Explain this question

Dear Student,
Meter bridge is based on the principle of wheatstone bridge. In this bridge one lateral pair of resistances is replaced by a straight wire of length 1m of uniform cross section. And other pair consist of one known and one unknown resistance.
At the balance point:
$\frac{R}{X}=\frac{\sigma l}{\sigma (100-l)}, where \sigma = resis\tan ce per unit of the metre bridge wire.$
So, if the resistances R and X are doubled, then $\frac{R}{X}$ will not change and hence the balance point will not change. But if A and X are interchanged then the ratio $\frac{R}{X}$ becomes $\frac{X}{R}$ so the balancing point will now be achieved at AJ' = 100 - AJ = 100 - l = 100 - 40 = 60cm.

Even if u interchange the battery and galvanometer, the balance point will not be affected.

Regards