1. Condition for just slipping

For a coin on a rotating table, the required centripetal force is provided by static friction.

At the point of just slipping, $f_{\max} = \mu_s N = \mu_s mg$

This maximum friction provides the centripetal force: $m\omega^2 r = \mu_s mg$

So, $\omega^2 r = \mu_s g$

or $r = \frac{\mu_s g}{\omega^2}$

Thus, $r \propto \frac{1}{\omega^2}$

2. Use the given initial condition

Initially, the coin just slips at $r_1 = 1\text{ cm}$ with angular velocity $\omega$.

So, $r_1 = \frac{\mu_s g}{\omega^2}$

3. When angular velocity is halved

New angular velocity: $\omega_2 = \frac{\omega}{2}$

Then the new slipping radius is $r_2 = \frac{\mu_s g}{\left(\frac{\omega}{2}\right)^2}$

$r_2 = \frac{\mu_s g}{\omega^2/4} = 4\frac{\mu_s g}{\omega^2} = 4r_1$

Since $r_1 = 1$ cm, $r_2 = 4\times 1 = 4\text{ cm}$

4. Check options

• A: $1$ cm ✗
• B: $8$ cm ✗
• C: $4$ cm ✓
• D: $2$ cm ✗

So the correct option is C.