Q.When two cells of emfs \varepsilon_1 and \varepsilon_2 are connected in series so as to assist each other, their balancing length on a potentiometer is found to be 2.7 m. When the cells are connected in series so as to oppose each other, the balancing length is found to be 0.3 m. Compare the emfs of the two cells.

[!TLDR] Using the sum-and-difference balancing lengths 2.7 m and 0.3 m with componendo-dividendo, $\varepsilon_1/\varepsilon_2 = (2.7+0.3)/(2.7-0.3) …

When the two cells assist each other (sum mode), the balancing length is $l_1=2.7$ m, so $\varepsilon_1+\varepsilon_2 = k\,l_1$; when they oppose (difference mode), the balancing length is $l_2=0.3$ m, so $\varepsilon_1-\varepsilon_2 = k\,l_2$. Using the sum-and-difference potentiometer result (Eq. 9.11), …

Apply the componendo-dividendo emf-ratio formula (Eq. 9.11) directly to …

Using the plain ratio $l_1/l_2$ (as in the individual-cells Method I) instead of the sum-difference formula $(l_1+l_2)/(l_1-l_2)$ …