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....

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.