# Mechanical Technology Grade 12 NSC November 2025 Memo (Eng)

> Grade 12 Mechanical Technology memorandum from the NSC November Exam 2025 (English), 27 pages. Read online or download the free PDF.

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- Updated: 2026-09-28
- Subject: Mechanical Technology
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: NSC November Exam
- Curriculum: CAPS
- Pages: 27

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## About this document

Mechanical Technology · Grade 12 · NSC November Exam · 2025. Memorandum, 27 pages. Read online or download the PDF.

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### Page 1

NATIONAL
                              SENIOR CERTIFICATE

                                    GRADE 12

           MECHANICAL TECHNOLOGY: FITTING AND MACHINING

                                  NOVEMBER 2025

                              MARKING GUIDELINES

MARKS: 200

                     These marking guidelines consist of 27 pages.

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Mechanical Technology: Fitting and Machining      2                       DBE/November 2025
                                      NSC – Marking Guidelines

 Instructions to marker: Mechanical Technology – Fitting and Machining

 1.        General marking instructions:

           1.1        During marking of multiple-choice questions, only ONE possible
                      answer can be accepted. If the candidate indicated two or more
                      responses, ONLY the first response will be recognised
                      /acknowledged and marked according to marking guideline.

           1.2        Where the number of responses of the candidate exceeds the
                      required number stated in the question, only the first required
                      number of responses will be accepted. E.g. if the question states,
                      'Name THREE …' and the candidate stated four different
                      responses, ONLY the first three will be accepted. The remaining
                      responses must be ignored with a line drawn through it.

           1.3        When the question clearly indicates what is expected of the
                      candidate:

                      (a) If the candidate is required to describe a process step by step
                          (e.g., a process in four steps), only the first four responses
                          should be considered.
                      (b) However, if the candidate is required to explain a process, it
                          should be noted that the response may be lengthy and not
                          necessarily well organized. In such cases, the marker must
                          evaluate the entire response to determine whether the
                          candidate has explained the required outcome satisfactorily
                          and allocate marks on merit.

           1.4        If question numbering of sub-questions is not correct according to
                      question paper sequence, the responses can be accepted if a
                      sequence pattern can be identified or followed.

           1.5        Attention must be given to mark allocation in questions where
                      two (2) marks were allocated to one response, e.g. (Any 1 x 2).

           1.6        ONE tick must be allocated for each mark awarded and no global
                      marking is allowed.

           1.7        Unanswered and incorrect responses of questions must be
                      indicated with a cross (X).

           1.8        All blank pages in an ANSWER BOOK must be crossed out to
                      indicate that the pages have been seen by the marker.

           1.9        Attention must be given to questions where the candidate did not
                      complete the answer and continued answer(s) on subsequent
                      pages in the answer book. Continue marking such answers and
                      award marks, if applicable. If a question is answered twice, mark
                      the first response.

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Mechanical Technology: Fitting and Machining      3                           DBE/November 2025
                                      NSC – Marking Guidelines

           1.10       During calculation marking, pay attention to the position of the tick
                      mark(s) in the marking guideline.

           1.11       If a candidate skipped a step where a tick mark is allocated in the
                      marking guideline and the final answer is correct, the total amount
                      of tick marks must be indicated by the marker to indicate full
                      marks awarded for the question correctly answered so that the
                      examination assistants can verify marks awarded.

           1.12       If the unit required for calculations is specified in the question, the
                      final answer can be considered correct without explicitly stating
                      the unit, provided that the numerical value matches the unit
                      specified or used in the question.

           1.13       If an incorrect unit is stated in the candidate's answer during
                      calculations, the answer will be marked incorrect even if the value
                      itself is correct.

           1.14       Attention must be given to calculation type questions when
                      different methods are used by candidates to obtain the correct
                      answer. Consider answers where candidates have rounded off in
                      sub-questions and it affects the final answer.

           1.15       Attention must be given to special marking instructions to the
                      marker which are stated for specific questions in the marking
                      guideline.

