A tube of length $l$ and radius R carries a steady flow of fluid whose density is $ρ$ and viscosity $η .$ The velocity v of flow is given by $v = v_{0} (1 - \frac{r^{2}}{R^{2}}),$ Where r is the distance of flowing fluid from the axis.

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The volume of fluid, flowing across the section of the tube, in unit time is $2 π v_{0} (\frac{R^{2}}{4})$

The kinetic energy of the fluid within the volume of the tube is $K . E . = π ρ l {v_{0}}^{2} (\frac{R^{2}}{6})$

The pressure difference at the ends of tube is $P = \frac{4 η l v_{0}}{R^{2}}$

The frictional force exerted on the tube by the fluid is $F = 4 π η k v_{0}$

answer is A, B, C, D.

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Detailed Solution

The volume of fluid flowing through this section per second $d v = (2 π r d r) v_{0} (1 - r^{2} / R^{2})$
total volume $V = \int_{0}^{R} (2 π r d r) v_{0} (1 - r^{2} / R^{2})$
$= 2 π v_{0} (R^{2} / 4)$
(ii). The kinetic energy of the fluid within the volume element of thickness dr
K.E of fluid within the tube is $= \frac{1}{2} (2 π l) ρ {v_{0}}^{2} \int_{0}^{R} (1 - r^{2} / R^{2}) r d r$
we get K.E $π ρ l {v_{0}}^{2} (R^{2} / 6)$
(iii) The viscous drag exerts a force on the tube $F = - η A {(\frac{d v}{d r})}_{r = R}$
Here ${(\frac{d v}{d r})}_{r = R} = v_{0} {(- 2 r / R^{2})}_{r - R} = - 2 v_{0} / R$
$∴ F = 4 π η l v_{0}$
(iv) $Δ P = P_{2} - P_{1} = P$
where $P_{1} = 0$ and $P_{2} = P$
$P = \frac{f o r c e (F)}{a r e a (π R^{2})}$
$= \frac{4 η l v_{0}}{R^{2}}$