In the given potentiometer circuit arrangement, the balancing length $\mathrm{AC}$ is measured to be $250 \mathrm{~cm}$. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes $400 \mathrm{~cm}$. The ratio of the emf of two cells, $\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is :
Correct Option: 1
$\mathrm{E}_{1}=\mathrm{k} \ell_{1}$...(1)
$\mathrm{E}_{1}+\mathrm{E}_{2}=\mathrm{k} \ell_{2}$...(2)
$\frac{\mathrm{E}_{1}}{\mathrm{E}_{1}+\mathrm{E}_{2}}=\frac{\ell_{1}}{\ell_{2}}=\frac{250}{400}=\frac{5}{8}$
$8 \mathrm{E}_{1}=5 \mathrm{E}_{1}+5 \mathrm{E}_{2}$
$3 \mathrm{E}_{1}=5 \mathrm{E}_{2}$
$\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{5}{3}$
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