in this example, we are given that in a bucket
shown in figure, a liquid of density 1 point
2 5 into 10 to power 3 kaygee per meter cube
is kept. we are required to find the speed
at which liquid will be coming out if the
bottom surface is removed, and its top level
is maintained by pouring the liquid through
the tap. in this situation we can consider
a point ay at the top level, and b at the
bottom level. so, in this situation, when
we remove the bottom, the top level will start
coming out with velocity v-a and, bottom level
liquid will start coming out with speed v
b. then in this situation we can directly
write by continuity equation, a-a v-a must
be equal to, a-b v b. so in this situation,
the speed of, liquid particles at point b,
can be written as in this situation this v-a,
multiplied by, this’ll be a-a upon a-b,
so this’ll be r 1 square by r 2 square.
or, vice versa we can write, v-a is equal
to, v b into r 2 square, by r 1 square. now
this we can use, in solution if we write,
by using bernawlli’s theorem, we can write,
bernawlli’s equation, at ay and b. then
here we can write, at end ay, the pressure
will be p atmospheric, plus, the ki-netic
energy of liquid particle will be, half ro
v-a square, plus compared to point b the gravitational
potential energy will be ro g h. this must
be equal to at point b we can write pressure,
p atmospheric plus, the ki-netic energy will
be half ro v b square. here p atmospheric
gets cancelled out, and in this situation,
the value of speed of liquid particles at
end b, we can find out by substituting the
value of v-a from here. the value of v-a when
we substitute we’ll get, v b square, 1 minus
r 2 square, by r 1 square, is equal to 2-
g-h as here rho also gets cancelled out. and
on simplifying, value of v b we get as root
of, 2 g h divided by, 1 minus r 2 square by,
r 1 square. when we substitute the values
it’ll be root of, 2 into g we can take as
10, into point 7 5, divided by 1 minus, this’ll
be zero point 8 whole square. so on simplifying
this result, we get, 6 point 4 5 meters per
second. that’ll be the answer to this problem.