           1.16       A red line must be drawn by the marker to indicate the end of each
                      question.

           1.17       Marks of sub-questions must be indicated in the outside margins
                      of the ANSWER BOOK pages and NOT in the inside margins.

           1.18       When sub-question marks are allocated in the outside margins
                      they must be in line with the sub-question and NOT be circled.

           1.19       The total marks awarded for the question must be written on the
                      red line drawn by the marker and circled to clearly indicate the
                      marks awarded for that particular question.

           1.20       Marker must transfer marks to cover page and ensure
                      correctness of transfer.

           1.21       Markers should familiarise themselves with the question and
                      answer before evaluating the responses of candidates.

           1.22       Always interpret the responses of the candidates within the
                      context of the question.

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Mechanical Technology: Fitting and Machining      4                        DBE/November 2025
                                      NSC – Marking Guidelines

           1.23       Mark what the candidate wrote and do not make assumptions
                      about or predict intended responses.

           1.24       Accept incorrect spelling in answers unless the spelling changes
                      the meaning of the answer.

           1.25       Pencil work is normally acknowledged as rough work and should
                      not be considered/marked.

           1.26       As a rule, marks are not awarded for formulas that are found in
                      the attached formula sheet. If a formula was correctly
                      manipulated, then only can a mark be awarded. Marks will also
                      be awarded for correct formulas given by candidates if the
                      particular formula is not found in formula sheet. Marks are
                      awarded for correct substitution.

           1.27       Incorrect values from previous calculations substituted at the right
                      place in the formula will be awarded marks.

           1.28       Markers must pay attention to calculations where an incorrect
                      answer from a previous step was correctly substituted in the next
                      calculation. In such cases, they must write “sub” next to the tick
                      mark on the script to indicate that the mark was awarded for
                      correct substitution only.

           1.29       If during the calculations of a candidate, the formula that appear
                      on the formula page was not indicated correctly but calculations
                      and steps were done correctly, full marks may be awarded.

           1.30       If, during a candidate’s calculations, the formula from the formula
                      sheet was transferred incorrectly, no marks should be awarded
                      for the values substituted in that calculation, as it renders the
                      solution mathematically incorrect.

 2.        Specific marking instructions:

           2.1        In Questions 9.3 and 9.4, the answers must be in context of the
                      subject.

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Mechanical Technology: Fitting and Machining      5              DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 1: MULTIPLE-CHOICE (GENERIC)

 1.1        D✓                                                                         (1)

 1.2        D✓                                                                         (1)

 1.3        A✓                                                                         (1)

 1.4        C✓                                                                         (1)

 1.5        A✓                                                                         (1)

 1.6        B✓                                                                         (1)
                                                                                       [6]

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Mechanical Technology: Fitting and Machining      6                    DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 2: SAFETY (GENERIC)

 2.1       Examination procedures:
           • Breathing irregularities. ✓
           • Abnormal position of limbs that indicate breakages. ✓
           • Any visible signs of bleeding. ✓
           • Level of consciousness. ✓
                                                                         (Any 2 x 1)          (2)

 2.2       Arc welding safety:
           • An operator must be adequately trained. ✓
           • A workplace must be effectively partitioned off. ✓
           • An operator uses protective equipment. ✓
           • Ensure that all equipment is in safe working conditions. ✓
           • Keep proper housekeeping to avoid fire hazards and electric shocks. ✓
                                                                        (Any 2 x 1)           (2)

 2.3       Portable angle grinder (PPE excluded):
           • Safety guard must be in place. ✓
           • Protective shields must be placed around the area where grinding is
              taking place. ✓
           • Use the correct disc. ✓
           • Ensure the grinder is in good and safe working order.
           • Make sure there are no cracks on the disc. ✓
           • Ensure the lockable switch is in the off position. ✓
           • Check electrical cables for any defects. ✓
           • Ensure that the surrounding area is dry. ✓
           • Ensure the grinder is in good and safe working order. ✓
           • Ensure the disc is securely fastened. ✓
           • Clamp work piece firmly. ✓
                                                                     (Any 2 x 1)              (2)

