$A$ uniform disc of radius $R$ is resting on a table on its rim. The coefficient of friction between the disc and the table is $\mu$. Now,the disc is pulled with a force $F$ applied at its center as shown in the figure. What is the maximum value of $F$ for which the disc rolls without slipping?

(D) Let $M$ be the mass of the disc. The forces acting on the disc are the applied force $F$ at the center and the friction force $f$ at the point of contact,acting in the opposite direction to $F$.
$1$. Translational motion equation: $F - f = Ma$ ... $(i)$
$2$. Rotational motion equation about the center of mass: $\tau = I\alpha$. Since the friction force $f$ acts at the rim,the torque is $\tau = fR$. The moment of inertia of a disc about its center is $I = \frac{1}{2}MR^2$. Thus,$fR = (\frac{1}{2}MR^2)\alpha$.
$3$. For pure rolling,the condition is $a = R\alpha$,which implies $\alpha = \frac{a}{R}$.
Substituting $\alpha$ into the torque equation: $fR = (\frac{1}{2}MR^2)(\frac{a}{R}) = \frac{1}{2}MaR$,which simplifies to $f = \frac{1}{2}Ma$ or $Ma = 2f$ ... $(ii)$.
$4$. Substitute $(ii)$ into $(i)$: $F - f = 2f$,which gives $F = 3f$.
$5$. For rolling without slipping,the friction force must satisfy $f \le \mu N$,where $N = Mg$ is the normal force. Therefore,$f \le \mu Mg$.
$6$. Since $F = 3f$,the maximum force $F$ is $F_{max} = 3f_{max} = 3\mu Mg$.