Velocity of a viscous liquid in long cylinder of radius R at a distance $R_{1}$ from centre is $V_{1}$ Find the velocity of the liquid as a function of the distance r from the axis of the cylinder,

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Solution

$v = \frac{p_{1} - p_{2}}{4 η L} (R^{2} - r^{2})$ (i)
$v_{1} = \frac{P_{1} - P_{2}}{4 η L} (R^{2} - R_{1}^{2})$ (ii)
Dividing Eq. (i) by Eq. (ii) we get
$v = (\frac{R^{2} - r^{2}}{R^{2} - R_{1}^{2}}) v_{1}$

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Q. A long cylinder of radius

R_{1}

is displaced along its axis with a constant velocity

v_{0}

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R_{2}

. The space between the cylinders is filled with viscous liquid. The velocity of the liquid as a function of the distance

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v = v_{0} \frac{ln \frac{k}{R_{2}}}{ln \frac{R_{1}}{R_{2}}}

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A long cylindrical rod of radius $R_{1}$ is placed co-axially inside a cylindrical tube of radius $R_{2} (> R_{1})$ filled withviscous liquid and pulled along the axis with a constant velocity. In steady state the velocity gradient $(\frac{d v}{d r})$ depends on the distance from the common axis 'r' as proportional to