(\frac{{\omega }_{1}}{{x}_{1}}=\frac{{\omega }_{2}}{{x}_{2}}=\frac{{\omega }_{3}}{{x}_{3}}=\frac{1}{60}=k\)

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