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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    The resistance of the meter bridge AB in given figure is \[4\Omega .\] With a cell of emf \[\varepsilon =0.5V\]and rheostat resistance \[{{R}_{h}}=2\Omega \]the null point is obtained at some point J. When the cell is replaced by another one of emf \[\varepsilon ={{\varepsilon }_{2}}\]the same null point J is found for\[{{R}_{h}}=6\Omega .\] The emf \[{{\varepsilon }_{2}}\] is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) 0.5 V                           

    B) 0.4 V

    C) 0.5 V   

    D)                  0.3 V

    Correct Answer: D

    Solution :

    In first case, current \[{{i}_{1}},\] flowing through secondary circuit\[{{i}_{1}}=\frac{6}{{{R}_{AB}}+2}=1A\] Now, once the balance point is achieved, \[{{\varepsilon }_{1}}={{R}_{AJ}}{{i}_{1}}\] \[\Rightarrow \]\[0.5={{R}_{AJ}}(1)\]                  ?.(i) Similarly, in the second case, \[{{\varepsilon }_{2}}={{R}_{AJ}}{{i}_{2}}\] \[\Rightarrow \]\[{{\varepsilon }_{2}}={{R}_{AJ}}\left( \frac{6}{4+6} \right)\Rightarrow {{\varepsilon }_{2}}=\frac{6}{10}{{R}_{AJ}}\]        ?.(ii) From eqn. (i) \[{{\varepsilon }_{2}}=\frac{6}{10}\times 0.5\Rightarrow \varepsilon =0.3V\]


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