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