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Haryana PMT Haryana PMT Solved Paper-2000

  • question_answer
    The electric field required to just balance a liquid drop of mass \[2.4\times {{10}^{-12}}kg\] and charge \[2.4\times {{10}^{-18}}\] coulomb, is:

    A)  \[9.8\times {{10}^{6}}N/C\]       

    B)  \[4.9\times {{10}^{6}}N/C\]

    C)  \[6.8\times {{10}^{6}}N/C\]       

    D)  none of these

    Correct Answer: A

    Solution :

                    For equilibrium of drop the force of gravity (mg) must be balanced by upward electric force So,          \[qE=mg\]                 \[E=\frac{mg}{q}\]                 \[=\frac{2.4\times {{10}^{-12}}\times 9.8}{2.4\times {{10}^{-18}}}\]                 \[=9.8\times {{10}^{6}}N/C\]


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