Question

We have two (narrow) capillary tubes T₁ and T₂. Their lengths are l₁ and l₂ and radii of cross-section are r₁ and r₂ respectively. The rate of flow of water under a pressure difference P through tube T₁ is 8cm³/sec. If l₁ = 2l₂ and r₁ =r₂, 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)

• A. (a) 4 cm³/sec
• A. (b) (16/3) cm³/sec
• A. (c) (8/17) cm³/sec
• A. (d) None of these

Answer 1

(b)

$V = \frac{\pi\Pr^{4}}{8\eta l} = \frac{8cm^{3}}{\sec}$

For composite tube $V_{1} = \frac{P\pi r^{4}}{8\eta\left( l + \frac{l}{2} \right)} = \frac{2}{3}\frac{\pi Pr^{4}}{8\eta l}$

$= \frac{2}{3} \times 8 = \frac{16}{3}\frac{cm^{3}}{\sec}$ $\left\lbrack \because\mspace{6mu}\mspace{6mu} l_{1} = l = 2l_{2}\mspace{6mu}\text{or}\mspace{6mu} l_{2} = \frac{l}{2} \right\rbrack$