in this example, we are given that water flows
through a tube which has 1 end ay at a height
of 20 centimeter above ground, and cross sectional
area is 1 centimeter square, and its other
end b at a height 15 centimeter above ground
and cross sectional area point 5 centimeter
square. the speed of water at ay is 10 centimeters
per second, and we are required to find the
speed at end b and pressure difference at
ends ay and b. so in the solution let us draw
the physical situation. here the situation
would be like this. say if this end is a,
and this end is b. and with respect to ground
the height of end ay is given to us as 20
centimeter, we take it as h 1. the height
of end b with respect to ground h 2 is given
as 15 centimeter. area at end ay is 1 centimeter
square, and at end b is point 5 centimeter
square. and speed of water which is flowing
at ay is 10 centimeter per second and here
we say velocity is v b. we are required to
find, the speed of flow, at point b. that
can be directly calculated by using, equation
of continuity. we can simply write a-1 v-1
is equal to a-2 v-2. so at end ay the cross
sectional area is 1 centimeter square, the
speed is 10 centimeter per second. area at
the other end is point 5 and speed is v b.
so, simply by, numerical calculation we can
get the value of v b to be 20 centimeter per
second. this is 1 answer to this problem which
we are required to find. in another part of
the problem we are required to find the pressure
difference at ay and b. let pressure at ay
is p-a and at b it is p b. that can be obtained
by using bernawlli’s equation. if we use
bernawlli’s equation at ends, ay and b,
we can see the relations that we write. it
is pressure at ay, plus half ro v-a square
plus, its gravitational potential energy is
ro g h 1, this should be equal to p b plus,
half ro v b square plus, ro g h 2. now in
this situation we can simply see, the difference
in pressures, if we just talk about point
b, and point a, here p b minus p a, can be
written as, ro g, h 1 minus h 2, plus, half
ro, v ay square minus, v b square. this we
can use. if we substitute the values, in this
situation, the density of fluid, in this situation
we can take as thousand as it is given that
water is flowing. so it is thousand multiplied
by, g we can take as 10 for numerical calculation,
h 1 minus h 2 is 5 centimeter, that is point
zero 5, plus, it is half multiplied by thousand,
multiplied by, if we talk about, v ay square
minus v b square, here it comes out to be
a negative value because we can see v ay is
less. so here v ay we can write as, zero point
1, v b we can write, zero point 2, so this
can be given as, on calculation, point zero
1 minus, zero point zero 4. on further simplifying
the values, this 1st term will be 500, 2nd
term will be negative, it can be given as,
point zero 3 multiplied by thousand, that’ll
be 30 by 2, it is 15, so pressure difference
here we can write as 485 newton, per meter
square. that’ll be the answer to this problem.
