AMU Medical AMU Solved Paper-2011

  • question_answer
    (The surface tension of the soap film is \[\text{3}\times \text{1}{{0}^{-\text{2}}}\text{ N}/\text{m}).\] A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is \[\text{2}.0\times \text{1}{{0}^{-\text{5}}}\text{ kg}/\text{ms},\] the terminal velocity of the drop will be

    A)  \[\text{1}.0\times \text{1}{{0}^{-\text{4}}}\text{ m}/\text{s}\]                

    B) \[2.0\times \text{1}{{0}^{-\text{4}}}\text{ m}/\text{s}\]

    C)  \[2.5\times \text{1}{{0}^{-\text{4}}}\text{ m}/\text{s}\]                             

    D)  \[5.0\times \text{1}{{0}^{-\text{4}}}\text{ m}/\text{s}\]               

    Correct Answer: C

    Solution :

    Radius r = 0.0015 mm                 \[=15\times {{10}^{-7}}m\]                 Coefficient of viscosity \[=2.0\times {{10}^{-5}}kg/ms\]                 Density of water \[=1\times {{10}^{3}}kgm/{{m}^{3}}\]                 Terminal velocity \[v=\frac{2}{9}\frac{{{r}^{2}}\rho g}{\eta }\]                 \[v=\frac{2}{9}\times \frac{{{(15\times {{10}^{-7}})}^{2}}\times 1\times {{10}^{3}}\times 10}{2\times {{10}^{-5}}}\]                 \[v=2.5\times {{10}^{-4}}m/s\] Note : - (The density of water\[=\text{1}.0\times \text{1}{{0}^{\text{3}}}\text{kg}/{{\text{m}}^{\text{3}}}\text{and}\,\text{g}=\text{1}0\text{m}/{{\text{s}}^{\text{2}}}\])


You need to login to perform this action.
You will be redirected in 3 sec spinner