## Physics Paper 3 Questions and Answers - Form 3 End Term 2 Exams 2022

QUESTION 1
You are provide with the following;

• Two dry cells
• One bulb
• Voltmeter (0 – 3 v or o – 5v)
• Ammeter  (0 – 2.5A)
• A mounted nichrome wire on millimetre scale;
• Switch
• Seven connecting wire at least two with crocodile clips.
• Micrometer screw gauge

Procedure as follows;

1.
1. Set up the circuit as shown in figure below; 2. With the crocodile clip at P take the voltmeter reading and the ammeter reading.  Record V and I. Repeat the readings for L = 80,60,40,20 and 0 cm respectively.
Complete the table below;
 Length L (cm) 100 80 60 40 20 0 Voltage V (v) Current, I (A)
(4mks)
3. What changes do you observe on the bulb as L decreases from P ? (1mk)
4. Plot a graph of the ammeter reading (y = axis) against voltmeter reading (5mks)
5. Determine the slope of your graph at  V = 1 volt (3mks)
6. What physical quantity is represented by the slope of the graph at any given point.?  (1mk)
2.
1. Given the apparatus in a (i) above, draw a diagram of the a circuit you would use to determine the current through the resistance wire and the potential difference across it. (1mk)
2. Set up the circuit you have drawn. Record the ammeter reading I and the voltmeter reading
V, when L = 100cm. (2mks)
3. Using a micrometer screwgauge, measure the diameter d of the wire. (1mk)
d = …………………………………m
4. Calculate the quantity,
p = 0.785(V/I)(d2/L) and give its units, where L is one metre. (2mks)

QUESTION 2.
You are provided with the following;

• A Complete retort stand
• A Stop watch/stop clock
• A Metre rule
• Two identical springs labeled R and P.
• A Weighing balance (to be shared)
• A Set of masses 10g, 20g, 50g and 100g
• A Pendulum bob

Proceed as follows:

1. Join springs R and P in parallel so that it has only one hook at one end and then arrange the apparatus as shown in the figure below. Note and record the initial pointer reading.
Initial pointer reading = ………………………………………………………….……cm mark.
(This mark should be maintained throughout the experiment)
2. Hang the 30g mass on the hook of the combined spring balance and record the final pointer reading. Hence calculate the extension, e, for m = 30g.
3. With mass, m = 30g, still suspended, slightly displace the mass vertically and time 20 complete oscillations
4. Repeat the experiment for m = 50, 70, 100, 120 and 150g and record your results in the table below.
 Mass m (g) Extension e (cm) e (m) Time, t, for 20 complete oscillations Periodic time, T (s) T2(s2) 30 50 70 100 120 150
5.
1. On the grid provided, plot a graph of e(x – axis) against T2 (5 mks)
2. Determine the slope, S, of the graph. (2 mks)
6. If the experiment obeys the law T = e/k   where k is a constant, determine the value of k (= 3.142) (2 mks
7. Weigh and record the mass of the pendulum bob provided.
Mass, m of pendulum bob = ……………..………………..g = ……………………….…….kg (1 mk)
8. Suspend the pendulum bob on the combined spring balance and note the extension produced.
extension e1= ………………………………….………..cm = ……………………..………..m (1 mk)
9. If v = mg/e =   where m = mass of the pendulum bob and e is the extension produced, find the value of v where v is the elastic constant of the springs. (2 mks) ## MARKING SCHEME

1.
1.
1.
 Length L (cm) 100 80 60 40 20 0 Voltage V (v) 0.25 0.45 0.55 0.75 1.15 1.60 Current, I (A) 0.12 0.14 0.15 0.16
2. Brightness increase
3. Axes – 1
Scale - 1
Plotting – 2 at least four each ½ mark
Curve - 1
4. Tangent at the point
Slope = 0.16 - 0.14 1 correct intervals
1.4 - 0.3
=0.02
1.1
= 0.018Ω-1   1 correct evaluation
5. Reciprocal of resistance
2.
1. NB – ammeter in cell switch in series voltmeter parallel to wire
2. V = 1.8 V
I = 0.14 A
3. d = 3.6  x  10-4m correct substition
= 1.308 x 10-6Ωm 1 evaluation and  correct units
2.
1. Initial pointer reading =   “ Candidates” cm mark
 Mass m (g) Extension e (cm) e (m) Time, t, for 20 complete oscillations Periodic time, T (s) T2(s2) 30 2.0 2.0 6.63 0.3317 0.11 50 3.2 3.2 8.49 0.4243 0.18 70 6.3 6.3 12.00 0.6 0.36 100 7.5 7.5 12.96 0.6481 0.42 120 9.0 9.0 14.97 0.7483 0.56 150 12 12 16.61 0.8307 0.69
(7mks)

2. Labelling- correct quantity  + units 1 mark
Scale – simple and uniform 1 mark
Plotting- ½ x 4 2 marks
Line 1 marks
5 marks
3. S =   0.561 - 0.113
(9.0 - 2.0) x 10-2
= 9.6
Correct internal  1mk
Correct evaluation  1 mk
4. T= e/k
T2 =2 e
k
Slope, S = T22
e        k
k =   = 4 x 3.1422
S            9.6
= 4 . 113 2mks
5. m = 26 g  =  0.026  kg
6. e=0.42 cm  =4.2 x 10-3
7. v = mk
0.026 x 4.113
4.2 x 10-3
= 6. 190
Correct  substitution
Correct evaluation

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