Question

Find the time period of meeting of the minute hand and second hand of a clock. (Initial positions are shown in the figure below)

• A. $\frac{59}{60}\text{min}$
• B. $\frac{60}{59}\text{min}$
• C. $\frac{3}{4}\text{min}$
• D. $\frac{3}{8}\text{min}$

Answer 1

The correct option is B $\frac{60}{59}\text{min}$


Angular velocity of minute hand, ${\omega }_{min}=\frac{2\pi }{60}{\text{rad min}}^{-1}$
Angular velocity of second hand,
${\omega }_{sec}=\frac{2\pi }{1}{\text{rad min}}^{-1}$
For second and minute hand to meet again
${\theta }_{sec}-{\theta }_{min}=2\pi$
$\Rightarrow ({\omega }_{sec}-{\omega }_{min})T=2\pi$
$\Rightarrow 2\pi (1-\frac{1}{60})T=2\pi$
$\Rightarrow T=\frac{60}{59}\text{min}$