In an experiment with a potentiometer, `V_(B)=10V`. R is adjusted to the `50Omega` (figure). A student wanting to measure voltage `E_(1)` to a battery (approx. 8V) finds no null point possible. He then diminishes R to `10Omega` and is able to locate the null point ont he last (4th) segment of the potentiometer. find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Explanation

Let `R `be the resistance of potentiometer wire , when `R = 50 Omega` in series with potentiometer with circuit current through potentiometer
wire is `I = (V_(B))/(R + R') = (10)/(50 + R)`
potential driop across potentiometer wire `= IR' = (10)/((50 + R)) xx R'`
As no moll is found on petntiometer wire while measurting voltage `E_(1) ( = 8V)` so
`(10R')/((50 +R') lt 8 `or `10R' lt 400 + 8R'` or `R" lt 2000 Omega `
when `R = 10 Omega ` in series of potentiometer wire , then `(10R')/(10 + R') gt 8 `or` 2R' gt 80` or `R' gt 40 Omega`
If so deffection in galvanometer in on the third segment of the potentiometer wire , then
`(10 xx (3//4)R')/(10 + R') lt 8 `or `7.5 R' lt 80 + 8R'` or `- 0.5 R' lt 80 or R' gt 80//0.5 `or `R' gt 160 Omega`
it means less any value of resistance of potentimeter wire between `160 Omega and 200 Omega ` will help ti it achieving the no deflection on fourth wire of potentimeter circuit
Here petential across `4m` of wire `gt 8V` drop across `3m` of wire `lt 8V` if `phi` is the potential then `4 phi gt 8V`...(i) and `3 phi gt 8V`...(ii)
From (i) `phi gt 8V//4 m = 2V//m`
from (ii) `3 phi lt 8V` or `phi = (8)/(3) V//m` or `phi lt 2 (2)/(3)V//m`