Water is leaking out of an inverted conical tank at a rate of 1100.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 15.0 meters and the diameter at the top is 5.0 meters. If the water level is rising at a rate of 300 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

To solve this problem, we need to use the concept of related rates. We'll consider the rates at which the water level is changing and the volume of water in the tank is changing.

Let's define some variables:
h - height of the water in the tank (in meters)
V - volume of water in the tank (in cubic meters)
r - radius of the water surface in the tank (in meters)

First, let's find an equation that relates h and r. Since the tank is conical, the height and radius are related by similar triangles. When the height is 15 meters, the radius is 2.5 meters (half the diameter), so we can write:

h / r = 15 / 2.5

Next, let's express V in terms of r. The volume of a cone can be calculated using the formula V = (1/3) * π * r^2 * h. Substituting the above equation relating h and r, we get:

V = (1/3) * π * r^2 * (15/2.5)

Now, let's differentiate both sides of this equation with respect to time t to find the rates at which V and r are changing with time:

dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * d(h)/dt)
Note: dV/dt represents the rate at which the volume is changing, dr/dt represents the rate at which the radius is changing, and d(h)/dt represents the rate at which the height is changing.

We're given that dV/dt = pumped-in rate - leaked-out rate = constant rate of water being pumped in - 1100 (since water is leaking out at a rate of 1100 cm^3/min).

We're also given that d(h)/dt = 300 cm/min (since the water level is rising at a rate of 300 cm/min when the height of water is 2.5 meters).

r is related to h by the equation h / r = 15 / 2.5. To find dr/dt, we need to differentiate this equation with respect to time t as well:

d(h)/dt / r - h * (dr/dt) / r^2 = 0

Simplifying this equation, we get:

dr/dt = (d(h)/dt * r) / h

Now, we have expressions for dV/dt and dr/dt in terms of known values and variables. Substituting these expressions into the derivative equation, we can solve for the constant rate at which water is being pumped into the tank.

I'll calculate the value for the rate at which water is being pumped into the tank for you.