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 the potentiometer wire.
Effective resistance of potentiometer and variable resistor `(R=50Omega)` is given by `=50Omega+R'` Effective voltage applied across potentiometer=10V.
the current through the main circuit,
`I=(V)/(50Omega+R)=(10)/(50Omega+R)`
Potential difference across wire of potentiometer
`IR'=(10R')/(50Omega+R)`
Since with `50Omega` resistor, null point is not obtained it's possible only when
`(10xxR')/(50+R)=(10)/(50Omega+R)`
Potential difference across wire of potentiometer,
`IR'=(10R')/(50Omega+R)`
Since with `50Omega` resistor, null point is not obtained it's possible only when
`(10xxR')/(50+R) lt 8`
`implies10R lt 400+8R'`
`2R' lt 400` or `R' lt 200Omega`
Similarly with `10Omega` resistor, null point is obtained its possible only when
`(10xxR')/(10+R') gt 8`
`implies2R' gt 80`
`R' gt 40`
`(10xx(3)/(4)R')/(10+R') lt 8`
`implies7.5R' lt 80 + 8R'`
`R'gt160`
`implies160 lt R' lt 200`
Any R' between `160Omega and 200Omega` will achieve.
Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop this imply that potential gradient
`kxx400cm gt 8V`
or `kxx4m gt 8V`
Similarly, potential drop across 300 cm wire `lt 8V`
`kxx100cm lt 8V`
or `kxx103mlt 8V`
`k lt 2(2)/(3)V//m`
Thus, `2(2)/(3)V//m gt k gt 2 V//m`