Question

Water flows in a stream line manner through a capillary tube of radius a, the pressure difference being P and the rate flow Q. If the radius is reduced to $\frac{a}{2}$ and the pressure is increased to 2P, the rate of flow becomes(a) 4Q(b) Q (c) $\frac{Q}{2}$(d) $\frac{Q}{8}$

Answer 1

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Answer 2

Answer:

D) Q/8

Explanation:

Rate of flow in the tube = Q (Given)

Radius of the tube = a (Given)

Pressure difference in the tube = P (Given)

Length of the tube = l

Coefficient of viscosity = л

From Poiseuille’s equation - Q= λpa²/ 8πl (1)

Now, let the new rate of flow be Q1, 

Radius = a/2 

Pressure difference = 2p 

Thus,

Q1= λ 2p (a/2)²/8πl (2)

Dividing eq 1 and 2 we will get -

Q1/ Q= 1/8

Q1= Q/8

Therefore, if the radius is reduced and the pressure is increased to 2P then the rate of flow becomes Q/8