# Mechanical Technology Grade 12 NSC November 2018 Memo (Eng)

> Grade 12 Mechanical Technology memorandum from the NSC November Exam 2018 (English), 21 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mechanical-technology-nov-2018-fitting-machining-final-memo-eng/40d94883-0687-412a-b074-581e9681770d
- Updated: 2026-09-27
- Subject: Mechanical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: NSC November Exam
- Curriculum: CAPS
- Pages: 21

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 2018 Memo (Eng)](https://www.hlayiso.com/library/mechanical-technology-nov-2018-fitting-machining-final-memo-eng/40d94883-0687-412a-b074-581e9681770d)

## About this document

Mechanical Technology · Grade 12 · NSC November Exam · 2018. Memorandum, 21 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: FITTING AND MACHINING

                                  NOVEMBER 2018

                               MARKING GUIDELINES

MARKS: 200

                      These marking guidelines consist of 21 pages.

Copyright reserved                                                    Please turn over

### Page 2

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      2                       DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 1: MULTIPLE-CHOICE QUESTIONS (GENERIC)

1.1       A                                                                                    (1)

1.2       C                                                                                    (1)

1.3       A                                                                                    (1)

1.4       B                                                                                    (1)

1.5       D                                                                                    (1)

1.6       A                                                                                    (1)

                                                                  TOTAL QUESTION 1:             [6]

Copyright reserved                                                           Please turn over

### Page 3

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      3                       DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 2: SAFETY (GENERIC)

2.1       Angle grinder: (Before using)
            • The safety guard must be in place before starting. 
            • Protective shields must be placed around the object being grinded
                to protect the people around. 
            • Use the correct grinding disc for the job. 
            • Make sure that there are no cracks in the disc before you start. 
            • Protective clothing and eye protection are essential. 
            • Check electrical outlets and cord/plugs for any damages. 
            • Ensure that lockable switch is disengaged. 
            • Ensure that the disc and the nut are well secured. 
            • Ensure that the removable handle is secured. 
                                                                       (Any 2 x 1)              (2)

2.2       Welding goggles:
            • To protect your eyes against sparks 
            • To protect your eyes against heat 
            • To be able to see where to weld 
            • To protect your eyes from UV rays 
                                                                          (Any 2 x 1)           (2)

2.3       PPE for Hydraulic Press:
            • Overall 
            • Safety shoes / boots
            • Safety goggle 
            • Leather gloves 
            • Face shield 
                                                                          (Any 2 x 1)           (2)

2.4       Workshop layouts:
             • Process layout 
             • Product layout                                                                  (2)

2.5       Employer’s responsibility regarding first-aid:
            • Provision of first-aid equipment 
            • First aid training 
            • First-aid services by qualified personnel 
            • Any first aid procedures / treatment 
            • Display first aid safety signs 
            • First aid personnel must be identified by means of arm bands or
               relevant personal signage 
                                                                   (Any 2 x 1)                  (2)

                                                                  TOTAL QUESTION 2:        [10]

Copyright reserved                                                           Please turn over

### Page 4

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      4                       DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 3: MATERIALS (GENERIC)

3.1       Bending test:
            • Ductility 
            • Malleability 
            • Brittleness 
            • Flexibility 
                                                                          (Any 1 x 2)           (2)

3.2       Heat-treatment:

          3.2.1       Annealing:
                        • To relieve internal stresses 
                        • To soften the steel 
                        • To make the steel ductile 
                        • To refine the grain structure of the steel 
                        • To reduce the brittleness of the steel 
                                                                          (Any 2 x 1)           (2)

          3.2.2       Case hardening:
                        • To require a wear resistant surface  and it must be
                            tough enough internally  at the core to withstand the
                            applied loads.
                        • Hard case  and tough core. 
                                                                       (Any 1 x 2)              (2)

3.3       Tempering process:
            • To reduce  the brittleness  caused by the hardening process.
            • Relieve  strain  caused during hardening process.
            • Increase  the toughness of the steel. 
                                                                     (Any 1 x 2)                (2)

3.4       Factors for heat-treatment processes:
             • Heating temperature / Carbon content 
             • Soaking (Time period at temperature) / Size of the work piece 
             • Cooling rate / Quenching rate                                                   (3)

