Question

What is the angular velocity in rad `s^(-1)` of the hour minute and second hand of a clock ?

Answer 1

Correct Answer - `(pi)/(21600) rad s^(-1) ; (pi)/(1800) rad s^(-1) ; (pi)/(30) rad s^(-1)` .
(i) The hour hand of a clock completes one revolution in `12` hours i.e
`T = 12 h = 12 xx 60 xx 60 s`
`omega = (2pi)/(T) = (2pi)/(60 xx 60) = (pi)/(1800) rad s^(-1)`
(ii) For minutes hand of the clock,
`T =60 min = 60 xx 60 s`
`:. omega = (2pi)/(T) = (2pi)/(60 xx60) = (pi)/(1800) rds^(-1)`
(iii) For second s hand of the clock `T = 60 sec`
`:. omega = (2pi)/(T) = (2pi)/(60 xx60) = (pi)/(30) rds^(-1)` .

Comments

Wrong calculations