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PP-DSE-PHY 1A–17
17
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PP-DSE-PHY 1A–18
18
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List of data, formulae and relationships
Data
R = 8.31 J mol 1 K 1
NA = 6.02 × 1023 mol 1
g = 9.81 m s 2 (close to the Earth)
G = 6.67 × 10 11 N m2 kg 2
c = 3.00 × 108 m s 1
e = 1.60 × 10 19 C
me = 9.11 × 10 31 kg
0 = 8.85 × 10 12 C2 N 1 m 2
0 = 4 × 10 7 H m 1
u = 1.661 10−27 kg
(1 u is equivalent to 931 MeV)
AU = 1.50 1011 m
ly = 9.46 1015 m
pc = 3.09 1016 m = 3.26 ly = 206265 AU
 = 5.67 10−8 W m−2 K−4
h = 6.63 × 10 34 J s
molar
gas
constant
Avogadro
constant
acceleration due to gravity
universal gravitational constant
speed of light in vacuum
charge of electron
electron rest mass
permittivity of free space
permeability of free space
atomic mass unit
astronomical unit
light year
parsec
Stefan constant
Planck constant
Rectilinear motion
Mathematics
For uniformly accelerated motion :
Equation of a straight line
y = mx + c
Arc length
=rθ
Surface area of cylinder
= 2rh + 2r2
Volume of cylinder
= r2h
Surface area of sphere
= 4r2
Volume of sphere
=
v = u + at
1
s = ut + at 2
2
v2 = u2 + 2as
4 3
r
3
For small angles, sin  ≈ tan  ≈  (in radians)
U=
GMm
r
a itati
te a
al
te tial e e
k
t
ate
k
la
e e
ta
e
c
t e mal t a mitta ce
cti
al e
le e ect
0
0
vma
= f
ma im m
Ei tei

t elect ic e ati
Ra lei
d
e
e e
=
h
h
=
p mv

eB
le el e
lie
ati
e at m
= c
m la
c ite i
e l i
e
PP-DSE-PHY 1A–19
I
Z
Z
=
I
Z +Z
I=I e
d
19
ac
x
i
c ite i
t
i e
e l i
e
a le
i te it le el
=
Ra lei
e
e
B
tic im e a ce
i te it
e lecti
c e icie t
t a mitte i te it t
a me i m
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Q1Q2
A1. E = mc ∆T
energy transfer during heating
and cooling
D1.
F=
A2. E = l ∆m
energy transfer during change
of state
D2.
E=
pV = nRT
equation of state for an ideal gas
D3.
V=
pV =
kinetic theory equation
D4.
E=
molecular kinetic energy
D5.
I = nAvQ
D6.
R=
A3.
Nmc 2
3
3RT
A5. EK =
2N A
A4.
v
=
t t
B1. F = m
4π 0 r 2
Q
4π 0 r
electric field strength due to
a point charge
4π 0 r
electric potential due to
a point charge
d
electric field between parallel plates
(numerically)
general current flow equation
l
A
resistance and resistivity
D7. R = R1 + R2
Force
Coulomb’s law
B2. moment = F d moment of a force
D8.
1
R
B3. EP = mgh
gravitational potential energy
D9.
P = IV = I 2 R
kinetic energy
D10. F = BQv sin 
force on a moving charge in a
magnetic field
mechanical power
D11. F = BIl sin 
force on a current-carrying
conductor in a magnetic field
centripetal acceleration
D12. V =
Hall voltage
Newton’s law of gravitation
D13. B =
mv 2
B4.
EK =
B5.
P = Fv =
2
t
2
B6.
B7.
v
a=
=  2r
r
Gm1 m2
F=
r2
D
C1.
y =
C2.
d sin  = n
1
R1
nQt
0
2πr
 0 NI
D14. B =
l
Φ
D15.  = N
t
V
N
D16. s  s
fringe width in
double-slit interference
diffraction grating equation
p
C3.
