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JEE Main & Advanced Sample Paper JEE Main Sample Paper-42

  • question_answer
    A rain drop of radius 0.2 cm is falling through air with a terminal velocity of 8.7 m/s. The viscosity of air in SI units is [Take \[{{\text{ }\!\!\rho\!\!\text{ }}_{\text{water}}}\] = 1000 kg/m3 and \[{{\text{ }\!\!\rho\!\!\text{ }}_{\text{air}}}=1\text{kg/}{{\text{m}}^{\text{3}}}]\]

    A)  \[\text{1}{{\text{0}}^{-4}}\]poise             

    B)  \[\text{1}\times \text{1}{{\text{0}}^{-3}}\]poise

    C)  \[8.6\times \text{1}{{\text{0}}^{-3}}\]poise          

    D)  \[1.02\times \text{1}{{\text{0}}^{-3}}\]poise

    Correct Answer: B

    Solution :

    We have,\[6\pi \eta rv=\frac{4}{3}\pi {{r}^{3}}\rho g-\frac{4}{3}\pi {{r}^{3}}\sigma g\] where\[\rho \to {{\rho }_{water}}\]and\[\sigma \to {{\rho }_{air}}\] \[\Rightarrow \]\[\eta =\frac{2g{{r}^{2}}(\rho -\sigma )}{9v}\] \[=\frac{2\times 9.81\times {{(0.2\times {{10}^{-2}})}^{2}}\times 999}{9\times 8.7}\] \[=1\times {{10}^{-3}}\]poise


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