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NEET 2024
Exam
NEET
Year
2024
Subject
Physics
Chapter
Oscillations

NEET 2024 · Physics · Oscillations

Let ω_1, ω_2 and ω_3 be the angular speed of the second hand,...

Let ω1,ω2\omega_1, \omega_2 and ω3\omega_3 be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If x1,x2x_1, x_2 and x3x_3 are their respective angular distances in 1 minute then the factor which remains constant (k)(k) is

Options

  1. Aω1x1=ω2x2=ω3x3=k\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=k
  2. Bω1x1=ω2x2=ω3x3=k\omega_1 x_1=\omega_2 x_2=\omega_3 x_3=k
  3. Cω1x12=ω2x22=ω3x32=k\omega_1 x_1^2=\omega_2 x_2^2=\omega_3 x_3^2=k
  4. Dω12x1=ω22x2=ω32x3=k\omega_1^2 x_1=\omega_2^2 x_2=\omega_3^2 x_3=k

Correct answer

A. ω1x1=ω2x2=ω3x3=k\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=k

Step-by-step solution

ω1=2π60;x1=2π60×60=2π\omega_1=\frac{2 \pi}{60} ; \quad x_1=\frac{2 \pi}{60} \times 60=2 \pi ω2=2π3600;x2=2π3600×60=2π60\omega_2=\frac{2 \pi}{3600} ; \quad x_2=\frac{2 \pi}{3600} \times 60=\frac{2 \pi}{60} ω3=2π3600×12;x3=2π3600×12×60=2π720\omega_3=\frac{2 \pi}{3600 \times 12} ; \quad x_3=\frac{2 \pi}{3600 \times 12} \times 60=\frac{2 \pi}{720} ω1x1=ω2x2=ω3x3=160=k\frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=\frac{1}{60}=k
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