In the arrangement shown in the figure,the ma | vedclass

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(C) Let $M_C$ be the mass of block $C$. The acceleration $a$ of the system is given by $a = \frac{M_C g}{M_C + M_B} = \frac{M_C g}{M_C + 2m}$.For bloc

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$\frac{3m}{\mu-1}$

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(C) Let $M_C$ be the mass of block $C$. The acceleration $a$ of the system is given by $a = \frac{M_C g}{M_C + M_B} = \frac{M_C g}{M_C + 2m}$.For block $A$ to remain stationary with respect to $B$,the pseudo force acting on $A$ (in the frame of $B$) must be balanced by the friction force.The pseudo force on $A$ is $F_p = ma$ (directed horizontally).The normal force on $A$ is $N = ma$ (since $A$ is pressed against the vertical face of $B$ due to acceleration $a$).The maximum static friction force is $f_{max} = \mu N = \mu ma$.For equilibrium of $A$ in the vertical direction,the friction force must balance its weight: $f = mg$.Thus,$mg \le \mu ma$,which implies $g \le \mu a$,or $a \ge \frac{g}{\mu}$.Substituting the expression for $a$: $\frac{M_C g}{M_C + 2m} \ge \frac{g}{\mu}$.$\mu M_C \ge M_C + 2m \implies M_C(\mu - 1) \ge 2m$.Assuming the configuration implies $A$ is on the vertical face,the acceleration $a$ causes $N=ma$. If $A$ were on a horizontal surface,$N=mg$. Given the diagram,$A$ is on the vertical face,so $N=ma$. The condition for $A$ not to slip down is $f \ge mg$. Thus $\mu ma \ge mg \implies a \ge g/\mu$.Using $a = \frac{M_C g}{M_C + 2m}$,we get $\frac{M_C}{M_C + 2m} \ge \frac{1}{\mu} \implies \mu M_C \ge M_C + 2m \implies M_C \ge \frac{2m}{\mu-1}$.Re-evaluating the standard problem type: If $A$ is on the vertical face,$N=ma$. If $A$ is on a horizontal face,$N=mg$. The diagram shows $A$ on the vertical face. However,checking the options,if $N=mg$,then $f_{max} = \mu mg$. For $A$ to be stationary,$ma \le \mu mg \implies a \le \mu g$. This doesn't match the options. If $N=ma$,then $f_{max} = \mu ma$. For $A$ to be stationary,$mg \le \mu ma \implies a \ge g/\mu$. This also doesn't match. Let's assume the force $N$ is provided by acceleration such that $N=ma$. If the system accelerates such that $A$ stays,$a = \frac{M_C g}{M_C + 2m + m} = \frac{M_C g}{M_C + 3m}$. Then $mg = \mu N = \mu ma \implies g = \mu a \implies a = g/\mu$. Then $\frac{M_C}{M_C + 3m} = \frac{1}{\mu} \implies \mu M_C = M_C + 3m \implies M_C = \frac{3m}{\mu-1}$. This matches option $C$.

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