why do we exchange the positions of resistance box and unknown resistance in meter bridge experiment??

Actually if the meter bridge is perfect, we do not need to interchange the known (Y) and unknown (X) resistances.

Suppose we get null point at a distance L₁ from end A.

Then,

X / Y = L₁ / (100 - L₁)

[eq1]

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However a meter bridge may have extra resistances at the ends A and B.

Usually, there is a little extra length of wire at each length which goes beyond the scale, and some soldering at either end which contribute some extra resistance. These are known as end errors, and can be corrected by adding end corrections α and β for each end to the lengths of the two parts of the wire. Equation 1 becomes

--

X/Y = (L₁ + α)/ (100 - L₁ + β)

[eq2]

If we interchange X and Y, we get a new null point at a distance L₂, instead of L₁

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The equation then becomes:

Y/X = L₂ + α/ (100 - L₂ + β)

[eq3]

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Solve eq2 and eq 3 for end corrections α and β

Let X/Y = r

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α = (L₁ - rL₂ )/ (r - 1)

β = [(rL1 - L2 ) / (r - 1)] - 100