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(C) For the block to be in equilibrium,the weight of the block must be equal to the total buoyant force exerted by the oil and water.Weight of block =

Vedclass · Accepted Answer

$760$

Vedclass · Answer

(C) For the block to be in equilibrium,the weight of the block must be equal to the total buoyant force exerted by the oil and water.Weight of block = Weight of displaced oil + Weight of displaced water$mg = V_{oil} ho_{oil} g + V_{water} ho_{water} g$$m = V_{oil} ho_{oil} + V_{water} ho_{water}$Given that the block is a cube of side $10 \ cm$,the total height is $10 \ cm$. Since $4 \ cm$ is submerged in water,the remaining $6 \ cm$ is in the oil.$V_{oil} = 10 \ cm 	imes 10 \ cm 	imes 6 \ cm = 600 \ cm^3$$V_{water} = 10 \ cm 	imes 10 \ cm 	imes 4 \ cm = 400 \ cm^3$Density of oil $ho_{oil} = 0.6 \ g/cm^3$ and density of water $ho_{water} = 1 \ g/cm^3$.$m = (600 	imes 0.6) + (400 	imes 1)$$m = 360 + 400 = 760 \ gm$.

$A$ cubical block of wood with a side length of $10 \ cm$ floats at the interface between oil and water,with its lower surface horizontal and $4 \ cm$ below the interface. The density of oil is $0.6 \ g/cm^3$. The mass of the block is ......... $gm$.

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