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JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1
Two cells A  and  BA\,\,and\,\,B are connected in the secondary circuit of a potentiometer one at a time and the balancing length are respectively 400cm and 440cm\mathrm{400 cm }\,\mathrm{and}\,\mathrm{ 440 cm}. The emf of the cell AA is 1.08  V\mathrm{1}\mathrm{.08}\,\,\mathrm{V}. The emf of the second cell BB is volt is
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Correct option: B
1.188
Option B (1.188) is correct because the emf of cell B is calculated using the potentiometer principle, where the ratio of emfs is equal to the ratio of balancing lengths.

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