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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2005

  • 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 15 ohms is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohms is:

    A)  3

    B)  6

    C)  9

    D)  12

    Correct Answer: C

    Solution :

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


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