 2.4      Handling gas cylinders:
          • Never stack cylinders on top of one another. ✓
          • Cylinders must not be stored in direct sunlight. ✓
          • Chain/handle cylinders in upright position. ✓
          • Valve guards must be on cylinders. ✓
          • Don't work or bang on cylinders. ✓
          • Never allow cylinders to fall. ✓
          • Do not allow oil or grease to come into contact with oxygen fittings. ✓
          • Do not interfere with the structure (no modification) of the gas
             cylinders. ✓
          • Store full and empty cylinders separately. ✓
          • Store acetylene and oxygen cylinders separately. ✓
          • Keep cylinders away from flammable substances. ✓
          • Do not drag/roll cylinders. ✓
          • Cylinders’ valves should be closed properly. ✓
                                                                          (Any 2 x 1)         (2)

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Mechanical Technology: Fitting and Machining      7                   DBE/November 2025
                                      NSC – Marking Guidelines

 2.5       Disadvantage of product layout:
           • Lack of flexibility. ✓
           • Optimum use of equipment is not possible. ✓
                                                                        (Any 1 x 1)          (1)

 2.6       Employee responsibilities:
           • Take reasonable care of the health and safety of himself and others. ✓
           • To cooperate with employer to comply with any duty or responsibility. ✓
           • Carry out any lawful order to him/her as to obey the health and safety
             rules. ✓
           • Report any unsafe conditions. ✓
           • Report any unhealthy conditions. ✓
           • Report any incident immediately. ✓
           • Report any injury immediately if possible. ✓
                                                                         (Any 1 x 1)         (1)
                                                                                              [10]

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Mechanical Technology: Fitting and Machining      8                        DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 3: MATERIALS (GENERIC)

 3.1       Bend test:

           3.1.1      Bends easily ✓                                                              (1)

           3.1.2      • Hard to bend ✓
                      • It will snap/break easily ✓
                      • Tends to crack/fracture easily✓
                                                                             (Any 1 x 1)          (1)

 3.2       Heat treatment:

           To change the properties ✓ of metals.                                                  (1)

 3.3       Quenching media:

           •   Water ✓
           •   Brine (salt and water) ✓
           •   Oil ✓
           •   Molten metal salts ✓
           •   Nitrogen ✓
           •   Air ✓
           •   Sand ✓
           •   Molten lead ✓
           •   Ash ✓
                                                                             (Any 3 x 1)          (3)

 3.4       Sound test:

           •   Tap with a hammer ✓
           •   Drop on the floor ✓                                                                (2)

 3.5       Purpose of case hardening:
           It is to produce a hard case ✓ over a tough core. ✓                                    (2)

 3.6       Case hardening methods:
           • Carburising ✓
           • Nitriding ✓
           • Cyaniding ✓
                                                                             (Any 2 x 1)          (2)

 3.7       Normalising:

           It is to relieve the internal stresses ✓ caused by machining/forging/welding. ✓        (2)
                                                                                                  [14]

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Mechanical Technology: Fitting and Machining      9              DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 4: MULTIPLE-CHOICE (SPECIFIC)

 4.1       B✓                                                                          (1)

 4.2       D✓                                                                          (1)

 4.3       B✓                                                                          (1)

 4.4       C✓                                                                          (1)

 4.5       A✓                                                                          (1)

 4.6       B✓                                                                          (1)

 4.7       C✓                                                                          (1)

 4.8       A✓                                                                          (1)

 4.9       C✓                                                                          (1)

 4.10      D✓                                                                          (1)

 4.11      B✓                                                                          (1)

 4.12      D✓                                                                          (1)

 4.13      A✓                                                                          (1)

 4.14      C✓                                                                          (1)
                                                                                       [14]

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Mechanical Technology: Fitting and Machining     10                              DBE/November 2025
                                      NSC – Marking Guidelines

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

 5.1       Disadvantages of set-over of the tailstock:
           • As the centres are not in line, a correct bearing surface on the centred
              work piece cannot be allowed for. ✓
           • Big cuts cannot be taken. ✓
           • Only small angled tapers can be turned. ✓
           • This method can only be used for external tapers. ✓
           • Can only use a faceplate and a centre, not a chuck. ✓
           • The work must be driven by using a dog clamp. ✓
           • Only practical when doing mass production. ✓
           • Set-over must be recalculated for different lengths but same included
              angle. ✓
                                                                          (Any 2 x 1)                   (2)