3.5       Hardening of steel:
             • Steel is heated to 30 – 50°C above the higher critical temperature.
               (AC 3 ) 
             • It is then kept at that temperature to ensure (soaking) that the whole
               structure is Austenite. 
             • The steel is then rapidly cooled by quenching it in clean water, brine
               or oil.                                                                         (3)

                                                                  TOTAL QUESTION 3:        [14]

Copyright reserved                                                           Please turn over

### Page 5

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      5                       DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 4: MULTIPLE-CHOICE QUESTIONS (SPECIFIC)

4.1       C                                                                                    (1)

4.2       A                                                                                    (1)

4.3       D                                                                                    (1)

4.4       A                                                                                    (1)

4.5       B                                                                                    (1)

4.6       A                                                                                    (1)

4.7       B                                                                                    (1)

4.8       B                                                                                    (1)

4.9       D                                                                                    (1)

4.10      C                                                                                    (1)

4.11      B                                                                                    (1)

4.12      D                                                                                    (1)

4.13      D                                                                                    (1)

4.14      C                                                                                    (1)

                                                                  TOTAL QUESTION 4:        [14]

Copyright reserved                                                           Please turn over

### Page 6

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      6                     DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 5: TERMINOLOGY (LATHE AND MILLING MACHINE) (SPECIFIC)

5.1       Advantages of using the tailstock to cut an external taper:
            • Long an accurate taper can be cut. 
            • The automatic feed can be used which result in a good finish.                   (2)

5.2       Calculate the compound slide set-over:
                  θ D−d        
              Tan =
                  2     2L
                 θ 60 - 28
             Tan =             
                 2     2 × 85  
                    = 0,188
                 θ
                    = 10,66 °  
                 2

                                                  OR

                       28                           60

                                            Ɵ
                                            2     X
                                       85

                  D−d
              X=                                                   
                    2                                  θ 16
                                                   Tan =
                  60 - 28                             2 85
                =                                                   
                     2                                θ
                                                       = 10,66 °
                = 16 mm                               2                                        (5)

5.3       Centre gauge:
            • To measure the form and angle of the screw cutting tool angle while
                grinding the tool 
            • To set the screw cutting tool square/perpendicular to the axis of the
                work piece                                                                    (2)

5.4       Parallel key:
          Length:
          Length = 1,5 × diameter 
                  = 1,5 × 42      
                  = 63 mm                                                                     (3)

Copyright reserved                                                          Please turn over

### Page 7

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      7                       DBE/November 2018
                                       NSC – Marking Guidelines

5.5       Advantages of up-cut milling:
            • Deeper cuts can be made as the cutting pressure on the cutter is
               lower than down cut milling. 
            • The process enables hard steel to be cut, because the total cutting
               pressure is absorbed by the material at the back of the edge. 
            • Metal with hard scale, such as castings or forgings, the cut is started
               under the scale where the material is softer which extends the life of
               the cutter. 
            • A quicker/course feed can be used. 
            • The strain on the cutter and arbour will be less. 
            • Vibration is limited 
            • Good finish 
            • Low noise level 
                                                                         (Any 2 x 1)            (2)

5.6       Disadvantage of down-cut milling:
             • Vibration in the arbour is unavoidable. 
             • A fine feed must be used. 
             • When milling a material with hard scale the milling cutter will be
                damaged. 
             • Process takes time because of slower feed. 
             • Noisy process. 
             • Bad finish because of vibration. 
                                                                          (Any 2 x 1)           (2)

5.7       Methods of centring a milling cutter:
            • Square and ruler method. 
            • Set-over method by milling machine dial. 
            • Dial indicator method 
            • Using reference points on digital read out equipment 
                                                                          (Any 2 x 1)           (2)

                                                                  TOTAL QUESTION 5:        [18]

Copyright reserved                                                           Please turn over

### Page 8

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      8                             DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 6: TERMINOLOGY (INDEXING) (SPECIFIC)

6.1       Spur gear:
          Chordal tooth thickness:
                         90°                                   90°
            t c = Tmsin                          t c = PCDsin              
                          T                                     T
                            90°                            90°             
                = 50 × 3sin                or      = 150sin
                            50                               50
                = 50 × 3(0,03141)                 = 150 × 0,03141          
                = 4,71 mm                         = 4,71 mm                
                                                                                                      (4)

