Two capillary tubes A and B of radii $r_{a}$ and $r_{b}$ and lengths $l_{a}$ and $l_{b}$ respectively are held horizontally. The volume of water flowing per second through tube A is $Q_{a}$ when the pressure difference across its ends is maintained at P. When the same pressure difference is maintained across tube B, the volume of water flowing per second through it is $Q_{b}$ . The ratio $Q_{a} / Q_{b}$ is

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$\frac{l_{b}}{l_{a}} (\frac{r_{a}}{r_{b}})$

$\frac{l_{b}}{l_{a}} {(\frac{r_{a}}{r_{b}})}^{2}$

$\frac{l_{b}}{l_{a}} {(\frac{r_{a}}{r_{b}})}^{3}$

$\frac{l_{b}}{l_{a}} {(\frac{r_{a}}{r_{b}})}^{4}$

answer is D.

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Detailed Solution

The volume of a liquid flowing per second through a capillary tube of length l and radius r, when a pressure difference P is maintained across its ends is given by
$Q = \frac{π P r^{4}}{8 η l}$
where $η$ is the coefficient of viscosity of the liquid. Since P is kept the same for tubes A and B.

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