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BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    A potential of 500 V is observed near the surface C with \[V=0\] at infinity. If two such drops of the same charge and radius combine to form a single spherical drop, the potential at the surface at the new drop?

    A)  590V                                    

    B)  690V

    C)  790V                                    

    D)  890V

    Correct Answer: C

    Solution :

                     After combination of drops \[{{(R')}^{3}}=2{{R}^{3}}\] \[R'={{2}^{1/3}}R\] \[q'=2q\] Thus,      \[V'=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q'}{R'}\]                 \[=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{2q}{{{2}^{1/3}}R}\]                 \[={{2}^{1/3}}V\]                 \[={{2}^{1/3}}(500)=790V\]


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