In an experiment with a potentiometer,$V_B = 10 \, V$ and the variable resistance is adjusted to $R = 50 \, \Omega$ (figure). $A$ student wanting to measure the voltage $E_1$ of a battery (approx. $8 \, V$) finds no null point. He then diminishes $R$ to $10 \, \Omega$ and is able to locate the null point on the last $(4^{th})$ segment of the potentiometer wire. Find the resistance of the potentiometer wire and the potential drop per unit length.

(D) Let the resistance of the potentiometer wire be $R'$ and its total length be $L = 4 \, m$ (assuming $4$ segments of $1 \, m$ each).
$1$. When $R = 50 \, \Omega$,the potential drop across the wire is $V_{wire} = \frac{10 \times R'}{50 + R'}$. Since no null point is found for $E_1 \approx 8 \, V$,the potential drop across the entire wire must be less than $E_1$.
$\frac{10 R'}{50 + R'} < 8 \Rightarrow 10 R' < 400 + 8 R' \Rightarrow 2 R' < 400 \Rightarrow R' < 200 \, \Omega$.
$2$. When $R = 10 \, \Omega$,the null point is on the $4^{th}$ segment,meaning the balance length $l$ is between $3 \, m$ and $4 \, m$. The potential drop across the wire is $V'_{wire} = \frac{10 \times R'}{10 + R'}$.
The condition for the null point to be on the $4^{th}$ segment is that the potential drop across $3 \, m$ is less than $8 \, V$ and the potential drop across $4 \, m$ is greater than $8 \, V$.
$\frac{3}{4} V'_{wire} < 8 < V'_{wire} \Rightarrow \frac{3}{4} \left( \frac{10 R'}{10 + R'} \right) < 8 < \frac{10 R'}{10 + R'}$.
From $8 < \frac{10 R'}{10 + R'}$,we get $80 + 8 R' < 10 R' \Rightarrow 2 R' > 80 \Rightarrow R' > 40 \, \Omega$.
From $\frac{7.5 R'}{10 + R'} < 8$,we get $7.5 R' < 80 + 8 R' \Rightarrow -0.5 R' < 80$ (always true for positive $R'$).
Combining with the first condition $R' < 200 \, \Omega$,the resistance $R'$ is in the range $40 \, \Omega < R' < 200 \, \Omega$.
The potential gradient $\phi = \frac{V'_{wire}}{4} = \frac{10 R'}{4(10 + R')} \, V/m$.