Uttarakhand PMT Uttarakhand PMT Solved Paper-2009

  • question_answer
    \[n\]identical mercury droplets charged to the same potential \[V\] coalesce to form a single bigger drop. The potential of new drop will be

    A)  \[\frac{V}{n}\]              

    B)  \[nV\]

    C)  \[n{{V}^{2}}\]           

    D)  \[{{n}^{2/3}}V\]

    Correct Answer: D

    Solution :

     Suppose we have\[n\]identical drops each having radius r, capacitance C, charge q and potential V. If these drops are combined to form a big drop of radius R, capacitance C charge Q and potential V will become: Charge on big drop \[Q=nq\] Capacitance of big drop \[C={{n}^{1/3}}C\] Hence potential of big drop \[V=\frac{Q}{C}=\frac{nq}{{{n}^{1/3}}C}\] \[={{n}^{2/3}}V\]


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