$8*.$ Consider a free liquid surface meeting a plane vertical wall. If the liquid wets the wall, that is$,$ has a

contact angle the liquid level will rise as the wall is approached, meeting the wall at its contact

angle $.$ The liquid is at rest; thus, the pressure inside the liquid varies linearly with depth below the

free surface. The pressure just above the free surface is the atmospheric pressure. As the horizontal

coordinate, $x,$ measured from the wall, approaches infinity, the free surface becomes horizontal.

Determine the height, $h_0,$ to which the liquid climbs at the vertical wall. Ans.: $h_0=\sqrt[\phantom{2}]{?}$

Hint: Show that the shape, $z=f(x),$ of the interface is described by the equation,

$\frac{f^{\text{'}\text{'}}}{[1+(f^{\text{'}}{)}^2{]}^{\frac{3}{2}}}=\frac{{}_{l}g}{}f$

Multiply both sides of this equation by $f^{\text{'}}$ and integrate.