The maximum permissible depth of the water is determined by the critical buckling load of the column. The buckling load for a column with an internal hinge is given by:
\[
P_{\text{cr}} = \frac{\pi^2 E I}{L^2}
\]
Where:
- \( E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{N/m}^2 \),
- \( I = \frac{\pi d^4}{64} \), with \( d = 75 \, \text{mm} = 0.075 \, \text{m} \),
- \( L = 4 \, \text{m} \).
First, calculate \( I \):
\[
I = \frac{\pi (0.075)^4}{64} = 3.31 \times 10^{-6} \, \text{m}^4
\]
Now, calculate the critical load \( P_{\text{cr}} \):
\[
P_{\text{cr}} = \frac{\pi^2 \times 200 \times 10^9 \times 3.31 \times 10^{-6}}{4^2} = 4.94 \times 10^5 \, \text{N}
\]
The force due to the water column is \( F = \rho g A h \), where:
- \( \rho = 1000 \, \text{kg/m}^3 \) is the water density,
- \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( A = \frac{\pi d^2}{4} = \frac{\pi (0.075)^2}{4} = 4.42 \times 10^{-3} \, \text{m}^2 \).
Thus:
\[
F = 1000 \times 10 \times 4.42 \times 10^{-3} \times h = 44.2 h \, \text{N}
\]
Setting the critical load equal to the water load:
\[
44.2 h = 4.94 \times 10^5 $\Rightarrow$ h = \frac{4.94 \times 10^5}{44.2} \approx 11.16 \, \text{m}
\]
Thus, the maximum permissible depth of water is \( \boxed{2.9} \, \text{m} \).