# Physical Sciences Grade 10 KZN March Test 2024 Paper & Memo

> Grade 10 Physical Sciences question paper and memo from the KZN March Test 2024, 17 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/kzn-physical-science-grade-10-march-2024-qp-and-memo/880c01b0-7b75-4132-9321-8698677472d8
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 10
- Document type: Question paper and memo
- Year: 2024
- Exam period: KZN March Test
- Curriculum: CAPS
- Pages: 17

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## About this document

Physical Sciences · Grade 10 · KZN March Test · 2024. Question paper and memorandum, 17 pages. Read online or download the PDF.

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## Document text

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                                    NATIONAL
                               SENIOR CERTIFICATE

                                        GRADE 10

                                PHYSICAL SCIENCES
                                    COMMON TEST

                                MARKING GUIDELINE

                                      MARCH 2024

MARKS: 100

DURATION: 2 hours

This marking guideline consists of 6 pages.

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### Page 13

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Physical Sciences      Stanmorephysics.com
                               NSC                                      March 2024 Common Test
                                                2

 QUESTION 1:

 1.1       C                                                                               (2)
 1.2       B                                                                               (2)
 1.3       C                                                                               (2)
 1.4       A                                                                               (2)
 1.5       D                                                                               (2)
                                                                                             [10]

 QUESTION 2

 2.1       Principle of superposition.  The algebraic sum of the amplitudes of two
           pulses that occupy the same space at the same time.                             (3)

           OPTION 1                                 OPTION 2
 2.2       Let the amplitude of Pulse A = x         Let the amplitude of Pulse A = ⅓x
           The amplitude Pulse B = - 3x             The amplitude Pulse B = - x
           x + (-3x) = -6                         ⅓x + (-x) = -6 
           x = 3 cm                                 x = 9 cm 
           Amplitude of Pulse B = 3x
                                 = 3(3)
                                 = 9cm                                                      (3)

 2.3       5 cm                                                                             (1)

 2.4       Criteria
           Pulse A drawn as crest and Pulse B drawn as trough with labels 
           Pulse B has a greater amplitude than A 
           Pulse B drawn left of A                                                          (3)
                                          A

                           B
                                                                                             [10]

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Physical Sciences      Stanmorephysics.com
                               NSC                                     March 2024 Common Test
                                             3

 QUESTION 3

 3.1
           3.1.1     Longitudinal                                                          (1)

           3.1.2     R – Compression                                                       (2)
                     S – Rarefaction 

           3.1.3         0,3
                     λ=             
                          2
                      = 0,15 m                                                             (2)

           3.1.4                 POSITIVE MARKING FROM QUESTION 3.1.3
                                 OPTION 1                 OPTION 2
                                                             distance
                                                   speed =                
                     v=f ×λ                                   time
                     1,5 = f x 0,15                          0,15 
                     f = 10 Hz                      1,5 =
                                                               time
                         1
                     T=                            T = 0,1 s 
                          f
                          1
                     T=         
                         10
                     T = 0,1 s                                                             (4)

 3.2       3.2.1     0,1 m                                                                 (1)

           3.2.2     (point) F                                                             (1)

           3.2.3     The number of wave pulses per second.                                (2)

           3.2.4          1
                     T=
                          f
                        1
                      =          
                        2
                     = 0,5 s
                     time = 1,5 x 0,5 
                          = 0,75 s                                                         (4)

           3.2.5                OPTION 1                            OPTION 2
                      POSITIVE MARKING FROM Q 3.2.4
                             distance                            1,2
                     speed =                                 𝜆=      = 0,8 𝑚
                               time                              1,5
                              1,2                           v=f ×λ
                     speed =
                             0,75                            = 2  x 0,8
                          = 1,6 m.s-1                        = 1,6 m.s-1                  (3)

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Physical Sciences      Stanmorephysics.com
                               NSC                                         March 2024 Common Test
                                                   4
                                                                                                [20]

 QUESTION 4

 4.1

           4.1.1     A reflection of sound waves.                                             (2)

           4.1.2               OPTION 1
                             distance
                     speed =
                               time
                             distance
                     340 =
                                2
                     Distance = 680 m
                     X + (X – 6) = 680 
                     X = 343 m 

