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v1 Measurement, STD2 M1 2021 HSC 16

The surface area, `A`, of a sphere is given by the formula

`A = 4 pi r^2,`

where `r` is the radius of the sphere.

A satellite dish resembles the inner surface of the lower half of a sphere with a radius of 1.5 meters.

 

Find the surface area of the satellite dish in square metres, correct to one decimal place.   (2 marks)

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`14.1\ text{m}^2`

Show Worked Solution
`A` `= frac{1}{2} times 4 pi r^2`
  `= 2 pi r^2`
  `= 2 pi times (1.5)^2`
  `= 2 pi times 2.25`
  `= 14.137…`
  `= 14.1\ text{m}^2\ \text{(1 d.p.)}`

Filed Under: Perimeter, Area and Volume (Std2-X) Tagged With: Band 4, num-title-ct-pathb, num-title-qs-hsc, smc-4235-60-Spheres, smc-798-50-Volume (Circular Measure)

v1 Measurement, STD2 M1 2019 HSC 16

A decorative light fixture is in the shape of a hollow hemisphere with a diameter of 24 cm.
 

The inside of the fixture is to be coated with reflective paint.

What is the area to be painted on the inside surface? Give your answer correct to the nearest square centimetre.   (2 marks)

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`905\ \text{cm}^2`

Show Worked Solution
`A` `= 2 pi r^2`
  `= 2 × pi × 12^2`
  `= 2 × pi × 144 = 905.0…`
  `≈ 905\ \text{cm}^2`

Filed Under: Perimeter, Area and Volume (Std2-X) Tagged With: Band 4, num-title-ct-pathb, num-title-qs-hsc, smc-4235-60-Spheres, smc-798-50-Volume (Circular Measure)

Volume, SMB-017

A deep ocean submarine is constructed in the shape of a sphere. 
 

If the volume of the sphere is 12.1 cubic metres, calculate its diameter in metres, correct to two decimal places.   (2 marks)

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`2.85\ text(m)`

Show Worked Solution
`text{Volume}` `=4/3 xx pi xx r^3`  
`12.1` `=4/3 xx pi xx r^3`  
`r^3` `= \frac{3 xx 12.1}{4 xx pi}`  
  `=2.888`  
`r` `=1.424…\ text{m}`  

 
`:. text{Diameter}\ = 2 xx 1.242… = 2.85\ text{m (2 d.p.)}`

Filed Under: Volume Tagged With: num-title-ct-pathb, smc-4235-60-Spheres

Volume, SMB-007

A cannon ball is made out of steel and has a diameter of 23 cm.

Find the volume of the sphere in cubic centimetres (correct to 1 decimal place).  (2 marks)

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`6370.6\ text{cm}^3`

Show Worked Solution

`text(Radius)= 23/2 = 11.5\ text(cm)`

`text(Volume)` `= 4/3pir^3`
  `= 4/3 xx pi xx 11.5^3`
  `= 6370.626…`
  `= 6370.6\ text{cm}^3\ text{(to 1 d.p.)}`

Filed Under: Volume Tagged With: num-title-ct-pathb, smc-4235-60-Spheres

Volume, SMB-006

A concrete water pipe is manufactured in the shape of an annular cylinder. The dimensions are shown in the diagrams.
 


 

Find the approximate volume of concrete needed to make the water pipe, giving your answer in cubic metres correct to two decimal places.   (3 marks)

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`0.70\ text(m)^3`

Show Worked Solution
`text(Volume)` `= text(Area of annulus) xx h`
  `= (piR^2 – pir^2) xx 2.8`
  `= (pi xx 0.45^2 – pi xx 0.35^2) xx 2.8`
  `= 0.7037…`
  `= 0.70\ text(m)^3`

Filed Under: Volume Tagged With: num-title-ct-pathb, smc-4235-60-Spheres

Volume, SMB-005

Two identical spheres fit exactly inside a cylindrical container, as shown.
 

The diameter of each sphere is 12 cm.

 What is the volume of the cylindrical container, to the nearest cubic centimetre?   (3 marks)

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`2714\ text{cm³}`

Show Worked Solution

`text(S)text(ince diameter sphere = 12 cm) `

`=>\ text(Radius of cylinder = 6 cm)`

`text(Height of cylinder)` `= 2 xx text(diameter of sphere)`
  `= 2 xx 12`
  `= 24\ text(cm)`
   
`:.\ text(Volume cylinder)` `= pi r^2 h`
  `= pi xx 6^2 xx 24`
  `= 2714.336…`
  `= 2714\ text{cm³}`

Filed Under: Volume Tagged With: num-title-ct-pathb, smc-4235-60-Spheres

Measurement, STD2 M1 2021 HSC 16

The volume, `V`, of a sphere is given by the formula

`V = frac{4}{3} pi r^3,`

where `r` is the radius of the sphere.

