A tube of length $l$ and radius R carries a steady flow of a liquid of density $ρ$ and viscosity $η$ . The velocity of flow of the liquid depends on the radial distance ‘r’ from the axis of the tube as $v = v_{0} (1 - \frac{r^{2}}{R^{2}})$ . If the pressure difference at the two ends of the tube is $k (\frac{η l v_{0}}{6 R^{2}})$ . The value of k is………

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answer is 24.

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Detailed Solution

Let the pressure difference at the two ends of the tube is $Δ P$ The viscous force acting on the cylindrical volume of the liquid of radius ‘r’ is

$F = - η 2 π r l \frac{d v}{d r}$
$F = - η 2 π l v_{0} (- \frac{2 r}{R^{2}})$
$F = 4 π η l v_{0} (\frac{r^{2}}{R^{2}})$ ……(i)
For the steady flow of the liquid
$Δ P π r^{2} = F$
$Δ P π r^{2} = 4 π η l v_{0} (\frac{r^{2}}{R^{2}})$
$Δ P = \frac{4 η l v_{0}}{R^{2}}$
Hence, $k = 24$