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(C) In case $(i)$,the system consists of masses $4m$ and $2m$ connected by a string over a pulley. The acceleration $a = \frac{(4m - 2m)g}{4m + 2m} =

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$F_1 = F_2$

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(C) In case $(i)$,the system consists of masses $4m$ and $2m$ connected by a string over a pulley. The acceleration $a = \frac{(4m - 2m)g}{4m + 2m} = \frac{2mg}{6m} = \frac{g}{3}$. The tension $T_1$ in the string is $T_1 = 4m(g - a) = 4m(g - g/3) = 4m(2g/3) = \frac{8mg}{3}$. The force on the pulley is $F_1 = 2T_1 = \frac{16mg}{3}$.In case $(ii)$,the system consists of masses $4m$ and $(m + m) = 2m$ connected by a string over a pulley. This is identical to case $(i)$ in terms of the masses hanging on either side of the pulley. Therefore,the acceleration $a$ and the tension $T_2$ in the string are the same as in case $(i)$. Thus,$T_2 = T_1 = \frac{8mg}{3}$. The force on the pulley is $F_2 = 2T_2 = \frac{16mg}{3}$.Comparing the two,we find $F_1 = F_2$.

In the arrangement shown in the figure,the pulley is smooth and massless,and all the strings are light. Let $F_1$ be the force exerted on the pulley in case $(i)$ and $F_2$ be the force in case $(ii)$. Then:

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