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(C) Let $L$ be the side of the cube in $cm$. The density of water is $\rho_w = 1 \,g/cm^3$.In the first case,the cube with the $200 \,g$ mass is just

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$10$

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(C) Let $L$ be the side of the cube in $cm$. The density of water is $ho_w = 1 \,g/cm^3$.In the first case,the cube with the $200 \,g$ mass is just floating,meaning the total weight is balanced by the buoyant force on the entire volume of the cube:$(M_{cube} + 200)g = (L^3 	imes ho_w)g$In the second case,the cube alone is floating,meaning its weight is balanced by the buoyant force on the submerged volume:$M_{cube}g = (L^2 	imes (L - 2) 	imes ho_w)g$Subtracting the second equation from the first:$(M_{cube} + 200)g - M_{cube}g = (L^3 	imes ho_w)g - (L^2(L - 2) 	imes ho_w)g$$200 = L^3 - L^2(L - 2)$$200 = L^3 - L^3 + 2L^2$$200 = 2L^2$$L^2 = 100$$L = 10 \,cm$.

$A$ wooden cube just floats inside water with a $200 \,g$ mass placed on it. When the mass is removed,the cube floats with its top surface $2 \,cm$ above the water level. The side of the cube is ......... $cm$.

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