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JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1
In a potentiometer experiment, when three cells A,B\mathrm{A, B} and C\mathrm{C} care connected in series the balancing length is found to be 740  cm\mathrm{740}\,\,\mathrm{cm}. if A\mathrm{A} and B\mathrm{B} are connected in series balancing length is 540  cm\mathrm{540}\,\,\mathrm{cm}. then the emf of EA,  EB  and  EC{E_A},\,\,{E_B}\,\,and\,\,{E_C} are respectively (in volts)
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Correct option: A
1,1.2  and  1.5\begin{array}{llllllllllllllllllll} {\mathrm{1,1}\mathrm{.2}\,\,\mathrm{and}\,\,\mathrm{1}\mathrm{.5}} \end{array}
Option A (1, 1.2 and 1.5) is correct because the emf values are proportional to the balancing lengths, allowing us to solve for the individual emfs.

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