A potentiometer wire of length `10 m` and resistance `30 Omega` is connected in series with a battery of emf `2.5 V`, internal resistance `5 Omega` and an external resistance `R`. If the fall of potential along the potentiometer wire is `50 mu V mm^(-1)`, then the value of `R` is found to be `23 n Omega`. What is `n`?

Answer

`(5)`

Explanation

Potential gradient, `k = (I R_(P))/(L) = (E R_(P))/((R + R_(P) + r)L)`
(where `R_(P)` is resistance of the wire)
or `50 xx 10^(-3) = (2.5 xx 30)/((R + 30 + 5) xx 10)` or R = 115 Omega`