Potentiometer

For a potentiometer, $E \propto \ell$: $\frac{E_1}{E_2} = \frac{\ell_1}{\ell_2}$ $\frac{1.5}{2.5} = \frac{36}{\ell_2} \Rightarrow \ell_2 = 36 \times \frac{5}{3} = 60$ cm

Potential gradient $x = 1$ mV/cm $= \frac{10^{-3}}{10^{-2}}$ V/m $= 0.1$ V/m
$x = \frac{E}{R + R'}\times\frac{R}{L}$
$0.1 = \frac{2}{8 + R'}\times\frac{8}{4}$
$8 + R' = 40 \Rightarrow R' = 32\ \Omega$

Current in the wire: $i = \frac{E_0}{r + r_1}$
Potential gradient: $x = \frac{ir}{L} = \frac{E_0r}{(r + r_1)L}$
$E = xl = \frac{E_0r}{(r + r_1)}\cdot\frac{l}{L}$

Internal resistance $r = \left(\frac{E - V}{V}\right)R = \left(\frac{l_1 - l_2}{l_2}\right)R$
$r = \left(\frac{3 - 2.85}{2.85}\right)(9.5) = 0.5\ \Omega$

The current through R and X in the secondary circuit is I = 1.0 A.
With the key between 1 and 2, the potential difference across R alone is balanced: $V_R = IR = kl_1$
Since I = 1 A, $R = kl_1$ ohm
With the key between 1 and 3, the potential difference across R and X in series is balanced: $I(R + X) = kl_2$
$R + X = kl_2$, so $X = kl_2 - kl_1$
$X = k(l_2 - l_1)$ ohm
So R and X are $kl_1$ and $k(l_2 - l_1)$ respectively.

In a potentiometer the potential difference is proportional to the balancing length.
On open circuit the cell balances at the longer length, so emf $E \propto l_1$ with $l_1 = 110$ cm.
When the cell is shunted by $R = 10\,\Omega$, its terminal voltage $V = \dfrac{E}{R + r}\cdot R \propto l_2$ with $l_2 = 100$ cm.
$\dfrac{E}{V} = \dfrac{R + r}{R} = \dfrac{l_1}{l_2}$
$\Rightarrow r = \left(\dfrac{l_1}{l_2} - 1\right)R = \left(\dfrac{l_1 - l_2}{l_2}\right)R$
$r = \dfrac{110 - 100}{100} \times 10$
$r = 1\,\Omega$
(The larger balancing length always belongs to the open-circuit emf, since $E \gt V$.)

Potential gradient $x = \dfrac{V}{l} = \dfrac{IR}{l} = \dfrac{I\rho l}{Al} = \dfrac{I\rho}{A}$
$x = \dfrac{0.1 \times 10^{-7}}{10^{-6}} = 10^{-2}$ V/m

A potentiometer measures voltage by a null (balance) method. At the balance point, no current flows through the galvanometer, i.e. no current is drawn from the cell being measured.
So it measures the potential difference as in an open circuit, i.e. the true emf of a cell, without the drop due to internal resistance.
A voltmeter draws some current, so it measures a slightly lower terminal voltage.
So the potentiometer is best because it measures p.d. as in an open circuit.