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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2000

  • question_answer
    A rain drop of radius 0.3 mm has a terminal velocity of 1 m/s in air. The viscosity of air is \[18\times {{10}^{-5}}\]poise. The visious forces on the drop will be:

    A) \[16.95\times {{10}^{-9}}N\]

    B)  \[1.695\times {{10}^{-9}}N\]

    C)  \[10.17\times {{10}^{-9}}N\]

    D)  \[101.73\times {{10}^{-9}}N\]

    Correct Answer: D

    Solution :

     The formula for viscous force is \[F=6\pi \,\eta \,r\,\upsilon \] Given : Here, viscosity \[\,\eta =18\times {{10}^{-5}}poise=18\times {{10}^{-6}}kg/m\sec \] \[Radiusr=0.3mm=0.3\times {{10}^{-3}}m\] Velocity \[\upsilon =1\text{ }m/s\] So, \[F=6\times 3.14\times 18\times {{10}^{-6}}\times 0.3\times {{10}^{-3}}\times 1\] \[=101.73\times {{10}^{-9}}N\]


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