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BHU PMT BHU PMT (Mains) 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 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)  400, 90                                

    B)  300, 7.50

    C)  200, 60                                

    D)  100, 30

    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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