Question

Find the time period of meeting of minute hand and second hand of a clock.

• A.
• B.
• C. t = 59 min
• D. t = 60 min

Answer 1

The correct option is B

${\omega }_{min}$ = $\frac{2\pi }{60}$rad/min, ${\omega }_{sec}$ = $\frac{2\pi }{1}$rad/min

${\theta }_{sec}-{\theta }_{min}$ = $2\pi$ (for second and minute hand to meet again)

$({\omega }_{sec}-{\omega }_{min})t$ = $2\pi$

$2\pi (1-\frac{1}{60})t$ = $2\pi$ $\Rightarrow t$ = $\frac{60}{59}$min.