MINISTRY OF EDUCATION
FIJI SEVENTH FORM EXAMINATION – 2004
EXAMINER’S REPORT
PHYSICS
GENERAL COMMENTS
PAPER I
The format for 2004 FSF Physics Examination paper has been similar to 2003 .
The examination paper on the whole had tested knowledge and concepts expected to be learned by seventh form physics students in Fiji, particularly :
- familiarity with basic concepts ; - problem solving techniques ;
- knowledge of practical situations encountered during practical lessons ; and - coverage of the FSF prescription
A total of 1280 candidates had registered to this examination. There is a decrease by 6.64% in the number of candidates registered for the examination, compared to the year 2003.
No grace marks were given as the examination was a valid one and proof read several times.
However, there was an inadequacy in question 7, part (iv) of section C. The value of R was not stated. This inadequacy should not have affected the thinking of the candidates as the answer should have been expressed in terms of R.
During the duration of the marking of the scripts, it has been revealed that candidates did not follow instructions given on the paper. All the instructions were very clearly stated on page 1 of the Question paper and Answer Book. Individual sections and questions and questions had instructions which candidates had to follow.
The setting out of the answers was generally good. Candidates should be well advised to list their information clearly, write down the correct formula and show their substitution to an incorrect calculation (answer), partial marks are awarded for correct substitution and consistent working.
Many candidates have poor understanding of the units especially in momentum, torque and magnetism.
Mathematical errors, especially those dealing with algebraic fractions and powers of 10 were quite common.
e.g. ɑ + ɑ/2
2.
Finally, I wish to call upon the teachers to encourage and motivate their students during the teaching and learning of Physics. Teachers should endeavour to maintain link between physical ideas and relevance to students lives in their teaching by making reference to real life experience and relevance of the concepts in real life. Upon request by seventh form Physics teachers solutions to 2004
examination paper is included in the report.
NOTE : Overall statistics for Paper I.
Parameter Score
Highest Score Lowest Score
Median Range
IQR Mean
94 5 45 89 31 42.4
This report is based on statistics from a random sample of raw scores. The distribution of the raw marks was approximately normal. It showed a more realistic spread, the paper discriminating very well between the able and weak candidates.
SECTION A
This section contained 30 multiple-choice questions, each worth 1 mark.
In general this section was not very well done. Analysis of scripts show that marks range from 5 – 26, with an average of 25, which is 50%.
Several candidates did not make a choice on some of the multiple-choice questions. They should be encouraged to do so.
3.
The table given below is taken from a random sample of 100 scripts. The correct answer is marked by an asterisk *.
Q.No. A B C D Q.No. A B C D
1 1 7 19 73* 16 27* 8 58 7 2 7 6 41 46* 17 5 80* 6 9 3 8 67* 7 18 18 11 32 53* 4 4 6 8 39 47* 19 8 67* 6 19 5 14* 9 63 15 20 9 33* 8 50 6 8 67* 8 17 21 21 2 13* 64 7 5 8 33* 54 22 47* 4 30 19 8 47* 7 3 43 23 10 3 14 73* 9 6 0 73* 21 24 9 21 20* 50 10 7 70 53* 10 25 33* 7 47 13
11 6 67* 22 5 26 8 33* 58 1
12 5 4 58 33* 27 6 53* 31 10
13 47* 10 41 2 28 7 22 67* 4 14 7 18 8 67* 29 10 65* 22 3 15 15 47* 6 32 30 4 23 67* 6
SECTION B
This section contained ten questions and all are compulsory. Each question was worth 3 marks. Analysis shows that this section was difficult to most candidates. The marks ranged from 0 – 28.5 (out of a possible 30), with an average of 11.4.
A handful of candidates did not attempt this section at all and as a result, they lowered the mean score of this section.
The following remarks relate to individual questions.
QUESTION 1
The question was very poorly done. It was very discouraging to see that students at Form 7 level lack the understanding of synthesis the word problems.
4.
The correct way to problem solve this question is :
2m TCOS 60°
A B 1m C
100N
100N
280N
(a) Taking A as pivot :
Clockwise moment = Anti clockwise moment
∑
∑
(280N x 1m) + (100N x 1m) + (100N x 2m) = 2mx T COS 60°
580N =T
(b) FAX = T sin 60° = 502.3 N
Some students calculated FAY = 19ON.
The question had very clearly stated calculate the horizontal component of force excited by the wall at A.
