To determine the internal resistance of a cell by using a potentiometer, the null point is at \(1 \mathrm{~m}\) when the cell is shunted by \(3 \Omega\) resistance and at a length \(1.5 \mathrm{~m}\) when cell is shunted by \(6 \Omega\) resistance. The internal resistance of the cell is.

Consider the following diagram:

The potential drop across the external shunt resistance \(R\) is equal to the potential drop across the battery of emf \(E\) and internal resistance \(r\).
\(\begin{aligned} & \Delta V=E-i r=i R \\ & \Rightarrow i=\frac{E}{(R+r)} \\ & \therefore \Delta V=i R=\frac{E R}{(R+r)}\end{aligned}\)
\(\because \Delta V \propto L\), i.e., the balance length is directly proportional to the potential drop.
Therefore,
\(\frac{E R}{(R+r)} \propto L\)
In case of internal resistance measurement by potentiometer, considering two balance lengths \(\left(L_1, L_2\right)\) corresponding to shunt resistors \(\left(R_1, R_2\right)\) respectively,
\(\Rightarrow \frac{V_1}{V_2}=\frac{L_1}{L_2}=\frac{\left\{E R_1 /\left(R_1+r\right)\right\}}{\left\{E R_2 /\left(R_2+r\right)\right\}}=\frac{R_1\left(R_2+r\right)}{R_2\left(R_1+r\right)}\).
Here, \(L_1=1 \mathrm{~m}, L_2=1.5 \mathrm{~m},
\)
\(\Rightarrow \frac{1}{1.5}=\frac{3(6+r)}{6(3+r)}\) \(
6+2r =9+1.5r
\)
\(\Rightarrow r=6 \Omega\)