00:01
High.
00:02
Here in this given problem, this is the pipe through which the water is flowing.
00:10
Its area of cross -section is going on reducing and its height is going on increasing.
00:17
Height above the ground.
00:23
Its right end is at a height h relative to the left end.
00:32
Here pressure p1 here the pressure p2, radius r1 here and radius is r2 here.
00:44
These values are p1 is equal to 1 .80 into 10 x to the power 5 possible.
00:51
R1 is 3 .20 centimeter.
00:57
Then p2, this is 1 .10 into 10 dash to the power 5 pascal and r2 here, this is 1 .20 this height of the right end relative to the left end is 2 .75 meters.
01:22
In the first part of the problem we have to find velocity of the water flow v1.
01:28
This is v1 here at the left end.
01:31
It will be v2 here at the right end.
01:33
So first of all, in the first part of the problem, first of all using equation of continuity at the left end and at the right end.
01:55
It says product of area of cross -section with the velocity remains constant.
02:02
And throughout the pipe.
02:04
So a1 v1 will be equal to a2 v2.
02:06
Means this is for a 1, pi r1 square into v1 is equal to pi r2 square into v2...