An inverted cone has a depth of 10cm and a base of radius 5cm. Water is poured into it at the rate of 3/2 c.c. per minute. Find the rate at which the level of water in the cone is rising when the depth is 4cm.

Explanation

`{H}=10 {cm}`


`{R}=5 {cm}`


In general,


`r=h . tan alpha=frac{h}{2}`


`frac{d V}{d t}=1.5 {min}^{-1}`


`V=frac{pi}{3} r^{2} h`


`=frac{pi}{12} h^{3}`


`Rightarrow frac{d V}{d t}=frac{pi}{4} h^{2} frac{d h}{d t}`


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