A solid cylinder (radius = 0.190 m, height = 0.190 m) has a mass of 21.4...

A solid cylinder (radius = 0.190 m, height = 0.190 m) has a mass of 21.4 kg. This cylinder is floating in water. Then oil (ρ = 867 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?

Expert Solution

Without the drawing, I will assume that the axis of revolution of the cylinder is pointing up. That makes the volume of the submerged cylinder proportional to the height submerged.

There is a bouyant force due to the oil, and another buoyant force due to the water. Their sum is equal to the weight of the cylinder.

Bo + Bw = W
po*g*Vo + pw*g*Vw = pc*g*Vc
po*Vo + pw*Vw = pc*Vc
po*(Vo/Vc) + pw*(Vw/Vc) = pc

where:
Vo, Vw, Vc = volume of cylinder in oil, in water, total volume of cylinder
po, pw, pc = density of oil, water, cylinder

Further, the portions in oil and water make up the whole, so we have

(Vo/Vc) + (Vw/Vc) = 1

We now have a system of equations, and the solution is

volume of cylinder pi *r^{2 *} h = 3.14 * (0.190)² * 0.190 = 0.0215m³

density pc=mass/volume = 21.4 kg/0.0215m³ =993.62 Kg/m3 ,pw=1000kg/m3

(Vo/Vc) = (pc - pw)/(po - pw)= 0.0479
(Vw/Vc) = (po - pc)/(po - pw)=0.952

Thus height of cylinder is in the oil =Vo/Vc * height = 0.0479* 0.19 = 0.0091 m is in the oil.

genius_generous answered 1 year ago

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