Question

water flows in a streamline manner through a capillary tube of radius a. the pressure difference Being p and the rate of flow is Q. if the radius is reduced to a/2 and the pressure is increased to 2p, then the rate of flow becomes :

Answer 1

In the first tube
 Rate of flow = Q
Radius = a
Pressure difference = P
length of the tube = l
coefficient of viscosity = л

From Poiseuille’s equation;
Q= λpa²/ 8лl

Now, let the new rate of flow be Q1, 
 radius = a/2 
pressure difference = 2p 

New Poiseulle’s equation;
Q1= λ 2p (a/2)²/8лl

dividing both eq we get

Q1/ Q= 1/8
Q1= Q/8

Answer 2

Explanation:

In the first tube

 Rate of flow = Q

Radius = a

Pressure difference = P

length of the tube = l

coefficient of viscosity = л

From Poiseuille’s equation;

Q= λpa²/ 8лl

Now, let the new rate of flow be Q1, 

 radius = a/2 

pressure difference = 2p 

New Poiseulle’s equation;

Q1= λ 2p (a/2)²/8лl

dividing both eq we get

Q1/ Q= 1/8

Q1= Q/8