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.
Show Hint
Use tau = mu times (v/h) for simple Couette flow between the plates, and remember to convert 2 mm to metres.
Step 1: Write viscosity directly as force times gap over area times velocity.
Rearrange Newton's law of viscosity $\tau=\mu\dfrac{v}{h}$ to solve for $\mu$ in one line:
\[ \mu=\frac{\tau h}{v} \]
Step 2: Plug in the numbers, keeping the gap in millimetres converted to metres.
Here $\tau=5$ N/m$^2$, $h=2$ mm $=2\times10^{-3}$ m, and $v=4$ m/s.
\[ \mu=\frac{5\times(2\times10^{-3})}{4} \]
Step 3: Work out the numerator first.
\[ 5\times2\times10^{-3}=10\times10^{-3}=1.0\times10^{-2} \]
Step 4: Divide by the velocity.
\[ \mu=\frac{1.0\times10^{-2}}{4}=0.25\times10^{-2}=2.5\times10^{-3} \text{ N.s/m}^2 \]
Step 5: Read off the coefficient asked for.
Since the question wants the answer as a multiple of $10^{-3}$ N.s/m$^2$, the required number is just the coefficient.
