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(D) Given: Side of the cube $a = 10 \ cm$, depth in water $x = 3 \ cm$, density of oil $\delta_1 = 0.9 \ g \ cm^{-3}$, density of water $\delta_2 = 1

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$930$

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(D) Given: Side of the cube $a = 10 \ cm$, depth in water $x = 3 \ cm$, density of oil $\delta_1 = 0.9 \ g \ cm^{-3}$, density of water $\delta_2 = 1 \ g \ cm^{-3}$.Since the cube is floating, the weight of the cube is equal to the total buoyant force exerted by the oil and water.Weight of cube $W = mg$.Buoyant force $F_B = (	ext{Volume in oil} 	imes \delta_1 	imes g) + (	ext{Volume in water} 	imes \delta_2 	imes g)$.Volume of cube $V = a^3 = 10^3 = 1000 \ cm^3$.Volume in water $V_2 = a^2 	imes x = 10^2 	imes 3 = 300 \ cm^3$.Volume in oil $V_1 = a^2 	imes (a - x) = 10^2 	imes (10 - 3) = 700 \ cm^3$.By the law of flotation, $mg = V_1 \delta_1 g + V_2 \delta_2 g$.$m = V_1 \delta_1 + V_2 \delta_2$.$m = (700 	imes 0.9) + (300 	imes 1)$.$m = 630 + 300 = 930 \ g$.

$A$ wooden cube of side $10 \ cm$ floats at the interface between water and oil with its lower surface $3 \ cm$ below the interface. If the density of oil is $0.9 \ g \ cm^{-3}$, the mass of the wooden cube is (in $g$)

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