The resistance per unit length of potentiometer wire of uniform cross section is  $\left(\frac{3R_{0}x}{{\mathcal{l}}^2}\right),$  where x is measured from end A. Balance length is also measured from A. Select correct option(s):

 

When S is open:

$$\begin{gathered} R_{AB}=\underset{0}{\overset{\mathcal{l}}{\int }}\frac{3R_{0}xdx}{{\mathcal{l}}^2}=\frac{3R_0}{2} \\ {I}_0=\frac{E_0}{R_0+\frac{3R_0}{2}}=\frac{2E_0}{5R_0} \\ \frac{E_0}{2}=I_0\times R_{AJ}=\frac{2E_0}{5R_0}\times \frac{3R_0}{{\mathcal{l}}^2}\underset{0}{\overset{\mathcal{l}}{\int }}xdx \\ x=\mathcal{l}\sqrt{\frac{5}{6}} \end{gathered}$$

When S is closed:
Current in secondary circuit
$i=\frac{\frac{E_0}{2}}{R_0+\frac{R_0}{4}}=\frac{2E_0}{5R_0}$ 
Terminal voltage of battery $=[{\epsilon }_0-\frac{2{\epsilon }_0}{5}]=\frac{3{\epsilon }_0}{5}$
$\frac{3{\epsilon }_0}{5}=I_0\times R_{AJ}=\frac{2{\epsilon }_0}{5R_0}\times \frac{3R_0}{{\mathcal{l}}^2}\times \frac{x^2}{2}\Rightarrow x=\mathcal{l}$