If the potential gradient of two potentiometer wire is ${X_1}$ and ${X_2}$, the resistance per unit length is equal and the current flowing through them are ${I_1}$ and ${I_2}$. Then ${I_1}:{I_2}$ is:-
(A) ${X_1}:{X_2}$
(B) ${X_2}:{X_1}$
(C) ${X_1}^2:{X_2}^2$
(D) ${X_1}^3:{X_2}^3$

Hint: Potentiometer is a device used to compare the emfs of two cells.(or) to find the emf of a cell (or) to find the internal resistance of a cell (or) to measure potential difference.
Construction: A potentiometer consists of a uniform wire of length 10m arranged between A and B as 10 wires each of length 1m on a wooden board. The wire has specific resistance and low temperature coefficient of resistance. A meter scale is arranged parallel to the wires to measure the balancing length. The resistance of the total wire of the potentiometer is about 5Ω. A Jockey J can be moved on the wire. The balancing length is measured from the end which is connected to the positive terminal of the battery.
Principle: In null deflection position,
Unknown potential difference = Known potential difference.
When steady current passes through the uniform wire of the potentiometer, the potential difference across any part of the wire is directly proportional to the length of the wire.

Complete step by step answer:
1. Potential gradient for wire 1 is ${X_1} = {I_1}{R_1}$.
2. Potential gradient for wire 2 is ${X_2} = {I_2}{R_2}$
 But here, resistance per unit length is equal that ${R_1} = {R_2}$
 So, The ratios are:
$\Rightarrow$ $
  \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\dfrac{{{X_1}}}{{{R_1}}}}}{{\dfrac{{{X_2}}}{{{R_2}}}}} = \dfrac{{{X_1}}}{{{X_2}}} \\
  \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{X_1}}}{{{X_2}}} \\
 $

hence, correct option is (a)

Note: Potential Gradient is the decrease in potential per unit length. It is calculated as V / L, where V is the potential difference between two points and L is the distance between two points. The longer the wire the lesser the potential gradient and the greater the sensitivity of the potentiometer.