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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    A potentiometer PQ is set up to comapre two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key \[{{K}_{3}}\] is open. The balance point is at a length \[{{l}_{1}}\]cm from P when two way key \[{{K}_{3}}\] is plugged in between 2 and 1, while the balance point is at a length \[{{l}_{2}}\] cm from P when key \[{{K}_{3}}\] is plugged in between 3 and 1. The ratio of the two resistance \[\frac{{{R}_{1}}}{{{R}_{2}}},\] is found to be -                                 [JEE Online 08-04-2017]

    A) \[\frac{{{l}_{1}}}{{{l}_{1}}-{{l}_{2}}}\]                      

    B) \[\frac{{{l}_{2}}}{{{l}_{2}}-{{l}_{1}}}\]

    C) \[\frac{{{l}_{1}}}{{{l}_{1}}+{{l}_{2}}}\]                     

    D) \[\frac{{{l}_{1}}}{{{l}_{2}}-{{l}_{1}}}\]

    Correct Answer: A

    Solution :

                    When key is at point \[{{V}_{1}}=i{{R}_{1}}=x{{l}_{1}}\] when key is at (3) \[{{V}_{2}}=i({{R}_{1}}+{{R}_{2}})=x{{l}_{2}}\] \[\frac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}-{{l}_{1}}}\]


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