Question

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

Answer 1

Correct Answer - c
Let `epsilon_(1)` and `epsilon_(2)` be the emits of two cells and `K` be the potential gradient of potentiometer wire
case (i), `epsilon_(1) + epsilon_(2) = K xx 50`….(i)
case (ii), `epsilon_(1) - epsilon_(2) = K xx 10`….(ii)
Dividing (i) and (ii), we get
`(epsilon_(1) + epsilon_(2))/(epsilon_(1) - epsilon_(2)) = (50)/(10) = 5`
or `epsilon_(1) + epsilon_(2)= 5epsilon_(1) - 5 or epsilon_(2) or 4epsilon_(1) = 6 or epsilon_(2)`
or `(epsilon_(1))/(epsilon_(2)) = (6)/(4) = (3)/(2)`