Question

A Newtonian fluid fills the clearance between the shaft and the sleeve. When a force of $800\text{N}$ is applied to the shaft, parallel to the sleeve, the shaft attains a speed of $1.5\text{cm/sec}$. If a force of $2.4\text{kN}$ is applied instead, the shaft would move with a speed of

• A. $1.5\text{cm/sec}$
• B. $13.5\text{cm/sec}$
• C. $4.5\text{cm/sec}$
• D. None

Answer 1

The correct option is C $4.5\text{cm/sec}$

Assuming the linear velocity profile.


By Newton's law of viscosity,
$\tau =\eta \frac{dv}{dy}$
Viscous force, $F=\eta A\frac{dv}{dy}$
Situation 1: $F=800\text{N},v=1.5\text{cm/sec}$
$\Rightarrow 800=\eta A\frac{1.5}{x}.....\left(i\right)$
Situation 2: $F=2400\text{N},v=?$
$2400=\eta A\frac{v}{x}....\left(ii\right)$
Dividing $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{1}{3}=\frac{1.5}{v}$
$\Rightarrow v=4.5\text{cm/sec}$