First slide

Viscosity

Question

Two capillary tubes AB and BC are joined end to end at B, AB is 16cm long and of diameter 4 mm whereas BC is 4 cm long and of diameter 2 mm. the composite tube is held horizontally with A connected to a vessel of water giving a constant head of 3 cm and C is open to the air. Calculate the pressure difference between B and C

Moderate

A

2.4 cm
B

2.8 cm
C

1.2 cm
D

1.8 cm

Solution

For tube AB ; l₁ = 16 cm, r₁ = 2 mm =0.2
For tube BC: l₂=4cm, r₂ = 1mm = 0.1 cm

Let h be the head of pressure at B over C, then it will be (3-h) at A over B. Rate of flow through AB,
$\large {Q_1} = \frac{{\pi {p_1}r_1^4}}{{8\eta {l_1}}} = \frac{{\pi (3 - h){{(0.2)}^2}}}{{8\eta \times 16}}c{m^3}{s^{ - 1}}$
Rate of flow through BC,
$\large {Q_2} = \frac{{\pi {p_2}r_2^4}}{{8\eta {l_2}}} = \frac{{\pi h{{(0.1)}^2}}}{{8\eta \times 4}}c{m^3}{s^{ - 1}}$
But Q₁ = Q₂
$\large \frac{{\pi (3 - h){{(0.2)}^4}}}{{8\eta \times 16}} = \frac{{\pi h{{(0.1)}^4}}}{{8\eta \times 4}}$
or 12-4h = h or 5h = 12 or h = 2.4 cm

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