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(C) Let the acceleration of the system be $a$ towards the left. For the mass $4M$ on the ice slab,the forces acting are the pulling force $2Mg$ to the

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$6$

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(C) Let the acceleration of the system be $a$ towards the left. For the mass $4M$ on the ice slab,the forces acting are the pulling force $2Mg$ to the left and the tension $T$ to the right. Applying Newton's second law: $2Mg - T = 4Ma$ (Equation $1$). For the hanging mass $M$,the forces acting are the tension $T$ upwards and its weight $Mg$ downwards. Since the system accelerates towards the left,the mass $M$ moves upwards. Applying Newton's second law: $T - Mg = Ma$ (Equation $2$). Adding Equation $1$ and Equation $2$: $(2Mg - T) + (T - Mg) = 4Ma + Ma$. $Mg = 5Ma$,which gives $a = \frac{g}{5}$. Substituting the value of $a$ in Equation $2$: $T = Mg + M(\frac{g}{5}) = Mg + \frac{Mg}{5} = \frac{6Mg}{5}$. Comparing this with the given expression $\frac{x}{5}Mg$,we get $x = 6$.

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