Angular velocity of hour hand of a watch is

Question

Angular velocity of hour hand of a watch is (A) (π/43200)rads-1 (B) (π/30)rads-1 (C) (π/21600)rads-1 (D) (π/1800)rads-1

Tardigrade Admin · Accepted Answer

Angular velocity = (angle traced / time taken) Time period of hour hand is 12 h . So, ω = (2π/12) rad h-1 = (2π/12 × 60 × 60)rad s-1 = (π/21600)rad s-1

Q. Angular velocity of hour hand of a watch is

$\frac{π}{43200} r a d s^{- 1}$

$\frac{π}{30} r a d s^{- 1}$

$\frac{π}{21600} r a d s^{- 1}$

$\frac{π}{1800} r a d s^{- 1}$

Angular velocity = $\frac{an g l e t r a ce d}{t im e t ak e n}$
Time period of hour hand is 12 h . So, $ω = \frac{2 π}{12} r a d h^{- 1}$
= $\frac{2 π}{12 \times 60 \times 60} r a d s^{- 1} = \frac{π}{21600} r a d s^{- 1}$

Q. Angular velocity of hour hand of a watch is

43200π​rads−1

30π​rads−1

21600π​rads−1

1800π​rads−1

Angular velocity = timetakenangletraced​ Time period of hour hand is 12 h . So, ω=122π​radh−1 = 12×60×602π​rads−1=21600π​rads−1

$\frac{π}{43200} r a d s^{- 1}$

$\frac{π}{30} r a d s^{- 1}$

$\frac{π}{21600} r a d s^{- 1}$

$\frac{π}{1800} r a d s^{- 1}$

Angular velocity = $\frac{an g l e t r a ce d}{t im e t ak e n}$
Time period of hour hand is 12 h . So, $ω = \frac{2 π}{12} r a d h^{- 1}$
= $\frac{2 π}{12 \times 60 \times 60} r a d s^{- 1} = \frac{π}{21600} r a d s^{- 1}$