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MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    The length of a wire of a potentiometer is 100 cm, and the emf of its stand and cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5\[\Omega \]If the balance point is obtained at \[l\] = 30 cm from the positive end, the emf of the battery is

    A) \[\frac{30E}{100.5}\]                                      

    B) \[\frac{30E}{100-0.5}\]

    C)  \[\frac{30(E-0.5i)}{100}\]where \[i\]is the current in the potentiometer wire

    D)  \[\frac{30E}{100}\,\]

    Correct Answer: D

    Solution :

                    (d )As    \[V\propto l\] \[\therefore \]  \[\frac{V}{E}=\frac{l}{L}\]where, \[l=\]balance point, \[L=\]length of potentiometer wire or               \[V=\frac{l}{L}E\] \[V=\frac{30\times E}{100}=\frac{30}{100}E\]


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