2. Solved Papers
  3. BHU PMT

BHU PMT BHU PMT (Screening) Solved Paper-2011

  • 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\[15\,\Omega \]is connected in series with the smaller of the two. The null point shift to 40 cm. The value of the smaller resistance in Ohm 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 meter bridge \[S=\left( \frac{100-l}{l} \right)R\] \[=\left( \frac{100-20}{20} \right)R=4R\] Again,  \[S=\left( \frac{100-l}{l} \right)(R+15)\]                 \[=\left( \frac{100-40}{40} \right)(R+15)\]                 \[4R=\frac{3}{2}(R+15)\] \[8R-3R=45\] \[5R=45\] \[R=9\Omega \].


You need to login to perform this action.
You will be redirected in 3 sec spinner