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+### Statement
+
+$4.5.12.$ [Insert the problem statement]
+
+### Solution
+
+The sum of the forces acting on the volume of liquid highlighted in the figure is zero,so let's write force balance to the $x$ axis:
+
+\begin{equation}
+\sigma L-\sigma L\sin\theta=P_{avg}Lh
+\end{equation}
+
+where $P_{avg}$ is an average pressure of the liquid.As pressure linearly increasing with depth we get:
+
+
+\begin{equation}
+P_{avg}=\frac{1}{2}\rho gh
+\end{equation}
+
+
+
+Plugging eq(2) into eq(1) we get rise height of the liquid:
+
+
+\begin{equation}
+h=\sqrt{\frac{2\sigma(1-\sin\theta)}{\rho g}}
+\end{equation}
+
+
+In the case of complete wetting, the contact angle is zero,so:
+
+
+\begin{equation}
+h=\sqrt{\frac{2\sigma}{\rho g}}
+\end{equation}
+
+
+For the water $\sigma=0.073 \frac{N}{m}$,$\rho=1000\frac{kg}{m^3}$ and $g=9.8\frac{m}{s^2}$,so:
+
+
+
+\begin{equation}
+h \approx 3.9mm
+\end{equation}
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]