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CMC Medical CMC-Medical Ludhiana Solved Paper-2014

  • 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 :

                    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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