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Two cells of E.M.F. E 1 and E 2 E 2 > E 1 are connected in series in the secondary circuit of a potentiometer experiment for determination of E.M.F. The balancing length is found to be 825 c m . Now when the terminals of cell E 1 are reversed, then the balancing length is found to be 225 c m . The ratio of E 1 and E 2 is

Options

  1. A2 : 3
  2. B4 : 7
  3. C7 : 4
  4. Dnone of these

Correct answer

B. 4 : 7

Step-by-step solution

E 2 + E 1 = x 825 E 2 – E 1 = x 225 Dividing above equations, we get E 2 + E 1 E 2 - E 1 = 11 3 Appling componendo and dividendo, we get E 1 E 2 = 4 7

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