In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells $$\frac{\varepsilon_1}{\varepsilon_2}$$ is:

We need to find the ratio of the electromotive forces (emfs) of two cells, $$\frac{\varepsilon_1}{\varepsilon_2}$$, based on their balancing lengths in a potentiometer circuit.

1. Understand the Circuit States

From the problem , the balancing length is proportional to the total potential drop across the balanced part of the loop ($$V \propto l$$):

• Connection at Point (1):
  When the galvanometer is connected to terminal (1), only the first cell ($$\varepsilon_1$$) is included in the balancing loop. The corresponding balancing length is $$l_1 = 250\text{ cm}$$.

  $$\varepsilon_1 = \phi \cdot l_1 = 250\phi$$

  (where $$\phi$$ represents the potential gradient of the potentiometer wire)
• Connection at Point (2):
  When the connection shifts to terminal (2), both cells are connected in series aiding each other ($$\varepsilon_1 + \varepsilon_2$$) in the circuit loop. The new balancing length increases to $$l_2 = 400\text{ cm}$$.

  $$\varepsilon_1 + \varepsilon_2 = \phi \cdot l_2 = 400\phi$$

2. Calculate the Ratio $$\frac{\varepsilon_1}{\varepsilon_2}$$

Divide the second condition by the first condition to eliminate the potential gradient variable $$\phi$$:

$$\frac{\varepsilon_1 + \varepsilon_2}{\varepsilon_1} = \frac{400\phi}{250\phi}$$

Simplify the fraction on the right-hand side:

$$1 + \frac{\varepsilon_2}{\varepsilon_1} = \frac{8}{5}$$

Isolate the ratio $$\frac{\varepsilon_2}{\varepsilon_1}$$:

$$\frac{\varepsilon_2}{\varepsilon_1} = \frac{8}{5} - 1 = \frac{3}{5}$$

Invert the fraction to get the required ratio of $$\frac{\varepsilon_1}{\varepsilon_2}$$:

$$\frac{\varepsilon_1}{\varepsilon_2} = \frac{5}{3}$$

Conclusion

The ratio of the emfs of the two cells is $$\frac{5}{3}$$, which matches the highlighted choice A.