Question

31.Two liquids are allowed to flow through two capillary tubes of lengths are in the ratio 1:2 and radii in the ratio 2:3 under the same pressure difference.If the volume rates of flow of liquids are in the ratio 8:9, the ratio of their coefficient of viscosity is

Answer 1

Dear Student,

Given,
$\frac{l_1}{l_2}=\frac{1}{2},\frac{V_1}{V_2}=\frac{8}{9},\frac{r_1}{r_2}=\frac{2}{3},\frac{P_1}{P_2}=1;$
where,
$\frac{l_1}{l_2}$ is the ratio of the lengths of the capillary tube
$\frac{V_1}{V_2}$ is the ratio of rate of volume flow
$\frac{r_1}{r_2}$ is the ratio of radius of the capillary tubes
$\frac{P_1}{P_2}$ is the ratio of pressure difference
coefficient of viscosity is given by,
$$\begin{gathered} \eta =\frac{{\mathrm{\pi Pr}}^4}{8lV} \\ \Rightarrow \frac{{\eta }_1}{{\eta }_2}=\frac{\frac{\pi P_1{r_1}^4}{8l_1{V}_1}}{\frac{\pi P_2{r_2}^4}{8l_2{V}_2}}=\frac{P_1{r_1}^4{l}_2{V}_2}{P_2{r_2}^4{l}_1{V}_1}=1\times {\left(\frac{2}{3}\right)}^4\times 2\times \left(\frac{9}{8}\right)=\frac{4}{9} \end{gathered}$$
Regards