A uniform solid cylinder of mass m and radius R is released from rest on a rough incline of angle θ with tan θ=3/4, its axis horizontal. The…

For pure rolling of a solid cylinder, $a=\dfrac{g\sin\theta}{1+I/mR^2}=\tfrac23 g\sin\theta=0.4g$ (since $\sin\theta=0.6$), and the friction required follows from $mg\sin\theta-f=ma$, giving $f=\tfrac13 mg\sin\theta=0.2\,mg$ acting up the slope. Rolling without slipping is possible only if $f\le\mu mg\cos\theta$, i.e. $\mu\ge\tfrac13\tan\theta=0.25$. For $\mu=0.5$ this holds, so (A) and (B) are true. For $\mu=0.2<0.25$ the cylinder slips and kinetic friction $\mu mg\cos\theta=0.16\,mg$ acts, so $a=g(\sin\theta-\mu\cos\theta)=0.44g$ and $\alpha=\dfrac{(\mu mg\cos\theta)R}{\frac12 mR^2}=\dfrac{2\mu g\cos\theta}{R}=0.32\,g/R$; (C) is true. (D) is false because rolling is impossible when $\mu<0.25$. Always compare the friction required for rolling with its limiting value before choosing the formula, because that decides whether the no-slip relation $a=R\alpha$ may be used at all in this problem.