Sunday, May 24, 2009

Irodov Problem 1.323

Suppose that at some instant the height of water in the cylinder is h. At the bottom of the cylinder, just before the water jets out, the pressure is given by , here Po is the atmospheric pressure. We assume that water here is still i.e. its velocity is 0.

As soon as the water jets out it enters the atmosphere where the pressure is Po. Let the velocity of the water shooting out be v. Then using Bernoulli's equation for water at points just before the orifice and just outside the orifice we have,

The rate (volume per unit time) at which water flows out is given by sv, where s is the area of the small hole. As water flows out of the cylinder, the height of the water also drops. The rate of change of volume in the cylinder must be equal to the rate at which water flows out of the cylinder. Since the area of the circular section of the cylinder is S, the volume of water contained in the cylinder is given by Sh. Hence, we can find the time taken for the tank to be empty as follows,

1 comment:

pranjal daga said...

hey buddy u r doing a great job
GO ON

I am Pranjal Daga in class 12th looking for solutions of IRODOV of the class 12th topics like electrostatics etc. plz help me out as soon as possible as I am preparing for IIT JEE .

e- mail: pranjaldaga@gmail.com

also visit :pranjaldaga.blogspot.com