 5.2       Taper calculations:

           5.2.1     Small diameter:
                                                                             x
                                                                   tan 3  =
                                      θ D−d                                 510
                                tan     =
                                      2   2l                           x = 510 tan 3✓
                                    6 102 − d ✓                        x = 26,72796743 mm 
                             tan      =
                                    2   2  510 ✓
                                                            OR          d = D − 2x
                      1020  tan 3 = 102 − d                           d = 102 − 2 ( 26,73 ) ✓
                     53,4559348 7 = 102 − d                             d = 48,54 mm ✓
                                        d = 102 − 53,4559348 7 ✓
                                        d = 48,54 mm ✓                                                  (4)

           5.2.2     Set-over of the tailstock:
                                  L (D − d)
                     Set − over =
                                    2l
                                  670 (102 − 48,54) ✓
                                =
                                        2  510 ✓
                                = 35,12 mm ✓

                                         OR

                                       X
                          Tan       =
                                2       L
                             6     X
                         Tan     =     ✓
                             2 670
                     670 Tan 3  = X ✓                     NOTE: X = Set-over
                                    = 35,12 mm ✓                                                        (3)

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Mechanical Technology: Fitting and Machining     11              DBE/November 2025
                                      NSC – Marking Guidelines

 5.3       Key ways:

           5.3.1      Width:
                                   D
                       Width =
                                   4

                                   210
                               =       ✓
                                    4

                               = 52,50 mm ✓                                            (2)

           5.3.2      Thickness:
                                           D
                       Thickness =
                                           6

                                           210
                                       =       ✓
                                            6

                                       = 35 mm ✓                                       (2)

           5.3.3      Length:
                      Length = 1,5  diameter of shaft
                               = 1,5  210 ✓
                               = 315 mm ✓                                              (2)

 5.4      Balancing:
          • Unnecessary bearing load. ✓
          • Excessive vibration. ✓
          • Work that is not perfectly round.
          • Poor finish/surface finish. ✓
          • Damage to the work piece. ✓
          • Clatter on the gear teeth. ✓
          • Tendency to bend the spindle. ✓
          • Damage to cutting tool. ✓
          • Danger to the operator. ✓
                                                                  (Any 2 x 1)          (2)

 5.5      Disadvantages of up-cut milling:
          • Tendency for cutter to lift the work piece. ✓
          • Poor finish. ✓
          • More vibrations occur on machine. ✓
                                                                  (Any 1 x 1)          (1)
                                                                                       [18]

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Mechanical Technology: Fitting and Machining     12                               DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 6: TERMINOLOGY (INDEXING) (SPECIFIC)

 6.1       Gear terminology:

           6.1.1      Circular pitch:
                      CP = m  
                           = 3 ✓
                           = 9,42 mm ✓                                                                   (2)

           6.1.2      Number of teeth:
                          PCD                                          CP  T
                      m=                                    PCD =
                             T                                            π
                          PCD                                          9,42  T
                      T =        ✓                           240 =                ✓
                            m                   OR                          
                          240                                         240  π
                      T =        ✓                                T=           ✓
                            3                                          9,42
                      T = 8 0 teeth ✓                             T = 80 teeth ✓                         (3)

           6.1.3      Dedendum:

                        Dedendum = 1,157  m                        Dedendum = 1,25  m
                                                      OR
                                    = 1,157  3 ✓                                 = 1,25  3 ✓
                                    = 3,47 mm ✓                                   = 3,75 mm ✓
                                                                                                         (2)

 6.2       Dovetails:

           6.2.1      Maximum width distance of dovetail: (W)

                      Calculate DE or y:

