Question

Oil having a density of 924 kg/m 3 floats on water. A rectangular block of wood 3.79 cm high and with a density of 970 kg/m3 floats partly in the oil and partly in the water. The oil completely covers the block. How far below the interface between the two liquids is the bottom of the block

Answer 1

x = 2.69 cm

Step-by-step explanation:

We are given ;

Density of oil; ρ_oil = 924 kg/m³

Density of wood; ρ_wood = 970 kg/m³

h = 3.79cm

Density of water ( ρ_water) has a constant value of 1000 kg/m³

At equilibrium position, we have;

ρ_wood•g•h - ρ_oil•g•(h - x) - ρ_water•g•x = 0

This is because the density of oil is lower than that of water while density of wood is higher than that of oil but lower than that of water.

x is the distance below the interface between the two liquids is the bottom of the block.

Thus, let's make x the subject;

x = [(ρ_wood - ρ_oil)/(ρ_water - ρ_oil)] x h

Plugging in the relevant values to get ;

x = [(970 - 924)/(1000 - 924)] x 3.79

x = (54/76) x 3.79 = 2.69cm