1
MINISTRY OF EDUCATION, HERITAGE & ARTS
FIJI YEAR 13 CERTIFICATE EXAMINATION 2020
PHYSICS
DETAILED
SOLUTIONS
2
STRAND 1
1.
L
B.
M
C.
P
D.
T
Weight
Tension
C.
D.
Static friction
Kinetic friction
linear
B.
orbital
C.
escape
D.
rotational
D.
parabolic.
According to Kepler, the paths of planets about the Sun are
A.
6.
Angle in radians
Angular acceleration
The smallest velocity a small body requires in order to overcome the gravitational
attraction of a more massive body is known as the ____________ velocity.
A.
5.
C.
D.
Which one of the following forces provides the centripetal force on an unbanked
road?
A.
B.
4.
Torque
Angular speed
The angular momentum of a moving body is represented by the symbol
A.
3.
[25 marks]
Which of these rotational quantities is analogous to distance in linear motion?
A.
B.
2.
MECHANICS
ecliptic.
B.
circular.
C.
straight.
A block of mass 4 kg is pulled along a horizontal surface by a force of 25 N at an
angle of 30º above the horizontal.
F = 25 N
30º
Fr = 10.5 N
4 kg
Calculate the coefficient of kinetic friction, µ, if a friction force of 10.5 N acts
between the block and the table top.
(2 marks)
3
Normal force (N) = (4 x 9.8 ) - (25 x Sin 300)
= 39.2 - 12.5
= 26.7 N
Fr
= µxN
10.5 N = µ x 26.7
µ
7.
= 0.39
A mass of 5 kg, on an inclined surface at an angle of 40º is being pulled by a force of
50 N which acts at an angle of 25o. The coefficient of friction between the surface and
the mass is 0.02. Calculate the acceleration of the mass up the slope.
F = 50 N
25º
40°
(3 marks)
Fr = µ(mgCosθ - 50 Sin 25)
Fr = 0.02 (37.54 -21.13)
Fr = 0.33 N
Fnett = 45.32 – (31.50 + 0.33)
Fnett = 45.32 – 31.17
Fnett = 14.15 N
Fnett = ma
14.15
a=
5
a = 2.83 ms2
4
8.
Show that the formula d = vit + ½at2 is dimensionally consistent, where d is the
distance travelled in time, t and vi is the initial velocity and a is the acceleration. (2 marks)

d has dimension of a length, L

vit: velocity x time has dimensions L  T =L ( L= length, T= time)
T
2
½at : acceleration x time has dimensions L2 T 2 =L
T