                                                                           AD
                                                                  tan  =
                                DE                                         DE
                        tan  =
                                AD                                         35
                                                                 tan60  =     ✓
                        DE = tan   AD ✓                                  DE
                                                 OR
                        = tan30   35 ✓                                     35
                                                                    DE =           ✓
                        = 2 0,21 mm ✓                                      tan60 
                                                                         = 20,21 mm ✓

                        W = 250 + 2(DE ) ✓
                          = 250 + 2(20,21 ) ✓
                          = 250 + 40,42
                          = 290,42 mm ✓                                                                  (6)

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Mechanical Technology: Fitting and Machining     13                            DBE/November 2025
                                      NSC – Marking Guidelines

           6.2.2      Distance over the rollers: (M)

                      Calculate AC or x:

                                   BC
                       Tan CAˆ B =
                                   AC                                        AC
                                                                 Tan CBˆ A =
                                     BC                                      BC
                            AC =           ✓
                                        ˆ
                                   Tan CAB            OR              AC = Tan CBˆ A  BC ✓
                                      8                                   = Tan 60   8 ✓
                                 =          ✓
                                   Tan 30                                = 13,86 mm ✓
                                 = 13,86 mm ✓

                      M = w + [2(AC) + 2(R)] ✓
                         = 250 + [2(13,86) + 2(8)] ✓
                         = 250 + (27,72 + 16)
                         = 293,72 mm ✓
                                 OR
                      M = w + 2(AC + R) ✓
                         = 250 + 2(13,86 + 8) ✓
                         = 250 + 27,72 + 16
                         = 293,72 mm ✓
                                OR
                      M= w + 2(AC) + 2(R) ✓
                         = 250 + 2(13,86) + 2(8) ✓
                         = 250 + 27,72 + 16
                         = 293,72 mm ✓                                                               (6)

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Mechanical Technology: Fitting and Machining     14                        DBE/November 2025
                                      NSC – Marking Guidelines

 6.3       Milling of spur gear:

           6.3.1      Indexing:
                                 40
                      Indexing =
                                  n
                                 40
                               =
                                 97
                                 40
                      Indexing =
                                  A
                                  40
                               =       ✓
                                 100
                                 2 5
                               = 
                                 5 5
                                 10
                               =     ✓
                                 25
                      Approximate indexing:
                               No full turns and 10 holes on a 25-hole circle   ✓
                                              OR
                               No full turns and 12 holes on a 30-hole circle ✓                    (3)

           6.3.2      Change gears:
                       Dr              40
                          = ( A − n) 
                       Dn              A
                       Dr                  40
                          = (100 − 97 )      ✓
                       Dn                 100
                                 40
                          =3 
                                100 ✓
                            120
                          =
                            100         ✓
                             12 4
                          =      
                             10 4
                       Dr 48 ✓
                          =
                       Dn 40 ✓                                                                     (5)

 6.4       CNC codes:
           G-codes ✓                                                                               (1)
                                                                                                   [28]

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Mechanical Technology: Fitting and Machining     15              DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 7: TOOLS AND EQUIPMENT (SPECIFIC)

 7.1       Determine hardness:
           • Resistance to penetration ✓
           • Elastic hardness ✓
           • Resistance to abrasion ✓
                                                                  (Any 1 x 1)          (1)

 7.2       Label indentation:
           A Material/work piece/test piece ✓
           B Indenter/ball ✓
           C Applied force/load ✓
           D Indentation diameter ✓                                                    (4)

 7.3       Depth of screw thread:
           Depth of screw thread = 0,613  Pitch (P)
                                 = 0,613  2,5 ✓
                                     = 1,53 mm ✓                                       (2)

 7.4       Label screw thread micrometer:
           A. Frame ✓
           B. Lock ✓
           C. Thimble ✓
           D. Ratchet ✓                                                                (4)

 7.5       Depth micrometer reading:
                                   ✓ ✓
           Reading = 7,00 + 0,46 = 7,46 mm                                             (2)
                                                                                       [13]

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Mechanical Technology: Fitting and Machining     16                            DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 8: FORCES (SPECIFIC)

           8.1.1      ∑ Horizontal component:

                        HC = 75cos0  + 120cos30  − 150cos60  − 45cos90 
                               ✓     ✓       ✓
                        HC = 75 + 103,92 − 75 − 0

