Under a constant pressure head, the rate of flow of liquid through a capillary tube is V. If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

The correct option is V / 32.

To find the new rate of flow, we use Poiseuille's law for the steady flow of a liquid through a capillary tube. According to Poiseuille's law, the volume flow rate $V$ of a liquid through a capillary tube of length $L$ and radius $r$ under a constant pressure head $P$ is given by the formula:

$V=\frac{\pi ·P·r^4}{8·\eta ·L}$

where $\eta$ is the coefficient of viscosity of the liquid.

Since the diameter of the bore is halved, the radius $r$ of the capillary tube is also halved because the radius is directly proportional to the diameter ($r=d/2$).
Let the new length be $L^{\prime }=2L$ and the new radius be $r^{\prime }=\frac{r}{2}$.

Substituting these new values into the flow rate equation, we get the new flow rate $V^{\prime }$ as:

$V^{\prime }=\frac{\pi ·P·{\left(\frac{r}{2}\right)}^4}{8·\eta ·\left(2L\right)}$

Simplifying the numerator and denominator:

$V^{\prime }=\frac{\pi ·P·\frac{r^4}{16}}{8·\eta ·2L}$

$V^{\prime }=\frac{1}{16·2}·\left(\frac{\pi ·P·r^4}{8·\eta ·L}\right)$

$V^{\prime }=\frac{1}{32}·V$

$V^{\prime }=\frac{V}{32}$

Therefore, the rate of flow becomes $\frac{V}{32}$.