Question

Two capillary tubes of same radius r but of lengths l₁ and l₂ are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the length of a single tube that can replace the two tubes so that the rate of flow is same as before?

• A. (a) $l_{1} + l_{2}$
• A. (b) $\frac{1}{l_{1}} + \frac{1}{l_{2}}$
• A. (c)$\frac{l_{1}l_{2}}{l_{1} + l_{2}}$
• A. (d) $\frac{1}{l_{1} + l_{2}}$

Answer 1

(c)

For parallel combination $\frac{1}{R_{eff}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

⇒ $\frac{\pi r^{4}}{8\eta l} = \frac{\pi r^{4}}{8\eta l_{1}} + \frac{\pi r^{4}}{8\eta l_{2}}$ ⇒ $\frac{\mathbf{1}}{\mathbf{l}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{l}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{l}_{\mathbf{2}}}\mathbf{\mspace{6mu}\mspace{6mu}\mspace{6mu}\mspace{6mu}}\mathbf{\therefore}\mathbf{l =}\frac{\mathbf{l}_{\mathbf{1}}\mathbf{l}_{\mathbf{2}}}{\mathbf{l}_{\mathbf{1}}\mathbf{+}\mathbf{l}_{\mathbf{2}}}$