A potentiometer wire has length 4 m and resistance 8 $\Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1mV per cm on the wire is

Given, length of potentiometer wire, L = 4 m = 400 cm
Resistance of wire, $\mathrm{R}=8\mathrm{\Omega }$
Accumulator voltage = V = 2 V
Let a resistance r be connected in series with the wire.
Effective resistance $=(8+r)\mathrm{\Omega }$
Let current flow be I.
From Ohm’s Law, V = IR,
where V is voltage
I is current
R is resistance.
(2 V) = I x (8 + r)
Current $I=\left(\frac{2}{8+r}\right)A$
Potential drop across the wire will be
$\begin{array}{l}{V}_{\text{wire }}=I\times R_{\text{wire }}\\ =\left(\frac{2}{8+r}\right)\times 8\mathrm{V}\\ =\frac{16}{8+r}V\end{array}$
Potential gradient is $\frac{V_{\text{wire }}}{L}$
$=\left(\frac{16}{8+r}\right)\times \frac{1}{400}{\mathrm{vcm}}^{-1}$
Given, potential gradient = 1 mV cm^-1
$\begin{array}{l}\Rightarrow \frac{16}{8+r}\times \frac{1}{400}={10}^{-3}\mathrm{v}{\mathrm{cm}}^{-1}\\ \Rightarrow r=32\mathrm{\Omega }\end{array}$