Given: The water is running into an inverted cone at the rate of π cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5 m.

To find how fast the water level is rising when the water stands 7.5 m below the base.

Let the height of the cone be H = 10m (given)

Let the radius of the base be R = 5m (given)

Let O’Y = r and CO’ = h

Now from the above figure,

ΔCOB~ΔCO’Q

So,

Let the volume of the water in the vessel at any time t be V

Then,

(from equation (i))

Now differentiate the above equation with respect to t, we get

But given the water is running at the rate of m³/min, i.e.,

So the above equation becomes

So when the water stands 7.5 m below the base

So h = 10 - 7.5 = 2.5m, the rate becomes

Hence the rate of water level rising when the water stands 7.5m below the base is 0.64 metres per min