Two cells of emf’s E₁ and E₂when placed in series produce null deflection at a distance of 204 cm in a potentiometer. When one cell is reversed they produce null deflection at 36 cm. If E₁=1.4V then E₂=

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0.098V

2.47V

980V

0.98V

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{E_{1}}{E_{2}} = \frac{l_{1} + l_{2}}{l_{1} - l_{2}}; \frac{1 .4}{E_{2}} = \frac{204 + 36}{204 - 36} \Rightarrow \frac{1 .4}{E_{2}} = 1 .42 \\ E_{2} = \frac{1 .4}{1 .42} = 0 .98 V \end{array}$

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