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JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Different Measuring Instruments

  • question_answer
    A potentiometer is used for the comparison of e.m.f. of two cells \[{{E}_{1}}\] and \[{{E}_{2}}\]. For cell \[{{E}_{1}}\] the no deflection point is obtained at 20\[cm\] and for \[{{E}_{2}}\] the no deflection point is obtained at 30\[cm\]. The ratio of their e.m.f.'s will be                                                                                            [MP PET 1984]

    A)                    2/3

    B)                                      1/2

    C)                    1

    D)                                      2

    Correct Answer: A

    Solution :

                 Ratio will be equal to the ratio of no deflection lengths i.e. \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{2}{3}\]


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