All the terms have the dimension of a length. The equation is therefore
dimensionally correct.
9.
An aircraft propeller rotates at a speed of 2500 revolutions per minute and has a
radius of 0.8 m.
Source: https://www.explainthatstuff.com/how-propellers-work.
Calculate the linear speed of the tip of the propeller blade.
(2 marks)
5
10.
11.
The area of a square microscope cover slip is (3.24  0.01) cm2.
Calculate the length of the cover slip.
A = kt2
Two quantities are related by the equation
(i)
(3 marks)
Using logarithms transform the above relation into a linear equation of the
form y = mx + c.
(2 marks)
A = kt
2
log A = log kt2
log A = 2 log t + log k
(ii)
Identify the y – intercept from the equation in (i) above.
(1 mark)
log k
(iii)
Identify the gradient of the graph from the equation in (i) above.
2
12.
To study the relationship between force and mass in the gravitational field, the
following experiment set up was used.
Trolley + masses
Ticker timer
Power supply
Scale pan
+ masses
Height
(1 mark)
6
Small masses are added to the scale pan at the start for dummy runs. What is the
purpose of these dummy runs.
(1 mark)
To compensate for friction
13.
An experiment was set-up to study the translational and rotational kinetic energy
of a solid cylinder.
Solid cylinder
Ticker timer
Power supply
Mass
Height
The following set of data was gathered from the above experiment.
Mass (m)
Height (h)
Radius of cylinder
Mass of cylinder (M)
Velocity of mass at impact
(i)
=
=
=
=
=
0.15 kg
0.65 m
0.03 m
0.80 kg
1.00 ms-1
Calculate the gravitational potential energy lost by the mass.
(1 mark)
Ep = mgh
Ep = 0.15 x 9.8 x 0.65
Ep = 0.96 J
(ii)
Calculate the rotational kinetic energy of the cylinder.
E p  EK R  EK T
mgh = EKR + ½mv2
0.15 x 9.8 x 0.65 = EKR + ½ x (0.15 + 0.8) x 12
0.9555 = EKR + 0.475
EKR = 0.48 J
(2 marks)
7
STRAND 2
1.
0.01
7.50
20.94
87.73
In a simple harmonic motion an object completes 10 vibrations in 2 seconds. The
angular frequency of the simple harmonic motion in rads-1 is
A.
B.
C.
D.
3.
[9 marks]
A 5 kg mass attached to a spring undergoes simple harmonic motion with a period of
1.5 s. The force constant of this spring in Nm-1 is
A.
B.
C.
D.
2.
OSCILLATORY MOTION
10
20
10 π
20 π
A 2.3 kg mass oscillates back and forth from the end of a spring with a force
constant of 120 Nm-1. Calculate the total energy at a displacement of 0.13 m where
the velocity is 3.4 ms-1.
(2 marks)
Total Energy = KE + U
= 13.294 + 1.014
= 14.31 J
4.
A simple pendulum of mass, m = 2 kg and length, l = 0.25 m, is held in an elevated
10º position then released, resulting in oscillatory motion.
Ceiling
100
m = 2 kg
8
Using an appropriate expression, calculate the period of oscillation.
5.
(1 mark)
The phenomenon of forced oscillations and resonance is investigated using the
experimental set-up shown. A graph of amplitude versus frequency is plotted.
Amplitude (m)
60
50
40
30
1
20
2
Driver pendulum
3
10
4
0
(i)
0.1
0.2
0.3
0.4 0.5
Frequency (Hz)
From the graph, determine the frequency at which resonance occurs.
(1 mark)
0.3 Hz
(ii)
As the frequency is increased beyond resonance, state whether the forced
pendulum swings in phase or out of phase with the driver pendulum.
(1 mark)
out of phase
(iii)
Sketch the graph of frequency (f) versus length (l) for the simple
pendulum.
(2 marks)
9
STRAND 3
1.
WAVES
[12 marks]
If the intensity of unpolarised incident light on the vertical polariser is , the
intensity of the ray transmitted through this polariser will be
When an unpolarised light passes through a polariser,
the intensity is reduced by a factor of
A.
x
2
C.
B.
2.
2
cosθ
The Doppler Effect is the change in observed frequency of a source due to the
A.
B.
C.
D.
3.
D.
x
type of wave.
type of the source.
medium through which the source travels.
relative motion between the source and the receiver.
A diffraction grating has 800 lines per millimetre. This grating is used with ultraviolet
light to form an interference pattern on a special phosphorescent screen.
Calculate the average slit spacing, d, of the diffraction grating.
(1 mark)
d
=
1
N
=
= 1.25 x 10-3 mm
or 1.25 x 10-6 m
4.
An air wedge between two microscope slides 7.5 cm long and separated at one end
by a paper of thickness 0.09 mm, is illustrated with red light of wavelength 663 nm.
Calculate the spacing between the dark fringes.
(2 marks)