1
1
u
v
1
f
p
resistors in parallel
power in a circuit
magnetic field due to a long
straight wire
magnetic field inside a long
solenoid
induced e.m.f.
ratio of secondary voltage to
primary voltage in a transformer
equation for a single lens
E1.
E2.
N = N0 e
t1 =
2
PP-DSE-PHY 1A–20
1
R2
resistors in series
20
kt
ln 2
k
law of radioactive decay
half-life and decay constant
E3.
A = kN
activity and the number of
undecayed nuclei
E4.
E = mc 2
mass-energy relationship
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(b)
When the oil flows through the pipe in the heater at a rate of 0.3 kg per minute, the temperature of
the oil rises from 25°C to 37°C. Determine the power absorbed by the oil.
Given : specific heat capacity of oil = 2500 J kg–1 °C–1
Answers written in the margins will not be marked.
(3 marks)
*(c)
If the wooden box is sealed and made air-tight, how would the air pressure inside change when
temperature increases ? Explain briefly in terms of kinetic theory. No mathematical derivation
is required.
(3 marks)
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PP-DSE-PHY 1B−3
3
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PP-DSE-PHY 1B−5
5
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*(c)
Ball Y lands on the ground at S which is at a horizontal distance of 1.2 m from the bench. Find
the height H of the bench.
(3 marks)
*(d)
Ball X is now released at Q such that ball Y moves off the bench horizontally with a smaller speed
after collision. Would the time of flight of ball Y change ? Explain briefly.
(2 marks)
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PP-DSE-PHY 1B−7
7
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4.
A communications satellite moves in a circular orbit around the Earth with a period of 24 hours and
remains above a certain place on the equator.
Given : radius of the Earth rE = 6400 km
*(a)
(i)
Find the orbital radius of the communications satellite.
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(3 marks)
*(ii)
Determine the orbital speed of the communications satellite.
(2 marks)
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PP-DSE-PHY 1B−8
8
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PP-DSE-PHY 1B−13
13
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(b)
(c)
(i)
Express tan  in terms of the electric force F acting on the ball and the weight W of the
ball.
(1 mark)
*(ii)
Given that the mass of the ball is 0.07 g. When the voltage between the plates is
4000 V,  = 2°. Estimate the magnitude of the charge carried by the ball. Assume that
the electric field between the plates is uniform.
(3 marks)
Using the setup in Figure 8.1, suggest a simple method to test whether the electric field between
the plates is uniform.
(3 marks)
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PP-DSE-PHY 1B−17
17
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(iii)
A 5 A fuse is installed in the power supply. Explain whether the fuse will blow if
another identical heater is connected in parallel with the original heater.
(2 marks)
*(b)
The heater is now connected to a sinusoidal a.c. power supply. The peak value of the voltage of
the a.c. power supply is 15 V. How would the output power of the heater change ?
(2 marks)
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(a)
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PP-DSE-PHY 1B−19
19
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(c)
Explain why thick wire should be used to construct the primary coil.
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(3 marks)
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PP-DSE-PHY 1B−21
21
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11.
The decay of radioactive isotope protactinium-238 ( 238 Pa) has a half-life of approximately 136 s. A
sample of 238 Pa is put in front of a GM tube and the initial count rate is 1000 counts per minute. The
background count rate is 50 counts per minute.
(a)
It is known that the decay of 238 Pa does not emit  radiation.
radiation from 238 Pa is  radiation but not  radiation.
Suggest a simple test to verify the
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(3 marks)
*(b)
238
Estimate the decay constant of
Pa.
(1 mark)
*(c)
Hence, or otherwise, estimate the time taken for the count rate to drop to 250 counts per minute. (3
marks)
END OF PAPER
Sources of materials used in this paper will be acknowledged in the Hong Kong Diploma of Secondary Education
Examination Practice Papers p bli ed by the Hong Kong E ami atio s and As essment A t ority at a later stage.
