A potentiometer wire is `100 cm` long hand a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obatined at `50 cm` and `10 cm` from the positive end of the wire in the two cases. The ratio of emfs is:

Options

A. `5 : 4`
B. `3 : 4`
C. `3 : 2`
D. `5 : 1`

Answer

C

Explanation

Let `E_(1) and E_(2)` be the emf of the two cells. The first reading is corresponding to the sum of emf.s and the second is corresponding to the difference of the emf.s
`E_(1) + E_(2) =k xx (50)`
`E_(1) -E_(2) =k xx (10)`
Dividing the above two equations we get the following:
`(E_(1) + E_(2))/(E_(1) - E_(2)) = (5)/(1) rArr (E_(1))/(E_(2)) = (5+1)/(5-1) = (6)/(4) = (3)/(2)` Hence option (c ) is correct