6. Most aircraft are held together with metal rivets.
(a) The rivet used on one type of aircraft is shown below. It can be thought of as a cylinder connected to a hemisphere.
Diagram not drawn to scale
The cylinder has a diameter of 6 mm and a length of 9 mm. The hemisphere has a diameter of 12 mm.
(i) Calculate the volume of the rivet. Give your answer in terms of π in its simplest form. [4]
(ii) The manufacturer plans to reduce the length of the cylindrical part of the rivet from 9 mm to 8 mm. Calculate the fractional reduction in the volume of a rivet this would produce. [2]
Fractional reduction in the volume of a rivet = ……………
(b) For another type of rivet, the manufacturer plans to reduce the volume by \(0{\cdot}16\dot{1}\) of its original volume. Write \(0{\cdot}16\dot{1}\) as a fraction. Give your answer in its simplest form. [3]
From the formula list: Volume of sphere \(= \dfrac{4}{3}\pi r^3\)
Allow values of π from 3.14 to 3.142 for M marks only
M1 for \(\dfrac{2}{3}\) × π × 63 (+ ….) Allow M1 for \(\dfrac{4}{3}\) × π × 63 + π × 32 × 9 or equivalent
= 225π (mm3)
A2
CAO A1 for sight of 144π OR A1 for an answer of 369π (from use of 4/3×π×63) OR A1 for a fully correct method but with an answer of 225 (omitting π)
If no marks awarded, SC2 for an answer of 1476π from consistent use of diameters as their radii OR SC1 for \(\dfrac{2}{3}\) × π × 123 + π × 62 × 9 or equivalent OR SC1 for ( \(\dfrac{1}{3}\) × π × 63 + π × 32 × 9 = ) 153π
(ii)
Scheme
Marks
Comments
(Reduction in volume =) 9π (mm3)
B1
from π × 32 × (9 – 8) OR π×32×9 − π×32×8 or 81π − 72π
(Fractional reduction =) \(\dfrac{9(\pi)}{225(\pi)}\) or \(\dfrac{1}{25}\) or equivalent
B1
Implies previous B1 ISW FT ‘their 225π’ and ‘their π×32×9 − π×32×8’ and both are multiples of π
If no marks awarded, SC1 for a numerator of ‘their 225π’ − 9π, which may be embedded in a final answer of e.g. 216/225 or 24/25 if correct work seen in part (a)(i) Must be from a reduction of 9π in their volume. ISW
Mark scheme (b)
Scheme
Marks
Comments
Complete method e.g. 100x = 16.111…, 1000x = 161.111… AND an attempt to subtract both sides
M1
\(\dfrac{145}{900}\) or \(\dfrac{1595}{9900}\) or \(\dfrac{16095}{99900}\) or equivalent
A1
Allow A1 for e.g. (10x – x =) 1.45/9, (100x – x =) 15.95/99 etc
\(\dfrac{29}{180}\)
A1
FT from M1A0 provided at least 1 stage of simplification required
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