Q. Two cells of emf E_{1} and E_{2} (E_{1} > E_{2}) are connected as shown in figure.

When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300cm. On connecting the same potentiometer between A and C, the balancing length is 100cm. The ratio E_{1}/E_{2} is

(a) 3 : 1

(b) 1 : 3

(c) 2 : 3

(d) 3 : 2

Ans: (d)

$ \displaystyle \frac{E_1 -E_2}{E_1} = \frac{100}{300} $

$ \displaystyle 1-\frac{E_2}{E_1} =\frac{1}{3}$

$ \displaystyle \frac{E_2}{E_1}=\frac{2}{3} $

$ \displaystyle \frac{E_1}{E_2}=\frac{3}{2}$