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(B) Given: Mass $M = 5\, kg$,Radius $R = 0.5\, m$,Force $F = 40\, N$. For a hollow cylinder,the moment of inertia about its central axis is $I = MR^2$

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$16$

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(B) Given: Mass $M = 5\, kg$,Radius $R = 0.5\, m$,Force $F = 40\, N$. For a hollow cylinder,the moment of inertia about its central axis is $I = MR^2$.Let $f$ be the friction force acting at the point of contact in the backward direction.The equation for linear motion is: $F + f = Ma$,where $a = R\alpha$ for rolling without slipping.So,$40 + f = M(R\alpha) \quad (i)$The equation for rotational motion about the center of mass is: $(F 	imes R) - (f 	imes R) = I\alpha$.Substituting $I = MR^2$: $(40 - f)R = (MR^2)\alpha \implies 40 - f = MR\alpha \quad (ii)$Adding equations $(i)$ and $(ii)$:$(40 + f) + (40 - f) = MR\alpha + MR\alpha$$80 = 2MR\alpha$$80 = 2 	imes 5 	imes 0.5 	imes \alpha$$80 = 5\alpha$$\alpha = 16\, rad/s^2$.

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