Name___________________________________________ Index No.____________
Candidates signature ____________________ Date __________________
PHYSICS
PAPER 3
PRACTICAL
2 ½ HOURS
JOINT EXAMINATION TERM 3
INSTRUCTIONS TO CANDIDATES
FOR EXAMINER’S USE ONLY
| Question | Maximum Score | Candidate’s Score |
| 1. | 20 | |
| 2. | 20 | |
| TOTAL | 40 |
This paper consists of 7 printed pages. Learners should check to ensure that all pages are printed as indicated and that no questions are missing.
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Question 1
You are provided with the following;
Proceeds as follows;
L0=________________________________(cm) (1mk)
iii) Calculate the extension, e= L-L0 due to the mass of 20g and record the value in the table given below.
| Mass (g) | 0 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
| Weight (N) (Force) | ||||||||||
| Reading L (cm) | ||||||||||
| Extension e (cm) |
Question 2
You are provided with the following;
E=_____________________________________ (1mk)
Voltmeter reading, V=_______________________________ (1mk)
Ammeter reading, I=________________________________ (1mk)
______________________________________________________________________________________________________________________________________________________________________________________________________
V1=
(b)
d= ___________________________________mm
d=___________________________________m (2mks)
I=__________________________________ (1mk)
V2= ________________________________ (1mk)
________________________________________________________________________
JOINT EXAMINATION
FORM 3PP3 PHYSICS
TERM 3 MARKING SCHEME
½mk correct unit (cm)
| Mass (g) | 0 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
| Weight (force) N | ||||||||||
| Reading, L cm | ||||||||||
| Extension, e(cm) | 1.0 | 2.6 | 4.8 | 6.3 | 8.1 | 10.2 | 12.3 | 14.2 | 16.4 |
2mks@ Row (force) – correct evaluation and substitution
F = mg
L to 1 d.p
e correct (L-L0) to 1 d.p
Correct labelling of axis – 1mk
Scale – simple and uniform – 1mk
Plotting (2mks) (7-10 points) – 2mks
(5-6 points) – 1mk
(0-4 points) – 0 mk
Line – straight line passing through the origin – 1mk
Clear points from the graph – 1mk
Correct evaluation – 1mk
Answer – ½mk
Correct unit – ½mk
vii) Correct substitution of k=se – 1mk
Correct evaluation – 1mk
Answer – ½mk
Correct unit – ½mk
Correct unit – ½mk
I = 0.20A 0.02 2 d.p (1mk) and correct unit
Correct unit – ½mk
Correct substitution – ½mk
Correct evaluation – ½mk
Answer in 4s.f – ½mk
Correct unit – ½mk
b (i) d=0.36mm
d= in metres – correct conversion to 5 d.p (1mk)
(iii) I = 0.10A Correct reading – ½mk
Correct unit – ½mk
iv)V2 = 1.9v 0.1 Correct reading – ½mk
Correct unit – ½mk
Correct evaluation – 1mk
Answer in 4s.f – ½mk
Correct unit – ½mk
Answer – ½mk 4s.f
Correct unit – ½mk (Ωm-1)
Answer in 4s.f – ½mk
Correct unit – ½mk (Ωm)
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