# Mechanical Technology Grade 12 NSC November 2012 Memo (Eng)

> Grade 12 Mechanical Technology memorandum from the NSC November Exam 2012 (English), 17 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mechanical-technology-nov-2012-memo-eng/0cd2c7e3-61d6-4a21-80c4-178cd05b4dad
- Updated: 2026-09-27
- Subject: Mechanical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2012
- Exam period: NSC November Exam
- Curriculum: CAPS
- Pages: 17

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mechanical Technology](https://www.hlayiso.com/library/grade-12/mechanical-technology) › [Mechanical Technology Grade 12 NSC November 2012 Memo (Eng)](https://www.hlayiso.com/library/mechanical-technology-nov-2012-memo-eng/0cd2c7e3-61d6-4a21-80c4-178cd05b4dad)

## About this document

Mechanical Technology · Grade 12 · NSC November Exam · 2012. Memorandum, 17 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Downloaded from hlayiso.com

                                NATIONAL
                           SENIOR CERTIFICATE

                                   GRADE 12

                         MECHANICAL TECHNOLOGY

                                NOVEMBER 2012

                                 MEMORANDUM

MARKS: 200

                       This memorandum consists of 17 pages.

Copyright reserved                                             Please turn over

### Page 2

Downloaded from hlayiso.com
Mechanical Technology                2          DBE/November 2012
                             NSC – Memorandum

QUESTION 1: MULTIPLE-CHOICE QUESTIONS

1.1       B                                                      (1)

1.2       A                                                      (1)

1.3       D                                                      (1)

1.4       B                                                      (1)

1.5       A or B                                                 (1)

1.6       B                                                      (1)

1.7       A                                                      (1)

1.8       C                                                      (1)

1.9       D                                                      (1)

1.10      A                                                      (1)

1.11      D                                                      (1)

1.12      B                                                      (1)

1.13      B                                                      (1)

1.14      C                                                      (1)

1.15      A                                                      (1)

1.16      C                                                      (1)

1.17      D                                                      (1)

1.18      A                                                       (1)

1.19      C                                                      (1)

1.20      D                                                      (1)
                                                                  [20]

Copyright reserved                                  Please turn over

### Page 3

Downloaded from hlayiso.com
Mechanical Technology                        3                            DBE/November 2012
                                     NSC – Memorandum

QUESTION 2: TOOLS AND EQUIPMENT

2.1      Causes of low compression:
         •    Worn cylinders                                                           
         •    Worn compression rings                                                   
         •    Worn piston                                                              
         •    Worn valves                                                              
         •    Worn head gasket                                                         
                                                                              (Any 3 x 1)         (3)

2.2      Torsion:
         Torsion is the twisting action in a member                                     
         caused by two opposing moments along the longitudinal axis.                             (2)

2.3      Multimeter:
         A – LCD display screen                                                         
         B – Range selector switch                                                      
         C – 10 A DC terminal socket (Input terminal)                                   
         D – VΩmA terminal socket (Input Terminal)                                      
         E – Common terminal socket                                                              (5)

2.4      Cylinder leakage tester:
         •     Listen at the carburettor and/or inlet manifold for hissing noise.       
               (inlet valve is leaking)                                                 
         •     Listen to the exhaust pipe or exhaust manifold for a hissing noise.      
               (exhaust pipe is leaking)                                                
         •     Listen for hissing noise in the dipstick hole.                           
               (piston rings worn)                                                      
         •     Remove the filler cap on the tappet cover and listen for hissing noise. 
               (rings are worn)                                                         
         •     If you see bubbles in the radiator water,                                
               (the cylinder head gasket is blown or the cylinder block is cracked)     
                                                                               (Any 3 x 2)        (6)

2.5      MAGS/MIGS – meaning
         •       MAGS: Metal Arc Gas Shielded                                           
         •       MIGS: Metal Inert Gas Shielded                                                  (2)

