Question

Two capillary tubes of the same length but different radii r₁ and r₂ are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as before?

• A. (a) $r_{1} + r_{2}$
• A. (b) $r_{1}^{2} + r_{2}^{2}$
• A. (c) $r_{1}^{4} + r_{2}^{4}$
• A. (d) None of these

Answer 1

(d)

Sol. $V = V_{1} + V_{2}$

⇒ $\frac{\pi Pr^{4}}{8\eta l} = \frac{\pi Pr_{1}^{4}}{8\eta l} + \frac{\pi Pr_{2}^{4}}{8\eta l}$⇒ $r^{4} = r_{1}^{4} + r_{2}^{4}$

∴ $r = (r_{1}^{4} + r_{2}^{4})^{1/4}$