Physics Paper 1, MAy/June. 2010
 Questions: 1 2 3 4 5 6 Main
Weakness/Remedies
Strength
Alternative a
Question 3

(a)  Diagram

You are provided with a voltmeter V, a chemical cell/battery E; two standard resistors R1and R2; a potentiometer AB; a key K; a jockey and other necessary materials.
(i) Set up a circuit as shown in the diagram above.
(ii) Close the key K.
(iii) Make contact with the potentiometer wire AB using the jockey at a point C such that AC = x = 20 crn.
(v) Evaluate x-1 and V-1.
(vi) Repeat the procedure for other values of x = 30, 40,50,60 and 80 cm.
(viiiPlot a graph with V-1 on the vertical axis and x-1on the horizontal axis, starting both axes from the origin (0, 0).
(ix) Determine the:
(A) slope, s, of the graph;
(B) intercept, e, on the vertical axis.
(x) State two precautions taken to ensure 'accurate results.

(b)(i) State two devices in which Ohm's law does not apply.
(ii) A current of l A is supplied to two resistors of resistances 2Ω and 3 Ω connected in parallel. Calculate the current in each resistor.

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Observation

You are provided with a voltmeter V, a chemical cell/battery E; two standard resistors R1and
R2; a potentiometer AB; a key K; a jockey and other necessary materials.
(i) Set up a circuit as shown in the diagram above.
(ii) Close the key K.
(iii) Make contact with the potentiometer wire AB using the jockey at a point C such that AC = x = 20 crn. (iv)Read and record the voltmeter reading, V.
(v)Evaluate x-1and V-1.
(vi)Repeat the procedure for other values of x = 30, 40,50,60 and 80 cm.
(viii Plot a graph with V-1on the vertical axis and x-1on the horizontal axis, starting both axes from the origin (0, 0).
(ix) Determine the:
(A) slope, s, of the graph;
(B) intercept, e, on the vertical axis.
(x) State two precautions taken to ensure 'accurate results.

(b)(i) State two devices in which Ohm's law does not apply.
(ii)A current of l A is supplied to two resistors of resistances 2Ω and 3 Ω connected in parallel. Calculate the current in each resistor.
This question was popular among the candidates and they gave good accounts of themselves. Some of them could not evaluate the required reciprocals of V and X and those that did, could not express the results to required number of decimal places. Determination of the intercept C on the vertical axis was a problem because of poor scale used by most candidates.
The part b (i) and (ii) were poorly attempted.
OBSERVATIONS [10]
(i) Six values of x read and recorded in cm to at least 1 decimal place
(ii) Six values of V read and recorded in volts to at least 1 decimal place and in trend, Yz mark each.
[Trend: V increases as x increases]
(iii)Six values of X-1 correctly evaluated to at least 3 decimal places.
(iv) Six values of V-1 correctly evaluated to at least 3 decimal places
(v) Composite table showing x, V, X-1 and V-1.
GRAPH [06]
(i) Axes correctly distinguished 1/2 mark each.
(ii)Reasonable scales,1/2 mark each.
(iii)Six points correctly plotted,1/2mark each.
(iv) Line of best fit
Note: If origin not shown, deduct 1/2 mark for d.i.
SLOPE [02l
(i)Large right angled triangJe
(ii) ∆V-1 correctly determined
(iii)∆x -1 correctly determined
(iv) (∆v-1)/(∆x-1) correctly evaluated

INTERCEPT [01]
Intercept on the vertical axis:
correctly shown
correctly recorded
ACCURACY [01]
Based on teacher's value of resistance per unit length ofpotentiometre wire slope = s = resistance per unit length ± 10%
PRECAUTIONS [02]
Award 1 mark each for any two correct precautions
e.g
Tight connections/clean terminals was ensured
Key removed when reading are not being taken
Parallax error avoided in reading voltmeter/rule
Zero error noted and corrected for in the voltmeter/rule
Dragging jockey along wire avoided
(Accept any other valid precautions).

(b)(i) Award 1 mark each for any two correct devices stated:
e.g. - Diode (p-n)
- Valves
- Transistors, metal rectifier
(Accept any other correct devices)
(ii) 1/R= 1/R1 + 1/R2
= ½ + 1/3 =     5/6
R = 6/5 Ω
V = IR = 1 x 6/5 = 6/5 V
I2Ω = 6/5 ÷ 2 =         3/5 A
I3Ω = 6/5 ÷ 3 =           2/5 A
Alt. Method
I1 = 2/5 × 1 =0.4 A
I2= 3/5 ×1= 0.6A