Water is running into an inverted cone at the rate of `pi` cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5m. How fast the water level is rising when the water stands 7.5 m below the base.

Explanation

Here, radius of the cone `(r)= 5m`

Height of the cone ` (h) = 10m`

`:. r/h = 5/10 = 1/2 => r = h/2`

Now, Volume of the cone `(V) = 1/3pir^2h`

`=> V = 1/3pi(h/2)^2h = pi/12h^3`

`:. (dV)/dt = pi/12(3h^2)(dh)/dt`
`=>(dV)/dt = pi/4h^2(dh)/dt->(1)`

We have to find the rate when water stands `7.5m` below the base.

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