Question

Two capillary of length L and 2L and of radius R and 2R are connected in series. The net rate of flow of fluid through them will be (given rate of the flow through single capillary, $X = \pi PR^{4}/8\eta L)$

• A. (a) $rac{8}{9}X$
• A. (b) $rac{9}{8}X$
• A. (c) $rac{5}{7}X$
• A. (d) $rac{7}{5}X$

Answer 1

(a)

Sol. Fluid resistance is given by $R = \frac{8\eta l}{\pi r^{4}}.$

When two capillary tubes of same size are joined in parallel, then equivalent fluid resistance is

$R_{e} = R_{1} + R_{2} = \frac{8\eta L}{\pi r^{4}} + \frac{8\eta \times 2L}{\pi(2R)^{4}} = \left( \frac{8\eta L}{\pi r^{4}} \right) \times \frac{9}{8}$

Equivalent resistance becomes $\frac{9}{8}$times so rate of flow will be $\frac{8}{9}X$