QUESTION 2
This question was also very poorly done. Many students lacked the understanding of 2 – dimensional momentum especially when Pythagoras theorem and/or sine rule is not to be used. A few students were able to calculate the magnitude of B’s final velocity, however, forgot or overlooked the direction of B’s final velocity.
To tackle such question, students could have used components of horizontal and vertical momentum and conservation of momentum to determine the magnitude ot B’s final velocity. Students should use their mathematical skills and their understanding on Geometry to determine the direction (θ).
5.
The correct way to problem solve this question is using the method :
before collision = after collision
∑
∑
• x
0.2kg x 1.50m/s COS 45° = 0.2kg x 1ms cos 25° + BX BX = 0.031 kg m/s
• x
0.2kg x – 1.50 m/s sin 45° = 0.2kg x 1 m/s sin 25° + BY
BY = - 0.297 kg m/s
B = √ BX² + BY²
= 0.3 kg m/s
VB = RB MB
VB = 0.75 m/s
θ = tan-1 RBY RBX
θ = 84°
Method 2
Using cosine rule : :
a² = b² + c² - 2bc cos A to calculate RB. VB = RB/MB
6.
QUESTION 3
This question was very well answered by majority of the students. Analysis showed that students had a thorough understanding of the concept of conservation of energy. This question was very similar to Question 3 of 2003 however the formula for I was changed to I = ½ mr2.
The correct working to derive the expression = √4gh is 3
Loss in Ep = Gain in Ek
Ep = Ek trans + Ek rot
mgh = ½ mv² + ½ Iw²
= ½ mv² = ½ (½ mr²) (v/r)²
= ½ mv² + ¼ mv²
mgh = ¾ mv²
v² = 4gh 3
v = √ 4gh 3
QUESTION 4
This question was generally well done showing that students had good understanding of the concept of thin film interference, however a few students had difficulty in conversions : especially mm to m and nm to m.
The correct working for answering this question is :
θ = ½ λ
x
t = ½ (360 x 10-9m) 0.25 x 10-3 m
θ θ = 7.2 x 10-4 5cm
θ = t/5cm
t = (7.2 x 10-4) (0.05m) t = 3.6 x 10-5m
Another method that could be used to calculate the thickness t is using the formula dx = (n-½) where n = 1
7.
QUESTION 5
This question was very poorly done. The question as a whole was very simple, testing students understanding of the concept on SHM. Analysis of scripts showed that students had difficulty interpreting the reference circle. Teachers and students are advised to take note that reference circle is included in the seventh form Physics prescription.
The correct working for answering this question is :
(a) y = A sin wt
= 20cm sin ( 16s 2 ) (6s)
y = 14.14cm
common error : students swapped T and τ
(b) Vm = VP cos θ
= WA cos θ
Vm = 7.38cm/s
Common error : students wrote Vm = WA
(c) αmax = W²A α = 3.08cm/s²
common error : stating the correct formula.
QUESTION 6
This question was very well answered by many students, however, a few students had difficulty in conversion of area : 2cm² to m². Most of the students were able to state the correct formula and calculate the required parameters. Consistency marking was done for consistent working.
The correct working for answering this question is :
(a) C = ∈oA/d
= (8.85 x 10-12 Fm-1) (2 x 10-4 m²) 5 x 10-3m
8.
(b) Q = CV
Q = 3.54 x 10-11C
(c) E = V/d
E = 2 x 104 Vm-1
QUESTION 7
This question had 2 parts ; (a) and (b) for part (a), students were required to calculate the magnitude of force and state the direction of the force. Many students overlooked the second part of (a) i.e. the direction.
The concept force between two current carrying wires placed at a distance (d) apart, is a repeat of form 6 work and students are expected to know the concept. It was very disappointing to see that students do not know the conventions used for currents into the page and out of page. This concept is
introduced at form 5 level physics under the topic electromagnetism.
The correct working for answering this question is :
(a) F = k I1 I2 l d
= (2 x 10-7 NA-2) (3A) (3A) (20 x 10-2m) 0.1m
F = 3.6 x 10-6 N
Direction : Repulsion
F F
Common error : - students used μo instead of K. - overlooked direction
(b)
x
9.
QUESTION 8
This question was generally well done. Majority of the students were able to correctly answer this question. This question had tested the concept of magnetic field of a soleroid and in particular the hand note used to determine the direction of magnetic field, the effect on magnetic field strength when (i) diameter is halved (ii) the length is reduced by half, while keeping the number of turns constant and calculating magnetic field strength.
