A coin placed on a rotating turntable just slips if it is placed at a distance of 16cm from the center. If the angular velocity of the turntable is doubled, it will just slip at a distance of-
A. 1cm
B. 2cm
C. 4cm
D. 8cm

Hint: We will measure the centripetal force which will act on the coin in the solution as well as the frictional force. From the figure as well as the condition given, we will understand that the frictional force as well as the centripetal force will be equal to each other. Refer to the solution below.
Formula used: $f = \mu N$,

Complete step-by-step answer:
Since the coin is rotating, there will be a centripetal force working on it in the outwards direction. Let that centripetal force be equal to $m{\omega ^2}r$. If the angular velocity at which the coin is rotating is $\omega $.
As it is given in the question that the coin is slipping, there will be a frictional force working on it. Let that frictional force be $f$.

The force working downwards on the coin will be $mg$which will be equal and opposite to the force working upwards on it which is $N$.
The formula of friction is $f = \mu N$ (equation 1)
According to the question, the coin slips. So, for the slipping to occur the frictional force will be equal to $f = m{\omega ^2}r$(refer to the figure as well).
The above formula can be said as equation 2.
$ \Rightarrow f = m{\omega ^2}r$ (equation 2)
Observing equation 1 and equation 2, we can say that-
$ \Rightarrow \mu N = m{\omega ^2}r$
As we already know that N is equal to mg -$N = mg$. So, putting this value in the above equation we get-
$
   \Rightarrow \mu N = m{\omega ^2}r \\
    \\
   \Rightarrow \mu mg = m{\omega ^2}r \\
    \\
   \Rightarrow \mu g = {\omega ^2}r \\
$
Let this be equation 3.
 Since the coin is placed at 16cm from the center, the radius will be calculated as 16cm.
Initial angular velocity of the coin was $\omega $.
The final angular velocity will be $\omega '$.
Since we doubled the angular velocity, we can say that-
$ \Rightarrow \omega ' = 2\omega $
The initial radius was r. Let the final radius be r’.
 Comparing the initial and the final conditions through equation 3, we can say that-
$ \Rightarrow {\omega ^2}r = \omega {'^2}r'$
Putting the values of $\omega '$ and r, we get-
$
   \Rightarrow {\omega ^2} \times 16 = {\left( {2\omega } \right)^2} \times r' \\
    \\
   \Rightarrow {\omega ^2} \times 16 = 4{\omega ^2} \times r' \\
    \\
   \Rightarrow 16 = 4r' \\
    \\
   \Rightarrow r' = \dfrac{{16}}{4} \\
    \\
   \Rightarrow r' = 4cm \\
$
Thus, we can say that option C is the correct option.

Note: Angular velocity is the velocity rate at which an object or a particle in a given time span rotates around a center or a particular point. This is sometimes referred to as rotational velocity. Angular velocity is measured by angle per unit time or radians per second (rad / s).