Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2008

  • question_answer A charged oil drop is suspended in uniform field of \[3\times {{10}^{4}}V/m\] so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge \[=9.9\times {{10}^{-15}}kg\] and \[g=10m/{{s}^{2}}\])

    A)  \[3.3\times {{10}^{-18}}C\]                        

    B)  \[3.2\times {{10}^{-18}}C\]

    C)  \[1.6\times {{10}^{-18}}C\]                        

    D)  \[4.8\times {{10}^{-18}}C\]

    Correct Answer: A

    Solution :

    Key Idea In steady state electric force on drop balances the weight of the drop. In steady state, electric force on drop                 = weight of drop                 \[\therefore \]  \[qE=mg\]                 \[\Rightarrow \]               \[q=\frac{mg}{E}\]                                 \[=\frac{9.9\times {{10}^{-15}}\times 10}{3\times {{10}^{4}}}\]                                 \[=3.3\times {{10}^{-18}}C\]


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