Consider $R^{\prime}$ be the resistance of the potentiometer wire and given variable resistor $(R=50 \Omega)$
Then the effective resistance of potentiometer
$$
=50 \Omega+\mathrm{R}^{\prime}
$$
Effective voltage applied across potentiometer $=10 \mathrm{~V}$ The current through the main circuit,
$$
\begin{aligned}
\mathrm{I} &=\frac{\mathrm{V}}{50 \Omega+\mathrm{R}^{\prime}} \\
&=\frac{10}{50 \Omega+\mathrm{R}^{\prime}}
\end{aligned}
$$
Potential difference across wire of potentiometer,
$$
I^{\prime}=\frac{10 R^{\prime}}{50 \Omega+R^{\prime}}
$$
As, with $50 \Omega$ resistor, null point can not obtained by 8 volt it's possible only when
$$
\frac{10 \times \mathrm{R}^{\prime}}{50+\mathrm{R}^{\prime}} < 8
$$
So, $\quad 10 \mathrm{R}^{\prime} < 400+8 \mathrm{R}^{\prime}$
$$
2 \mathrm{R}^{\prime} < 400 \text { or } \mathrm{R}^{\prime} < 200 \Omega
$$
Similarly with $10 \Omega$ resistor, null point is obtained at $V^{\prime}>8$ its possible only when
$$
\frac{10 \times R^{\prime}}{10+R^{\prime}}>8
$$
So, $\quad 2 R^{\prime}>80, \quad R^{\prime}>40$
As the null point is obtained on 4th segment or at $\left(\frac{3}{4}\right)$ of total length so at $\left(\frac{3}{4} R^{\prime}\right)$.
$$
\begin{aligned}
\frac{10 \times \frac{3}{4} R^{\prime}}{10+R^{\prime}} & < 8 \\
7.5 R^{\prime} & < 80+8 R^{\prime} \\
R^{\prime} &>160 \\
160 < R^{\prime} < 200
\end{aligned}
$$
So,
(At balance point)
Any R' between $160 \Omega$ and $200 \Omega$ will achieve.
Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop across $400 \mathrm{~cm}$ of wire $>8 \mathrm{~V}$.
This imply that potential gradient
or
$$
\begin{aligned}
\mathrm{k} \times 400 \mathrm{~cm} &>8 \mathrm{~V} \quad \text { (at balance point) } \\
\mathrm{k} \times 4 \mathrm{~m} &>8 \mathrm{~V} \\
\mathrm{k}>2 \mathrm{~V} / \mathrm{m}
\end{aligned}
$$
Similarly, potential drop across $300 \mathrm{~cm}$ wire $ < 8 \mathrm{~V}$
$$
\mathrm{k} \times 300 \mathrm{~cm} < 8 \mathrm{~V}
$$
$$
\begin{aligned}
\mathrm{k} \times 3 \mathrm{~m} & < 8 \mathrm{~V} \\
\mathrm{k} & < \frac{8}{3} \mathrm{~V} / \mathrm{m}
\end{aligned}
$$
Hence, $\frac{8}{3} V / m>k>2 V / m$