Question

A solid cylinder (radius = 0.150 m, height = 0.120 m) has a mass of 7.36 kg. This cylinder is floating in water. Then oil (ρ = 851 kg/m³) is poured on top of the water until the situation shown in the drawing results. How much of the height of the cylinder is in the oil?

Answer 1

you need to use Arquimedes :
There is a force that the water and the oil produced to the cylinder, so the weight of the cylinder must be equal to the plus of those two forces :

Force by the water : 1000*9.8*V'
Force by the oil : 851*9.8*V''
V' = Volume of the cylinder in water
V'' = Volume of the cylinder in oil

1000*9.8*V' + 851*9.8*V'' = 7.36*9.8

In other words, the push by the water + the push by the oil must be equal to the weight of the cylinder. Now then : The volumes can be calculated by :

V' = pi*(0.150)^2*a
V'' = pi*(0.150)^2*b

where : a + b = height = 0.12

Replacing on the first equation :
1000*9.8*V' + 851*9.8*V'' = 7.36*9.8
1000V' + 851V' = 7.36
1000*pi*(0.150)^2*a + 851*pi*(0.150)^2*b = 7.36
22,5*pi*a + 19.15*pi*b = 7.36

a + b = 0.12
we need to find "b" , then a = 0.12 - b
8.48 - 70.68b + 60.16*b = 7.36
1.12 = 10.52*b
b = 0.1065 meters
b = height of oil

Hope that helps