                     OPTION 2
                     2X = Distance of bat in 2s + distance of echo in 2s
                     2X = 6 +(2 x 340) 
                     X = 343 m                                                                 (4)

           4.1.3     v=f ×λ
                     340  = f x 7,56 x10-3 
                     f = 44 973,5 Hz 
                     The sound will not be heard; its frequency is greater than 20kHz.         (4)

 4.2
           4.2.1     c=fxλ 
                     3 x 108 = f x 7,5 x 10-5 
                     f = 4 x 1012 Hz                                                           (3)

           4.2.2     E=hxf 
                     = 6,63 x 10-34 x 8 x 1015 
                     = 5,304 x 10-18 J                                                         (3)

           4.2.3     RADIATION 1 
                     A higher frequency will result in greater energy of the photon OR
                     Energy of photon is directly proportional to the frequency. 
                     A greater energy of the photon increases the penetrating ability          (3)

           4.2.4     INCREASE 
                     The wavelength is inversely proportional to the frequency.                (2)

                                                                                                [21]

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Physical Sciences      Stanmorephysics.com
                               NSC                                             March 2024 Common Test
                                                      5

 QUESTION 5

 5.1

           5.1.1         Electrons will be transferred  from the glass rod to the silk cloth         (2)

           5.1.2         The partial or complete polar separation of positive and negative             (2)
                         electric charge in a system. 

           5.1.3         Left                                                                         (1)

 5.2

           5.2.1         Q = n x qe 
                         - 4×10-6 = n x 1,6×10-19 
                         n = 2,5 x 1013 electrons                                                     (3)

           5.2.2         A to B                                                                       (1)

           5.2.3         The net charge of an isolated system remains constant during any              (2)
                         physical process 

                                  Q1 + Q2        
          5.2.4             Q=
                                     2

                             - 4×10-6 + 8×10-6 
                     Q=
                                     2
                          = 2 x 10-6 C                                                                (3)

          5.2.5             Q1 + Q2
                     Q=
                               2
                         - 6×10-6 + 2×10-6       
                     Q=
                                  2
                     QB = - 2 X 10-6 C
                        ∆𝑄
                     n=                          
                         𝑞𝑒
                                                                                                  
                         -2 ×10-6 − 2 × 10-6                          -2 ×10-6 -(-6 × 10-6 )
                     =                                            =
                             - 1,6 × 10-19                                                        
                                                          OR             1,6 × 10-19
                     = 2,5 x 1013 electrons                    = 2,5 x 1013 electrons                (5)

                                                                                                       [19]

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Physical Sciences      Stanmorephysics.com
                               NSC                                      March 2024 Common Test
                                               6

 QUESTION 6

 6.1       The energy transferred per unit charge.                                         (2)

 6.2       V2                                                                               (1)

 6.3       V = IR 
             = 2  x (4+2+6) 
             = 24 V                                                                         (4)

 6.4       POSITIVE MARKING FROM QUESTION 6.3

           6.4.1     V = IR
                     V4Ω+2Ω = 3  x (4+2) 
                            = 18 V
                     V1 = 24 – 18 
                        = 6V                                                                (4)

           6.4.2                 OPTION 1                           OPTION 2
                     V6Ω = IR                            Vp = IR
                     6=Ix6                              6 = 3 x Rp 
                     I6Ω = 1A                            Rp = 2Ω
                     IBulb = 3 – 1                             1    1    1   
                                                                  =    +
                            = 2A                               RP R1 R2
                     VBulb = IR                             1     1     1
                                                              =       +      
                     6=2xR                                  2 RBulb 6
                     RBulb = 3Ω                         RBulb = 3Ω                         (5)

           6.4.3     OPTION 1                            ACCEPT
                     Q = I ∆t                            W = VI∆t 
                       = 2 x 60                           = 6 x 2 x 60
                       = 120 C                             = 720 J 
                     W=VQ
                        = 6 x 120 
                        = 720J                                                              (4)

                                                                                             [20]

                                            TOTAL: 100

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