In this illustration, we'll analyze, a water
jet crossing a wall. here the figure shows
a container in which there is a small orifice
of area ay on its side wall at a depth h,
bellow the top. and at t equal to zero liquid
is filled in it. we are required to find the
time after which the ejecting water jet from
orifice will cease to cross the wall. and
we are given, that the cross sectional area
of tank is ay and that of orifice is very
less compare to ay, and the wall distance
from, the orifice is given as x and depth
is given as y. now in this solution here we
can write, when. liquid level. is at a height.
h, above the orifice. the flux velocity. of
water is. this flux velocity we can write
by using torricelli's theorem, we already
studied earlier in concept videos that is
root 2 gee h. divided by 1 minus, ay square
by, capital ay square, and as ay is very small
compare to ay then this can be approximated
as root 2 gee h only. and, in this situation
with this velocity water jet is coming out,
so if we wish to calculate the height of water
level. when it'll just crossing the wall then
we can write, water jet. just crosses. the
wall when. we can write the range, of water
jet is equal to, if flux velocity multiplied
by time taken to fall. by a distance y that
can be written as root 2 y by, gee. this implies,
in this situation if we substitute the value
of v as root 2 gee h and we square it this
gives us x square. is equal to 2 gee h multiplied
by. 2 y by, gee, here g gets cancelled out.
and, we get the value of h is equal to x square
upon, 4 y. so when the height above the orifice
of water level x square upon 4 y the water
jet will just be crossing the wall. so we
are required to find the time taken in falling
the level of water from capital h to, this
small h. so we can use from, equation of continuity.
here, we can write if the water level is coming
down, at a rate of d h by d t we can write
d h by d t. multiplied by the cross sectional
area, ay, and this should be minus d h by
d t as, h is decreasing, should be equal to
small ay multiplied by, root 2 gee h. this
is the flow rate at which the, water is coming
out from the orifice. so here rearranging
the terms this will give us, d h by root h.
is equal to small ay by capital ay multiplied
by root 2 gee, multiplied by d t. and we can
integrate these terms at t equal to zero.
the height will be, height was given as h.
and at time t this reaches to small h which
is giving us the x square by 4 y when water
jet, here just be crossing the wall. now if
we integrate this term we can see this will
give us twice of, root h. and we put the limits
from, h to small h. with a negative sign is
equal to ay by capital ay, root 2 gee multiplied
by t. and here, simplifying this it'll give
us, the value of time is equal to you can
substitute the limits. and this will give
us time is equal to ay by small ay and multiplied
by root 2 by gee. and, here, value of small
h we can substitute as square upon 4 y. so
this will be root of capital h minus root
of x square upon. 4 y. that is the result
of this problem.
