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

Answer

B

Explanation

Suppose two cells have emfs `epsilon_(1) and epsilon_(2)` (also `epsilon_(1)gtepsilon_(2)`). Potential difference per unit length of the potentiometer wire = k (say)
When `epsilon_(1) and epsilon_(2)` are in series and support each other then `epsilon_(1)+epsilon_(2)=50xxk" "...(i)`
When `epsilon_(1)andepsilon_(2)` are in opposite direction
`epsilon_(1)-epsilon_(1)=10xxk`
On adding eqn. (i) eqn. (ii)
`2epsilon_(1)=60kimplies epsilon_(1)=30k` and
`epsilon_(2)=50k-30k=20k`
`therefore (epsilon_(1))/(epsilon_(2))=(30k)/(20k)=(3)/(2)`