Name___________________________________________ Index No.____________
Candidates signature ____________________ Date __________________
PHYSICS
PAPER 3
PRACTICAL
2 ½ HOURS
JOINT EXAMINATION TERM 3
INSTRUCTIONS TO CANDIDATES
- Answer all questions in both sections in the spaces provided in the question paper.
- Read the whole paper carefully before commencing your work.
- Marks are given for clear record of the observations actually made, their suitability and accuracy and the use of them.
- Mathematical tables and electronic calculators may be used.
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;
- Masses; 10g, 20g(2), 50g and 100g
- A helical spring with a pointer
- Metre rule or half metre rule
Proceeds as follows;
- a) Suspend the helical spring vertically, alongside the clamped half metre rule as shown in figure 1 below. Determine the length, L0 of the spring before loading it.
L0=________________________________(cm) (1mk)
- i) Load the spring with a mass of 20g and determine the new reading on the metre rule (L).
- ii) Record this in the table below.
iii) Calculate the extension, e= L-L0 due to the mass of 20g and record the value in the table given below.
- Repeat step b(i-iii) for other masses and complete the table. (8mks)
| Mass (g) | 0 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
| Weight (N) (Force) | ||||||||||
| Reading L (cm) | ||||||||||
| Extension e (cm) |
- Plot a graph of Force(N) against extension e (cm). (5mks)
- Determine the slope S of the graph. (3mks)
- Given that K=Se, determine the value of K where S is the slope of the graph and e is the extension of the spring when loaded with a 50g mass. (3mks)
Question 2
You are provided with the following;
- A Nichrome wire mounted on a millimeter scale YZ.
- A carbon resistor R (10Ω)
- 8 connecting wires
- Ammeter
- Voltmeter
- 2 dry cells
- Cell holders
- Switch
- (a) Set up the circuit as shown below.
- Record the voltmeter reading when the switch is open.
E=_____________________________________ (1mk)
- Close the switch and record the voltmeter and ammeter readings.
Voltmeter reading, V=_______________________________ (1mk)
Ammeter reading, I=________________________________ (1mk)
- Explain why V is less than E. (2mk)
______________________________________________________________________________________________________________________________________________________________________________________________________
- Now connect the voltmeter across the carbon resistor R and record voltmeter reading V1 when the switch is on. (1mk)
V1=
- Determine R given that; R= (3mks)
(b)
- Measure and record the diameter, d of the resistance wire (YZ) using a micrometer screw gauge.
d= ___________________________________mm
d=___________________________________m (2mks)
- ii) Set-up the circuit as shown in figure;
- Determine the current I flowing in the circuit.
I=__________________________________ (1mk)
- Now connect the voltmeter across wire (YZ) and record the p.d, V2 across wire YZ.
V2= ________________________________ (1mk)
- Determine the resistance R of the wire YZ. (3mks)
- Determine k, the resistance per metre of wire YZ. (2mks)
- Determine Q given that Q=(where d is in metres) (2mks)
________________________________________________________________________
JOINT EXAMINATION
FORM 3PP3 PHYSICS
TERM 3 MARKING SCHEME
- a) L0 – ½mk answer to 1d.p
½mk correct unit (cm)
- ii) (8mks)
| 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
- v) Graph of Force(N) vs extension, e (cm). (5mks)
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
- vi) Slope, s== 0.8N/cm (3mk)
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
- a)
- E = 3.0- ½mk
Correct unit – ½mk
- V = 2.5v1 1 d.p (1mk) and correct unit
I = 0.20A 0.02 2 d.p (1mk) and correct unit
- V is less than E due to the lost voltage; voltage against the internal resistance of the cell. (2mk)
- V1 = 1.4V1 Correct reading to 1 d.p – ½mk
Correct unit – ½mk
- R=1
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
- v) R= Correct substitution – 1mk
Correct evaluation – 1mk
Answer in 4s.f – ½mk
Correct unit – ½mk
- K=Correct evaluation – 1mk
Answer – ½mk 4s.f
Correct unit – ½mk (Ωm-1)
- Q= Correct evaluation – 1mk
Answer in 4s.f – ½mk
Correct unit – ½mk (Ωm)
