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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    In Millikans oil drop experiment, a charged drop of mass \[1.8\times {{10}^{-13}}\]kg is stationary between its plates. The distance between its plates is 0.9 cm and potential difference is 2000 V. The number of electrons in the drop is

    A)  500                                       

    B)  50

    C)  5                            

    D)         10

    Correct Answer: B

    Solution :

    Drop is stationary,\[ie\], weight of drop = electric force \[ie\],    \[mg=qE\] or            \[mg=q\frac{V}{d}\] or            \[mg=ne\frac{V}{d}\] \[\therefore \]  \[n=\frac{mgd}{eV}\]                 \[=\frac{1.8\times {{10}^{-13}}\times 10\times 0.9\times {{10}^{-2}}}{1.6\times {{10}^{-19}}\times 2000}\]                 \[n=50\]


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