∆x = L  
 2t 
9


∆x = 0.075  66310 3 
 20.0910 
∆x = 2.76 x 10-4 m
10
5.
A string of linear mass density (mass per unit length) 0.4 kgm-1, is under a tension
of 40 N. A wave of frequency 150 Hz and amplitude 3 mm travels down the string.
Calculate the rate, at which the wave transports energy using the formula,
1
(2 marks)
P  μvω 2 A 2 .
2
µ = 0.4 kgm-1
v =√ = √
= 10
ω = 2πf
= 2 x π x 150
= 942.48
P =
P =
= 15.99 W
6.
The diagram shows a vibrating steel wire emitting a frequency of 200 Hz.
1.5 m
(i)
Identify the harmonic shown in the wire.
(1 mark)
3rd harmonic
(ii)
Calculate the wavelength of the standing wave formed in the wire.
(1 mark)
(iii)
Calculate the speed of the wave along the wire.
(1 mark)
v = fλ
v = 200 x 1
v = 200 ms-1
11
(iv)
Calculate the speed of the wave along the wire if the length of the wire is
reduced to 0.75 m.
(2 marks)
3
2
3
0.75 =
2
λ = 0.5 m
L=
v = fλ
v = 200 x 0.5
v = 100 ms-1
STRAND 4
1.
The amount of work done in bringing unit positive charge from infinity to a point
in an electric field is known as
A.
B.
C.
D.
2.
ELECTROSTATICS
resistance.
electric potential.
electrostatic force.
gravitational force.
A capacitor is a device used to
A.
B.
C.
D.
vary the current.
vary the resistance.
store magnetic energy.
store electrical energy.
[13 marks].
12
3.
Three point charges are located at the corners of a right-angled triangle.
- 4 µC
5 cm
3 cm
+
+
7 µC
8 µC
Calculate the net electric force on the 7 µC charge.
F
k q 1q 3
r
2
910 410 710 
6
9
F
6
0.032
F  280N
F
(3 marks)
upwards
k q 2q 3
r
2
910 710 810 
F
6
9
6
0.04 2
F  315N
4.
towards theleft
A charged particle (q = 1.4 mC) moves a distance of 0.4 m from a surface of potential
10 V to 20 V. Calculate the work done by the field during this motion.
(1 mark)
W = ∆Vq
W = 10 x 1.4 x 10-3
W = 1.4 x 10-2 J
13
5.
An air filled capacitor consists of two parallel plates, each with an area of 5 cm2 and
is separated by a distance of 1.5 mm. If a potential difference of 20 V is applied to
1
these plates, calculate the energy density using the formula U E   o E 2 .
(2 marks)
2
V = Ed
E=
=
E = 13,333.33 Vm-1
U =
U =
U = 7.87 x 10-4 Jm-3
6.
The plates of a capacitor are 0.005 m apart and filled with a dielectric constant of
κ = 5. The plates have an area of 0.002 m2 and are connected to a battery of voltage
30 V.
Calculate the
(i)
capacitance of the capacitor.
(1 mark)
C =
C =
C = 1.77 x 10-11 F
(ii)
charge on the plates of the capacitor.
(1 mark)
Q = CV
Q = 1.77 x 10-11 x 30
Q = 5.31 x 10-10 C
7.
A combination of two capacitors with capacitances 4 µF and 7 µF are arranged in
series across a potential difference of 18 V.
18 V
4 µF
7 µF
14
Calculate the
(i)
total capacitance of the circuit.
=
+
=
1
1
+
6
4 x10
7 x10 6
(1 mark)
= 3.93 x 105
CT = 2.55 x 10-6 F
(ii)
total charge in the circuit.
(1 mark)
Q = CV
Q = 2.55 x 10-6 x 18
Q = 4.58 x 10-5 C
(iii)
potential difference across the 4 µF capacitor.
(1 mark)
Q = CV
V =
4.58 x10 5
V =
4 x10 6
V = 11.45 V
STRAND 5
1.
[13 marks]
When a heating coil is connected to a power supply, a current of 2.5 A is recorded.
The charge that flows through the heating coil in 5 minutes is
A.
2.
DIRECT CURRENT
0.5 C
B.
2C
C.
Wheatstone bridge consists of 4 resistive arms.
12.5 C
D.
750 C
15
3.
Resistance of a conductor increases with
A.
B.
C.
D.
4.
increasing current.
constant temperature.
increase in temperature.
decrease in temperature.
Calculate the resistance of a copper rod which is 2 m long and 8 mm in diameter.
The resistivity of the material is 1.756 x 10-8 Ωm.
(2 marks)
A = πr2
A = π x (4 x 10-3)2
A = 5.03 x 10-5 m2
R = ρ
R = 1.756 x 10-8
R = 6.99 x 10-4 Ω
5.
A galvanometer has a resistance of 12 mΩ. A current of 150 mA produces full scale
deflection of the galvanometer. A shunt is connected in parallel to convert the
galvanometer into an ammeter. The resistance of the shunt is 70 µΩ. Calculate
the greatest current that the ammeter can measure.
(2 marks)
IG = 150 x 10-3 A
VG = IGRG
= 150 x 10-3 x 12 x 10-3
= 1.8 x 10-3 V
VS = 1.8 x 10-3 V
IS =
I = IS + IG
I = 25.71 A + 150 x 10-3 A
I = 25.86 A
= 25.71 A
16
6.
The circuit shows two voltage sources, three resistors and an ammeter of negligible
resistance.
3Ω
4Ω
I1
I2
2Ω
ε
9V
Loop 1
(i)
A 3A
Calculate the value of I1 and I2.
Loop 2
(2 marks)
9 - 3I1 - 6 = 0 using loop 1 as shown above.
3 = 3I1
I1 = 1 A
I1 + I2 = 3
12 = 3 - I1
I2 = 3 - 1
I2 = 2 A
(ii) Calculate the value of the unknown voltage, ε.
(2 marks)
3x2 + 4I2 - ε = 0 using loop 2 as shown above.
6 + 4 (2) - ε = 0
6 + 8 - ε = 0
ε = 14 V
7.
Differentiate between conductors and insulators.
Conductors
- allows electricity to pass through
- contain large number of free electrons
- electrical resistances are very low
- valance and conduction bands are overlapped
(2 marks)
Insulators
- does not allow electricity to pass through
- do not contain free electrons
- electrical resistances are very high
- valance and conduction bands are separated
17
STRAND 6
MAGNETIC FIELD
[8 marks]