The answer(s) for this question is as follows :-
(a)
Use R.H Grip Rule in reverse manner for soleroids.
(b) (i) half d ; no effect B = 0.5T
(ii) B ½ haL l ; ß doubles
B = 2 x 0.5T = 1 T
(c) B =
μ
o NI l10.
QUESTION 9
Overall, this question was very poorly answered. Approximately about 90% of the students do not know how to derive an expression for the resonant frequency and define the term reactance. It is a pity to see that students hide behind the formulae and use symbols (notations) for defining a term. Analysis reveals that students haven’t learnt the concept, they have however claimed the formula for resonant Frequency.
The answer(s) for this question is/are as follows :-
(a) At resonance Vl = Vc
I X L = I Xc
2πf L = 1 4πf C
f² = 1 4π² L C
1
f = √ 4π² L C
1
f = 2π√ L C
(b) Reactance is defined as : the ratio of voltage to current for a capacitor or inductor in an a.c circuit.
QUESTION 10
This question was also very poorly done. Students failed to comprehend magnetic field at an angle of 30° from the vertical shown on the diagram, and as a result were unable to correctly calculate magnetic flux. The other fact picked up was that students had difficulty in conversions; cm to m, ms to s.
Majority of students could not calculate area of the square loop given the dimension in centimeters Faradays law is used to calculate the induced emf in the loop. Many students used B l to calculate emf.
The correct working for answering this question is :
O = BA cos θ ; A = l²
= (4 x 10-2m)²
11.
O = D.ST (1.6 x 10-3 m²) cos 30°
= 6.93 x 10-4 Wb
є
= Δ OΔ t
= 6.93 x 10-4 Wb 20 x 10-3 s
є
= 3.46 x 10-2 VCommon error : Rounding off. Some students wrote 4 x 10-2V; as a result, they were penalized.
SECTION C
This section of the paper contained seven questions, out of which, any four questions were to be answered. Each question was worth 10 marks. The following table presents some statistics on this section.
PARAMETERS RAW SCORES
HIGHEST SCORE LOWEST SCORE RANGE
MEAN SCORE
39.5 0 39.5 16
The table shown below indicates the mean score per question.
Question No. 1 2 3 4 5 6 7
Mean Score 6.5 3.5 4.5
This section was fairly done. A very many candidates who did not do well in Section A and B managed to pull through this section.
A few candidates were able to pass the overall paper due to this section.
The following remarks relate to individual questions :
QUESTION 1
This question was very popular, and generally very well answered by majority of the students. Analysis of scripts portrayed the following :
% attempting this question : 85
12.
Specific Comments
(a) All the parts (i) – (iv) were done satisfactorily. The correct answers for these were :
(i) = Fr = 0.08 Nm
(ii) α = T/I
= 0.27 rad s²
(iii) Δ W = αt
= 0.54 rad/s
(iv) L = IW
= 0.16 kg m²/s
(b) This question was generally well done, however, many students failed to indicate the direction of the force in (i). Some students had difficulty in re-arranging equations. Many were able to clearly state qBυ = Mυ²/r, but cannot find r. H is advisable that weaker students substitute first and then solve. A few students substituted incorrect mass (9.11 x 10-31 kg.
Mass, of proton = 1.673 x 10-27kg. This value is given in the physical data on page 3 of the answer book.
(i) F = qBυ
F = 2.88 x 10-12N
direction : to the left
(ii) r =
r = 0.52m
(c) This question was fairly done. It has been observed that students are very good at questions involving calculations, but when it comes to explanation, they cannot explain the concept tested. The marking criteria for parts (i) and (ii) were such that though students could not phase the sentence properly, key words such as out of phase and destructive interference for (i) and absorb other colours and allow red light to pass for (ii) were looked for while marking.
The correct answer for this question was :
(i) Light waves arrive out of phase at B and interfere destructively forming a dark band. (ii) Filter will absorb other wavelength of lights and will only allow red light to pass
= 6 x 10-7m
13.
QUESTION 2
In general, many candidates did not choose to answer this question. Analysis of scripts portray the following
Highest Score : 8.5 Lowest Score : 0.5
Range : 8
Mean Score : 3.5 % attempting this question : <10%
Specific Comments
(a) Many students had difficulty with vector related question on the gravitational field.
