Physical Sciences Grade 10 KZN March Test 2024 Paper & Memo

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Downloaded from Stanmorephysics.com 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. Copyright reserved

Downloaded from 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] Copyright reserved Please turn over

Downloaded from 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) Copyright reserved Please turn over

Downloaded from 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] Copyright reserved Please turn over

Downloaded from 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] Copyright reserved Please turn over

Downloaded from 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 Copyright reserved Please turn over