This strand has 5 questions.
Show necessary working for questions 3-5 as partial marks will be awarded for correct
working.
1.
Charged particles, P1 (proton) and P2 (electron) enter a region of constant magnetic
field with same velocity. If the magnetic field is perpendicular to the velocity, their
path of travel would be
A.
B.
2.
C.
D.
linear.
parabolic.
A law that relates the net magnetic field along a closed loop to the electric current
passing through the loop is best described as
A.
B.
3.
cubic.
circular.
Ohm’s law.
Gauss’s law.
C.
D.
Faraday’s law.
Ampere’s law.
A velocity selector in an electron microscope uses cross electric field, E and
magnetic field, B to allow positively charged protons of very high velocity to
pass through without bending.
proton
x
x
x
x
x B into the page
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
E down page
If the electric field strength is 4.8 x 106 Vm-1 and the proton passes straight
through with a velocity of 2.8 x 107 ms-1, calculate the strength of magnetic
field, B.
Eq = Bvq
B =
=
B = 0.17 T
(2 marks)
18
4.
A solenoid of length 0.15 m, has 200 turns in it and carries a current of 4 A.
l = 0.15 m
current in
current out
Source: https://www.arborsci.com/products/solenoid-air-core
Calculate the magnetic field strength
(i)
at the centre of the solenoid.
(1 mark)
B=
B=
B = 6.70 x 10-3 T
(ii)
5.
outside the solenoid.
zero
The following are the components of a DC motor.
Source: https://courses.lumenlearning.com/physics
(1 mark)
19
Calculate the maximum torque on a 100 turns square loop of wire 10 cm on a side
that carries 15 A of current in a magnetic field of 2 T.
(2 marks)
A =l x l
A = 0.1 x 0.1
A = 0.01 m2
τ = BANI cosθ
τ = 2 x 0.01 x100 x 15 x cos0
τ = 30 Nm
STRAND
1.
[12 marks]
increase
not change
C.
D.
reverse
decrease
An inductor of inductance, 30 mH is connected to a 60 Hz alternating current
source. The inductive reactance, XL, of the inductor is
A.
3.
ELECTROMAGNETIC INDUCTION AND
ALTERNATING CURRENT THEORY
Increasing the angular frequency of the coils in an electric generator will
_______________ the voltage output.
A.
B.
2.
7
0.09 Ω
B.
8.84 Ω
C.
11.31 Ω
D.
11,310 Ω
A circular loop of radius, r = 10 cm is placed in a uniform magnetic field of
magnitude, B = 2.0 T where the face of the loop is perpendicular to the direction
of the magnetic field.
B = 2.0 T (out of page)
Calculate the voltage induced in the loop if the magnetic field is reduced to
zero in 0.2 s.
(2 marks)
20
A = πr2
A = π x 0.1 x 0.1
A = 0.03 m2
φ B initial  BA
 20.03
φ B final  0
 0.06Wb
  N
  B final   B initial 
d B

