Q.Why should not the jockey be slided along the potentiometer wire?

The two main sources of error in a meter bridge are end resistance at the copper strips and non-uniformity of the bridge wire; both are corrected by calibration/measurement technique rather than by circuit change.

(1) End resistance (end correction): The resistance of the copper strips at the two ends of the meter bridge wire, and of the connections/jockey contact, adds an unknown extra resistance not accounted for by the measured length alone, causing a small error in the balance-point reading.

Minimisation: Determine the end correction experimentally (by interchanging known resistances in the two gaps and comparing the calculated vs. actual balance lengths), and add this correction to every observed balancing length; alternatively, keep the unknown and known resistances comparable so the balance point falls near the middle of the wire, where end effects influence the percentage error least.

(2) Non-uniform cross-section / material of the wire: If the meter-bridge wire does not have a strictly uniform cross-sectional area or uniform resistivity along its length, the assumption $R \propto l$ (resistance proportional to length) breaks down, introducing error.

Minimisation: Use a wire of uniform diameter (checked by a screw gauge at several points) made of a single, homogeneous material (e.g. constantan/manganin), and take the mean of several observations obtained by interchanging the positions of the known and unknown resistances.

Meter bridge balance condition: $\dfrac{X}{R} = \dfrac{l}{100-l}$.

In a meter bridge, with unknown resistance $X$ in the left gap and known resistance $R$ in the right gap, if the null (balance) point is at length $l$ (cm) from the left end, the balance (Wheatstone bridge) condition gives

$\frac{X}{R} = \frac{l}{100-l}$

Given $R = 50\ \Omega$, $l = 40\ \text{cm}$:

$X = R\times\frac{l}{100-l} = 50\times\frac{40}{60} = 50\times\frac{2}{3} = 33.33\ \Omega$