# In a U tube manometer, the height of water in one arm is 10.0 cm and the height of oil in the other arm is 20.0 cm. The density of water is 1.0. Find the density of oil.Please answer immediately....

In a U tube manometer, the height of water in one arm is 10.0 cm and the height of oil in the other arm is 20.0 cm. The density of water is 1.0. Find the density of oil.

Please answer immediately. Please also shw process or explanation.

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The basic principle behind this problem is for the weight of the oil and the weight of the water to be equal. When the weights are equal, the liquids are in static equilibrium and stop moving up and down in the tube. If one knows the weight then one can determine the mass and density.

However the problem as stated does not provide enough information to determine the mass directly because we do not know the volume of the water or oil. Therefore we must determine it indirectly.

We will use a lower case 'w' to label the variables associated with the water and a 'o' for the oil.

Wo = Ww so, MoXg = MwXg where g is the acceleration due to gravity. We can cancel the g, so

Mo = Mw

We can now use the definition of density to get an expression for the masses: D = M/V which gives M = DV

Mo = DoVo and Mw = DwVw Therefore

DoVo = DwVw which gives

Do = DwVw/Do

Volume of the cylinder of water in the tube is given by V = HxPiR^2

where H is the height of the liquid and R is the radius of the tube.

So

Do = Dw (HwxPiR^2)/(HoxPiR^2)

The PiR^2 cancels out in denominator and numerator leaving

Do = DwHw/Ho substituting values into the equation:

Do = 1.0g/cm^3 x 10cm/20cm

Do = 0.5 g/cm^3

The density of the oil is half the density of the water. This answer should "feel" right considering the relationship between the heights of the liquids.