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BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    A galvanometer of resistance\[50\,\Omega \]is connected to a battery of 3 V along with a resistance of\[2950\,\Omega \]in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions the resistance in series should be

    A)  \[4450\,\Omega \]                        

    B)  \[5050\,\Omega \]

    C)  \[5550\,\Omega \]                        

    D)  \[6050\,\Omega \]

    Correct Answer: A

    Solution :

                     The current through the galvanometer \[=\frac{3}{2950+50}={{10}^{-3}}A\] To reduce deflection from 30 divisions to 20 divisions, the required current \[=\frac{20}{30}\times {{10}^{-3}}=\frac{2}{3}\times {{10}^{-3}}A\] The required resistance R is given by                 \[\frac{3}{R+50}=\frac{2}{3}\times {{10}^{-3}}\] \[R=4450\,\Omega \]


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