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BVP Medical BVP Medical Solved Paper-2014

  • question_answer
    In Millikans oil drop experiment, a charged drop of mass \[64V,\,\,2\Omega \] is stationary between the plates. The distance between the plates is 0.9 cm and potential difference between the plates is 2000 V. The number of electrons on the oil drop is                 

    A)  \[15min\]                                          

    B)  \[20\text{ }min\]                 

    C)  \[7.5min\]                                         

    D)  \[25min\]                

    Correct Answer: B

    Solution :

                    \[qE=mg\] \[E=\frac{V}{d}\] \[q=ne\] \[q=\frac{mg}{E}\] \[ne\frac{mg}{V}d\]     (\[\therefore \]\[q=ne\]) \[n=\frac{mgd}{Ve}\] \[=\frac{18\times {{10}^{-14}}\times 9.8\times 0.9\times {{10}^{-2}}}{2000\times 1.6\times {{10}^{-19}}}\] \[n=4.96=5\]     


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