Question

There are two (narrow) capillary tubes T₁ and T₂ with lengths l₁ and l₂ and radii of cross-section r₁ and r₂ respectively.The rate of water under a pressure difference P through tube T₁ is 8cm³/sec.What will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before(= p) if l₁ = 2l₂ and r₁ = r₂?

• A. 4 cm³/sec
• B. (16/3) cm³/sec
• C. (8/17) cm³/sec
• D. None of these

Answer 1

Answer: (B)

$\text{V} = \frac{\pi \text{P} \text{r}^{4}}{8 η \text{l}} = \frac{8 \text{cm}^{3}}{\text{sec}}$
For composite tube
$\left(\text{V}\right)_{1} = \frac{\text{P} \pi \left(\text{r}\right)^{4}}{8 η \left(\right. \text{l} + \frac{\text{l}}{2} \left.\right)} = \frac{2}{3} \frac{\left(\pi \text{P} \text{r}\right)^{4}}{8 η \text{l}} = \frac{2}{3} \times 8 = \frac{1 6}{3} \frac{\left(\text{cm}\right)^{3}}{\text{sec}} \left[∴ \left( \text{l}\right)_{1} = \text{l} = 2 \left(\text{l}\right)_{2} \text{or} \left(\text{l}\right)_{2} = \frac{\text{l}}{2}\right]$