A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire, is $$k\,volt/cm$$   and the ammeter, present in the circuit, reads $$1.0\,A$$  when two way key is switched off. The balance points, when the key between the terminals (a) 1 and 2 (b) 1 and 3, is plugged in, are found to be at lengths $${l_1}\,cm$$  and $${l_2}\,cm$$  respectively. The magnitudes, of the resistors $$R$$ and $$X$$ in ohm, are then equal, respectively to

Solution :
The balancing length for $$R$$ (when 1, 2 are connected) be $${l_1}$$ and balancing length for $$R + X$$  (when 1, 3 is connected) is $${l_2}.$$
$$\eqalign{ & {\text{Then,}}\,\,iR = k{l_1}\,{\text{and}}\,i\left( {R + X} \right) = k{l_2} \cr & {\text{Given,}}\,\,i = 1A \cr & \therefore R = k{l_1}\,.......\left( {\text{i}} \right) \cr & R + X = k{l_2}\,.......\left( {{\text{ii}}} \right) \cr & {\text{Also, subtracting Eq}}{\text{.}}\,\left( {\text{i}} \right){\text{ fom Eq}}{\text{. }}\left( {{\text{ii}}} \right){\text{, we get}} \cr & X = k\left( {{l_2} - {l_1}} \right) \cr} $$