Question

A solid cylinder (radius = 0.150 m, height = 0.120 m) has a mass of 7.20 kg. This cylinder is floating in water. Then oil (ρ = 725 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 in meters.

Picture:
https://general.physics.rutgers.edu/gifs/CJ/11-16.gif

Answer 1

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 cyl
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

(Vo/Vc) = (pc - pw)/(po - pw)= .593
(Vw/Vc) = (po - pc)/(po - pw)= .407

Thus almost 60% of the cylinder is in the oil. Or
.593*.12 = 0.071m is in the oil.

Answer 2

@tfguss there you go ^ ^

Answer 3

Thanks!

Answer 4

welcome! :)

Answer 5

A solid cylinder (radius = 0.150 m, height = 0.120 m) has a mass of 6.50 kg. This cylinder is floating in water. Then oil (ρ = 725 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? what would the answer to this be??