Let the volume of the block be V and the density D. The mass of the block is D*V.

In an oil-water interface with oil having a density of 800 kg/m^3 floating on water with a density of 1000 kg/m^3, the block floats such that 1/4th of its volume is...

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Let the volume of the block be V and the density D. The mass of the block is D*V.

In an oil-water interface with oil having a density of 800 kg/m^3 floating on water with a density of 1000 kg/m^3, the block floats such that 1/4th of its volume is in oil and 3/4th is in water. When an object is submerged in a fluid it is buoyed up with a force that is equal to the weight of the fluid displaced.

The block displaces a volume of oil equal to V/4 and a volume of water equal to 3*V/4. The mass of oil displaced is 800*V/4 and the mass of water displaced is equal to 1000*3*V/4. The forces acting on the block when it floats are the weight of the block equal to D*V*g that acts in a direction vertically downward and a force equal to (200*V + 750*V)*g in a direction vertically upwards. The two are equal.

This gives D*V*g = (200*V + 750*V)*g

=> D = 200 + 750 = 950

**The density of the block is 950 kg/m^3.**