6.2       Calculate simple indexing:

                              40
          Simple Indexing =
                              N
                              40
                            =             
                              13
                                1         
                            =3
                               13
                                1 3
                            =3    ×
                               13 3
                                3         
                            =3
                               39
                                          
                             3 full turns and 3 holes in a 39 hole circle                             (4)

Copyright reserved                                                                 Please turn over

### Page 9

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining      9                      DBE/November 2018
                                       NSC – Marking Guidelines

6.3       Differential indexing:

          6.3.1       Indexing required:
                                 40 40
                      Indexing =    =
                                  n 127
                                 40 40 5                          
                               =    =     ÷
                                  A 125 5
                                  8                               
                               =
                                 25
                      Indexing = 8 holes on the 25 hole circle                                 (3)

          6.3.2       Change gears required:
                      Dr A - n 40       
                        =       ×
                      Dn A        1
                          125 - 127 40 
                        =           ×
                             125      1
                             2 40
                        =      ×
                          125 1
                          - 80 5        
                        =      ÷
                          125 5
                          - 16 4
                        =      ×
                           25 4
                          - 64
                        =               
                          100                                                                  (5)

          6.3.3       Direction of rotation of index plate:
                      The index plate will turn the opposite  direction as the index
                      crank handle.                                                             (1)

Copyright reserved                                                           Please turn over

### Page 10

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     10                              DBE/November 2018
                                       NSC – Marking Guidelines

6.4       Calculate distance “x” between rollers:

          " x" = 150 + 2(AB ) − 2(CD) − 2r
                                                          B                      A
                   θ = BC
                 tan             
                       AB                                                  60°
                        BC
                  AB = θ
                       tan
                         35
                     =           
                       tan 60°                                                   35
                     = 20,207 mm 
                     = 20,21 mm

                                                          C

                  θ = DE
                tan
                      CD                                                        E
                       15
                 CD = θ                                                              15
                      tan               
                        15                                        30°
                    =                                 C
                      tan30°                                                     D
                    = 25,98 mm          

          " x" = 150 + 2(AB ) − 2(CD) − 2r           
             = 150 + 2(20,21) − 2(25,98 ) − 2(15 )
             = 150 + 40,42 − 51,96 − 30              
             = 108,454 mm
             = 108, 45 mm                            
                                                                                                         (9)

6.5       Reasons for balancing work piece on a centre lathe:
            • Prevent unnecessary bearing loads 
            • Prevent excessive vibration 
            • To obtain a good finish 
            • To prevent clatter on the gear teeth 
            • To prevent the spindle from bending 
                                                                                 (Any 2 x 1)             (2)

                                                                        TOTAL QUESTION 6:           [28]

Copyright reserved                                                                    Please turn over

### Page 11

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     11                       DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 7: TOOLS AND EQUIPMENT (SPECIFIC)

7.1       Hardness testers:
             • Brinell-hardness tester 
             • Rockwell-hardness tester 
             • Vickers 
                                                                          (Any 2 x 1)           (2)

7.2       Moment tester:
          To determine the reactions  on either side of a simply loaded beam.                 (2)

7.3       Tensile test:
          A piece of material is subjected to an increasing axial load  while
          measuring  the corresponding elongation  of the material.                           (3)

7.4       Depth micro-meter:
                                         
          Reading = 100 + 11,00 + 0,50 + 0,09
                      
                  = 111,59 mm                                                                   (5)

7.5       Measure depth:
          Vernier calliper                                                                     (1)

                                                                  TOTAL QUESTION 7:         [13]

Copyright reserved                                                           Please turn over

### Page 12

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     12                      DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 8: FORCES (SPECIFIC)

8.1       Forces:

                                                                                           (15)

       Horizontal                                        Vertical
                               Magnitudes                               Magnitudes
     Components                                     Components 
   1150cos0°                     1150N            1150sin0°                 0N
   800cos150°                   -692,82N          800sin150°             400N 
   650cos200°                   -610,80N          650sin200°           -222,31N 
   550cos270°                       0N             550sin270°             -550N 
              TOTAL:            -153,62N                      TOTAL:   -372,31N 

Copyright reserved                                                          Please turn over

### Page 13

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     13                       DBE/November 2018
                                       NSC – Marking Guidelines

8.2       Moments:

           2800 N                                                 700 N
                                                           “x”

                            O
                     1m                                    5m

          Calculate “x”:
          Take moments about O.

