Water is being pumped into an inverted conical tank at a rate of 7.8 cubic meters per min.
The tank has height 11 meters and the diameter at the top is 4 meters.
Find the rate at which the water level is rising when the height of the water is 3.5 meters.
\[
\frac{\mathrm{m}}{\min }
\]

Step 1 ：The volume \(V\) of a cone is given by the formula \(V = \frac{1}{3} \pi r^2 h\), where \(r\) is the radius and \(h\) is the height.

Step 2 ：Given that the diameter of the tank is 4 meters, the radius \(r\) is 2 meters.

Step 3 ：The height \(h\) and radius \(r\) of the water in the tank are related by the similar triangles in the cone and its water part, which gives \(r = \frac{2h}{11}\).

Step 4 ：Substitute \(r = \frac{2h}{11}\) into the volume formula, we get \(V = \frac{1}{3} \pi \left(\frac{2h}{11}\right)^2 h = \frac{4}{363} \pi h^3\).

Step 5 ：Differentiate both sides with respect to time \(t\), we get \(\frac{dV}{dt} = \frac{4}{121} \pi h^2 \frac{dh}{dt}\).

Step 6 ：We know that \(\frac{dV}{dt} = 7.8\) cubic meters per min, and we want to find \(\frac{dh}{dt}\) when \(h = 3.5\) meters.

Step 7 ：Substitute \(\frac{dV}{dt} = 7.8\), \(h = 3.5\) into the equation, we get \(7.8 = \frac{4}{121} \pi (3.5)^2 \frac{dh}{dt}\).

Step 8 ：Solve the equation for \(\frac{dh}{dt}\), we get \(\frac{dh}{dt} = \frac{7.8 \times 121}{4 \pi (3.5)^2} = 0.2\) meters per min.

Step 9 ：Therefore, the rate at which the water level is rising when the height of the water is 3.5 meters is \(0.2 \frac{m}{min}\).

Step 10 ：\boxed{0.2}