A Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of $800$ $N$ is applied to the shaft, parallel to the sleeve, the shaft attains a speed of $1.5$ $cm/sec$. If a force of $2.4$ $kN$ is applied instead, the shaft would move with a speed of ......... $ cm/sec$
Diffcult
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Using the relation $\tau=\mu \frac{d u}{d y} .$ Here $\tau$ is the shear stress between the layers of the liquid which is equal to $\frac{F}{A} . d u$ is the change of velocity $=u-0=u$ and $d y$ is the clearance which is equal to $y$
Thus we get
$\frac{F}{A}=\mu \frac{u}{y}$
or
$F=\frac{A \mu u}{y} \propto u$
Thus we get
$\frac{F_{1}}{u_{1}}=\frac{F_{2}}{u_{2}}$
Substituting the values we get
$\frac{800}{1.5}=\frac{2400}{u_{2}}$
Thus we get $u_{2}=4.5 \mathrm{cm} / \mathrm{s}$
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