A similar question on resultant electric fluid between two point charges was tested in the year 2003.A few candidates were able to state the correct formula, but were unable to correctly substitute the Universal Gravitational constant.
Common error : G = 0 x 109 instead of G = 6.67 x 10-11 Nm2 kg-2
The correct working for this question is :
g = GM/r²
g1 = 3.34 x 10-6 m/s²
g2 = 3.34 x 10-6 m/s² gR √ g12 + g2²
gR = 4.72 x 10-6 m/s2
14.
The correct answers for this question are as follows :
(i)
3μf C1 6μf
1/C = 1/4μf + 1/53μf
C = 20/9 μf
C1 = 2μf + 20/9μf = 4.22μf
1/CT = 1/3μf + 1/C1 + 1/6μf
CT = 1.36 x 10-6F
(ii) Q = CV
Q = 2.45 x 10-5C
(iii) Vyz = Q/C = 4.08 V
(iv) E = ½ CV² = 4.99 x 10-5J
Common error : V = 18V was substituted instead of 4.08V.
(c) Overall, this question was very poorly done. Some candidates had difficulty in re-arranging equations. A few candidates were able to state qBυ = mυ2/r, but were unable to make υ the subject of the formula in part (i). Quite a few did not cancel hence making the calculation cumbersome. Candidates did not know the concept of circular motion. They also lack the profound. Understanding of the concept Ek = eV. The correct answer(s) for this question is :
(i) υ = rqB m
υ = 1.72 x 10-6 m/s
(ii) T = 2π/ υ
T = 1.1 x 10-7 S
15.
Common error : T = 1.1 x 10-7 s. It seems candidates did not read the question properly. The question very clearly states in bold letters “one-half” of a revolution.
(iii) V = Ek/e = ½ υ m² e
V = 3.1 x104V
Many candidates used mass of electron instead of the mass of deutron which was stated in the question.
QUESTION 3
This question was also attempted by many candidates. Analysis portray the following :
Highest : 10 Lowest : 0 Range : 10 Mean Score : 3.6
% attempting this question : N 75%
Specific Comments
(a) Overall this question was not well answered well by most candidates. Many candidates do not know the concept of diffraction Grating. It seems this concept is inadequately covered by teachers. The diffraction grating is a valuable tool for analyzing the light given off by atoms which are excited by heat or electrical discharges.
Students had difficulty in calculating the slit spacing d, given the number of slits per centimeter (N).
Due to large number of slits, the fringes are bright. Hence the formula d θsin = λn should be used. Many candidates used d θsin = (n-½)λ which is wrong.
Candidates had to calculate the angular separation (θ) between the ∝ and δ lines in the second – order spectrum (n = 2). The correct working for this question is :
d = 1/N d = 2 x 10-6 m d sin θ = ∩λ/d)
θ∝ = 41°
θδ = 24.2°
θ = θ∝ - θδ
θ = 16.8°
16.
(b) Instructions for this part was not followed. Those candidates who did not follow instructions were penalized. The correct answers for this question is :
X - IV
Y - II
(c) This part was based on the subtopic projectile motion. Many candidates just stated the answers as “ball passes over the crossbar” without justifying. To justify the once the following working was required :
t = R/Vx = 50m 20 m/s
t = 2.5s
s = ut - ½ gt²
= 15 m/s (2.5 s) -½ (10 m/s²) (2.5s)² = 6.25 m
At t = 2.5s, s = 6.25 m, so the ball passes over the crossbar which is 4 m high.
(d) This question was generally well answered by many candidates. However, a few candidates were unable to answer (iv) of the question. The correct answer to this question is :
(i) T = 2π √ m/k
T = 0.04s
Common error : conversion of mass to kilogram – 10g = 10 x 10-3 kg.
(ii) w = 2π/T = 157 rad/s V = WA V = 47.1 m/s
Consistency of T and W was taken into consideration while marking for V.
(iii) ET = ½ KA²
= 10.8J
(iv) t = ¼ T
= 0.01 s
17.
QUESTION 4
Analysis of scripts portray the following for this question.
