Question

A coin placed on a rotating turntable just slips. If it is placed at a distance of 4 cm from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of 

विकल्प

• 1 cm

• 2 cm

• 4 cm

•  8 cm

Answer 1

8 cm

Answer 2

1 cm 

Let the force of friction between the coin and the rotating turntable be F.
For the coin to just slip ,

we have : \[\text{ F = m }\omega^2 \text{r}\]

Here, \[\text{ m } \omega^2 \text{ r}\] is the centrifugal force acting on the coin.

For constant F and m, we have : \[\text{r} \propto \frac{1}{\omega^2}\]

Therefore,

\[\frac{\text{r}'}{\text{r}} = \left( \frac{\omega}{\omega'} \right)^2 \]

\[ \Rightarrow \text{ r' = 1 cm }\]