Find the time interval between two successive overlaps of the hour hand and the minute hand of a clock.

Explanation

Let the time interval after which the two hands meet each other be t hours.
in the case of the hour hand. The angular velocity
`omega = (2pi)/(12) = (pi)/(6) "rad"cdot h^(-1)`.
Hence, in t hours, the angular displacement of the hour hand, `theta = omega t = (pi)/(6) t " rad"` .
In the case of the minute hand, the angular velocity `omega. = (2pi)/(1) = 2pi " rad " cdot h^(-1)` .
Hence, the angular displacement of the minute hand, `theta. = omega. t = 2 pi` t rad.
After meeting each other, when the hands coincide again after t hours, the minute hand completes one revolution more than the hour hand, i.e., the angular displacement of the minute hand is `2pi` rad more than that of the hour hand.
So,`" " theta. - theta = 2pi " " or, 2pi t - (pi t)/(6) = 2pi " " or, " " t xx (11)/(6) = 2`
or,`" "t = (12)/(11)` h = 1 h 5 min 27 s (approx).