Form 3 Physics paper 3 Exams and Marking Schemes Free

By Editorial Team

February 21, 2026

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

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)