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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • 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 :

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


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