Highest : 9.5 Lowest : 1.0 Range : 8.5 Mean Score : 3.9
Specific Comments
(a) This question was well answered by many candidates. This shows that candidates had good understanding of progressive wave. The correct answer to this question is :
(i) W = 2π (50) = 100π
f = w/2π
f = 50Hz
(ii) k = 4π
λ = 2π/k
λ = 0.5m v = fλ
v = 25 m/s
Consistency from (i) was noted while making the scripts. (iii) Y = 0.02 sin 2π (0.5 x + 50t)
(b) This question was fairly done. Most candidates were able to correctly recall the formula J = I/A for part (i). Part (ii) of this question was very poorly done. Almost all candidates wrote p = EJ. There is a confusion here especially with the formula. The correct formula is E = pJ. For part (iii) most of the candidates were able to get the correct answer for velocity.
The correct answer to this question is :
(i) J = I/A
= sA 1 x 10-6m² J = 2 x 106 A/m²
Common error : conversion of cross-sectional area from mm² to m²
(ii) E = pJ
18.
(iii) ν = I/neA
ν = 1.25 x 10-4 m/s
The formula ν = J/ne can also be used to calculate velocity.
Marks were awarded for consistent working and correct substitution.
(c ) Only very few able candidates were able to correctly answer this question. Students need to study the laws in electro magnetism thoroughly. A large percentage could not work with SI units correctly. Candidates had to use Flemmings left hand Rule to determine the direction in which the coil with rotate.
The correct answer(s) to this question is :
(i) Direction : anticlockwise.
(ii) λ = BANI
λ = 1.28 Nm
Common error : Conversion and calculation of area - S.I. Unit for torque.
(iii) λ = BANI cosθ
= 0 Nm
Note : When θ = 90° ; there is minimum torque.
QUESTION 5
This was another popular question. Analysis portray the following :
Highest : 10 Lowest : 3.5 Range : 6.5 Mean Score : 4.5
% attempting this question : 80
Specific Comments
(a) This question was very well done. Many students were able to write the loop equations correctly but had some difficulty solving it simultaneously. It must be noted that substitution for I3 should be done while solving the equations. A few students wrote answer for I1 as – 33/28 but used the positive value of I1 in further substitution, hence getting incorrect value of I2 and I3.
19.
The correct answers were :
(i)
∑
current arriving a junction =∑
current leaving a junctionI1 + I2 = I3
(ii) 1. For loop ABEFA ;
2I3 + 4 I1 + 11 = 0
2. For loop CDEBC 11ν - 4I1 + BI2 = 0
(iii) I1 = -1.18A , I2 = -1.97A & I3 = -3.15A
(b) The performance in this question was satisfactory, however a few points that were picked up while marking was that students do not set out working in the required manner for deriving an expression.
(i) Some students were able to write = Iα but were unable to aquate this to, Te
The correct working for deriving T = ½ ma is :
= Iα
T.r = ½ mr2 . α
T = ½ mr α ; a = 2α
T = ½ m a
(ii) The diagram was drawn to aid students to answer this question. The common error made was that students wrote T –n mg = ma instead of mg – T = ma. A few students could not add a + ½a. This reflects their mathematical skills and needs to be improved on. Consistency of T from (i) was noted while marking. The correct working for deriving an expression for the acceleration in terms of g is :
T
mg - T +ma • mg - ½ ma = ma
20.
(c) Parts (i) and (ii) of this question was fairly well done, however students made mistake in substituting the value for fo. They used 1.2 x 1015 Hz instead of fo = 9.014 Hz. Part (iii) was very poorly done. Many students failed to answer this question. A few students answered the effect of increasing the frequency of light. The question very precisely states “Explain the effect of increasing the intensity of the incident light on the cut-off voltage.
The correct answers for these parts are :
(i) O = hfo
= s.97 x 10-19 J
(ii) EK = E - O = hf - O
= 1.99 x 10-19J or 1.24 eV
(iii) Stopping potential (cut-off voltage) is independent of intensity of light. Ans: No effect
QUESTION 6
The statistics for this question are as follows :
Highest : 9.5 Lowest : 0 Range : 9.5 Mean Score : 4.2
Specific Comments
(a) Generally this question was well answered by many candidates, however, teachers should consistently remind students to use the correct diagrammatic symbols of electric
components. A few candidates used the obsolete (old) symbol for resistor, as a result they were penalized. The electrical symbols that are to be used, is introduced at form 5 level and this is very dearly given on the last page of FSLC prescription. Part (ii) of this question had tested a similar concept as that of 2003 FSFE Physics paper. Due to very low If.s.d, Vg was low. Thus students had used the formula R = Vg/10 to get the value of Resistance (R = 0.05Ω) which is incorrect. For this, the working was looked at while marking.
