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(C) The potential drop per unit length of the potentiometer wire is $k = \frac{V_{AB}}{L}$,where $L = 10\, m = 1000\, cm$.For the first battery $E_{1}

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$3$

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(C) The potential drop per unit length of the potentiometer wire is $k = \frac{V_{AB}}{L}$,where $L = 10\, m = 1000\, cm$.For the first battery $E_{1}$,the balancing length $l_{1}$ is measured from point $A$. The wire consists of $10$ segments of $1\, m$ each. $J_{1}$ is at $20\, cm$ on the second segment,so $l_{1} = 100 + 20 = 120\, cm$.For the second battery $E_{2}$,the balancing length $l_{2}$ is measured from point $A$. $J_{2}$ is at $60\, cm$ on the eighth segment,so $l_{2} = 700 + 60 = 760\, cm$.Using the principle of the potentiometer,$E_{1} = k l_{1}$ and $E_{2} = k l_{2}$.Therefore,$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} = \frac{120}{760} = \frac{12}{76} = \frac{3}{19}$.Given $\frac{E_{1}}{E_{2}} = \frac{a}{b}$,we have $a = 3$ and $b = 19$.

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