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# Find the time period of the meeting of the minute-hand and the second-hand of a clock.A) $60{\text{min}}$B) $\dfrac{{59}}{{60}}{\text{min}}$C) $\dfrac{{60}}{{59}}{\text{min}}$D) ${\text{59min}}$

Last updated date: 20th Jun 2024
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Hint: Here the minute-hand and the second-hand execute circular motion along the same circular path but with different angular velocities. Their periods will differ too. The second-hand takes one minute to complete one rotation while the minute-hand takes 60 minutes to complete one rotation. So we can say that the minute-hand is moving relative to the second-hand and the relative angular displacement of the two hands as they meet will be $2\pi$ .

Formula Used:
1) The relative angular displacement of two bodies moving along the same circular path is given by ${\theta _{rel}} = {\theta _2} - {\theta _1}$ where ${\theta _1}$ and ${\theta _2}$ are the angular displacements of the two bodies.
2) The angular displacement of a body is given by, $\theta = \omega t$ where $\omega$ is the angular velocity of the body and $t$ is the time taken to undergo the angular displacement.

Step 1: List the parameters of the second-hand and the minute-hand.
Let ${\omega _{\sec }} = \dfrac{{2\pi }}{1}{\text{rad/min}}$ be the angular velocity of the second-hand and let ${\omega _{\min }} = \dfrac{{2\pi }}{{60}}{\text{rad/min}}$ be the angular velocity of the minute-hand.
The relative angular displacement of the two hands as they meet again will be
${\theta _{rel}} = {\theta _{\sec }} - {\theta _{\min }} = 2\pi$ -------- (1).
Let $t$ be the period of the meeting of the two hands.
Step 2: Express equation (1) in terms of the angular velocities of the two hands to find $t$ .
Since $\theta = \omega t$ , equation (1) can be represented as $\left( {{\omega _{\sec }} - {\omega _{\min }}} \right)t = 2\pi$ ------ (2)
Substituting the values for ${\omega _{\sec }} = \dfrac{{2\pi }}{1}{\text{rad/min}}$ and ${\omega _{\min }} = \dfrac{{2\pi }}{{60}}{\text{rad/min}}$ in equation (2) we get, $\left( {\dfrac{{2\pi }}{1} - \dfrac{{2\pi }}{{60}}} \right)t = 2\pi$
$\Rightarrow \left( {1 - \dfrac{1}{{60}}} \right)t = 1$
$\dfrac{{59}}{{60}}t = 1 \Rightarrow t = \dfrac{{60}}{{59}}{\text{min}}$
Then the period of the meeting will be $\Rightarrow t = \dfrac{{60}}{{59}}{\text{min}}$ .

Hence the correct option is C.

Note: Alternate method
As the second hand completes one rotation, the minute-hand will have moved to the next minute. So to meet the minute-hand, the second-hand needs to travel a bit more. Hence the time period of the meeting of the two hands will be greater than one minute. Out of the given options, only option C is greater than one minute i.e., $\dfrac{{60}}{{59}} > 1{\text{min}}$ . So the correct option is C.