2019 A/L Physics Past Paper · Question 23 (English Medium)

Velocity of the middle layer of oil between two plates

The gap of 2 cm2\ \text{cm} between two horizontal plates of surface area 500 cm2500\ \text{cm}^2 is filled with an oil having the coefficient of viscosity 0.2 N s m−20.2\ \text{N s m}^{-2}. A horizontal force of 5 N5\ \text{N} is applied to the upper plate while keeping the lower plate at rest. What is the velocity of the middle layer of the oil, if the velocities of the oil layers vary linearly across the gap between the plates?

  1. (1) 2.5 m s−12.5\ \text{m s}^{-1}
  2. (2) 5 m s−15\ \text{m s}^{-1}
  3. (3) 10 m s−110\ \text{m s}^{-1}
  4. (4) 25 m s−125\ \text{m s}^{-1}
  5. (5) 50 m s−150\ \text{m s}^{-1}
Answer and explanation

Answer: (2) 5 m s−15\ \text{m s}^{-1}

Given information

Given:

The area of the plates A=500 cm2A = 500\ \text{cm}^2.

The gap d=2d = 2 cm.

The viscosity η=0.2 N s m−2\eta = 0.2\ \text{N s m}^{-2}.

The force F=5F = 5 N.

The lower plate is at rest.

The speed changes in a straight line across the gap.

What the question asks

We want:

The speed of the middle layer of the oil.

It is in the middle of the gap, at the height d2\dfrac{d}{2}.

Relevant theory

Newton law of viscosity:

F=ηAdvdxF = \eta A\dfrac{dv}{dx}

The speed changes in a straight line, so dvdx=vd\dfrac{dv}{dx} = \dfrac{v}{d}.

Here vv is the speed of the upper plate.

The upper plate moves at a steady speed.

So the force we apply is equal to the viscous force.

The coefficient of the viscosity of a fluid is defined as the tangential force on a unit area under a unit velocity gradient of the fluid in uniform stream line motion.

Required equations

F=ηAvdF = \eta A \frac{v}{d}

v=FdηAv = \frac{Fd}{\eta A}

vmid=v2v_{mid} = \frac{v}{2}

Applying it to this question

The plan:

First change the units to SI.

Then find the speed of the upper plate.

The middle layer has half of it.

Reasoning and calculation

0.050.05

A=500×10−4A = 500 \times 10^{-4}

0.020.02

d=2×10−2d = 2 \times 10^{-2}

v=FdηA=5×0.020.2×0.05v = \frac{Fd}{\eta A} = \frac{5 \times 0.02}{0.2 \times 0.05}

1010

v=0.10.01v = \frac{0.1}{0.01}

55

vmid=v×d/2d=102v_{mid} = v \times \frac{d/2}{d} = \frac{10}{2}

The speed gradient is the same across the whole gap:

vd=100.02=500 s−1\dfrac{v}{d} = \dfrac{10}{0.02} = 500\ \text{s}^{-1}

At the height d2=0.01\dfrac{d}{2} = 0.01 m the speed is 500×0.01=5 m s−1500 \times 0.01 = 5\ \text{m s}^{-1}.

Checks

Unit check:

N⋅mN s m−2⋅m2=m s−1\dfrac{\text{N} \cdot \text{m}}{\text{N s m}^{-2} \cdot \text{m}^2} = \text{m s}^{-1}

We used SI units at every step.

Does it make sense?

The middle layer must move faster than the lower plate, 0.

And slower than the upper plate, 10.

55 is in between.

F∗d/(eta∗A)/2F * d / (eta * A) / 2

Conclusion

vmid=5 m s−1v_{mid} = 5\ \text{m s}^{-1}

The upper plate moves at v=FdηA=10 m s−1v = \dfrac{Fd}{\eta A} = 10\ \text{m s}^{-1}.

The middle layer has half of it, 5 m s−15\ \text{m s}^{-1}.

So the answer is option (2).

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