1. Given data

• Frequency of rotation: $f = 3.5\ \text{rev/s}$
• Distance of coin from axis: $r = 1.25\ \text{cm} = 0.0125\ \text{m}$
• Acceleration due to gravity: $g = 10\ \text{m/s}^2$

The coin remains at rest relative to the disc, so static friction provides the required centripetal force.

2. Angular speed

$\omega = 2\pi f = 2\pi(3.5) = 7\pi\ \text{rad/s}$

3. Required centripetal force

For a coin of mass $m$, the centripetal force needed is

$F_c = m\omega^2 r$

This is supplied by static friction. Maximum static friction is

$f_{\max} = \mu mg$

For the coin to remain at rest:

$m\omega^2 r \le \mu mg$

At the limiting condition,

$\mu = \frac{\omega^2 r}{g}$

4. Substitute values

$\mu = \frac{(7\pi)^2(0.0125)}{10}$

Using $\pi^2 \approx 9.8$,

$\mu = \frac{49\pi^2 \cdot 0.0125}{10}$

$\mu = \frac{49 \times 9.8 \times 0.0125}{10}$

$49 \times 9.8 = 480.2$

$480.2 \times 0.0125 = 6.0025$

$\mu = \frac{6.0025}{10} \approx 0.60$

5. Answer

$\mu \approx 0.6$

So the correct option is D.