A conical container with a diameter of 10 cm and a height of 8 cm is filled with water to a depth of 4 cm.

  1. (a)

    Show that the volume of the water in the container is 253π\frac{25}{3}\pi cm^3.

  2. (b)

    Find the surface area of the inside of the container that is in contact with the water.

    Answer ___ cm^2

  3. (c)

    The water in the cone is poured into a test tube.
    The test tube is made by joining together a cylinder of diameter 3 cm and a hemisphere of diameter 3 cm.
    Calculate the height of the water in the test tube.

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Worked solution

26(a)

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

The full cone has radius 55 cm and height 88 cm.

The water forms a smaller similar cone of height 44 cm.

r5=48\frac{r}{5}=\frac{4}{8}

r=48×5=2.5 cmr=\frac{4}{8}\times 5=2.5\text{ cm}

2. Calculate the volume of the water cone.

Using V=13πr2hV=\frac{1}{3}\pi r^2h,

V=13π(2.5)2(4)V=\frac{1}{3}\pi(2.5)^2(4)

=13π(6.25)(4)=\frac{1}{3}\pi(6.25)(4)

=253π cm3=\frac{25}{3}\pi\text{ cm}^3

Answer

Volume of water=253π cm3\text{Volume of water}=\frac{25}{3}\pi\text{ cm}^3

26(b)

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

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

Full cone: radius 55 cm, height 88 cm.

Water cone: height 44 cm.

r5=48\frac{r}{5}=\frac{4}{8}

r=2.5 cmr=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=42+2.52=16+6.25=22.25=892l=\sqrt{4^2+2.5^2}=\sqrt{16+6.25}=\sqrt{22.25}=\frac{\sqrt{89}}{2}

Curved surface area:

πrl=π(2.5)(892)\pi rl=\pi(2.5)\left(\frac{\sqrt{89}}{2}\right)

=5894π=\frac{5\sqrt{89}}{4}\pi

≈37.1 cm2\approx 37.1\text{ cm}^2

Answer

5414π cm2≈25.1 cm2\frac{5\sqrt{41}}{4}\pi\text{ cm}^2 \approx 25.1\text{ cm}^2

26(c)

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

The test tube has diameter 33 cm, so its radius is
r=32=1.5 cm.r=\frac{3}{2}=1.5\text{ cm}.
Volume of hemisphere:
12(43πr3)=23π(1.5)3=2.25π cm3.\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
253π cm3.\frac{25}{3}\pi\text{ cm}^3.
Volume in the cylindrical part:
253π−2.25π=253π−94π=7312π cm3.\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 hh cm.
π(1.5)2h=7312π\pi(1.5)^2h=\frac{73}{12}\pi
2.25h=73122.25h=\frac{73}{12}
h=7327 cm.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 1.51.5 cm.
Total height=7327+1.5\text{Total height}=\frac{73}{27}+1.5
=7327+32=146+8154=22754=\frac{73}{27}+\frac{3}{2}=\frac{146+81}{54}=\frac{227}{54}
≈4.20 cm.\approx 4.20\text{ cm}.

Answer

Height of water=2027+1.5=12154 cm≈2.24 cm\text{Height of water}=\frac{20}{27}+1.5=\frac{121}{54}\text{ cm}\approx 2.24\text{ cm}

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