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.
![](https://www.sarthaks.com/?qa=blob&qa_blobid=2961339929024683314)
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,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=2115572301932220890)
Let the volume of the water in the vessel at any time t be V
Then,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=15673465866424561822)
Now differentiate the above equation with respect to t, we get
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5877809047350023808)
But given the water is running at the rate of \(\pi\)m3/min, i.e., \(\frac{dv}{dt}=\pi\)
So the above equation becomes
![](https://www.sarthaks.com/?qa=blob&qa_blobid=16690976280956857950)
So when the water stands 7.5 m below the base
So h = 10 - 7.5 = 2.5m, the rate becomes
![](https://www.sarthaks.com/?qa=blob&qa_blobid=2996800601843379457)
Hence the rate of water level rising when the water stands 7.5m below the base is 0.64 metres per min