In this illustration we'll study about a water
level controller in a tank. here the figure
shows a tank in which a hole at its bottom
is opened or closed by a level controller,
which consist of a thin light circular plug
ay, of diameter 10 centimeter and a cylindrical
float b, of diameter 20 centimeter and mass
3 kilogram, here both ay and b are tied together
with a light, rigid rod of length 10 centimeter.
and here it is saying as water fills up and
level reaches to h 1.
plug opens, and we are required to find h
1 here. and also we are required to find the
level of water h 2 when plus closes again.
so, in this solution here we can analyze.
at a height.
h 1. if, plug opens.
then, h 1 is the level where the buoyancy
acting on beam minus the downward force on
ay will be balancing the total weight.
so here we can calculate if plug opens then,
the buoyancy, on b, is.
this is the upward force, f, 1, acting on
b in upward direction.
this can be written as submerged volume multiplied
by density multiplied by g. this is the volume
of the liquid displaced, as the rod is of
length, 10 centimeter, the submerged depth
of the cylinder would be, h 1 minus zero point
1 multiplied by, its cross sectional area
which is given as pie d square by 4 so we
can write pie, zero point 2 whole square by
4.
multiplied by water density we can take as
thousand and g to be 10. which the numerical
value here is, this given as hundred, pie
h 1, minus, 10 pie. this is the value of upward
force acting on block b. if we calculate the
downward force, on plug. is, say this is f
2 then its value can be written as bottom
pressure is, h 1 ro g multiplied by its cross
sectional area which is.
pie, its, radius is 10 by 2 so this is zero
point 1 whole square by 4.
so numerically we are getting again ro can
be taken as thousand, g to be 10, then, this
will be given as on simplifying 25 pie h 1.
and we can say.
plug will open.
when, the net upward force which is f 1 minus
f 2 will be equal to, mass of b into g that
is equal to 3 gee as, the plug is lightweight.
so we substitute the 2 values here this gives
us, hundred pie h 1.
minus 10 pie, minus 25, pie h 1, is equal
to 30 here i have substituted g as 10.
so simplifying this value we are getting the
value of h 1 is.
30 plus 10 pie, divided by 75 pie, or this
is twice of 3 plus pie divided by, 15 pie
that is the result of this problem the height
h 1. of water at which the plug opens.
now, once it opens, f 2 will no longer exist
because on this plug ay which is light weight
upward downward force becomes equal. and now
only force acting on this system will be the
buoyant force on b. so, we can say plug. will
close again.
when, f 1, will be equals to 3 gee when, the
buoyant force acting on b, will be equal to,
the weight of b then it'll be closing down.
so here rather then f 1 we use it, another
force, so force, numerically we can substitute
here.
but we change the value of h 1 to h 2, so,
this will give us hundred pie, h 2.
minus 10 pie, should be equal to 30. here,
h 2 is the, water level.
when plug closes.
so here the same force f 1 we have calculated
we have replaced h 1 by h 2.
so on simplifying this value we'll get the
value of h 2 is equal to.
this is 3, plus pie divided by 10 pie. and
that is the result of this problem.