             ∑ RHM = ∑ LHM
                                 
           700 × " x" = 2800 × 1
           700 × " x" = 2800     
                         2800       
                  " x" =
                          700
                  " x" = 4 m                                                                   (4)

8.3       Stress and Strain:

          8.3.1      Type of stress:
                     Compressive stress                                                    (1)

          8.3.2      Stress:
                         A=
                                (
                           π D2 - d2    )              
                                 4
                                (
                           π 0,04 2 − 0,03 2
                           =
                                               )
                                     4
                         A = 0,55 × 10 -3 m2           

                       σ =F                        
                           A
                            50 × 10 3
                         =
                           0,55 × 10 −3            
                       σ = 90,91× 10 6 Pa
                       σ = 90,91 MPa               

                                            (NO UNIT – NO MARK)                             (5)

Copyright reserved                                                           Please turn over

### Page 14

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     14                       DBE/November 2018
                                       NSC – Marking Guidelines

          8.3.3       Change in length:
                       σ
                      Eε=                        
                       σ
                     ε =
                          E
                          90,91× 10 6
                        =
                            90 × 10 9
                        = 1,01× 10 −3            

                                      (IF ANY UNIT IS GIVEN – NO MARK)
                     Δ
                   ε =L                              
                        L
                  Δ ε= ×L
                   L                                 
                           (            )
                     = 1,01× 10 −3 × (80 )
                         = 0,08 mm                                                         (5)

          8.3.4       Safety factor:
                                               Break stress
                               Safety factor =                     
                                            Safe working stress
                                            Break stress
                      Safe working stress =
                                            Safety factor
                                              600 × 10 6
                                            =                      
                                                  4
                                            = 150 × 10 6 Pa
                                            = 150 MPa                                      (3)

                                                                  TOTAL QUESTION 8:         [33]

Copyright reserved                                                           Please turn over

### Page 15

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     15                     DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 9: MAINTENANCE (SPECIFIC)

9.1       Lack of preventative maintenance:
             • Risk of injury or death. 
             • Financial loss due to damage suffered as a result of part failure and
                the waste of material. 
             • Loss of valuable production time.                                              (3)

9.2       Causes for the malfunctioning of chain drive systems:
            • Lack of or incorrect lubrication 
            • Lack of maintenance 
            • Overloading 
            • Misalignment of sprockets 
            • Incorrect chain tension 
            • Contamination of chain drive system such as dust or sand 
                                                                      (Any 2 x 1)              (2)

9.3       Procedures to reduce the physical wear on a belt drive system:
             • Check the belt alignment. 
             • Checking the belt tension. 
             • Prevent overloading of the system. 
             • Keep the pulleys and belt clean. 
             • Check that all covers are secure. 
                                                                     (Any 2 x 1)               (2)

9.4       Procedures to replace the belt on a belt drive system:
             • Ensure that the machine is switched off 
             • Release the tension on the belt 
             • Remove the belt from the pulleys 
             • Fit the correct size replacement belt onto the pulleys 
             • Check the pulley alignment 
             • Apply adequate tension according to specification and lock the
               system 
                                                                        (Any 5 x 1)            (5)

Copyright reserved                                                          Please turn over

### Page 16

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     16                       DBE/November 2018
                                       NSC – Marking Guidelines

9.5       Properties of materials:

          9.5.1      Poly vinyl chloride (PVC):
                        • Flexible 
                        • Rubber-like substance 
                        • Makes a dull sound when dropped 
                        • Tough 
                        • Act as an insulator 
                        • It is durable 
                        • Highly resistant to oxidative material 
                        • Oil, water and chemical resistant 
                                                                          (Any 1 x 1)           (1)

          9.5.2      Carbon fibre:
                        • Strong 
                        • Tough 
                        • Light weight 
                        • Good electrical conductor 
                                                                          (Any 1 x 1)           (1)

9.6       Difference between “Thermoplastic” and “Thermo                   hardened
          (thermosetting)” composites:
          Thermoplastics can be reheated and deformed. / Recyclable 
          Thermo hardened cannot be reheated. / Non-recyclable                                 (2)