2. 6     MAGS/MIGS gases
         •       Argon                                                                  
         •       CO 2                                                                   
         •       Helium                                                                 
                                                                              (Any 2 x 1)         (2)
                                                                                                 [20]

Copyright reserved                                                            Please turn over

### Page 4

Downloaded from hlayiso.com
Mechanical Technology                         4                          DBE/November 2012
                                      NSC – Memorandum

QUESTION 3: MATERIALS

3.1       Non-ferrous materials and composites:
          •     Non-ferrous Metals are metallic and composites are non-metallic. 
          •     Non-ferrous metals are original substances and composites are
                combinations of two or more materials.                           
                                                                         (Any 1 x 2)            (2)

3.2       Compressive strength test:

          3.2.1       Material A has the highest compressive strength.                         (1)

          3.2.2       The material that can resist a large compression force will have
                      little deformation or compression, and has a higher compressive
                      strength.                                                               (2)

3.3       Carbon steel:
          •    Low-carbon steel                                                        
          •    Medium-carbon steel                                                     
          •    High-carbon steel                                                               (3)

3.4       Carbon steel – properties:
          •    Greater hardness is obtained                                            
          •    Tensile strength is increased                                           
          •    Ductility is decreased                                                  
          •    Welding ability is decreased                                                    (4)

3.5       Uses and properties of engineering materials:

          3.5.1       Uses of Duralumin:
                      It is used to make the following:
                      •       Bars                                                 
                      •       Sheets                                               
                      •       Piston rods                                          
                      •       Tubes                                                
                      •       Rivets                                               
                      •       Motorcar and aircraft parts                          
                                                                            (Any 2 x 1)         (2)

                      Properties of Duralumin:
                      o     Lightweight                                            
                      o     High tensile strength                                  
                      o     Good resistance to corrosion                           
                      o     Hardens with age                                       
                                                                            (Any 2 x 1)         (2)

Copyright reserved                                                           Please turn over

### Page 5

Downloaded from hlayiso.com
Mechanical Technology                       5                             DBE/November 2012
                                    NSC – Memorandum

          3.5.2       Uses of PVC:
                      It is used to make:
                      •       Pipes and fittings                                   
                      •       Cable and services ducting                           
                      •       Roofing and ceiling systems and membranes            
                      •       Healthcare materials                                 
                      •       Automotive industry materials                        
                                                                             (Any 2 x 1)         (2)

                      Properties of PVC:
                      •    Lightweight                                             
                      •    Weather resistant                                       
                      •    Rigid or flexible                                       
                      •    Clear or coloured                                       
                      •    Good electrical insulator                               
                      •    Good resistance to corrosion                            
                                                                             (Any 2 x 1)         (2)
                                                                                                 [20]

Copyright reserved                                                            Please turn over

### Page 6

Downloaded from hlayiso.com
Mechanical Technology                        6                           DBE/November 2012
                                     NSC – Memorandum

QUESTION 4: SAFETY, TERMINOLOGY AND JOINING METHODS

4.1       Beam bending tester:
          •    Make sure that the object to be tested is firmly secured.             
          •    Make sure that all the holding devices are properly fitted.           
          •    Check components of tester for wear.                                  
          •    Check for leaks at the hydraulic pump ram and hose.                   
          •    Make sure that the tester is clean and free from oil and grease.      
                                                                            (Any 3 x 1)         (3)

4.2       Gas cylinders:
          •    Store full cylinders apart from empty cylinders.                      
          •    Keep in cool, dry place away from sunlight.                           
          •    Acetylene cylinders should be stored in an upright position.          
          •    Oxygen cylinders should be stored away from acetylene cylinders. 
          •    Do not allow cylinders to fall.                                       
          •    No oil and grease should come into contact with oxygen cylinders and
               fittings.                                                             
                                                                            (Any 3 x 1)         (3)