The answer(s) to this question is as follows :-
G
21.
(ii) IP = 10A - 0.01A = 9.99A
Vg = 0.01Z x 50 Ω = 0.5ν
Vg = Vp
0.SV = (9.99A) Rp
Rp = 0.05Ω
(b) This question was not satisfactorily done. Candidates are relatively weak at questions the involve derivation.
A few candidates equated Fe and Fg and as a result got an expression for the velocity of satellite.
The correct derivation for escape velocity of he object of mass m/from the earth’s gravitational field is :
Ek - Ep = 0 ½ mν² - Gmme/Re = 0
½ mν² - = Gmme Re
ν² = 2GMe Re
ν = √ 2GMe Re
(c) This question was generally well done, however, a few candidates did not convert 0.2mH to Henry (H). The correct answer to this question is :
(i) emf = - L di/dt
= (0.2 x 10-3H) (30A/S) emf = 6 x 10-3V
(ii) E = ½ LI² E = 0.25J
22.
(d) This question was very poorly answered. The word “tuned” meant the circuit is at resonance. A few candidates were able to correctly state the formula for resonant frequency, but could not make C the subject of the formula. Mathematical skills certainly have been a drawback for many candidates. As stressed, several times in the report, candidates did not convert 0.14mV
to 0.4 x 10-3V.
I call upon teachers to stress on this point to students, so that they prepare well for the future examinations.
The correct working to answer this question is as follows :
(i) fo = 1/2π √ LC 1
C = 4π ½ fo² L = 6.7 x 10-12 C = 1.49 x 10-13F
(ii) I = V/R
= 0.4 x 10-3V 5Ω
I = 8 x 10-5A
QUESTION 7
The statistics for this question are as follows :-
Highest : 9.0 Lowest : 0.5 Range : 8.5 Mean Score : 3.2
Specific Comments
(a) This question as a whole was unsatisfactorily answered. Majority of candidates had (i) and (ii) incorrect. A very common response for part (i) was Jump A, which is incorrect. Many
candidates had difficulty in recalling the correct formula part (ii) . A few who were able to write the correct formula did not know that as L →∞ ; 1/L → O . Mathematical skills are very poor. For part (iii), there was no need to convert the answer to Joules; Electron – volt is accepted.
23.
The correct answer for this question is :
(i) Jump C
(ii) 1/λ = R (1/s² - 1/L²)
1/λ = 1.097 x 1.097 x 107 (1/3² - 0)
λ = 8.2 x 10-7m.
(iii) ΔE = (-3.4 - -13.6 )ev = 10.2eV
(b) Overall this question was very poorly done. This confirms that candidates do not understand the concept is inadequately convened by the seventh form Physics teachers. For part (iv) of this question, the value of R was not stated, however, the answer should have been expressed in terms of R. Full marks was awarded for the correct expression for power in terms of R. A few candidates were confused with the term phase difference and phase angle. These two terms should be clarified to students. Phasor diagram would be of great help for better understanding.
The answer to the question is as follows :
(i) VL = √ VS² - VR²
= √ (240V)² - (100V)²
VL = 218.2V
(ii) Vs
VL
= tan –1 (VL/VR)
= 65.38°
VR
Note : phase difference is 90° (means VL leads VR)
(iii) P. F = cos = 0.42
(iv) P = VR Irms where Irms = VR/R or
P = VR²/R = (100) ²
24.
(c) This question was satisfactorily done, however, a few candidates gave the answer for beat frequency as negative. Part (ii) of the question tested the concept of using wave equation
ν = fλ
The answer to this question is
(i) fB = f1 - f2
= 550Hz - 540 Hz
fB = 10Hz
(ii) λ = v/f
λ1 = 1440 m/s λ2 = 1440 m/s 540Hz 550Hz
λ1 = 2.67m λ2 = 2.62m
(d) This question was very poorly done. For many candidates who attempted this question, it can be deduced that they lacked the basic knowledge and understanding of the concept tested. The principle of moment should be applied to calculate the value of x.
This concept is introduced at form five level.
The answer to this question is :
W1X = W2 (L – X)
W1X = W2 (L – X)
X = W2L W1 + W2
25.
PAPER II
GENERAL COMMENTS
The format for 2004 Fiji Seventh Form Physics Examination Paper II has been very similar to the 2003 paper.
