MCQ on Viscosity – Poiseuille’s Flow

Poiseuille’s formula for the volume ‘V’ of a liquid of density ‘ρ’ flowing in time ‘t’ through a capillary tube of length ‘L’ and radius ‘r’ under a pressure difference ‘P’ between the ends of the tube is

V = (πPr⁴t)/ (8Lη) where ‘η’ is the coefficient of viscosity of the liquid. [You should note that this formula holds good only if the flow is slow and steady (stream-lined)].

The rate of flow (volume flowing per second) ‘Q’ is given by

Q = (πPr⁴)/ (8Lη)

Here P = hρg where ‘h’ is the height of the liquid column which produces the pressure difference so that

Q = (π hρg r⁴)/ (8Lη)

You will often find questions based on Poiseuille’s formula in entrance tests for admitting students to various courses. Consider the following question:

A capillary tube of length ‘L’ and radius ‘r’ is joined to another capillary tube of length L/4 and radius r/2 A liquid flows through this series combination. If the pressure difference between the ends of the first tube is P, that between the ends of the second tube is

(a) P (b) 4P (c) 8P (d) 16P (e) P/4

Since the tubes are in series, the rates of flow through the tubes are equal so that from Poiseuille’s equation

πPr⁴/8Lη = πP'(r/2)⁴/[8(L/2)η], where P’ is the pressure difference between the ends of the second tube.

From this P' = 16P/2 = 8P.

Now, consider the following MCQ:

Under a constant pressure head, the volume of a liquid flowing per second through a capillary tube of radius 1 mm and length 16 cm is 4 cm³. If another tube of radius 0.5 mm and length 8 cm is connected in series with it and the same pressure head is applied across the combination, the volume of liquid flowing per second will be ( in cm³)

(a) 5/3 (b) 5/6 (c) 4/3 (d) 4/5 (e) 4/9

When tubes are connected in series, the net rate of flow (Q_net) under a given pressure head is given by the reciprocal relation,

1/Q_net = 1/Q₁ + 1/Q₂ + 1/Q₃ +…..etc. where Q₁, Q₂, Q₃.....etc. are the individual rates of flow when the tubes are connected separately to the same pressure heaed.

In the present case, since there are two tubes only, 1/Q_net = 1/Q₁ + 1/Q₂ so that

Q_net = Q₁Q₂/(Q₁+Q₂). Here Q₁ = 4 cm³.

Since the rate of flow through a tube is given by Q = (πPr⁴)/ (8Lη), we have Q α r⁴/L.

Since the radius of the second tube is half that of the first, the rate of flow is reduced to (½)⁴ = 1/16 of that through the first tube on account of this. Since the length of the second tube is half that of the first, the rate of flow is increased to twice that through the first tube on this account. Therefore, rate of flow through the second tube is given by

Q₂ = Q₁×(2/16) = Q₁/8 = 4/8 cm³ = 0.5 cm³.

The net rate of flow is therefore given by Q_net = (4×0.5)/(4+0.5) = 4/9 cm³ [Option (e)].