A tank consists of the bottom half of a sphere of radius 2 metres, as shown.
 

Find the volume of the tank in cubic metres, correct to one decimal place.   (2 marks)

Show Answers Only

`16.8\ text{m}^3`

Show Worked Solution
 `V` `= frac{1}{2} times frac{4}{3} pi r^3`
  `= frac{1}{2} times frac{4}{3} times pi times 2^3`
  `= 16.755…`
  `= 16.8\ text{m}^3\ \ text{(1 d.p.)}`

Filed Under: Perimeter, Area and Volume (Std 2), Perimeter, Area and Volume (Std2-2027) Tagged With: Band 4, num-title-ct-pathb, num-title-qs-hsc, smc-4235-60-Spheres, smc-6304-50-Volume (Circular Measure), smc-798-50-Volume (Circular Measure)

Measurement, STD2 M1 2019 HSC 16

A bowl is in the shape of a hemisphere with a diameter of 16 cm.
 

What is the volume of the bowl, correct to the nearest cubic centimetre?  (2 marks)

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`1072\ text(cm)^3`

Show Worked Solution
`V` `= 1/2 xx 4/3pir^3`
  `= 1/2 xx 4/3 xx pi xx 8^3`
  `= 1072.3…`
  `= 1072\ text{cm}^3\ text{(nearest cm}^3 text{)}`

Filed Under: Perimeter, Area and Volume (Std 2), Perimeter, Area and Volume (Std2-2027) Tagged With: Band 4, num-title-ct-pathb, num-title-qs-hsc, smc-4235-60-Spheres, smc-6304-50-Volume (Circular Measure), smc-798-50-Volume (Circular Measure)

Measurement, STD2 M1 2008 HSC 21 MC

A sphere and a closed cylinder have the same radius.

The height of the cylinder is four times the radius.

What is the ratio of the volume of the cylinder to the volume of the sphere?

  1. `2 : 1`
  2. `3 : 1`
  3. `4 : 1`
  4. `8 : 1`
Show Answers Only

`B`

Show Worked Solution

♦♦ Mean mark 33%.

`V_text(cylinder)` `: V_text(sphere)`
`pir^2h` `: 4/3pir^3`
`underbrace(pir^2 4r)_(h = 4r)` `: 4/3pir^3`
`4pir^3` `: 4/3pir^3`
`3` `: 1`

  
`=> B`

Filed Under: Perimeter, Area and Volume (Std 2), Perimeter, Area and Volume (Std2-2027) Tagged With: Band 5, num-title-ct-pathb, num-title-qs-hsc, smc-4235-60-Spheres, smc-6304-50-Volume (Circular Measure), smc-798-50-Volume (Circular Measure)

Measurement, STD2 M1 2017 HSC 30e

A solid is made up of a sphere sitting partially inside a cone.

The sphere, centre `O`, has a radius of 4 cm and sits 2 cm inside the cone. The solid has a total height of 15 cm. The solid and its cross-section are shown.
 


 

Using the formula  `V=1/3 pi r^2h`  where `r`  is the radius of the cone's circular base and `h` is the perpendicular height of the cone, find the volume of the cone, correct to the nearest cm³?  (3 marks)

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`113\ text{cm}^3`

Show Worked Solution

`V = 1/3 xx text(base of cone × height)`

`text(Consider the circular base area of the cone,)`

`text(Find)\ x\ \ text{(using Pythagoras):}`

`x^2` `= 4^2-2^2 = 16-4 = 12`
`x` `= sqrt12\ text(cm)`

 

`:. V` `= 1/3 xx pi xx (sqrt12)^2 xx (15-6)`
  `= 1/3 xx pi xx 12 xx 9`
  `= 113.097…`
  `= 113\ text{cm}^3\ text{(nearest cm}^3 text{)}`

Filed Under: Perimeter, Area and Volume (Std 2), Perimeter, Area and Volume (Std2-2027), Volume Tagged With: Band 6, num-title-ct-extension, num-title-qs-hsc, smc-4235-50-Pyramids/Cones, smc-4235-60-Spheres, smc-6304-50-Volume (Circular Measure), smc-798-50-Volume (Circular Measure)

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