In this illustration, we'll analyze, a vertically
floating glass tube. we are given that a thin
walled glass tube has diameter 4 centimeter
and weighs 30 gram. and the center of mass
of the tube is at a distance 10 centimeter
above the bottom. we are required to find
the amount of water which must be poured in
the tube so that it can vertically float in
a water tank. here, in this situation we need
to understand. body can float vertically in
water tank only when, its center of gravity,
is bellow the center of buoyancy, in the,
submerged part of body. like for example here
if we wish this tube to flow to vertically.
in water. then in this situation we'll consider,
liquid is filled in it to a height say y.
and if we consider x is the total depth which
is submerged in it. then the buoyancy force,
will act at a point which we call center of
buoyancy is at the center of the submerged
part. which is x by 2. so the center of mass
of this tube plus the water inside, must coincide
or will be bellow, the point x by 2. otherwise
the tube will become horizontal due to the
torque applied by, buoyancy force and the
force of gravity. so in this situation here
we can say if water. is poured. upto, a level.
y in tube which we can see i have drawn in
the figure. and x, is. the submerged. length
of tube. here, we can use. for equilibrium.
of tube. here we can rise the weight of tube
is 30 gram so we can solve it in, c g s units
also, this will be 30 plus if ay is the cross
sectional area, of the tube, the mass of water
in it will be, ay, y, ro, and this must be
equal to the mass of water displaced by, this
tube, that will be, ay x ro. and this gives
us the value of x is equal to 30. plus ay
y divided by, ay. and in this situation for,
vertical. floatation. of tube. we must. use.
the center of mass, of, tube, plus water.
should be equal to. the center of. buoyancy.
here we are talking about the height of center
of mass of tube plus water is equal to height
of center of buoyancy. so with the bottom
if we calculate, this, center of buoyancy
is located at a distance of x by 2. and if
we calculate the center of mass of tube plus
water this can be given by. the expression
m 1 x 1 plus m 2 x 2 upon m 1 plus m 2. so
this will write as 30 multiplied by 10. plus,
ay y ro multiplied by y by 2. divided by the
total mass is 30 plus, ay y, ro. and, we can
also substitute. the value of, x over here
which is. 30 plus, ay y divided by 2 ay, so
simplifying this relation it gives, we can
also cross multiplied, here we can consider
water density to be 1. in c g s. so this is
30 plus, ay y whole square. divided by, 2
ay is equal to. 300, plus, ay y square by,
2. if we simplify this relation. here, it
gives the value of y. which is 10 minus, 15
by, ay. i am leaving it as an exercise for
you to simplify this equation and verify.
the value of y will be getting is 10 minus
15 by ay. and from here we are given that
the glass tube has a diameter, 4 centimeter
then we can, now write, mass of, water poured.
in the tube is, equal to. ay y ro we can take
as 1 as we are resolving the problem in c
g s this can be written as ay multiplied by
10 minus 15 by ay, so this mass of water we
are getting is, 10 ay minus 15. and the value
of ay we can use as, pie r square and the
radius we can take as 2 centimeter, so this
will be 10 multiplied by. this pie, multiplied
by 2 square, minus, 15. if you are numerically
solve it this gives us, 40 pie minus 15. which
is equal to 110 point 6 gram that is the answer
for this problem.
