Question

Two capillary tubes of same radius r but of different lengths l1 and l2 ate fitted in parallel to the bottom of a vessel with pressure head p . The length of a single tube of same radius which can replace the two tubes so that the ratio of flow is same will be

Answer 1

we know, $V=\frac{\pi Pr^4}{8\eta l}$
where , P is pressure, r is the radius of capillary tube, $\eta$ is viscosity and l is length of capillary tube.

here, Two capillary tube are fitted in parallel to the bottom of a vessel with pressure head .
so, $V=V_1+V_2$

$\frac{\pi Pr^4}{8\eta l}=\frac{\pi Pr^4}{8\eta l_1} +\frac{\pi Pr^4}{8\eta l_2}$

$\frac{1}{l}=\frac{1}{l_1}+\frac{1}{l_2}$

$l=\frac{l_1l_2}{l_1+l_2}$