Ncert Exercises & Solutions

10.13 Glycerine flows steadily through a horizontal tube of the length of 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10^–3 kg s^–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = $1.3 \times 10^{3}$ kg m^–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Given,

l = 1.5 m, r = 1 cm = 0.01 m, d = 2r = 0.02 m

M = 4.0 × 10^–3 kg s^–1

ρ = 1.3 × 10³ kg m^–3

η = 0.83 Pa s

The volume of glycerine flowing per sec:

$V = \frac{M}{ρ}$
$= \frac{4.0 \times 10^{- 1}}{1.3 \times 10^{3}}$

= 3.08 × 10^–6 m³ s^–1

According to Poiseville’s formula,

$V = \frac{{πPr}^{4}}{8 ηl}$

$P = \frac{V 8 ηl}{{πr}^{4}} = \frac{3.08 \times 10^{- 6} \times 8 \times 0.83 \times 1.5}{π \times {(0.01)}^{4}}$

= 9.8 × 10² Pa

Reynolds’ number is given by:

$R = \frac{4 ρV}{πdη}$
$= \frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{- 6}}{π \times (0.02) \times 0.83}$

Reynolds’ number is about 0.3.

The flow is laminar.

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