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JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Cathode Rays and Positive Rays

  • question_answer
    In Millikan oil drop experiment, a charged drop of mass \[1.8\times {{10}^{-14}}kg\]is stationary between its plates. The distance between its plates is 0.90 cm and potential difference is 2.0 kilo volts. The number of electrons on the drop is [MP PMT 1994, 2003; MP PET 1997]

    A)            500

    B)            50

    C)            5    

    D)            0

    Correct Answer: C

    Solution :

                       \[QE=mg\Rightarrow Q=\frac{mg}{E}\Rightarrow n=\frac{mgd}{Ve}\]            \[\Rightarrow n=\frac{1.8\times {{10}^{-14}}\times 10\times 0.9\times {{10}^{-2}}}{2\times {{10}^{3}}\times 1.6\times {{10}^{-19}}}=5\]


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