 N


dt
t


  1
(0  0.06)
0.2
  0.3 V
4.
A thick copper rod is pushed in a uniform magnetic field along contacts which
are connected to a 24 Ω resistor. The length of the rod in the magnetic field is
0.5 m. The rod moves horizontally at a uniform speed of 4 ms-1. The strength
of the magnetic field is 0.5 T.
Calculate the
(i)
emf induced across the rod.
emf
emf
emf
(ii)
= Bvl
= 0.5 x 4 x 0.5
= 1V
power generated by the moving rod.
P
P
P
(1 mark)
V2
R
12
=
24
=
= 0.04 W
(1 mark)
21
5.
A LCR circuit at resonance has a 300 Ω resistor, 0.1 H inductor and a 100 µF
capacitor connected in series. The AC source has an rms voltage of 200 V at 50 Hz.
200 V
0.1 H
300 Ω
100 µF
Calculate the
(i)
(ii)
resonant frequency.
fo
=
fo
=
fo
= 50.33 Hz
(2 marks)
√
√
total impedance, Z.
(2 marks)
At resonance R = Z
Thus Z = 300 Ω
(iii)
(iv)
rms current.
V
I
= IZ
=
I
=
I
= 0.67 A
phase angle.
At resonance X L  X C , ϕ = 0
 X XC 

Tan ϕ =  L
R


ϕ = tan-1 0
ϕ = 0
(1 mark)
(1 mark)
22
STRAND


1.
8
MODERN PHYSICS
This strand has 4 questions.
Show necessary working for questions 3 & 4 as partial marks will be awarded for correct
working.
The de Broglie wavelength, using the formula,  
moving with a speed of 300 ms-1 is
A.
B.
C.
D.
2.
h
of a bullet of mass 0.02 kg
mv
2.21 x 10-36 m
2.20 x 10-36 m
1.11 x 10-34 m
3.32 x 10-32 m
The orbits in which electrons move according to Niels Bohr are
A.
B.
C.
D.
3.
[8 marks]
oval.
linear.
circular.
cylindrical.
The diagram shows possible jumps of the electron from a higher to a lower energy
state in the hydrogen atom.
n
∞
5
4
B
3
2
A
1
(i)
Calculate the energy of the photon, if the frequency of the photon emitted by
the electron making jump B is 6.71 x 1014 Hz.
(1 mark)
E = hf
E = 6.63 x 10-34 x 6.71 x 1014
E = 4.45 x 10-19 J
23
(ii)
Calculate the wavelength of the photon emitted by the electron in making
the energy jump B.
(1 mark)
v = fλ
λ =
λ =
= 4.47 x 10-7 m
(iii) Jump B is in the visible part of the electromagnetic spectrum. Name the
type of electromagnetic energy that is emitted by jump A. Explain your
answer.
(2 marks)
Jump A lies in the ultraviolet part of the spectrum
Jump A emits greater energy than B. As frequency is directly proportional to energy,
jump A emits a photon of greater frequency.
4.
Calculate the shortest possible wavelength at the end of the Paschen series
( n f  3 ).
For shortest wavelength in Paschen series n f  3 and ni  


1  R 1  1 
n 2 n 2 
λ
i 
 f
= 1.097 x 107 (
= 1.097 x 107 ( )
= 1,218,888.89
λ = 8.20 x 10-7 m
.
)
 1  0
  
(2 marks)
24
Formulae and Constants
Mechanics
Mechanics (continued)
1. v f  v i  at
20. v  2GM
r
2. d  v i t  1 at 2
2
3. v f  v i  2ad
2
21.  
2
d
r
Oscillatory Motion
1. y  A sint  
2. v max  A
2
2
3. v   A  y
22. ω  Δθ
Δt
4. a max  ω 2 A  k A
m
23. α  Δω
Δt
5. a  2 y
6. E p  mgh
24. a  r
6. y  A sin t
7. W  Fd
25. f  i  t
7. v  A cos t
1
8. E p  kx 2
2
1
26.   i t   t 2
2
4. F  ma
5. E k 
9. P 
1
mv 2
2
27. f 2  i 2  2
W
t
8. a  2 A sin t
9. F  kx
10. T  2
m
1 k
, f
k
2 m
11. T  2
l
1 g
, f
g
2 l
28.   F r
10. p  mv
29.   I
11. F mg
30. L  I
12. Ff  μ.N
31. L  mvr
12. E 
1
kA 2
2
32. EK r  1 Iω 2
2
13. E 
1
m2 A 2
2
13. v 
2r
T
2
33. f  1
T
14. a c  v
r
2
34. ω  2π f
2
35. E k 
15. Fc  mv
r
16. a c  4π 2 r
T
17. v  rgtanθ
GMm
18. Fg 
r2
19. v 
GM
r
T

