A cubical block of wood of edge 10 cm and mas | vedclass

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(D) Let $x_1$ be the length of the block submerged in oil and $x_2$ be the length submerged in water. The total length of the block is $L = 10$ $cm =

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$B$ and $C$ both.

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(D) Let $x_1$ be the length of the block submerged in oil and $x_2$ be the length submerged in water. The total length of the block is $L = 10$ $cm = 0.1$ $m$.The density of oil $ho_{oil} = 0.6 	imes 1000 = 600$ $kg/m^3$ and the density of water $ho_{water} = 1000$ $kg/m^3$.At equilibrium,the weight of the block equals the total buoyant force:$mg = F_{oil} + F_{water}$$0.92 	imes 10 = (A 	imes x_1 	imes ho_{oil} 	imes g) + (A 	imes x_2 	imes ho_{water} 	imes g)$where $A = 0.1 	imes 0.1 = 0.01$ $m^2$.$9.2 = 0.01 	imes x_1 	imes 600 	imes 10 + 0.01 	imes x_2 	imes 1000 	imes 10$$9.2 = 6x_1 + 10x_2$ (where $x_1, x_2$ are in meters).Since the total depth of the block is $10$ $cm$,$x_1 + x_2 = 10$ $cm = 0.1$ $m$.Substituting $x_1 = 0.1 - x_2$:$9.2 = 6(0.1 - x_2) + 10x_2$$9.2 = 0.6 - 6x_2 + 10x_2$$8.6 = 4x_2 \implies x_2 = 2.15$ $cm$ (This implies the block is mostly in water).Wait,re-evaluating the depth of oil: The oil layer is $4$ $cm$ deep. If $x_1$ is the part in oil,$x_1$ must be $\leq 4$ $cm$. Solving $9.2 = 6x_1 + 10(10 - x_1)$ (using $cm$):$9.2 = 6x_1 + 100 - 10x_1$$4x_1 = 100 - 92 = 8 \implies x_1 = 2$ $cm$.Then $x_2 = 10 - 2 = 8$ $cm$.Thus,$8$ $cm$ is in water and $2$ $cm$ is in oil.

$A$ cubical block of wood of edge $10$ $cm$ and mass $0.92$ $kg$ floats in a tank of water with a layer of oil of relative density $0.6$ to a depth of $4$ $cm$ above the water. When the block attains equilibrium with four of its side edges vertical,which of the following is true?

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