4.3       Milling operations:

          4.3.1       Upcut milling:
                      •     Less vibration occurs.                                     
                      •     Less strain on the cutter and arbor.                       
                      •     There is positive pressure on the feed screw spindle and its
                            nuts because of the direction of the cutter.               
                      •     A coarser feed may be used.                                
                                                                             (Any 2 x 1)        (2)

          4.3.2       Down-cut milling:
                      •    Deeper cuts can be made because the force of the cutter is
                           downwards.                                                
                      •    A finer finish is obtained.                                           (2)

Copyright reserved                                                           Please turn over

### Page 7

Downloaded from hlayiso.com
Mechanical Technology                           7                          DBE/November 2012
                                        NSC – Memorandum

4.4       Indexing:

          4.4.1                      40
                        Indexing   =                                                      
                                      A
                                     40
                                   =                                                      
                                     60
                                     4× 4                                                 
                                   =
                                     6× 4
                                     16 20 26 28 34 36 44
                                   =    or or or or or or                                 
                                     24 30 39 42 51 54 66

                      16 holes on the 24-hole circle or
                      20 holes on the 30-hole circle or
                      26 holes on the 39-hole circle or
                      28 holes on the 42-hole circle or
                      34 holes on the 51-hole circle or
                      36 holes on the 51-hole circle or
                      44 holes on the 44-hole circle                                           (4)

         4.4.2                                                                            
                                 = ( A − n )×                 = (N − n )×
                              Dr              40           Dr             40
                                                      OR
                              Dv               A           Dv             N
                                                                                          
                                 = (60 − 63)×                 = (60 − 63)×
                              Dr              40           Dr              40
                              Dv              60           Dv              60
                              Dr − 120                      Dr − 120                      
                                 =                             =
                              Dv   60                       Dv   60
                              Dr − 12 × 4                   Dr − 12 × 4
                                 =                             =                          
                              Dv   6 ×4                     Dv   6 ×4
                              Dr − 48 56 64                Dr − 48 56 64
                                 =    or or                   =    or or                   (5)
                              Dv   24   28 32              Dv   24   28 32

Copyright reserved                                                              Please turn over

### Page 8

Downloaded from hlayiso.com
  Mechanical Technology                         8                          DBE/November 2012
                                        NSC – Memorandum

  4.5       Cutting feed:

            V = πDN                                                                        
                 V
            N=                                                                             
                πD
                   20                                                                      
            N=
                π × 0,08
            N = 79,577 r/min                                                               

            f = f1 × T × N                                                                 
            f = 0,08 × 14 × 79,577                                                         
            f = 89,13 mm/min                                                                      (7)

  4.6       Ultrasonic test:

            •      A high frequency sound wave is send into the metal for a very short
                   period of 1 to 3 microseconds.                                      
            •      The same unit which was used to send the sound wave then acts as
                   a receiver to listen to the ultrasonic waves it reflected through the
                   metal.                                                              
            •      This cycle is repeated from one to five million times per second.
            •      The oscilloscope is calibrated only to pick up defects of a size that
                   would be considered harmful.                                        
            •      The oscilloscope wave pattern is also calibrated to show the
                   distance between the searching unit and any defects found.          
                                                                                                  (5)

4.7     Weld defects:

         4.7.1         Slag inclusion:
                       Causes:
                       •      Included angle too narrow                              
                       •      Rapid cooling                                          
                       •      Welding temperature too low                            
                       •      High viscosity of molten metal.                        
                       •      Welding second run without removing slag               
                                                                            (Any 2 x 1)           (2)

                       Prevention:
                       •    Increase the included angle.                             
                       •    Let the welded metal cool slowly                         
                       •    Pre-heat the metal                                       
                       •    Remove slag before welding a second run                  
                                                                            (Any 1 x 1)           (1)

  Copyright reserved                                                           Please turn over

### Page 9

Downloaded from hlayiso.com
  Mechanical Technology                         9                             DBE/November 2012
                                        NSC – Memorandum

