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If the length of the second’s hand of the clock is 10cm, the speed of its tip (in $cm{{s}^{-1}}$) is nearly
A.2
B.0.5
C.1.5
D.1

Answer
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Hint: To solve this problem we have to apply the concept of circular motion and of angular velocity. We will also apply the relationship of velocity with radius and angular velocity. The second’s hand will go through the circular path and make one complete circle in one revolution.

Formula used:
We will use the following given formulae to get the precise and accurate solution:-
$\omega =\dfrac{2\pi }{T}$ and $v=r\omega $

Complete step by step solution:
From the question above we have the length of the second hand which is also considered as the radius of its circular motion, $r=10cm$.
We will consider the full revolution to find its speed, $v$ for easy and sound calculation.
During one complete revolution, the time taken, $T=60s$.
To find the speed we will first find the angular velocity, $\omega $ for the second’s hand.
Using formula for angular velocity, $\omega =\dfrac{2\pi }{T}$ for this case we get
$\omega =\dfrac{2\pi }{60}$
$Rightarrow \omega =\dfrac{\pi }{30}rad/s$……………. $(i)$
Now, using the formula for the velocity of the second’s hand in terms of radius and angular velocity, $v=r\omega $for this case we have
$v=10\times \dfrac{\pi }{30}$
We have used the value of $\omega $ from equation $(i)$.
Solving more we get
$v=\dfrac{\pi }{3}$…………………. $(ii)$
We know that the value of $\pi =3.14$ and putting this value in $(ii)$ we get
$v=\dfrac{3.14}{3}$
$v=1.05cm{{s}^{-1}}$
From the options given above the nearest value of $v$ is $1$.

Hence, option (D) is the correct option.

Note:
We have found the linear velocity after finding angular velocity of the second’s hand of the clock because we had to find the velocity in $cm{{s}^{-1}}$ and it is linear velocity only which has the units as $cm{{s}^{-1}}$. Therefore in solving these types of problems we should take care of units also to know what actually we have to find. If the velocity is given $rad/s$ then we will find angular velocity.