The diagram shows a solid ornament in the shape of a cylinder with an upright cone cut out at one end and a hemisphere attached to the other end. The vertical heights of the cone and cylinder are xx cm and 2x2x cm respectively.

  1. (a)

    Find the ratio of the volume of the cone to that of the cylinder, expressing your answer as a fraction in the simplest form.

    Answer ...

  2. (b)

    If the volume of the cylinder is 345 cm3 and its height is 8 cm, calculate

    1. (i)

      the radius of the cylinder,

      Answer ... cm

    2. (ii)

      the curved surface area of the cone,

      Answer ... cm2

    3. (iii)

      the quantity of paint needed to paint the exterior of the ornament with a 0.2 mm thick coat of paint.

      Answer ... cm3

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Worked solution

5(a)

1. Form the ratio of the cone volume to the cylinder volume and simplify.

Volume of cone =13πr2x= \dfrac{1}{3}\pi r^2 x

Volume of cylinder =πr2(2x)= \pi r^2(2x)

Ratio=13πr2x2πr2x=16\text{Ratio} = \frac{\frac{1}{3}\pi r^2 x}{2\pi r^2 x}=\frac{1}{6}

Answer

16\dfrac{1}{6}

5(b)(i)

1. Substitute the given cylinder volume and height into the cylinder volume formula.

For a cylinder,
V=πr2hV=\pi r^2h
Given V=345V=345 and h=8h=8,
345=πr2(8)345=\pi r^2(8)

2. Solve for the radius.

r2=3458πr^2=\frac{345}{8\pi}
r=3458πr=\sqrt{\frac{345}{8\pi}}
r=3.704…r=3.704\ldots
So, to 3 significant figures,
r=3.70r=3.70

Answer

3.703.70 cm

5(b)(ii)

1. Find the cone height from the given relationship between cylinder height and cone height.

The cylinder height is 2x=82x=8.

Hence
x=4x=4

So the vertical height of the cone is 4 cm4\text{ cm}.

2. Find the slant height of the cone using Pythagoras' theorem.

Using the radius from part (b)(i), r=3.70 cmr=3.70\text{ cm}.

For the cone,
l=r2+h2l=\sqrt{r^2+h^2}
l=3.702+42l=\sqrt{3.70^2+4^2}
l=29.69=5.448…l=\sqrt{29.69}=5.448\ldots

3. Calculate the curved surface area of the cone.

Curved surface area of cone
=πrl=\pi r l
=π(3.70)(5.448…)=\pi(3.70)(5.448\ldots)
=63.3 cm2, to 3 s.f.=63.3\text{ cm}^2\text{, to 3 s.f.}

Answer

86.1 cm286.1\text{ cm}^2

5(b)(iii)

1. Convert the paint thickness into centimetres.

0.2 mm=0.02 cm0.2\text{ mm}=0.02\text{ cm}

2. Find the total exterior surface area to be painted.

Using r=3.70 cmr=3.70\text{ cm} from part (b)(i), cylinder height =8 cm=8\text{ cm}, and cone curved surface area =86.1 cm2=86.1\text{ cm}^2 from part (b)(ii),

curved surface area of cylinder=2πrh\text{curved surface area of cylinder}=2\pi rh
=2π(3.70)(8)=186 cm2 approximately=2\pi(3.70)(8)=186\text{ cm}^2\text{ approximately}

curved surface area of hemisphere=2πr2\text{curved surface area of hemisphere}=2\pi r^2
=2π(3.70)2=86.0 cm2 approximately=2\pi(3.70)^2=86.0\text{ cm}^2\text{ approximately}

Total area to be painted
=186+86.0+86.1=358.1 cm2 approximately=186+86.0+86.1=358.1\text{ cm}^2\text{ approximately}

3. Multiply surface area by thickness to find the volume of paint.

Volume of paint=area×thickness\text{Volume of paint}=\text{area}\times\text{thickness}
=358.1×0.02=358.1\times 0.02
=7.162=7.162

Using unrounded values gives approximately 7.17 cm37.17\text{ cm}^3.

Answer

7.17 cm37.17\text{ cm}^3

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