Two cells of emfs ε1 and ε2 (ε1 > ε2) are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio $\frac{ε_1}{ε_2}$is :
Answer & explanation
Correct answer: option 4
When potentiometer is connected between A and B, then it measures only ε1 and when connected between A and C, then it measures ε1 - ε2.
∴ $\frac{\varepsilon_1}{\varepsilon_1-\varepsilon_2}=\frac{l_1}{l_2}$, $\frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1}=\frac{l_2}{l_1}$
or $1-\frac{\varepsilon_2}{\varepsilon_1}=\frac{100}{300}$ or $\frac{\varepsilon_2}{\varepsilon_1}=1-\frac{1}{3}$ or $\frac{\varepsilon_2}{\varepsilon_1}=\frac{2}{3}$ or $\frac{\varepsilon_1}{\varepsilon_2}=\frac{3}{2}$