PHYS*1080 Sample Quiz 9

  1. (a) What is meant by a "Newtonian" fluid? (One sentence.)

    (b) A student measures the coefficient of viscosity to be \(7.8 \times 10^{-4} \; \text {N/m}^{-2}\; \text s\) at \(20^\circ \text C\). The accepted value is \(1.00 \times 10^{-3} \text{N/m}^{-2}\text s\). What is the \(\%\) difference between these two values? (2 marks)
    Part Solution Marks
    (a)

    A 'Newtonian' fluid will have constant viscosity (at a constant temp) regardless of the method used to measure it and regardless of the shear forces in the liquid.  

    		 (1) The two"regardless" are essentially redundant so either would do. "at a constant temp" is not required either.
    (b) \(\% \; difference = \frac{10 \times 10^{-4}- 7.8 \times 10^{-4}}{10 \times 10 ^{-4}}\times 100\% = 22\%\) (1)

     
  2. In exercise (a) of experiment 14, you used equation \(V = \frac{\pi r^+Pt}{8 \eta L}\) to determine the coefficient of viscosity of water. In a sentence or two, explain what "\(P\)" represents and how its numerical value was determined. (2 marks)
    Solution  Marks
    "P" is the pressure difference from one end of the capillary tube to the other. \(P = \rho gh\) where \(h\) is the measured height of the water column, \(\rho\) is the density of water and  \(g\) is the acceleration due to gravity.  (1) for what it is
    (1) for how it is measured.

     
  3. What flow per minute goes through the blood vessel for which this graph was obtained? The heart beats 150 times per minute and the vessel has a diameter of \(1.0 \times 10^{-3}\; \text m.\). (4.5 marks)
    graph of flow through a blood vessel
    Solution Marks
    To get flow - require area under curve.  (0.5)
    The 2 small areas represent equal forward and reverse flows and contribute no net flow.  (0.5)
    \(360^\circ \Rightarrow \text{time for 1 beat} = 60\; \text s/ 150 = 0.4\; \text s\) (0.5)
    \(\therefore\; \text{base of large triangle represents} \\ \quad 180/360 \times 0.4 \; \text s = 0.2 \; \text s\) (1.0)
    \(\text{height must be}\; Q = Av = \pi R^2v \\ \therefore \; Q = \pi \times (10^{-3}/2)^2 \times 0.5 = 3.9 \times 10^{-7}\; \text m^3/\text s\) (1.0)
    \(\begin {align} \text{Flow/beat} = \frac{1}{2} b \times h & = \frac{1}{2}\times 3.9 \times 10^{-7} \times 0.2\; \text s \\ & = 3.9 \times 10^{-8}\; \text{m}^3 \end {align}\) (0.5)
    \(\begin {align} \text{Flow/min} = 3.9 \times 10^{-8} \times 150 & = 5.9 \times 10^{-6}\; \text m^3 \\ & (\text{or}\; 5.9 \; \text{cm}^3) \end {align}\) (0.5)

     
  4. What is the tension in an arteriole wall if the mean blood pressure is \(50\; \text{mm Hg}\) and the radius is \(0.0080\; \text{cm}\) ? (1.5 marks)
    Solution Marks
    \(\begin {align} T &= \Delta PR = \rho ghR \\ & = 13, 600 \; \text{kg/m}^3 \times 9.80\; \text{m/s}^2 \times 0.050\; \text m \times 8.0 \times 10^{-5}\; \text m \end {align}\) (1.0)
    \(\quad = 0.53\; \text{N/m}\) (0.5)