ACSI 2019 Y3 IP Physics Paper1 Answers
Uploaded by skibidi · 20 September 2024
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Text from the first pagesEND-OF-YEAR EXAMINATION 2019 YEAR 3 INTEGRATED PROGRAMME PHYSICS PAPER 1 MarkScheme Monday 7 October 2019 1 hour 45 minutes INSTRUCTIONS TO STUDENTS Write your index number in the box provided on the top right corner of this page. Do not open this booklet until you are told to do so. Section A Answer all questions in the spaces provided in the paper. Section B Answer all questions in the spaces provided in the paper. INFORMATION FOR STUDENTS Candidates are reminded that all quantitative answers should include appropriate units. Candidates are advised to show their answers in a clear and orderly manner as more marks are awarded for sound use of physics than for correct answers. The number of marks is given in brackets [ ] at the end of each question or part question. Calculators are allowed for this paper. Take g = 10 N/kg. There are 18 pages. Marks Awarded Section Marks A B Penalty Sig. Fig. Units TOTAL SCORE Index Number
2019 FYE P1 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 A A D A C C B C D D Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 A D A B A D C A B C Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 D C A C B A A B B B Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 C B B C B B D D B A Note Q33: Negative voltage implies then that Liquid B must be at a higher temp. than Liquid A. Hence option A and B are both as likely. Next for a 30 degrees difference → 2 mV This means that 1 mV corresponds to 15 degrees. So the new temp will be 30 + 15 = 45 degrees. Q34: Energy supplied = Pt = 100 (1x60x60) = 360000 J (Using one hour time frame) Q needed to raise temp in 1 hour = mcΔθ = 0.50(4200)(50) = 105000J Hence energy loss in 1 hour = 255000J
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