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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    An unknown resistance R1 is connected in series with a resistance of \[10\,\Omega \]. This combination is connected to one gap 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;                \[\frac{{{P}_{1}}+10}{{{R}_{2}}}=\frac{50}{50}\] \[\Rightarrow \]               \[{{R}_{1}}+10={{R}_{2}}\]                                           ?. (1) 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                 \[{{R}_{1}}+10=\frac{60}{40}{{R}_{1}}\] \[\Rightarrow \]               \[\frac{60}{40}{{R}_{1}}-{{R}_{1}}=10\] \[\Rightarrow \]               \[\frac{20}{10}{{R}_{1}}=10\] \[\Rightarrow \]               \[{{R}_{1}}=\frac{10\times 40}{20}\] \[\therefore \]  \[{{R}_{1}}=20\,\,\Omega \]


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