                        HC = 103,92 N ✓

                      ∑ Vertical component:

                        VC = 75sin0  + 120sin30  + 150sin60  − 45sin90 

                                  ✓       ✓      ✓
                        VC = 0 + 60 + 129,90 − 45

                        VC = 144,90 N ✓

                                                        OR
      Force   θ         ∑HC/x = Fcosθ                             ∑VC/y = Fsinθ
      75 N    0º HC = 75cos0°      75 N     ✓               VC = 75sin0°     0N
     120 N   30º HC = 120cos30°    103,92 N ✓               VC = 120sin30°   60 N ✓
     150 N 120º HC = 150cos120° -75 N       ✓               VC = 150sin120° 129,90 N ✓
      45 N  270º HC = 45cos270°    0N                       VC = 45sin270°   -45 N ✓
                            Total 103,92 N ✓                                 144,90 N ✓              (8)

           8.1.2      Resultant:

                        R 2 = VC 2 + HC 2
                        R 2 = (144,9) 2 + (103,92) 2 ✓
                          R = 178,31 N ✓                                                             (2)

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Mechanical Technology: Fitting and Machining     17              DBE/November 2025
                                      NSC – Marking Guidelines

           8.1.3      Angle of resultant:
                      Angle:

                                         VC
                               tan  =
                                         HC
                                               144,90  ✓
                                    = tan-1          
                                               103,92 

                                          ✓        ✓
                                    = 54,35° OR 54° 21'

                                              OR

                                         HC
                               tan  =
                                         VC
                                                103,92 
                                tan  = tan -1         ✓
                                                144,90 

                                           ✓       ✓
                                    = 35,65° OR 35°39'

                      Angle and direction:

                      Direction:              ✓
                      R = 178,31 N 54,35 N from E
                       OR
                                               ✓
                       R = 178,31 N 35,65 E from N                                    (3)

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Mechanical Technology: Fitting and Machining     18                         DBE/November 2025
                                      NSC – Marking Guidelines

 8.2      Calculations, UDL and supports A and B:

           8.2.1      UDL:

                      UDL = 14 N/m  12 m ✓
                      UDL = 168 N ✓                                                               (2)

           8.2.2      Reaction in support A:
                      Take moments about B:
                                                    Σ LHM = Σ RHM

                            ✓           ✓              ✓
                       (155  12 ) + (168  6 ) + (90 1,5 ) = (A  12 )
                                       1860 + 1008 + 135 = 12A
                                                               3003
                                                           A=
                                                                12
                                                           A = 250,25 N ✓

                      Reaction in support B:
                      Take moments about A:
                       Σ LHM = Σ RHM
                                        ✓              ✓
                       (B  12 ) = (90  10,5 ) + (168  6 ) + (155  0 )
                           12B = 945 + 1008 + 0
                                 1953
                             B=
                                  12
                             B = 162,75 N ✓                                                       (7)

 8.3       Label A–C:
           A. Limit of proportionality ✓
           B. Elastic limit ✓
           C. Maximum stress ✓                                                                    (3)

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Mechanical Technology: Fitting and Machining     19                    DBE/November 2025
                                      NSC – Marking Guidelines

 8.4       Stress calculations:

           8.4.1      Diameter:

                             π d2
                       A=
                              4
                             π ( 0,045 )
                                           2

                       A=            ✓
                               4
                       A = 1,590431281  10 −3 m 2 ✓ OR 0,001590431281 m 2 ✓

                         F
                       σ =
                         A
                      F= A σ ✓
                             (                    ) (
                      F = 1,59043128 1  10 −3  28,56  10 6    ) ✓
                      F = 45422,72 N ✓                                                       (5)

           8.4.2      The original length:

                          ΔL
                       ε=
                          oL
                       ΔL = oL  ε ✓
                       ΔL = 350  0,0003 ✓
                       ΔL = 0,11 mm ✓                                                        (3)
                                                                                             [33]

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### Page 20

Mechanical Technology: Fitting and Machining     20              DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 9: MAINTENANCE (SPECIFIC)

