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CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2014 series 9702 PHYSICS 9702/21 Paper 2 (AS Structured Questions), maximum raw mark 60 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicat
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  ® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2014 series 9702 PHYSICS   9702/21 Paper 2 (AS Structured Questions), maximum raw mark 60 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2014 series for most Cambridge IGCSE ® , Cambridge International A and AS Level components and some Cambridge O Level components.  Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9702 21 © Cambridge International Examinations 2014 1 (a) temperature B1   current B1 [2]   (allow amount of substance and luminous intensity) (b) base units of force constant: kg   m   s –2   m –1  or kg   s –2 B1 base units of time and mass: s and kg C1 base units of C  :   s   (kg   s –2   /   kg) 1/2  cancelling to show no units B1 [3] 2 (a) pressure = force / area (normal to the force) [clear ratio essential] B1 [1]  (b) (i) P = mg /  A = (5.09 ×  9.81) /  A  C1    A = ( π d  2  / 4) = π   ×  (9.4 ×  10 –2 ) 2  / 4   (= 0.00694   m 2 ) C1   P = 49.93 / 0.00694 = 7200 (7195)   Pa (minimum of 2 s.f. required) A1 [3]  (ii) ∆ P /  P = ∆ m /  m +  2 ∆ d /  d   C1   = 0.01 / 5.09 + (2 ×  0.1) / 9.4 (= 0.0020 + 0.021 or 2.3%) C1   ∆ P    = 170 (165 to 167)   Pa A1 [3] (iii) P = 7200 ± 200   Pa A1 [1] 3 (a) random error (in the measurements) of the length OR resistance B1 [1] (b) gradient = (3.6 – 1.9 ) / (0.8 – 0.4) C1   = 4.25 A1   [2] (c) R =  ρ  l    /  A  C1    ρ   = gradient ×  area = 4.25 ×  0.12 ×  10 –6 C1 = 5.1(0) ×  10 –7   Ω   m A1 [3] (d) resistance decreasing with increasing area B1   correct shape with curve being asymptote to both axes B1 [2]  Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9702 21 © Cambridge International Examinations 2014 4 (a) (i) acceleration = ( v – u  ) / t or (12 – 0.5) / 4 C1   = (12 – 0.5) / 4 = 2.9 (2.875) (= approximately 3   m   s –2 ) M1 [2] (ii)  x   = ( u +   v  )   t / 2   = [(12 +  0.5) ×  4] / 2 C1   = 25   m A1 [2] (iii) line with increasing gradient M1   non-zero gradient at srcin A1 [2] (b) (i) weight down slope = 2 ×  9.81 ×  sin 25° = 8.29 / 8.3 M1 [1] (ii)   ( F = ma ) 8.3 – F  R  = 2 ×  2.9 C1   F  R   = 2.5 (2.3 if 3 used for a )   N A1 [2]  5 (a) (i) change in   kinetic energy = ½   mv  2  C1   = 0.5 ×  25 ×  (0.64) 2 = 5.1(2)   J A1 [2] (ii) zero A1 [1] (iii) (–)   5.1(2)   J A1 [1] (b) (i) PE = mgh C1   = 350 ×  0.64 ×  25 C1   = 5600   J A1 [3] (If full length used allow 1/3) (ii) P = Fv   or gain in PE   /   t  , E  P   /   t   or work done   /   t  , W   /   t   C1 = 350 ×  0.64 or 5600 / 25   = 220 (224)   W A1 [2]  6 melting: solid to liquid B1 at a specific   /   one temperature   /   at the melting point B1   evaporation: liquid to vapour    /   gas OR molecules escape from surface of liquid B1 at all temperatures B1 [4]  Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9702 21 © Cambridge International Examinations 2014 7 (a) due to the lost volts in internal resistance   /   cell or energy losses in the internal resistance   /   cell B1 [1] (b) (i) V = І  R C1   = 1.2 ×  6 = 7.2   V A1 [2]  (ii) p.d. across Y and internal resistance r   = 4.8   (V) [12 – 7.2] C1 resistance of Y + r   = 4.8 / 1.2 = 4   ( Ω ) C1   resistance of Y = 4 – 0.5 = 3.5 Ω  A1 [3]   or R  total  = 12 / 1.2 = 10 ( Ω ) (C1)   X + r   = 6.5 ( Ω ) (C1)   resistance of Y = 3.5   Ω  (A1)  (iii) P = І  2 r   C1   = (1.2) 2   ×  0.5 = 0.72   W    A1 [2]  (c) terminal p.d. increases as R   is increased current decreases so there are less lost volts B1 [1] 8 (a) two waves (of the same kind) travelling in opposite directions overlap B1 waves have same frequency   /   wavelength and speed B1   [2]  (b) (i) T = 0.8   (ms) C1   f   = 1 / (0.8 ×  10 –3 ) = 1250   (Hz) A1 [2]  (ii) microphone is moved from plate to loudspeaker or vice versa B1   wavelength is the twice the distance between adjacent maxima or minima (seen on c.r.o.) B1 [2] (iii) v = f  λ   C1 = 1250 ×  0.26 = 330 (325)   m   s –1  A1 [2]  
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