This paper was a valid one as it tested the practicals outlined in the seventh form Physics course by the Curriculum Development Unit of Fiji.
The aims and objectives outlined in the Physics Prescription and explanatory now associated with the prescription required the practical programme to be an integral part of the teaching programme. The experiments should give the pupils first hand experience of the ideas, and in many cases introduce the ideas to pupil before the theory teaching begins.
The marks obtained in Paper II are low. This portrays a low level of performance of practical word in seventh form Physics and the manner of answering question in this year’s Paper II seems to confirm that practical components are not adequately covered in school laboratory classes.
Candidates scored well on questions that relate more to theoretical in nature rather than the practical aspects of Paper II.
Given below is the statistics for this year’s Paper II.
Parameter Score
Highest Score Lowest Score
Median Range
IQR Mean
37 0 19 37 11 22.3
NOTE : All statistics is based on raw scores.
The following remarks relate to individual question.
Question No. 1 2 3 4 5 6 7 8
Mean Score 2.0 2.1 2.5 1.5 2.4 3 3 2.8
26.
QUESTION 1
Part (i) was well answered by most candidates.
Part (b) was unsatisfactorily answered for part (i) of this question, candidates did not precisely describe the motion of the rotating disc. The preferred answer is : moves with constant angular acceleration. Many candidates were unable to calculate the gradient skill seems to be a drawback.
QUESTION 2
(a) Part (ii) was generally well answered by many candidates however (iii) was very
unsatisfactorily done. Many candidates had difficulty in converting given units to s.I unit.
Recalling the formula v = T/μ seemed to be another problem for many candidates.
(b) This question was very poorly done. Basic electrical symbols are introduced at Form 5 level Physics. Some candidates drew obsolete symbols, as a result they were penalized. The
question very clearly states “Draw the new preferred symbols”.
QUESTION 3
This question tested the concept of SHM and the response from many candidates was fair.
However, only 0.4% of the candidates were able to correctly fill the column for X (displacement).
Furthermore, candidates had difficulty in plotting points and drawing a best fit line through the points. Candidates had difficulty in calculating the gradient of the graph and how the gradient of the graph is related to angular frequency of the pendulum. Part (v) of the question was generally well done.
QUESTION 4
Both parts (a) and (b) of this question was very poorly done. The mean score for this question is lowest out of all the eight questions.
(a) Majority of the candidates could not :
• write down the variables measured in this experiment ;
• write down the relationship between rise in temperature and the variable measured in (i) ;
• have knowledge that field remains same when – 40 cm and 80 cm length of wire are cut from same reel and the two lengths are joined in series. The ratio of field in the two conductors
is 1 : 1.
well done. This shows that < of of emphasis is placed on the use of mathematical formula rather than the understanding of the basic concepts.
27.
QUESTION 5
Part (a) of this question was well answered by many candidates however, a few common errors that were identified are :
• K = 0.3 (should be log k = p- 0.3)
• Swapping the values of k and n.
Part (b) of this question was unsatisfactorily use Kirchoff’s second law to write an equation relating
ℓi , E and K. The correct equation is : Kℓi - E = 0 or E = kℓi. Majority of candidates could not calculate the cross-sectional area of the nichrome wire given the diameter as 0.75 mm. A few were able to write the formula for calculating partial marks.
QUESTION 5
This question was generally well answered by many candidates. For part (a) candidate should have applied the basic concepts learnt in Form 5 Physics class. A few teachers complained that this practical should not be tested as it is out of syllabus. I wish to convey to these teachers that this practical on Mutual Inductance is very much in the syllabus and there hasn’t been any circular in black and white stating that this practical is out of syllabus.
Parts (b), (c) and (d) were generally well done.
QUESTION 7
This question was generally well done. Majority of the candidates were able to identify the a.s. and d.c. voltage source. A few candidates wrote Vmax instead of Vpp. Part (iv) of this question had a typographical error. The question should have read as “What is the supply voltage of the A.C. source ? (Hind calculate V.ms) Both the supply voltage of a.c. source and d.c. source were considered while marking (iv) and (v).
QUESTION 8
(a) This question was satisfactorily answered. It can be deduced that candidates understood the concept on Doppler effect.
(b) Majority of the candidates were able to state the purpose of the galvanometer (G) in the bridge circuit. Part (ii) of this question had an error. R3 was written instead of R2. This was taken
into consideration while making the script.
