in this example we are given that a cylindrical
vessel filled with water upto a height of
2 meter, stands on a rough horizontal plane.
at bottom in its side wall, a small hole is
opened and it is found that the vessel slides
on plane as water comes out. we are required
to find the diameter of hole, if friction
coefficient between vessel and plane is zero
point 4. and total mass of vessel with water
is hundred kaygee. if we draw the situation
here, we are given with, a rough ground, on
which, a vessel is kept. and total weight
of vessel in this situation is given as hundred
kaygee. so obviously on ground if we find
out the value of limiting friction it can
be written as, mu n that is mu m g, that’ll
come out to be zero point 4 into hundred multiplied
by, value of g we can take as 10 for calculations.
so the limiting friction here we can say it
is, 400 newton. now, the limiting friction
is 400 newton, and if we just, make a hole
near the bottom, and water starts coming out,
with some efflux velocity v, then we can say,
it’ll experience a reaction force in backward
direction. and we already studied about, reaction
force, due to, ejection of water is, this
force we can write as ro ay v square, that
we already studied. if this reaction force,
is more than the limiting friction, then we
can state, it’ll start moving. so we can
write, to slide, the vessel, ro a-v square,
in this situation has to be more than or atleast
equal to, the limiting friction that is 400
newton. and, the area of cross section of
this hole can be written as in terms of its
diameter, if it is taken as d, so we can write
it, ro pi d square by 4, v square is 400.
on simplifying we’ll get the diameter of
hole is equal to, under the root of, 4 multiplied
by 400, divided by, ro pi v square. the efflux
velocity v we can write as, root 2 g h. so
v square we write as, 2, g, h, height we are
given as 2 meter, so we put 2. in this situation,
we substitute the value of pi as well as the
density of water we take as thousand. and,
final expression we will have for dia-meter
of hole is, 4 multiplied by 400, divided by,
thousand multiplied by 3 point 1 4, multiplied
by, g we can take as 10 so it’ll be 20 into
2. on simplifying this expression, finally
we get, the diameter of hole is, zero point
1 1 4 meter. that’ll be the answer to this
problem.
