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BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer
    A charged dust particle of radius\[5\times {{10}^{-7}}m\]is located in a horizontal electric field having an intensity of\[6.28\times {{10}^{5}}V{{m}^{-1}}\]. The surrounding medium is air with coefficient of viscosity\[\eta =1.6\times {{10}^{-5}}N-s{{m}^{-2}}\]. If this particle moves with a uniform horizontal speed\[0.02\,m{{s}^{-1}}\]. Find the number of electrons on it.

    A)  10                                         

    B)  20

    C)  30          

    D)  40

    Correct Answer: C

    Solution :

                     The horizontal force on dust particle\[=qE\].                 If v is speed of particle in air, then                 Viscous force\[=6\pi \eta rv\](Stoke's law)                 For uniform speed v, we have                 \[qE=6\pi \eta rv\]                 If n is the number of excess electrons, then \[q=ne\]                 \[\therefore \]  \[neE=6\pi \eta rv\]                 Or           \[n=\frac{6\pi \eta rv}{eE}\]                 Substituting given values                 \[n=\frac{6\times 3.14\times 1.6\times {{10}^{-5}}\times 5\times {{10}^{-7}}\times 0.02}{1.6\times {{10}^{-19}}\times 6.28\times {{10}^{5}}}\]\[=30\]                          


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