Measuring resistance using potentiometer

[AdityaDev]

I don't think that would solve the problem. Because even then, the current in both would be different.

That is what we want. Now the potential difference across the resistors will be different.

 

[andrevdh]

Usually this experiment is done with the resistors connected in series to the cell, in which case the
current will be the same through both resistors.
For this particular setup I can only reduce it to two unknowns, the resistance X, and the internal resistance of the cell.

 

[Faris Shajahan]

That is what we want. Now the potential difference across the resistors will be different.

Yes, it will be different but still it would be impossible for us to calculate the value of $X$ as we don't have the value of $r$ given in the question.

Instead if they say there is a rheostat in series with ε in order to maintain constant current in both the cases, then we can calculate $X$ as
$X=R\frac{l_2}{l_1}$