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Punjab Medical Punjab - MET Solved Paper-2008

  • 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 \[80cm\] from left. When a \[10\,\,\Omega \] resistance is connected in parallel to \[X\] the balancing point is \[50cm\] from left. The values of \[X\] and \[Y\] respectively are

    A) \[40\Omega ,\,\,9\Omega \]                     

    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            \[10X=10Y+XY\] or            \[40Y=10Y+4{{Y}^{2}}\]  [from Eq. (i)] or                \[Y=7.5\Omega \] Putting in Eq. (i), we get                    \[X=30\Omega \]


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