 9.1       Thermoplastic definition:

           9.1.1      Reheated/Heated/Melted ✓                                         (1)

           9.1.2      Shaped/made/reformed/moulded ✓                                   (1)

           9.1.3      recycled ✓                                                       (1)

           9.1.4      heated ✓                                                         (1)

 9.2       Reason for conducting adjustments:

           •   To prevent faults ✓ from occurring.
           •   Ensure operations are at optimal level. ✓
                                                                  (Any 1 x 1)          (1)

 9.3       Thermo-hardened composites materials:

           •   Carbon fibre ✓
           •   Glass fibre/Fibre glass ✓
           •   Bakelite ✓
           •   Epoxy ✓
           •   Silicone ✓
           •   Vulcanized rubber ✓
                                                                  (Any 3 x 1)          (3)

 9.4       Thermoplastic composites materials:

           •   Nylon ✓
           •   Teflon ✓
           •   PVC ✓
           •   Vesconite ✓                                                             (4)

 9.5       Preventative maintenance procedures:

           •   Cleaning uncovered chain drives. ✓
           •   Check sprocket for wear. ✓
           •   Check link plates/chain for wear. ✓
           •   Ensure sufficient lubrication of chains. ✓
           •   Checking functioning of tensioning devices. ✓
           •   Inspecting chains regularly for elongation. ✓
           •   Check the alignment. ✓
                                                                  (Any 3 x 1)          (3)

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### Page 21

Mechanical Technology: Fitting and Machining     21              DBE/November 2025
                                      NSC – Marking Guidelines

 9.6       Type of composite material:

           9.6.1      •   Nylon/Polyurethane ✓
                      •   Vesconite ✓
                      •   Teflon ✓
                                                                   (Any 1 x 1)         (1)

           9.6.2      Teflon ✓                                                         (1)

           9.6.3      Bakelite ✓                                                       (1)
                                                                                       [18]

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### Page 22

Mechanical Technology: Fitting and Machining     22                      DBE/November 2025
                                      NSC – Marking Guidelines

 QUESTION 10: JOINING METHODS (SPECIFIC)

 10.1      Screw thread terminology:

           10.1.1     Metric ✓                                                                   (1)

           10.1.2     Major-/Crest-/Basic-/Nominal-/Outside diameter ✓                           (1)

           10.1.3     Pitch ✓                                                                    (1)

 10.2      Square thread:

           10.2.1     Pitch diameter:

                      Lead = Pitch  Number of starts
                                     Lead
                      Pitch =                     
                                 Number of starts
                               36
                             =     
                                2
                             = 18 mm 

                                   P
                      PD = OD −
                                   2
                                  18
                           = 70 −    
                                   2
                           = 61 mm                                                              (5)

           10.2.2     Helix angle of the thread:

                                 Lead
                       Tan θ =
                                 π PD
                                   36
                       Tan θ =          
                                 π  61
                           θ = tan −1 (0,18785501 48 ) 
                                             
                              = 10,64  OR 10  38'                                              (3)

           10.2.3     Leading angle:

                      Leading angle = 90° – (helix angle + clearance angle)
                                        = 90° – (10,64° + 3°) ✓
                                        = 76,36° ✓ OR 76° 22ꞌ ✓                                  (2)

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### Page 23

Mechanical Technology: Fitting and Machining     23                       DBE/November 2025
                                      NSC – Marking Guidelines

           10.2.4     Following angle:

                      Following angle = 90° + (helix angle – clearance)
                                        = 90° + (10,64° – 3°) ✓
                                        = 97,64° ✓ OR 97° 38ꞌ ✓                                 (2)

 10.3      V-thread terminology:
           A - Root ✓
           B - Pitch ✓
           C - Minor diameter/Root diameter/Core diameter ✓                                     (3)
                                                                                                [18]

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### Page 24

Mechanical Technology: Fitting and Machining     24                          DBE/November 2025
                                      NSC – Marking Guidelines

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

 11.1      Hydraulic systems:

           11.1.1     Area ram:
                           d2
                      A =
                            4
                              ( 0,1)
                                        2