         4.7.2         Undercutting:
                       Causes:
                       •    Faulty electrode manipulation                               
                       •    Arc length too long                                         
                       •    Current too high                                            
                       •    Welding speed too fast                                      
                                                                               (Any 2 x 1)            (2)

                       Prevention:
                       •     Use a uniform weaving movement in butt joint               
                       •     Use the correct electrode                                  
                       •     Use the correct current                                    
                       •     Weld slowly                                                
                                                                               (Any 1 x 1)            (1)

4.8     Milling Cutters:
        4.8.1 Spur gear – Involute cutter                                                 
        4.8.2 Groove – Side-and-face cutter or end mill or slot drill                     
        4.8.3 Rack – Involute cutter/Fly cutter                                           
        4.8.4 Blind hole – Flute-end mill                                                        (4)

4.9     Dividing head of a milling machine:
        A = Index plate                                                                   
        B = Index crank                                                                   
        C = Sector arms                                                                   
        D = Single-start worm                                                             
        E = Wormwheel/gear                                                                       (5)

4.10    Classification of milling cutters:
        Arbor cutters                                                                            (1)
        Examples: Plain cutter , side cutter, staggered-tooth cutter, slitting-saw
        cutter, angular cutter, profile/form cutters, side-and-face cutter, helical cutter 
                                                                                (Any 1 x 1)       (1)

        Shank cutters                                                                            (1)
        Examples: End mill; shell end mill; T-slot cutter and Woodruff keyseat cutter
                                                                           (Any 1 x 1)            (1)
                                                                                                  [50]

  Copyright reserved                                                               Please turn over

### Page 10

Downloaded from hlayiso.com
Mechanical Technology                       10                           DBE/November 2012
                                     NSC – Memorandum

QUESTION 5: MAINTENANCE AND TURBINES

5.1       Clutch

          5.1.1       Parts:
                      A.     Crankshaft                                                
                      B.     Clutch plate                                              
                      C.     Gearbox shaft                                             
                      D.     Diaphragm spring                                          
                      E.     Pressure plate                                            
                      F.     Flywheel                                                          (6)

          5.1.2       Functions:
                      •     To provide friction between the clutch and pressure plate
                                                                                     
                      •     To connect the flywheel to the gearbox shaft                      (4)

          5.1.3       Causes of slip:
                      •    Oil on the friction surfaces                               
                      •    Worn friction surfaces                                     
                      •    Lack of compressive force on the friction surfaces caused by
                           weak springs                                               
                      •    Lack of compressive force on the friction surfaces caused by
                           incorrect clutch settings                                  
                      •    Uneven friction surfaces                                   
                      •    Overheating                                                
                                                                             (Any 3 x 1)        (3)

          5.1.4       Clutch types:
                      •     Single-disc clutch                                        
                      •     Multi-disc clutch                                         
                      •     Centrifugal clutch                                        
                                                                             (Any 2 x 1)        (2)

5.2       Functions of lubricating oil:
          •     Provides lubrication between contact surfaces                         
          •     Resists oxidation                                                     
          •     Prevents rust                                                         
          •     Avoids foaming                                                        
          •     Resists carbon forming                                                
          •     Prevents corrosion                                                    
          •     Resists extreme pressure                                             
                                                                             (Any 3 x 1)        (3)

Copyright reserved                                                           Please turn over

### Page 11

Downloaded from hlayiso.com
Mechanical Technology                        11                            DBE/November 2012
                                      NSC – Memorandum

5.3       Cutting fluid:
          •     Avoid contamination of the cutting fluid by draining and regularly
                replacing it.                                                           
          •     Always clean metal cuttings from the machine’s splash tray after
                use.                                                                    
          •     Regularly wipe cutting fluid splashes off machine parts when machine
                is stationary.                                                          
          •     Ensure that the sump is topped up from time to time, and check that
                there is sufficient flow of cutting fluid to the cutting tool.          
          •     Check for correct ratio of cutting fluid to water.                      
                                                                               (Any 3 x 1)        (3)

