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Q.

A liquid is flowing through a narrow tube. The coefficient of viscosity of liquid is 0.1308 poise. The length and inner radius of tube are 50 cm and 1 mm respectively. The rate of flow of liquid is $360 m^{3} / m i n$ . Find the pressure difference between ends of tube.

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a

$1.8 \times 10^{4} dyne / {cm}^{2}$

b

$10^{4} dyne/cm 2$

c

$10^{6} dyne / {cm}^{2}$

d

$10 {dyne/cm}^{2}$

answer is A.

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Detailed Solution

From Poiseuille's equation

$Q = \frac{π (P_{1} - P_{2}) r^{4}}{8 η ℓ} \Rightarrow P_{1} - P_{2} = \frac{Q 8 η ℓ}{π r^{4}}$

$= \frac{360}{60} \times \frac{8 \times 0.1308 \times 50}{π \times (0.1)^{4}} = 10^{6} dyne$

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A liquid is flowing through a narrow tube. The coefficient of viscosity of liquid is 0.1308 poise. The length and inner radius of tube are 50 cm and 1 mm respectively. The rate of flow of liquid is 360m3/min. Find the pressure difference between ends of tube.