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JEE Main & Advanced Sample Paper JEE Main Sample Paper-36

  • 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 \[0.90\text{ }cm\] and potential difference is \[2.0\] kilovolts. The number of electrons on the drop is

    A)  500                             

    B)  50      

    C)  5                    

    D)  0

    Correct Answer: C

    Solution :

     \[ne\times E=m\times g\] or \[\frac{n\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}{0.9\times {{10}^{-2}}}=1.8\times {{10}^{-14}}\times 10\] \[n=\frac{1.8\times {{10}^{-14}}\times 10\times .9\times {{10}^{-2}}}{1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}\] or \[n\approx 5\]


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