When the frictional force is sufficient to provide the centripetal force the coin revolves with the disc. Coin would slip the record, if this force is not sufficient.
Force of friction, $f=\mu R=\mu \mathrm{mg}$,
Centripetal force, $m v^2 / r=m r \omega^2$
In order to prevent slipping, $\mu m g \geq m r \omega^2$, i.e. $\mu g \geq r \omega^2$
Case 1: Given, $r=4 \mathrm{~cm}=0.04 \mathrm{~m}$, $n=33 \frac{1}{3} \mathrm{rev} / \mathrm{min}$
$$
\begin{aligned}
&\omega=2 \pi n=2 \times(22 / 7) \times\left(33 \frac{1}{3} \mathrm{rev} / \mathrm{min}\right)=3.49 \mathrm{~s}^{-1} . \\
&\therefore r \omega^2=(.04)(3.49)^2=0.49 \mathrm{~ms}^{-2}
\end{aligned}
$$
and $\mu g=0.15 \times 9.8=1.47 \mathrm{~ms}^{-2}$
$\because \mu g \geq r \omega^2$, hence first coin will rotate with the record.
Case 2: Given, $r=14 \mathrm{~cm}=0.14 \mathrm{~m}$,
and $\omega=3.49 \mathrm{~s}^{-1}$.
$\therefore r \omega^2=(.14)(3.49)^2=1.7 \mathrm{~ms}^{-2}$ and $\mu g$
$=0.15 \times 9.8=1.47 \mathrm{~ms}^{-2}$
$\therefore \mu \mathrm{g}$ is less than $r \omega^2$, hence second coin will not rotate with the record.