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EAMCET Medical EAMCET Medical Solved Paper-2007

  • question_answer
    Two unknown resistances X and Y are connected to left and right gaps of a metre bridge and the balancing point is obtained at 80 cm from left. When a 10\[\Omega \] resistance is connected in parallel to X the balancing point is 50 cm from left. The values of X and Y respectively are

    A)  40\[\Omega \], 90\[\Omega \]                                

    B)  30\[\Omega \], 7.5\[\Omega \]

    C)  20\[\Omega \], 6\[\Omega \]                                   

    D)  10\[\Omega \], 3\[\Omega \]

    Correct Answer: B

    Solution :

                     Let \[l\] be the distance of balancing point from left gap, then \[\frac{X}{Y}=\frac{l}{100-l}=\frac{80}{20}=4\] or            \[X=4Y\]                                               ?(i) Again in parallel, the net resistance is \[X=\frac{10X}{10+X}\]                 So,          \[\frac{X}{Y}=\frac{50}{100-50}=1\]                 or            \[\frac{10X}{10+X}=Y\]                 or            \[10X=10Y+XY\]                 pr            \[40Y=10Y+4{{Y}^{2}}\]  [from Eq. (i)] or           \[Y=7.5\Omega \] Putting in Eq. (i), we get \[X=30\Omega \]


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