In this illustration we'll analyze 3 liquids
in a triangular tube. we are given that a
closed tube is in form of an equilateral triangle
of side l which consist of equal volumes of
3 liquids which do not mix. and the densities
of liquids are in the ratio 1 is to 2 is to
3. and the tube is kept in vertical plane
as shown, we are required to find the offset
distance x, in the figure.
here we can see as the liquids are considered
to be of densities ro 2-ro 3-ro, and the 1
which is, at the bottom must be the heaviest
liquid.
so this will be a having a density 3 ro this
would be 2 ro and a top 1 will be of, the
lighter density that can be taken as ro. here,
we can write, to find x. here, we write.
pascal's pressure equation, round the tube.
as. in this situation as, this is an equilateral
triangle these angles are all 60 degree.
so all vertical distances we can measure in
terms of x, and if the total length of the
tube is l this can be taken as l minus x.
so, we can write, if we start from point ay
pressure at ay is considered as p ay, and
throughout the horizontal length of the tube
pressure will remain constant.
so if we calculate the pressure at this point.
then we can write if we are going up, at this
point pressure will be less.
so we subtract it, h ro g and its height we
can write as x, sine 60 degree, so we can
write it x sine 60 degree.
multiplied by, 3 ro multiplied by g. and if
we move further then this density is ro, so
we can write minus.
l minus x. sine 60 degree.
multiplied by ro multiplied by g, then if
we further come down here we can write plus,
this is, x sine 60 degree.
multiplied by ro multiplied by g. and here,
if we come down from this point to c. again
it'll be plus and the density will be taken
as 2 ro, so this is, l minus x multiplied
by sine 60 degree.
multiplied by 2 ro, g. and this is equal to
pressure at point c, which is equal to the
pressure at point ay so these 2 can be cancelled
out.
here we can also mention, that, this pressure
is the pressure of, liquid column which is
density 3 ro, in the left arm of this triangular
tube, this is in the upper arm of tube this
is in the right arm of tube where the liquid
is of density ro and this is for, the liquid
which is of density 2 ro from this point to
this point.
here, we can also mark, points as, ay 1, b
1 and c 1, as a junctions, so we can write
this is the pressure, at ay 1, this is the
pressure at point b. then this is the pressure,
at, point b 1 and, again we are reaching to
c. so we can also write for clarity this is
the pressure at ay 1. this is the pressure,
at b, and this is the pressure, at point,
b 1.
now simplifying these values here g ro and
sine 60 degree can be canceled, and simplifying
this giving us minus 3 x. minus, l plus x,
plus, x. plus, this can be given as 2 l. minus
2 x. is equal to zero. and now we simplify
this, it is giving us x is equal to l by 3
that is a result of this problem.
