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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2004

  • question_answer
    An unknown resistance \[{{R}_{1}}\]is connected in series with a resistance of 10\[\Omega \]. This combination is connected to one gap of a metre bridge while a resistance \[{{R}_{2}}\]is connected in the other gap. The balance point is at 50 cm. Now, when the 10\[\Omega \] resistance is removed the balance point shifts to 40 cm. The value of \[{{R}_{1}}\] is (in ohms):

    A)  20

    B)  10

    C)  60

    D)  40

    Correct Answer: A

    Solution :

    The balance condition of a metre bridge experiment \[\frac{R}{S}=\frac{{{l}_{1}}}{(100-{{l}_{1}})}\] Here :    \[R={{R}_{1}},S={{R}_{2}}.\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{(100-{{l}_{1}})}\] Ist case:\[\frac{{{R}_{1}}+10}{{{R}_{2}}}=\frac{50}{50}\] \[\Rightarrow \] \[{{R}_{1}}+10={{R}_{2}}\] IInd case;\[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{40}{60}\] \[\Rightarrow \] \[{{R}_{2}}=\frac{60}{40}{{R}_{1}}\] ?(2) So, equations (1) and (2) give \[{{R}_{1}}+10=\frac{60}{40}{{R}_{1}}\] \[\Rightarrow \] \[\frac{60}{40}{{R}_{1}}-{{R}_{1}}=10\] \[\Rightarrow \] \[\frac{20}{40}{{R}_{1}}=10\] \[\Rightarrow \] \[{{R}_{1}}=\frac{10\times 40}{20}\] \[\therefore \] \[{{R}_{1}}=20\Omega \]


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