Two cells of $e.m.f.$ $E_1$ and $E_2$ are joined in series and the balancing length of the potentiometer wire is $625 \, cm$. If the terminals of $E_1$ are reversed,the balancing length obtained is $125 \, cm$. Given $E_2 > E_1$,the ratio $E_1 : E_2$ will be

  • A
    $2 : 3$
  • B
    $5 : 1$
  • C
    $3 : 2$
  • D
    $1 : 5$

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