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A coin placed on a rotating turntable just slips if it is at a distance of 40 cm from the centre if the angular velocity of the turntable is doubled, it will just slip at a distance of
A. 10 cm
B. 20 cm
C. 40 cm
D. 80 cm

Answer
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Hint: The coin will slip when the angular force will be equal to gravitational force. So to find the distance when it will slip just equate both the forces.

Complete step by step solution:
 In the above question given that
r = 40 cm
By using formula of angular force
$F = m{\omega ^2}r$
It slips when
$F = \mu mg$

Now,
$\mu mg = m{\omega ^2}r$
 $ \Rightarrow \mu g = {\omega ^2}r$
$ \Rightarrow \mu g = {\omega ^2} \times 40 \cdots \cdots \left( 1 \right)$

Now,
$\omega = 2\omega $
$ \Rightarrow \mu g = {\left( {2\omega } \right)^2}r$
$ \Rightarrow \mu g = 4{\omega ^2}r \cdots \cdots \left( 2 \right)$
From equation 1 & 2 we get,
${\omega ^2} \times 40 = 4{\omega ^2}r$
$ \Rightarrow r = 10cm$

Thus, option A is correct.

Additional Information: When ω is the constant angular velocity and is equal to v/R. From Newton's laws of motion it follows that the natural motion of an object is one with constant speed in a straight line, and that a force is necessary if the object is to depart from this type of motion. Whenever an object moves in a curve, a centripetal force is necessary. In circular motion the tangential speed is constant but is changing direction at the constant rate of ω, so the centripetal force along the radius is the only force involved.

Note: In circular motion the tangential speed is constant but is changing direction at the constant rate of ω, so the centripetal force along the radius is the only force involved.