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# Water is passed into an inverted cone of base radius 5 cm and depth 10 cm at the rate of (3/2) c.c/sec Find the rate at which the level of water is rising when depth is 4 cm.

Jitender Singh IIT Delhi
6 years ago
Ans:
Let ‘V’ be the volume of tank, ‘r’ be radius & ‘h’ be the height or depth of the tank.
Volume ‘V’:
$V = \frac{1}{3}\pi r^{2}h$
Relation b/w ‘h’ & ‘r’:
$h = 2r$
$V = \frac{1}{3}\pi (\frac{h}{2})^{2}h$
$V = \frac{\pi h^{3}}{24}$
$\frac{\partial V}{\partial t} = \frac{3\pi h^{2}}{24}.\frac{\partial h}{\partial t}$
$\frac{\partial V}{\partial t} = \frac{\pi h^{2}}{8}.\frac{\partial h}{\partial t} = \frac{3}{2}(\frac{cm^{3}}{sec})$
$h = 4cm$
$\frac{\partial V}{\partial t} = \frac{\pi (4)^{2}}{8}.\frac{\partial h}{\partial t} = \frac{3}{2}(\frac{cm^{3}}{sec})$
$\frac{\partial h}{\partial t} = \frac{3}{4\pi }(\frac{cm}{sec})$
Thanks & Regards
Jitender Singh
IIT Delhi