一个直径为 10 cm、高为 8 cm 的圆锥形容器装有水，水深为 4 cm。

1. Find the radius of the water surface using similar cones.

The full cone has radius $5$ cm and height $8$ cm.

The water forms a smaller similar cone of height $4$ cm.

$\frac{r}{5}=\frac{4}{8}$

$r=\frac{4}{8}\times 5=2.5\text{ cm}$

2. Calculate the volume of the water cone.

Using $V=\frac{1}{3}\pi r^2h$,

$V=\frac{1}{3}\pi(2.5)^2(4)$

$=\frac{1}{3}\pi(6.25)(4)$

$=\frac{25}{3}\pi\text{ cm}^3$

1. Find the radius of the water surface using similarity.

The water forms a smaller cone similar to the full container.

Full cone: radius $5$ cm, height $8$ cm.

Water cone: height $4$ cm.

$\frac{r}{5}=\frac{4}{8}$

$r=2.5\text{ cm}$

2. Find the curved surface area of the submerged inside surface.

The submerged inside surface is the curved surface of the smaller cone.

Its slant height is

$l=\sqrt{4^2+2.5^2}=\sqrt{16+6.25}=\sqrt{22.25}=\frac{\sqrt{89}}{2}$

Curved surface area:

$\pi rl=\pi(2.5)\left(\frac{\sqrt{89}}{2}\right)$

$=\frac{5\sqrt{89}}{4}\pi$

$\approx 37.1\text{ cm}^2$

1. Find the volume of the hemispherical bottom of the test tube.

The test tube has diameter $3$ cm, so its radius is
$r=\frac{3}{2}=1.5\text{ cm}.$
Volume of hemisphere:
$\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)=\frac{2}{3}\pi(1.5)^3=2.25\pi\text{ cm}^3.$

2. Use the remaining water volume to find the height of water in the cylindrical part.

From part (a), volume of water poured in is
$\frac{25}{3}\pi\text{ cm}^3.$
Volume in the cylindrical part:
$\frac{25}{3}\pi-2.25\pi=\frac{25}{3}\pi-\frac{9}{4}\pi=\frac{73}{12}\pi\text{ cm}^3.$
Let the height of water in the cylindrical part be $h$ cm.
$\pi(1.5)^2h=\frac{73}{12}\pi$
$2.25h=\frac{73}{12}$
$h=\frac{73}{27}\text{ cm}.$

3. Add the hemisphere radius to get the total height of water from the bottom of the test tube.

The total height of water includes the height of the hemisphere, which is its radius $1.5$ cm.
$\text{Total height}=\frac{73}{27}+1.5$
$=\frac{73}{27}+\frac{3}{2}=\frac{146+81}{54}=\frac{227}{54}$
$\approx 4.20\text{ cm}.$