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CMC Medical CMC-Medical VELLORE Solved Paper-2010

  • question_answer
    Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15Q is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohm is

    A)  3                                            

    B)  6

    C)  9                                            

    D)  12

    E)  None of these

    Correct Answer: C

    Solution :

                    Let S be the large and R be the smaller resistance. From formula for meter bridge \[S=\left( \frac{100-l}{l} \right)R\] \[=\left( \frac{100-20}{20} \right)R=4R\] Again,   \[S=\left( \frac{100-l}{l} \right)(R+15)\] \[=\left( \frac{100-40}{40} \right)(R+15)\] \[=\frac{3}{2}(R+15)\] \[\therefore \]  \[4R=\frac{3}{2}(R+15)\] or            \[8R-3R=45\] or            \[5R=45\] \[\therefore \]  \[R=9\,\Omega \]


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