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NEET NEET SOLVED PAPER 2015 (C)

  • question_answer
    A potentiometer wire has length 4 m and resistance \[8\,\Omega \]. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1 mV per cm on the wire is              

    A)                                                          32 \[\Omega \]            

    B)  40 \[\Omega \]

    C)  44 \[\Omega \]               

    D)  48\[\Omega \]

    Correct Answer: A

    Solution :

                    Given. 1 = 4 m, R = potentiometer wire resistance = \[8\Omega \] Potential gradient = \[\frac{dV}{dr}=1mV/cm\] So, for \[400cm,\Delta V=400\times 1\times {{10}^{-3}}=0.4V\] Let a resistor \[{{R}_{s}}\] connected in series, so as \[\Delta V=\frac{V}{R+{{R}_{s}}}\times R\]                 \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,0.4=\frac{2}{8+R}\times 8\Rightarrow 8+R=\frac{16}{0.4}=40\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,R=32\,\Omega \]


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