9.7       Examples of thermo hardened composites:
            • Carbon fibre or (Any application) 
            • Glass fibre or (Any application) 
            • Bakelite or (Any application) 
            • Teflon or (Any application) 
                                                                          (Any 2 x 1)           (2)

                                                                  TOTAL QUESTION 9:         [18]

Copyright reserved                                                           Please turn over

### Page 17

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     17                            DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 10: JOINING METHODS (SPECIFIC)

10.1      Square thread:

          10.1.1      The lead of the thread:
                      Lead = pitch × no of starts      
                           = 5×2
                           = 10 mm                                                                  (2)

          10.1.2      The helix angle of the thread:

                                   θ =               lead
                      Helix angle tan                                      
                                             pitch cercumfrence
                                                        10                 
                                         =
                                         π ×  outside dia − 1 pitch    
                                                             2        
                                                 10
                                         =
                                         π   × (82 − 2,5)                  
                                         = 0,0400
                                     θ = 2,29° / 2°17'24"                  

                                                        OR

                                    θ =           lead
                      Helix angle tan                             
                                             pitch diameter
                                           10                     
                                         =                        
                                        82 − 2,5
                                    θ = 7,17° / 7°10'12"                                            (5)

          10.1.3      The leading tool angle:

                      Leading tool angle = 90° − (helix angle + clearance angle ) 
                                         = 90° − (2,29° + 3°)
                                                                                  
                                         = 84,71° / 84°42'36"

                                                        OR

                      Leading tool angle = 90° − (helix angle + clearance angle ) 
                                         = 90° − (7,17° + 3°)
                                         = 79,83° / 79°49'48"                                       (2)

Copyright reserved                                                                Please turn over

### Page 18

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     18                          DBE/November 2018
                                       NSC – Marking Guidelines

          10.1.4      The following tool angle:

                      Following tool angle = 90° + (helix angle − clearance angle ) 
                                           = 90° + (2,29° − 3°)
                                           = 89,29° / 89°17'24"                     

                                                        OR

                      Following tool angle = 90° + (helix angle − clearance angle ) 
                                           = 90° + (7,17° − 3°)
                                           = 94,17° / 94°10'12"                                    (2)

10.2      Measurements of a screw thread :

          10.2.1      Metric screw thread                                                          (1)

          10.2.2      Crest / Major / External / Basic / Nominal / Outside diameter                (1)

          10.2.3      Pitch                                                                        (1)

10.3      Angles of a square thread cutting tool:
          A – Helix angle 
          B – Clearance angle 
          C – Leading tool angle 
          D – Following tool angle                                                                 (4)

                                                                  TOTAL QUESTION 10:           [18]

Copyright reserved                                                               Please turn over

### Page 19

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     19                                    DBE/November 2018
                                       NSC – Marking Guidelines

QUESTION 11: SYSTEMS AND CONTROL (DRIVE SYSTEMS) (SPECIFIC)

11.1      Advantages of a belt drive system compared to a chain drive system:
            • Silent operation 
            • Less expensive 
            • Drive can take place over a longer distance 
            • No lubrication needed 
                                                                    (Any 2 x 1)                              (2)

11.2      Hydraulics:

          11.2.1      Fluid pressure:
                         π 2
                      AA = d                                
                             4
                         π (0,032)2
                         =
                                4
                         = 0,8 × 10 −3 m2                   

                            FA
                      p=                                    
                            AA
                               120
                        =
                            0,8 × 10 −3
                        = 0,1492 × 10 6 Pa
                        = 0,15 MPa or 149207,76 Pa
                                                       
                                           (NO UNIT – NO MARK)                                               (4)

          11.2.2      Diameter of the ram:
                           F                               FB  F
                        p= A                                 = A                               
                          AA                               AB A A
                          F                                           A A × FB
                      AB = B                                AB =
                           p                      OR                     FA

                            =
                                 18 × 10 3                       =
                                                                      (0,8 × 10 )× (18 × 10 ) 
                                                                                 −3         3

                                0,15 × 10 6                                      120
                                                                             2
                            = 0,12 m   2
                                                                 = 0,12 m                      

                         π        2
                      AB = d                  
                            4
                             4A
                      d =π

