Question

A solid cylinder (radius = 0.160 m, height = 0.190 m) has a mass of 14.2 kg. This cylinder is floating in water. Then oil (ρ = 781 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

Bo = Bouyant force due to oil

Bw = Buoyant force due to water

Bo +Bw = W

(po*g*Vo) +(pw*g*Vw) = pc*g*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

(VO/VC) = (pc - pw)/(po - pw)

r = 0.16 m , h = 0.19 m , m =14.2 kg

pc = m/V = 14.2/(3.14*0.16^2*0.19) = 929.7 kg/m^3

(VO/VC) = (929.7 -1000) /(781 -1000) = 0.321

required height = 0.321*0.19 = 0.061 m