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

  • question_answer
    In the experimental set up of metre bridgre shown in the figure, the null point is obtained at a distance of 40 cm from A. If a \[10\Omega \]resistor is connected in series with \[{{R}_{1}},\] the null point shifts by 10 cm. The resistance that should be connected in parallel with \[({{R}_{1}}+10)\Omega \] such that the null point shifts back to its initial position is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) \[40\,\Omega \]

    B)                                       \[20\,\Omega \]

    C) \[60\,\Omega \]             

    D)   \[30\,\Omega \]

    Correct Answer: C

    Solution :

    Case I: \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{40}{60}=\frac{2}{3}\]                  ?(i) Case II: \[\frac{{{R}_{1}}+10}{{{R}_{2}}}=1\]or,\[{{R}_{1}}+10={{R}_{2}}\] or\[\frac{2}{3}{{R}_{2}}+10={{R}_{2}};{{R}_{2}}=30\Omega \] \[{{R}_{1}}=\frac{2}{3}\times 30=20\Omega \] Case III: Resistance R is connected in parallel with resistance \[({{R}_{1}}+10)\Omega \] \[\frac{(20+10)\times R}{\frac{30+R}{30}}=\frac{2}{3}\] \[3R=60+2R\] \[R=60\Omega .\]


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