Question

Two parallel plates, with Newtonian incompressible liquid in between, are 2 mm apart. The upper plate is stationary and the lower plate moves with a velocity of 4 m/s. A force per unit area of 5 N/m² is applied parallel to the lower plate to maintain its motion. The viscosity of the liquid (rounded off to two decimal places) is ______ \(\times10^{-3}\) N.s/m².

Hint:

Use tau = mu times (v/h) for simple Couette flow between the plates, and remember to convert 2 mm to metres.

Answer 1

Correct Answer: 2.4

Step 1: Recall Newton's law of viscosity.
For a Newtonian fluid sheared between two parallel plates, the shear stress (force per unit area) is proportional to the velocity gradient across the gap:
\[ \tau=\mu\frac{dv}{dy} \]
where \(\tau\) is the shear stress, \(\mu\) is the dynamic viscosity, and \(\frac{dv}{dy}\) is the rate of change of velocity across the gap between the plates.

Step 2: Note the given values.
Gap between the plates, \(h=2\) mm \(=2\times10^{-3}\) m.
Velocity of the moving (lower) plate relative to the stationary (upper) plate, \(v=4\) m/s.
Shear stress, \(\tau=5\) N/m\(^2\).

Step 3: Approximate the velocity gradient as linear.
Since the upper plate is stationary and the lower plate moves at 4 m/s across a small, fixed gap, the velocity varies linearly across the gap (simple Couette flow), so
\[ \frac{dv}{dy}=\frac{v}{h}=\frac{4}{2\times10^{-3}}=2000 \text{ s}^{-1} \]

Step 4: Solve for the viscosity.
\[ \mu=\frac{\tau}{dv/dy}=\frac{5}{2000} \]
\[ \mu=0.0025 \text{ N.s/m}^2 \]

Step 5: Convert to the units asked in the question.
\[ \mu=0.0025 \text{ N.s/m}^2=2.5\times10^{-3} \text{ N.s/m}^2 \]

Step 6: Final answer.
Rounded to two decimal places, the required value is
\[ \boxed{2.50} \]