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(D) Let $a_c$ be the acceleration of the center of mass and $f$ be the frictional force.From Newton's second law for translation: $F - f = ma_c$  $(1)

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$B, D$

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(D) Let $a_c$ be the acceleration of the center of mass and $f$ be the frictional force.From Newton's second law for translation: $F - f = ma_c$  $(1)$From the torque equation about the center of mass: $fR = I_c \alpha$  $(2)$For rolling without slipping: $a_c = \alpha R$,so $\alpha = \frac{a_c}{R}$.Substituting $\alpha$ in $(2)$: $fR = I_c \frac{a_c}{R} \implies f = \frac{I_c a_c}{R^2}$.Substituting $f$ in $(1)$: $F - \frac{I_c a_c}{R^2} = ma_c \implies a_c = \frac{F}{m + \frac{I_c}{R^2}}$.$(A)$ Incorrect: $a_c$ depends on $I_c$,which differs for solid $(I_c = \frac{1}{2}mR^2)$ and hollow $(I_c = mR^2)$ cylinders.$(D)$ Correct: For a thin-walled hollow cylinder,$I_c = mR^2$. Thus,$a_c = \frac{F}{m + \frac{mR^2}{R^2}} = \frac{F}{2m}$.$(C)$ Incorrect: The frictional force $f = \frac{I_c a_c}{R^2} = \frac{I_c F}{R^2(m + I_c/R^2)}$,which is not necessarily $\mu mg$.$(B)$ Correct: For rolling without slipping,$f \leq \mu mg$. Since $f = \frac{I_c a_c}{R^2}$,we have $\frac{I_c a_c}{R^2} \leq \mu mg \implies a_c \leq \frac{\mu mgR^2}{I_c}$. For a solid cylinder,$I_c = \frac{1}{2}mR^2$,so $a_c \leq \frac{\mu mgR^2}{0.5mR^2} = 2\mu g$.

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