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NEET NEET SOLVED PAPER 2014

  • question_answer
    A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of (i) infinity, (ii) \[9.5\Omega \] the 'balancing lengths', on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is [AIPMT 2014]

    A)  0.25 \[\Omega \]         

    B)  0.95\[\Omega \]

    C)  0.5 \[\Omega \]            

    D)  0.75 \[\Omega \]

    Correct Answer: C

    Solution :

    Given, e = 2V and \[l\]= 4m Potential drop per unit length \[\phi =\frac{e}{l}=\frac{2}{4}=0.5V/m\] For the first case, Þ \[e'=\phi {{l}_{1}}\,\,\,\,\,\,\,\,\,....(i)\] (\[e'\]® emf of the given cell) For the second case, \[V=\phi {{l}_{2}}\,\,\,\,\,\,\,\,\,....(ii)\] From Eqs. (i) and (ii), \[e'/V={{l}_{1}}/{{l}_{2}}\] \[e'=l(r+R)\] and \[V=lR\] for the second case So,  \[r=R\left( \frac{{{l}_{1}}}{{{l}_{2}}}-1 \right)=9.5\left( \frac{3}{2.85}-1 \right)=9.5(1.05-1)\] \[=9.5\times 0.05=0.475\simeq 0.5\Omega \]


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