GMm
2r
1. v 
GMm
r
2. f  1 T
2L μ
36. U  
37. g 
Waves
GM
r2
3. P  1 μvω 2 A 2
2
4. v  fλ
25
Waves (continued)
Electrostatics (continued)
 v  v0 
 f
5. f '  
v

v
s 

14. E p 
6. y  A sint  kx 
7. I  I o cos 2 
8.   2f
2

9. k 
10. PD  dsinθ  dx  nλ
L
Electrostatics
1.
F
k q 1q 2
r
2
1
QV
2
1
 CV 2
2
Q2

2C
Electromagnetic
Induction &
Alternating Current
Theory (continued)
16. Vrms 
Vmax
Electromagnetic
Induction & Alternating
Current Theory
17. I rms 
I max
1. V  BAN sint
Direct Current
2. VC  I C
1. I  nevA
2

Vrms
RT
2. Q  It
1
1

2fC C
3.  C 
2
3. V  IR
4. V  IZ
1
1
1


 ...
R p R1 R 2
2.
φ E  EAcosθ
5. VL  I L
4.
3.
F Eq
6.  L  2fL  L
5. R s  R 1  R 2  ...
4.
E
kQ
r2
5.
W  Vq
6.
V
kQ
r
7.
V  Ed
8.
Q  CV
9.
C
r 0A
d
10. C p  C1  C 2  ...
11.
1
1
1


 ...
C s C1 C 2
7. Z  R 2   L   C 
6. E  J
8. Pf  cosφ  R
Z
7. P  VI  I 2 R 
2
9. f o 
1
2 LC
10. tan  
L  c
R
11. V  Bvlsinθ
12.    N
d B
dt
13.   L
dI
dt
12.   RC
14. E L 
13. U E  1  0 E 2
2
15. V 
1 2
LI
2
 
t
8. J  I  nev
A
Magnetic Field
1. F  Bvq
2. V  E
B
3. F  BI l sin
4.   BANIcos 
5. F 
kI1 I 2 l
r
v2
R
2
26.
Magnetic Field(continued)
Constants
μ I
6. B  0  kI
2π r r
Speed of light,
c = 3 × 108 ms-1
Electronic charge,
e = 1.60 × 10-19 C
Electron rest mass,
me = 9.11 × 10-31 kg
Proton rest mass,
mp= 1.67 × 10-27 kg
Rydberg constant,
R = 1.097 × 107 m-1
Mass of Earth,
Me = 5.97 × 1024 kg
Modern Physics
Radius of Earth,
Re = 6.37 × 106 m
1. E  hf
Planck’s constant,
h = 6.63 × 10-34 Js
7. B 
8.
μ 0 NI
 μ 0 nI .
l
B
 0 NI
2 r
9. φ B  BAcosθ
10. r  mv
Bq
2. E k  eV
= 340 ms-1
3. E k  hf  
Speed of sound in air


4. 1  R  1 2  12 
n
λ
n i 
 f
Coulomb’s constant,
k = 9.0 × 109 Nm2C-2
Gravitational Acceleration
g= 9.8 ms-2
Magnetic field constant,
k = 2 × 10-7 Hm-1
Permeability of free space,
µo= 4π × 10-7 Hm-1
Gravitational constant,
G = 6.67 × 10-11 Nm2kg-2
Electric field constant,
o = 8.85 × 10-12 Fm-1
5. c  f
 13.6
eV
6. E n   hcR
2
2
n
n
7. λ  h
mv
8. L  nh
2π
9. v  nh
2π mr
3