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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