Sand is pouring from a pipe at the rate of 12 `c m^3//s` . The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when t

Explanation

Let at any time 't' , the height of cone be 'h' and radius be 'r'.
Now given, ` h = r/6`…(1)
Volume of cone
` V=1/3 pi r^(2) h`
`rArr V = 1/3 pi (6h)^(2)* h = 12 pi h^(3)`
` rArr (dV)/(dt) = 36 pi h^(2) * (dh)/(dt)`
Put ` (dV)/(dt) = 12 cm^(3)` / sec and h = 4 cm
` 12 = 36 pi (4)^(2) * (dh)/(dt)`
` rArr (dh)/(dt) = 1/(48 pi) `cm/sec