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A U-tube with a cross-sectional area of 1.00 cm2 is open to the atmosphere at both ends. Water is poured into the tube until the water rises part way along the straight sides, and then 5.00 cm3 of oil is poured into one end. As a result, the top surface of the oil ends up 0.550 cm higher than the surface of the water on the other side of the U. What is the density of the oil?

Solution 1

Height of oil in the tube = 5/1 = 5 cm
Pressure at oil water junction = 5cm* rho(oil) *g
Pressure at the same level on the other side of the u tube = (5-0.55)cm * 1 g/cm3 *g
We know that pressure at oil water junction = Pressure at the same level on the other side of the u tube
Density of oil (rho) = 4.45/5 g/cm3 = 0.89 g/cm3

Solution 2

in the tube
density of water,ρw = 1g/cm^3
density of oil =ρo
A = 1 cm^2
volume of oil = 5 cm^3
since V = Ah => h = 5 cm = height of oil column
now oil is 0.55 cm higher
thus, at the interface of oil with water pressure is equal
ρogh = ρwg(h-0.550)
putting values gives
ρo = 1(5-0.55)/5 = 0.89 g/cm^3=density of oil

Solution 3

P(atm) + p(water)gh(1) = P(atm) + p(oil)gh(2)
Therefore, h(1) = [p(oil) \ p(water)] * h(2)
h = h(2) - h(1)
h = h(2) - [p(oil) \ p(water)] * h(2)
h(2) - h = [p(oil) \ p(water)] * h(2)
[h(2) - h] \ h(2) = [p(oil) \ p(water)]
p(water) * [h(2) - h] \ h(2) = p(oil)
1000 * [5.00cm3 - 0.550cm3] \ 5.00cm = p(oil)
1000 * .89cm3 = 890 kg/m3
So the oil density is 890 kg/m3