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## Homework Statement

A cylinder of radius ##{R}## is partially filled with water. There is an inverted cone of cone angle 90° and base radius ##{\frac{R}{2}}## which is falling in it with a constant speed ##v=30~cm/s##.

Find the rate of increase of water level (in ##cm/s##) when half the height of cone is immersed in water.

The given answer is: ##2~cm/s##

## Homework Equations

Let the height of the cone be h, and height of cylinder be H

Volume of cone ## = \frac{1}{3}\pi (\frac{R}{2})^2 h = \frac{1}{12} \pi R^2 h##

Volume of cylinder = ##\pi R^2 H##

## The Attempt at a Solution

Now for the cone ##\frac{dh}{dt}=v=30~cm/s##

Let the increase in cylinder's water level be ##dH## when ##dV## of cone's volume is immersed in it.

So

##dV = \pi R^2 dH##

⇒ ## \frac{dH}{dt} = \frac{\frac{dV}{dt}}{\pi R^2}##

⇒ ## \frac{dH}{dt} = \frac{\frac{d}{dt}(\frac{1}{12} \pi R^2 h)}{\pi R^2}##

⇒ ## \frac{dH}{dt} = \frac{\frac{1}{12} \pi R^2~\frac{dh}{dt}}{\pi R^2}##

⇒ ## \frac{dH}{dt} = \frac{1}{12} × 30 = 2.5~cm/s##

Please help.

Thank you.