Question:

In a potentiometer experiment, two cells of emf E1E_1 and E2E_2 are used in series in the secondary circuit. The balancing length is found to be 58 cm. If the polarity of E2E_2 is reversed, the balancing length becomes 29 cm. The ratio E1/E2E_1/E_2 of the emfs of two cells is :

Updated On: May 10, 2024
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The Correct Option is B

Solution and Explanation

Using, V l\propto \, l we get E1+E2E1E2=l1+l2l1l2=58+295829=31\frac{E_1 + E_2}{E_1 - E_2} = \frac{l_1 + l_2}{l_1 - l_2} = \frac{58 + 29}{58 - 29} = \frac{3}{1}
i.e.i.e. E1E2=21\frac{E_1}{E_2} = \frac{2}{1} (E1+E2+E1E2E1+E2(E1E2)=3+131)\left( \, \because \, \frac{E_1 + E_2 + E_1 - E_2 }{E_1 + E_2-(E_1-E_2)} = \frac{3 + 1}{3 - 1} \right)
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