Question

In a clock, what is the time period of meeting the minute hand and the second hand?
A) $59s$
B) $\dfrac{{60}}{{59}}s$
C) $\dfrac{{59}}{{60}}s$
D) $\dfrac{{3600}}{{59}}s$

Answer 1

Hint- In the clock, the second hand, minute hand and the hour hand are rotating in the angular velocity. Here, we have to consider only the second hand and the minute hand. The second hand and the minute hand meet every second in an hour. Angular velocity is defined as the amount of time taken for the angular displacement. The angular velocity changes with the angle in which the object resides at a time $t$.

Formula used:
$\omega = \dfrac{{2\pi }}{T}$
Where,
$\omega$=the angular velocity
$T$=Time
$2\pi$=angular displacement

Complete step by step answer:
(i) Let's consider the one of the minutes in an hour, the second hand and the minute hand meet. We know $3600$ seconds in an hour and the $60$ seconds in a minute.
(ii) Let ${\omega _1}$ and ${\omega _2}$ are the angular velocities of the second hand and the minute hand and ${T_1}$, ${T_2}$ are the time period of second hand and the minute hand.
${\omega _1} = \dfrac{{2\pi }}{{3600}}$; ${\omega _2} = \dfrac{{2\pi }}{{60}}$
(iii) \omega \Rightarrow $$$${\omega _2} - {\omega _1}
$\Rightarrow \dfrac{{2\pi }}{{60}} - \dfrac{{2\pi }}{{3600}}$
$\omega \Rightarrow \dfrac{{2\pi \left( {59} \right)}}{{3600}}$
(iv)To find the time of the known angular velocity,
$\omega = \dfrac{{2\pi }}{T} \Rightarrow$ $T = \dfrac{{2\pi }}{\omega }$
Applying the value of $\omega$ in the above time equation,
$T = \dfrac{{2\pi \times 3600}}{{59 \times 2\pi }}$
Cancelling $2\pi$, we get
$T = \dfrac{{3600}}{{59}}s$

Hence the correct option is D.

Additional information:
(i) The velocity is the amount of linear displacement travelled in a period of time $t$. And its unit is metre per second. The angular velocity is the total amount of angular displacement travelled in a period of time $t$. ‘Radians per second’ is the unit of the angular velocity.

Note: The second and the minute hand meet every $60$ seconds. We can also say that the minute hand and the second hand meet every minute. We found the answer approximately equal to $60$seconds. We find this time period by evaluating the angular velocity. This describes how fast an object is moving in an orbit.