                                 4 × 0,12     
                            =     π
                            = 0,39088 m       
                            = 390,88 mm                                                                      (6)
Copyright reserved                                                                        Please turn over

### Page 20

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     20                                DBE/November 2018
                                       NSC – Marking Guidelines

11.3     Hydraulic symbols: One-way valve

                                                       

                                                                                                          (1)

11.4      Belt drives:
          Rotation frequency of the drive pulley:
          NDRD DR = NDND DN      
                    N × D DN
             NDR = DN            
                       D DR
                    80 × 240     
                  =
                       75
                  = 256 r/min                                                                            (4)

11.5      Gear drives:

          11.5.1 Rotation frequency of the output:
                 N A Product of Driven gears
                    =
                 ND Product of Driver gears
                     ND TA × TC                
                        =
                     N A TB × TD
                         T × TC × N A          
                     ND = A
                            TB × TD
                          20 × 25 × 3000       
                         =                     
                              35 × 30
                          1428,57 r/min
                     ND =                      
                                60
                        = 23,81 r/sec          

                                                      OR
                                                                  NB = NC = 1714,29 r/min
                     NB × TB = NA × TA         
                                 NA × TA                   ND × TD = NC × TC             
                         NB =
                                   TB                                    NC × TC
                                                                  ND =
                               3000 × 20                                   TD
                             =
                                  35                                   1714,29 × 25      
                                                                     =
                             = 1714,29 r/min                               30
                                                                       1428,57 r/min     
                                                                     =
                                                                            60
                                                                                         
                                                                     = 23,81r/sec                         (6)

Copyright reserved                                                                     Please turn over

### Page 21

Downloaded from hlayiso.com
Mechanical Technology: Fitting and Machining     21                        DBE/November 2018
                                       NSC – Marking Guidelines

          11.5.2 Gear ratio:
                                 Product of the number of teeth on driven gears 
                     Gear ratio =
                                 Product of the number of teeth on driver gears
                                 35 30
                               =    ×                                           
                                 20 25
                               = 2,1 : 1                                                        (3)

11.6      Work done:
          Work done = F × s                      
                        = 250 × 15
                        = 3750 Joule or N.m                                                     (2)

                                                                  TOTAL QUESTION 11:         [28]

                                                                             TOTAL:          200

Copyright reserved                                                            Please turn over

## Related documents

- [Mechanical Technology Nov 2018 (Welding Metalwork) FINAL Memo Afr](https://www.hlayiso.com/library/mechanical-technology-nov-2018-welding-metalwork-final-memo-afr/f10b5a6d-e8cc-4439-b3a7-3af568e5497b) — Mechanical Technology · Memo · 2018 · Afr
- [Mechanical Technology Nov 2018 (Fitting Machining) FINAL Memo Afr](https://www.hlayiso.com/library/mechanical-technology-nov-2018-fitting-machining-final-memo-afr/8fd509f9-c09f-4420-ab1a-3185bd4f5b41) — Mechanical Technology · Memo · 2018 · Afr
- [Mechanical Technology Nov 2018 (Automotive) FINAL Memo Afr](https://www.hlayiso.com/library/mechanical-technology-nov-2018-automotive-final-memo-afr/caf8f192-6dc9-4b09-9b6d-55e098484396) — Mechanical Technology · Memo · 2018 · Afr
- [Mechanical Technology Nov 2018 (Fitting Machining) Eng](https://www.hlayiso.com/library/mechanical-technology-nov-2018-fitting-machining-eng/ed9ad8c2-0759-4c20-a789-0e5dd65fb5fc) — Mechanical Technology · QP · 2018 · Eng
- [Mechanical Technology Nov 2018 (Welding Metalwork) Afr](https://www.hlayiso.com/library/mechanical-technology-nov-2018-welding-metalwork-afr/a3b6bb12-7bae-437f-9fd5-c43dec461d29) — Mechanical Technology · QP · 2018 · Afr
- [Mechanical Technology Nov 2018 (Fitting Machining) Afr](https://www.hlayiso.com/library/mechanical-technology-nov-2018-fitting-machining-afr/578c96a1-0a1e-4ea4-9237-a60a84bd962e) — Mechanical Technology · QP · 2018 · Afr

## 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.
