Two tubes of radii $r_{1}$ and $r_{2},$ and lengths $l_{1}$ and $l_{2}$ , respectively, are connected in series and a liquid flow through each of them in streamline conditions. $P_{1}$ and $P_{2}$ are pressure differences across the two tubes. If $P_{2}$ is $4 P_{1}$ and $l_{2}$ is $l_{1} / 4,$ then the radius $r_{2}$ will be equal to

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$r_{1}$

$2 r_{1}$

$4 r_{1}$

$\frac{r_{1}}{2}$

answer is D.

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Detailed Solution

For viscous fluid flowing in a capillary tube we use flow rate $\frac{d Q}{d t} = \frac{π r^{4}}{8 η l} Δ P$
As tubes in this case are in series, so we use
${\frac{d Q}{d t} |}_{1} = {\frac{d Q}{d t} |}_{2}$

$\Rightarrow \frac{π r_{1}^{4}}{8 η l_{1}} P_{1} = \frac{π r_{2}^{4}}{8 η l_{2}} P_{2}$
$\Rightarrow \frac{r_{1}^{4}}{l_{1}} P_{1} = \frac{r_{2}^{4}}{l_{2} / 4} (4 P_{1})$
$\Rightarrow r_{1}^{4} = 16 r_{2}^{4}$
$\Rightarrow r_{2} = {(\frac{r_{1}^{4}}{16})}^{1 / 4} = \frac{r_{1}}{2}$