Two cells of emf `epsilon_(1)` and `epsilon_(2)(epsilon_(2)lt epsilon_(1))` are joined as shown in figure :

When a potentiometer is connected between X and Y it balances for 300 cm length against `epsilon_(1)`. On connecting the same potentiometer between X and Z it balances for 100 cm length against `epsilon_(1)` and `epsilon_(2)`. Then the ratio `(epsilon_(2))/(epsilon_(1))` is :

Options

A. `(1)/(3)`
B. `(3)/(4)`
C. `(1)/(4)`
D. `(2)/(3)`

Answer

D

Explanation

`epsilon_(1)=300 alpha`....(i)
`-epsilon_(2)+epsilon_(1)=100 alpha` ....(ii)
where `alpha` is the potential gradient
`:. (epsilon_(2))/(epsilon_(1))=(2)/(3)`