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JEE Main & Advanced AIEEE Solved Paper-2003

  • 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 \[1=30\] cm from the positive end, the emf of the battery is           AIEEE  Solved  Paper-2003

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

    B)                       \[\frac{30\,E}{100-05}\]

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

    D) \[\frac{30\,E}{100}\]

    Correct Answer: D

    Solution :

    Potential difference of wire                     \[V\propto l\] \[\therefore \]      \[\frac{V}{E}=\frac{l}{L}\] where, \[l=\] length   of   balance   point   of potentiometer-wire     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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