In the given potentiometer circuit arrangement, the balancing length ${AC}$ is measured to be $250$ ${cm}$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram, the balancing length becomes $400\, {cm}$. The ratio of the emf of two cells, $\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -
JEE MAIN 2021, Diffcult
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$E_{1}=k l_{1} \ldots .$ (i)
$E_{1}+E_{2}=k l_{2} \ldots$ (ii)
$\frac{E_{1}}{E_{1}+E_{2}}=\frac{l_{1}}{l_{2}}=\frac{250}{400}=\frac{5}{8}$
$8 E_{1}=5 E_{1}+5 E_{2}$
$3 E_{1}=5 E_{2}$
$\frac{E_{1}}{E_{2}}=\frac{5}{3}$
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