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(D) $1$. Let the system consist of blocks $M$,$m_0$,and $m$. Since the string is inextensible,all three blocks move with the same acceleration $a$.$2$

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all the above

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(D) $1$. Let the system consist of blocks $M$,$m_0$,and $m$. Since the string is inextensible,all three blocks move with the same acceleration $a$.$2$. The driving force for the system is the weight of block $m$,which is $mg$.$3$. The total mass of the system is $M + m_0 + m$.$4$. Applying Newton's second law to the whole system: $mg = (M + m_0 + m)a$,which gives $a = \frac{mg}{M + m_0 + m}$. Thus,option $A$ is correct.$5$. Since there is no friction between $M$ and $m_0$,and no external horizontal force acts on $m_0$,its acceleration is zero. Thus,option $B$ is correct.$6$. For block $m$,the equation of motion is $mg - T = ma$. Substituting $a$,we get $T = mg - m(\frac{mg}{M + m_0 + m}) = mg(1 - \frac{m}{M + m_0 + m}) = mg(\frac{M + m_0}{M + m_0 + m})$.$7$. Since $\frac{M + m_0}{M + m_0 + m}  0$,$T > 0$. The condition $Mg $8$. Since $A$,$B$,and $C$ are correct,the correct option is $D$.

$A$ block of mass $M$ is placed on a horizontal surface and it is tied with an inextensible string to a block of mass $m$,as shown in the figure. $A$ block of mass $m_0$ is also placed on $M$. If there is no friction between any two surfaces,then:

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