JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Self Evaluation Test - Current Electricity

  • question_answer
    125 cm of potentiometer wire balances the emf of a cell and 100 cm of the wire is required for balance, if the poles of the cell are joined by \[2\Omega \]resistor. Then the internal resistance of the cell is

    A) \[0.25\Omega \]

    B) \[0.5\Omega \]

    C) \[0.75\Omega \]

    D) \[1.25\Omega \]

    Correct Answer: B

    Solution :

    [b] \[r=\frac{{{\ell }_{1}}-{{\ell }_{2}}}{{{\ell }_{2}}}\times R\Omega \] Here, \[{{\ell }_{1}}=125cm,{{\ell }_{2}}=100cm,R=2\Omega .\] \[\therefore r=0.5\Omega \]


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