in this example we are given that water flows
through the tube shown in figure. the areas
of cross section of wide and narrow sections,
of this tube are 5 centimeter square and 2
centimeter square. the rate of water flow
in the tube is 500 centimeter cube per second.
we are required to find the difference in
mercury levels in the u tube. water is flowing
in the tube, and, we have taken mercury in
the u tube, and we are required to find this
level difference h. we know well that, in
the narrow section, the flow velocity increases
due to which here pressure is decreased. and
as here pressure is more, we can say u tube
will, experience more pressure at this end
compared to this end. so mercury level here
will go down and here it will be raised up.
we are required to find the value of this
h. now, this can be easily obtained by using,
bernawlli’s theorem. if we use bernawlli’s
theorem we can write, according to. bernawlli’s
theorem, at points ay and b, where we can
take ay point to be here and, b point to be
here. in this situation if these are the points
ay and b, we can simply write down as these
are at same horizontal level, pressure at
ay plus, half ro v-a square, is equal to pressure
at b plus, half ro v b square, this we can
use. in this situation we’ll simply get
the pressure difference p-a minus p b is equal
to, half ro, we can also, substitute the value
of speeds, according to continuity equation
let’s take it as, v b square minus, v-a
square. and in this situation if we apply
the continuity equation, then continuity equation
gives us, area at point ay, multiplied by
speed at point ay is equal to, area at b into
speed at b which is given as, 500 centimeter
cube per second. so if we substitute the areas
which are given as to be 5 and 2 centimeters,
so if we put ay as 5 then we can see, v-a
is equal to hundred centimeter per second,
and this flow velocity at point b will be
500 by 2 it is 2 50 centimeter per second.
so we substitute the relation here, we’ll
get the pressure difference and, between these
2 we can also write, p-a minus p b is equal
to, h, ro of mercury minus, ro of water into
g as, the difference in liquid column, pressure
can be given as 1 side it is water other side
it is mercury, so it must be, h ro mercury
g minus h ro g. if we put the values here
it’ll be h, density of mercury will be taken
as, 13 thousand 6 hundred minus, density of
water to be thousand multiplied by g we can
take as 10 is equal to, half ro, v b square
minus v-a square, this v b we can take as,
2 point 5 meter per second, so 2 point 5 square
minus, v-a as 1 square. and ro can be taken
as thousand, so this thousand gets cancelled
out, this is 13 point 6. on solving we get
value of h as, 2 point 5 square is 6 point
2 5 minus 1 is, 5 point 2 5. so it’ll be,
5 point 2 5 divided by, 2 multiplied by here
it is 13 point 6 minus 1, 12 point 6 into
10 it’ll be 1 26. on simplifying we are
getting the value of h as, zero point zero,
2 zero 8 meter. or it can be written as, 2
point zero 8 centimeter. that’ll be the
answer to this problem.
