2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-3

  • question_answer
    In Millikan oil drop experiment, a charged drop of mass \[1.8\times {{10}^{-14}}\text{ }kg\]is stationary between its plates. The distance between its plates is 0.90 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            \[\frac{n\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}{0.9\times {{10}^{-2}}}=1.8\times {{10}^{-14}}\times 10\]


You need to login to perform this action.
You will be redirected in 3 sec spinner