A layer of oil with density 724 kg m - 3 floats on water of density 1000 kg m - 3 . A block floats on the oil-water interface with 1/6 of its volume in oil and 5/6 of its volume in water, as shown in the figure. What is the density of the block?

Step-by-step solution

Given, Density of oil ρ 1 = 724 kg / m 3 Density of water ρ 2 = 1000 kg / m 3 According to Archimedes' principle, Up thrust = Weight of the liquid displaced ⇒ Vρ b g = V o ρ 1 g + V w ρ 2 g ( V o and V w are volumes of oil and water displaced respectively) Vρ b g = V 6 724 g + 5 V 6 1000 g , where ρ b =density of block. and, V o = V 6 and V w = 5 V 6 ⇒ Vρ b g = 724 V 6 + 5000 V 6 g ⇒ Vρ b g = 724 V + 5000 V 6 g ⇒ Vρ b g = 5724 6 Vg ρ b = 954 kg / m 3