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Two cells of emf epsilon(1) and epsilon(...

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_(0)`. On conntecting 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 :

A

`(1)/(3)`

B

`(3)/(4)`

C

`(1)/(4)`

D

`(2)/(3)`

Text Solution

Verified by Experts

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