If you are provided with two tubes ($A$ and $B$),where one is narrow and the other is relatively wider,and if both are immersed in a beaker containing water as shown in the figure. Why does $B$ show higher water rise than $A$?

(N/A) The rise of water in a capillary tube is governed by the principle of capillarity.
According to the formula for capillary rise,$h = \frac{2T \cos \theta}{r \rho g}$,where $h$ is the height of the liquid column,$T$ is the surface tension,$\theta$ is the contact angle,$r$ is the radius of the tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
From this relationship,it is clear that the height of the water column $(h)$ is inversely proportional to the radius of the tube $(r)$,i.e.,$h \propto \frac{1}{r}$.
Since tube $B$ is narrower (has a smaller radius) than tube $A$,the water rises to a greater height in tube $B$ compared to tube $A$ due to stronger capillary action.