Q 12-03-208JEE MainJEE Main 2021 (25 Jul, Shift 2)Medium
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:
Answer: (A) $\frac{5}{3}$
With the galvanometer at point (1), only cell $\varepsilon_1$ is in the galvanometer loop: $\varepsilon_1 \propto 250$.
At point (2), both cells (aiding each other) are in the loop: $\varepsilon_1 + \varepsilon_2 \propto 400$, so $\varepsilon_2 \propto 150$.
$$\frac{\varepsilon_1}{\varepsilon_2} = \frac{250}{150} = \frac{5}{3}$$
Solution by Sreeraj P, M.Sc Physics