BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    Two cells of emfs \[{{E}_{1}}\] and\[{{E}_{2}}({{E}_{1}}>{{E}_{2}})\] are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 on. The ratio\[\frac{{{E}_{1}}}{{{E}_{2}}}\]is

    A)  3 : 1                                      

    B)  1 : 3

    C)  2 : 3                      

    D)         3 : 2

    Correct Answer: D

    Solution :

    When potentiometer is connected between A and B, then it measures only \[{{E}_{1}}\] and when connected between A and C, then it measures \[{{E}_{1}}-{{E}_{2}}.\] \[\therefore \]  \[\frac{{{E}_{1}}}{{{E}_{1}}-{{E}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\]                 \[\frac{{{E}_{1}}-{{E}_{2}}}{{{E}_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\]                 \[\Rightarrow \]               \[1-\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{100}{300}\]                 \[\Rightarrow \]               \[\frac{{{E}_{2}}}{{{E}_{1}}}=1-\frac{1}{3}\]                 \[\Rightarrow \]               \[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{2}{3}\] \[\Rightarrow \]               \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{3}{2}\]


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