Two capillary tubes of 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

The rate of flow in a capillary tube is given by the Hagen-Poiseuille equation:

$\mathrm{Q}=\frac{{{\mathrm{\pi Pr}}_1}^4}{8\mathrm{\eta l}}+\frac{\mathrm{\pi P}{{\mathrm{r}}_2}^4}{8\mathrm{\eta l}}\mathrm{\_\_\_\_\_\_\_\_}\left(1\right)$

If two capillary tubes of different radii r₁ and r₂ are fitted in parallel to the bottom of a vessel, the rate of flow through the tubes is

$\mathrm{Q}=\frac{{{\mathrm{\pi Pr}}_1}^4}{8\mathrm{\eta l}}+\frac{\mathrm{\pi P}{{\mathrm{r}}_2}^4}{8\mathrm{\eta l}}=\frac{\mathrm{\pi P}({{\mathrm{r}}_1}^4+{{\mathrm{r}}_2}^4)}{8\mathrm{\eta l}}$

To find the radius of a single tube that can replace the two tubes so that the rate of flow is the same as before, we can set the rate of flow through the single tube equal to the rate of flow through the two tubes and solve for r:

 $\mathrm{Q}=\frac{\mathrm{\pi }\times {\mathrm{r}}^4\times \mathrm{P}}{8\mathrm{\eta l}}$

$\mathrm{Q}=\frac{\mathrm{\pi }\times ({{\mathrm{r}}_1}^4+{{\mathrm{r}}_2}^4)\times \mathrm{P}}{8\mathrm{\eta l}}$

${\mathrm{r}}^4=({{\mathrm{r}}_1}^4+{{\mathrm{r}}_2}^4)$

$\mathrm{r}={({{\mathrm{r}}_1}^4+{{\mathrm{r}}_2}^4)}^{\frac{1}{4}}$

Hence the correct answer is ${({{\mathrm{r}}_1}^4+{{\mathrm{r}}_2}^4)}^{\frac{1}{4}}.$