5.4       Supercharger

          5.4.1       Centrifugal type                                                           (1)

          5.4.2       Parts:
                      A.     Inlet port                                                  
                      B.     Outlet port                                                 
                      C.     Rotor                                                       
                      D.     Vane                                                                (4)

          5.4.3       Operation:
                      •    The engine drives the rotor                                   
                      •    Air is drawn in behind the rotor                              
                      •    The air is forced around into a decreasing volume             
                      •    This raises the pressure of the air                           
                      •    The air is forced into the inlet manifold and into the cylinders
                                                                                                 (5)

          5.4.4       Advantages of a supercharger:
                      •    More power is developed compared to a similar vehicle
                           without a supercharger                                  
                      •    Supercharged engines are more economical per given
                           kilowatt output                                         
                      •    Less fuel is used compared to engine mass               
                      •    Power loss above sea level is eliminated                
                                                                          (Any 3 x 1)             (3)

Copyright reserved                                                             Please turn over

### Page 12

Downloaded from hlayiso.com
Mechanical Technology                     12                           DBE/November 2012
                                   NSC – Memorandum

5.5       Steam turbines:
          •    Condensing turbines                                                  
          •    Non-condensing turbines                                              
          •    Reheat turbines                                                      
          •    Extracting turbines                                                  
          •    Induction turbines                                                   
                                                                           (Any 3 x 1)        (3)

5.6       Advantages of gas turbine:
          •    High power output from a given weight of engine.                      
          •    The torque output permits a notable simplification of the transmission
               system.                                                               
          •    Smooth vibrationless running due to absence of reciprocating parts. 
          •    No rubbing parts such as piston so that internal friction and wear are
               almost eliminated.                                                    
          •    Easy starting.                                                        
          •    Can use a wide range of fuels without expensive anti-knock
               additives.                                                            
          •    Low lubricating oil consumption.                                      
          •    No water cooling system needed.                                       
          •    Non-poisonous exhaust gases gives very little trouble with pollution. 
          •    Requires little routine maintenance.                                  
                                                                           (Any 3 x 1)        (3)
                                                                                              [40]

Copyright reserved                                                         Please turn over

### Page 13

Downloaded from hlayiso.com
Mechanical Technology                                13                             DBE/November 2012
                                              NSC – Memorandum

QUESTION 6: FORCES, SYSTEMS AND CONTROL

6.1       Hydraulics:

        6.1.1    Fluid pressure:

                 Rounded off                                     Fully extended
                        πD 2                                           πD 2
                   AA =                                         AA =                              
                          4                                              4
                         π × (0,038 )                                  π × (0,038 )
                                     2                                             2
                      =                                              =
                               4                                             4
                                                                                                  
                      = 1,13 × 10 −3 m 2                            = 0,001134114 m 2
                        F                                             F
                    P= A                                         P= A                              
                        AA                                           AA
                                300                                      300
                        P=                                      P=
                             1,13 × 10 −3                            0,001134114                   
                           = 0,27 MPa                             = 264523,45 Pa
                                                                                                          (5)

        6.1.2    Load lifted by piston B:

                 Rounded off                                     Fully extended

                          πD 2                                        πD 2
                     AB =                                       AB =                              
                            4                                           4
                           π × (0,175 )                                π × (0.175 )
                                       2                                           2
                        =                                           =
                                 4                                           4
                        = 24,05 × 10 −3 m 2                        = 0,024052818 m 2              

                         FB                                           FB
                 P=                                             P=                                
                         AB                                           AB
                 FB = AB × P                                    FB = AB × P                       
                                                                  = (0,024052818 × 264523,45      
                       = (24,05 × 10 −3 ) × 0,27 × 10 6
                       = 6,49 kN                                  = 6362,54 N                     
                                                                                                           (6)

Copyright reserved                                                                      Please turn over

### Page 14

Downloaded from hlayiso.com
Mechanical Technology                            14                              DBE/November 2012
                                          NSC – Memorandum

