Two capillary tubes of the 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 correct option is None of these.

To understand why, let us derive the relationship for the rate of flow of a liquid through capillary tubes connected in parallel.
According to Poiseuille's law, the volume flow rate (rate of flow of liquid) $V$ through a capillary tube of length $l$, radius $r$, under a pressure difference (pressure head) $P$, and liquid viscosity $\eta$ is given by:

$V=\frac{\pi P{r}^4}{8\eta l}$

Here, the term $\frac{8\eta l}{\pi r^4}$ is analogous to electrical resistance, known as the liquid resistance or fluid flow resistance, $R$:
$R=\frac{8\eta l}{\pi r^4}$

Thus, we can write the flow rate as:
$V=\frac{P}{R}$

When two capillary tubes of the same length $l$ but different radii $r_1$ and $r_2$ are connected in parallel to the bottom of the vessel, they both experience the same pressure head $P$.
The total flow rate $V$ will be the sum of the flow rates through the individual tubes, $V_1$ and $V_2$:
$V=V_1+V_2$

Substituting Poiseuille's formula for each tube:
$V=\frac{\pi P{r}_1^4}{8\eta l}+\frac{\pi P{r}_2^4}{8\eta l}=\frac{\pi P}{8\eta l}(r_1^4+r_2^4)$

Now, let us assume these two tubes are replaced by a single capillary tube of the same length $l$ and radius $r$ under the same pressure head $P$ such that the flow rate remains the same. The flow rate through this replacement tube is:
$V=\frac{\pi P{r}^4}{8\eta l}$

Equating the flow rates, we get:
$\frac{\pi P{r}^4}{8\eta l}=\frac{\pi P}{8\eta l}(r_1^4+r_2^4)$

Canceling out the common terms $\frac{\pi P}{8\eta l}$ on both sides:
$r^4=r_1^4+r_2^4$

Taking the fourth root of both sides, the equivalent radius $r$ is:
$r={(r_1^4+r_2^4)}^{1/4}$

Since the value ${(r_1^4+r_2^4)}^{1/4}$ is not present in the options "r₁ +r₂", "r₁² +r₂²", or "r₁⁴ +r₂⁴", the correct choice is None of these.