A potentiometer is used for the comparison of e.m.f. of two cells `E_(1)` and `E_(2)`. For cell `E_(1)` the no deflection point os obtained at `20 cm` and for `E_(2)` the no deflection point is obtained at `30 cm`. The ratio of their e.m.f.'s will be

Options

A. `2//3`
B. `1//2`
C. 1
D. 2

Answer

A

Explanation

(a) Ratio will be equal to the ratio of no deflection lengths. Hence, `(E_(1))/(E_(2)) = (l_(1))/(l_(2)) = (2)/(3)`