6.2     Stress and Strain:

        Tensile force:

        Rounded off                                          Fully extended
            ΔL                                                  ΔL                             
        ε=                                                   ε=
           OL                                                    OL
            0,2                                                  0,2
         =                                                    =
            300                                                  300
         = 0,66 × 10 −3                                      = 0,0006666                       

             σ                                                   σ
         E=                                                 E=                                 
             ε                                                   ε
         σ = E×ε                                             σ = E×ε
            = 245 × 10 9 × 0.66 × 10 −3                       = 245 × 10 9 × 0,000666666       
                                                              = 163333333,3 Pa                 
            = 161.7 × 10 Pa
                          6

            F                                                   F
         σ=                                                  σ=
            A                                                  A                               
         F =σ×A                                              F =σ×A
                                                                                               
           = 161,7 × 10 × 2,2 × 10
                          6          −6
                                                              = 163333333,3 × 2,2 × 10 −6      
           = 355,74 N                                         = 359,33 N                       
                                                                                                        (9)

6.3     Belt drives:

        6.3.1    Rotation frequency of the driven pulley:
                 driven pulley:
                  N A × DA = N B × DB
                             N × DA        
                       NB = A
                                DB
                                1000 × 0,25
                              =
                                   0,35             
                              = 714,29 rpm
                                       r/min
                                                                                                       (3)

Copyright reserved                                                                   Please turn over

### Page 15

Downloaded from hlayiso.com
Mechanical Technology                             15                              DBE/November 2012
                                           NSC – Memorandum

        6.3.2    Power transmitted:

                 P=
                        (T1 − T2 )πDN
                           60                         

                 P=
                     (200 − 90 )× π × 0,25 × 1000     
                                  60
                 P = 1439,90 Watts
                 P = 1,44 kW                                                                            (3)

        6.3.3    Belt speed:

                     πDN                              
                 v=
                      60
                                                      
                      π × 0,25 × 1000
                   =
                             60                       
                 v = 13,09 m.s −1
                                                                                                         (3)

6.4     Gears:

        6.4.1    Rotation frequency of the output shaft:

                     N F TA × TC × TE               N F Product of driven gears
                        =                      or      =                                          
                     N A TB × TD × TF               N A Product of driver gears
                          TA × TC × TE × N A              24 × 20 × 42 × 1440                     
                 NF =                                 =
                            TB × TD × TF                     40 × 48 × 90                         
                          24 × 20 × 42 × 1440
                 NF =                                  = 168 rpm
                                                             r/min
                             40 × 48 × 90
                                                                                                  
                        = 168 rpm
                              r/min                                                                      (4)

        6.4.2    Velocity ratio:

                          NA
                 VR =                                 
                          NF
                      1440
                 VR =
                       168
                 VR = 8.57 : 1                        
                                                                                                         (2)

Copyright reserved                                                                    Please turn over

### Page 16

Downloaded from hlayiso.com
Mechanical Technology                       16          DBE/November 2012
                                     NSC – Memorandum

6.5     Differential wheel and axle:

        6.5.1    Mechanical advantage:

                      W                        
                 MA =
                      F
                      2400
                 MA =
                       400
                 MA = 6                                                       (2)

        6.5.2    Velocity ratio:

                          2D
                 VR =                          
                        d1 − d 2
                        2(210 )                
                 VR =
                      160 − 140
                      420
                 VR =
                       20
                 VR = 21 : 1                                                  (3)

        6.5.3    Mechanical efficiency:

                        MA
                 ηmech =    × 100%             
                        VR
                        6
                      = × 100%
                        21
                     = 28.57%                                                 (2)

Copyright reserved                                          Please turn over

### Page 17

Downloaded from hlayiso.com
Mechanical Technology                            17          DBE/November 2012
                                          NSC – Memorandum

