PHYS*1080 Sample Quiz 8

  1. A vertical hollow steel tube is \(2.5\; \text m\) long. Its internal diameter is \(5\; \text {cm}\) and its external diameter is \(6\; \text{cm}\). Calculate the increase in length of the tube when a load of \(1.5 \times 10^4 \; \text{kg}\) is suspended from the lower end. For steel Young's modulus is \(21 \times 10^{10} \; \text {N m}^{-2}\) and the shear modulus is \(8.3 \times 10 ^{10} \text{N m}^2\). (3 marks)

    diagram of cross section of steel tube

    Solution Marks
    \(F/A = Y(\Delta \ell /\ell) \\ A - \pi (0.030^2 - 0.0250^2) = 8.64 \times 10^{-4}\; \text m^2\) (1.0) area
    \(\Delta\ell = F \ell /YA = 1.5 \times 10^4 \times 9.80 \times 2.5 \; \text m/[21 \times 10^{10} \times 8.64 \times 10^{-4}]\) (1.5)
    \(\therefore\; \Delta \ell = 2.0 \times 10^{-3}\text m\; (2\; \text {mm})\) (0.5)

     
  2. In mammals, brain volume is proportional to the surface area of the mammal. A small mammal of \(1.0\; \text{kg}\) body mass has a brain mass of \(10 \; \text g.\) What will be the brain mass of a large mammal of mass \(1000\; \text{kg}\) ? (4 marks)
    Solution Mark
    \(V(\text{large})/V(\text{small})= 1000 = L^3\) (1.0)
    \(\therefore \; L = 10\) (1.0)
    \(A(\text{large})/A(\text{small})= L^2 = \text{Brain}(\text{large})/\text{Brain}(\text{small})\) (1.0)
    \(\therefore \; \text{Brain}(\text{large})= 10\; \text g \times 10^2 = 10^3 \; \text g = 1 \; \text{kg}\) (1.0)

     
  3. A biophysicist wants to determine the surface tension of ethyl alcohol. Its density is \(790\; \text{kg m}^{-3}\), and its contact angle with glass is the same as that for pure water. Inside a glass capillary tube, water "climbs" 2.7 times higher than ethyl alcohol. What is the surface tension of ethyl alcohol? (3 marks)
    Solution Marks
    \(2.7 = \frac{h_w}{h_a} = \frac{\frac{2 \gamma_w \cos \theta_w}{\rho_wgr}}{\frac{2 \gamma_a \cos \theta_a}{\rho _a gr}} = \frac{\gamma _w\rho_a}{\gamma _a \rho _w}\) (1.5)
    \(\gamma _a = \frac{h_a}{h_w}\frac{\gamma_w\rho_a}{\rho_w}= \frac{1}{2.7} \frac{(7.28 \times 10^{-2})(790)}{1000}\) (1.0)
    \( = 2.1 \times 10^{-2}\; \text{N/m} \; [\text{J/m}^2]\) (0.5) answer