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(D) Let $a$ be the acceleration of the system $(M_0 + M + m)$ towards the right. Then $F = (M_0 + M + m)a$,so $a = F / (M_0 + M + m)$.For blocks $M$ a

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Both $(A)$ and $(B)$.

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(D) Let $a$ be the acceleration of the system $(M_0 + M + m)$ towards the right. Then $F = (M_0 + M + m)a$,so $a = F / (M_0 + M + m)$.For blocks $M$ and $m$ to remain stationary with respect to $M_0$,they must have the same acceleration $a$ as $M_0$.For block $M$ (on the horizontal surface): The only horizontal force is the tension $T$ in the string. Thus,$T = Ma$.For block $m$ (on the vertical surface): The vertical forces are gravity $mg$ downwards and tension $T$ upwards. For it to be stationary relative to $M_0$,it must not move vertically,so $T = mg$.Equating the two expressions for $T$: $Ma = mg$,which implies $a = g$.Substituting $a = g$ into the equation for $F$: $F = (M_0 + M + m)g$.Since $a = g$ is a fixed value required for equilibrium,there is only one unique value of $F$ that satisfies this condition. Thus,statement $(B)$ is correct.If $F = 0$,then $a = 0$. For $M$ and $m$ to be stationary,we need $T = Ma = 0$ and $T = mg$. This implies $mg = 0$,which is impossible. Thus,if $F = 0$,the blocks cannot remain stationary. Statement $(A)$ is also correct.Therefore,the correct option is $(D)$.

Consider a situation in which both the horizontal and vertical surfaces of the block $M_0$ are smooth,as shown in the adjoining figure. $A$ force $F$ is applied to $M_0$. Mark the correct statement$(s)$.

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