6.6     Clutches:

        6.6.1    Diameter of clutch plate:

                 T = μWnR                             
                       T
                 R=                                   
                     μWn
                          336                         
                 R=
                     0,4 × 3500 × 2
                 R = 0,12 m
                 D = 2 × 0,12                         
                 D = 0,24 m
                   = 240 mm                           
                                                                           (5)

        6.6.2   Power transmitted at
                3500 rpm in kW:

                     2π NT
                 P=                               
                        60
                     2 × π × 3200 × 336
                 P=                               
                              60
                 P = 112,59kW                                            (3)
                                                                            [50]

                                                                   TOTAL:        200

Copyright reserved

## Related documents

- [Mechanical Technology Nov 2021 (Fitting Machining) MG Afr](https://www.hlayiso.com/library/mechanical-technology-nov-2021-fitting-machining-mg-afr/545abfde-8923-4811-aed9-fe51d7f86ad6) — Mechanical Technology · Memo · 2021 · Afr
- [MT Nov 2020 (Automotive) Memo Eng](https://www.hlayiso.com/library/mt-nov-2020-automotive-memo-eng/62bbe2de-9d39-485e-a55f-c51567c03422) — Mechanical Technology · Memo · 2020 · Eng
- [MT Nov 2020 (Fitting Machining) Memo Afr](https://www.hlayiso.com/library/mt-nov-2020-fitting-machining-memo-afr/c4d9ece6-0cba-42c1-a1e6-01ad8f18a69c) — Mechanical Technology · Memo · 2020 · Afr
- [MT Nov 2020 (Automotive) Memo Afr](https://www.hlayiso.com/library/mt-nov-2020-automotive-memo-afr/3f9a8d78-97b6-4d96-a459-e2cd2e0f4240) — Mechanical Technology · Memo · 2020 · Afr
- [MT Nov 2020 (Welding Metalwork) Memo Eng](https://www.hlayiso.com/library/mt-nov-2020-welding-metalwork-memo-eng/1c2f46e0-7366-454c-93bb-25a7a72400b1) — Mechanical Technology · Memo · 2020 · Eng
- [MT Nov 2019 (Welding) Memo Eng](https://www.hlayiso.com/library/mt-nov-2019-welding-memo-eng/5d6b5e2a-2e34-47b1-b41f-207fe9bd9f6f) — Mechanical Technology · Memo · 2019 · Eng

## More from Grade 12 Mechanical Technology

- [Gr 12 Weld Met ATP 2023-24](https://www.hlayiso.com/library/gr-12-weld-met-atp-2023-24/8ecc8de5-c625-4611-9b94-1f3fd325e05d) — ATP
- [Mechanical Technology Nov 2024 (Automotive) Eng](https://www.hlayiso.com/library/mechanical-technology-nov-2024-automotive-eng/5d2e2bd0-b02f-4218-849f-e7933bc6c4d1) — QP · 2024 · Eng
- [Mechanical Technology Nov 2024 (Fitting and Machining) Afr](https://www.hlayiso.com/library/mechanical-technology-nov-2024-fitting-and-machining-afr/1e7205e4-aef1-4cc4-afdf-aac93562ae67) — QP · 2024 · Afr
- [Mechanical Technology May-June 2024 (Welding and Metalwork) MG Afr](https://www.hlayiso.com/library/grade-12-mechanical-technology-2024-nsc-june-exam-memorandum-afrikaans/552ebeca-d294-4c37-b3fb-cded79fa438f) — Memo · 2024 · Afr
- [Mechanical Technology May-June 2024 (Automotive) Eng](https://www.hlayiso.com/library/grade-12-mechanical-technology-2024-nsc-june-exam-question-paper-english/10859467-be3e-41fb-aee0-b797c385437d) — QP · 2024 · Eng
- [MECH TECH GR12 MEMO SEPT2023 Welding Afrikaans](https://www.hlayiso.com/library/mech-tech-gr12-memo-sept2023-welding-afrikaans/f45c553c-63a1-4e1c-b9c6-d426a3dfef26) — Memo · 2023 · Afr

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