                       A =           ✓
                              4
                       A = 7,85  10 -3 m 2 ✓                                                      (2)

           11.1.2     Mass lifted:

                         F
                      P=
                         A
                      F =P A ✓
                             (              ) (
                      F = 6,36  10 6  7,85  10 -3 ✓)
                      F = 49926 N ✓

                              49926 ✓                                   49926 ✓
                      Mass =                                     Mass =
                                10                                       9,81
                                                      OR
                       Mass = 4992,60 kg ✓                       Mass = 5089,30 kg ✓
                                                                                                   (5)

 11.2      Components in hydraulic system:
           • Motor ✓
           • Pump ✓
           • Filter ✓
           • Pressure gauges ✓
           • Non-return valves/check valves/Direction control valve ✓
           • Reservoir ✓
           • Pipes ✓
           • Bleeding valve ✓
           • Levers ✓
           • Actuator ✓
           • Accumulator ✓
           • Flow control valve ✓
           • Pressure relief valve ✓
           • Lubricator ✓
                                                                              (Any 2 x 1)          (2)

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### Page 25

Mechanical Technology: Fitting and Machining     25                       DBE/November 2025
                                      NSC – Marking Guidelines

 11.3      Hydraulic pump designs:
           • Radial pump ✓
           • Rotary pump ✓
           • Axial pump ✓
                                                                           (Any 1 x 1)          (1)

 11.4      Belt drives:

           11.4.1     Rotational frequency on the driven pulley in r/s:

                      NDN  D DN = NDR  D DR
                              NDR  D DR
                      NDN =              ✓
                                D DN
                            22  85
                      NDN =          ✓
                              375
                      NDN = 4,99 r/s ✓                                                          (3)

           11.4. 2    Power transmitted:

                                  T1
                      Ratio =
                                  T2
                                  450
                          2,5 =       ✓
                                   T2
                                450
                           T2 =
                                 2,5
                           T2 = 180 N ✓

                           P =     (T1 − T2 ) π  D  N
                                     ✓
                           P = (450 − 180 )  π  0,085  22 ✓
                           P = 1586,19 W
                           P = 1,59 kW ✓

                           OR

                           P =     (T1 − T2 )  π  D  N
                                   ✓
                           P = (450 − 180 )  π  0,375  4,99 ✓
                           P = 1587,25 W
                           P = 1,59 kW ✓                                                        (5)

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### Page 26

Mechanical Technology: Fitting and Machining     26                             DBE/November 2025
                                      NSC – Marking Guidelines

 11.5      Gear drive:

           11.5.1     Rotational frequency of the output shaft NF in r/min:

                       Ninput        Product of the number of teeth on driven gears
                                 =
                       Noutput       Product of the number of teeth on driver gears
                          NA TB  TD  TF
                            =
                          NF TA  TC  TE
                           95 50  70  85 ✓
                             =
                           NF 30  25  60 ✓
                                     30  25  60  95
                           NF =                        ✓
                                       50  70  85
                                     4 275 000
                           NF =
                                      297 500
                           NF = 14,37 r/min ✓                                                            (4)

           11.5.2     Speed ratio:

                                           Ninput
                       Speed ratio =
                                           Noutput

                                           95 ✓
                      Speed ratio =
                                          14,37 ✓

                      Speed ratio = 6,61 : 1 ✓
                                          OR
                                         Product of teeth on driven gears
                      Gear ratio =
                                         Product of teeth on driver gears

                                         50  70  85 ✓
                      Gea r ratio =
                                         30  25  60 ✓

                      Gear ratio = 6,61 : 1 ✓                                                            (3)

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### Page 27

Mechanical Technology: Fitting and Machining     27              DBE/November 2025
                                      NSC – Marking Guidelines

 11.6      Torque:

           Torque = Force  Radius
                    Torque
           Radius =         ✓
                     Force
                    712,5
           Radius =        ✓
                     750
           Radius = 0,95 m ✓                                                         (3)
                                                                                     [28]

                                                                      TOTAL:         200

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