(ignore viscosity of air)
$\sqrt{2 gh }=\frac{2}{9} \frac{ r ^{2} g }{\eta}\left(\rho_{\ell}-\rho\right)$
$\Rightarrow h =\frac{2}{81} \frac{ r ^{4} g \left(\rho_{\ell}-\rho\right)^{2}}{\eta^{2}}$
$\Rightarrow h \propto r ^{4}$





$A$ and $B$ are on the axis of tube $............\,Pa$