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PP-DSE-PHY 1B−22
提供
22
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PP-DSE-PHY 1B−23
23
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PP-DSE-PHY 1B−24
24
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PP-DSE-PHY 2−5
5
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Q.2: Structured question
(a)
In studying the photoelectrons emitted from sodium, it was found that no photoelectrons were
emitted when the wavelength of the incident light was longer than 5.27  10–7 m.
(i)
Explain why the wave model of light cannot account for this phenomenon.
(2 marks)
(ii)
Determine the work function for sodium.
Express your answer in electron-volts.
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(3 marks)
(iii)
What is the physical meaning of work function ?
(1 mark)
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PP-DSE-PHY 2−10
10
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Section D : Medical Physics
Q.4: Multiple-choice questions
4.1
The far point of Phoebe’s eye is at 60 cm and its lens-to-retina distance is 2.5 cm.
the corrective lens that she should wear ?
A.
B.
C.
D.
4.2
Answers written in the margins will not be marked.
A
D
A
(1) and (2) only
(1) and (3) only
(2) and (3) only
(1), (2) and (3)
B
C
D
The sound intensity of a machine is 0.01 W m 2. By adding a noise barrier, the sound intensity is
reduced to 6  10 6 W m–2. Find the reduction in the sound intensity level.
Given : threshold of hearing = 1  10 12 W m 2
A.
B.
C.
D.
A
32 dB
68 dB
88 dB
100 dB
(For questions 4.4 and 4.5)
different media X, Y and Z.
B
C
D
The following table shows the acoustic impedances and the densities of three
Medium
X
Y
Z
Acoustic impedance /  106 Rayl
1.48
1.63
1.66
Density / kg m–3
1000
1058
1060
Arrange the speed of sound in the three media, vX, vY and vZ, in descending order.
A.
B.
C.
D.
4.5
C
The middle ear is filled with air.
The ear bones form a lever system.
The area of the ear drum is larger than that of the oval window.
A.
B.
C.
D.
4.4
B
Which of the following features of the middle ear amplify the pressure ?
(1)
(2)
(3)
4.3
– 2.0 D
– 1.7 D
1.7 D
2.0 D
What is the power of
A
vX > v Y > v Z
vY > v X > v Z
vZ > vX > v Y
vZ > v Y > v X
B
C
D
What is the intensity reflection coefficient between medium X and medium Z for normal incidence ?
A.
B.
C.
D.
7.97  10–4
8.48  10–4
3.29  10–3
5.73  10–2
A
C
D
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PP-DSE-PHY 2−16
B
16
4.6
Which of the following statements about a fibre optic endoscope is incorrect ?
A.
B.
C.
D.
Tissue samples can be obtained at the same time for further examination.
Coherent bundle fibres are used for image transport.
The cladding of the optical fibre must have a smaller refractive index than the glass fibre
inside.
The critical angle of the optical fibre should be as large as possible.
A
4.7
(3)
Answers written in the margins will not be marked.
C
D
Which of the following statements about radionuclide planar imaging is/are incorrect ?
(1)
(2)
Radionuclide planar images provide functional information about the organ.
Radiation of radionuclide planar imaging stays shorter inside our body than that of X-ray
radiographic imaging.
Tracers are used to absorb radiation.
A.
B.
C.
D.
4.8
B
A
(1) only
(2) only
(1) and (3) only
(2) and (3) only
B
C
D
Technetium-99m has a half-life of 6 hours, and a biological half-life of 3 hours. A patient is given an
injection of technetium-99m at 12:00 noon. Which of the following statements is/are correct ?
(1)
(2)
(3)
At 3:00 pm, the number of undecayed technetium nuclei in the patient’s body is less than half of
the initial value.
All the technetium nuclei will be removed from the body by biological processes by 6:00 pm.
All the technetium nuclei removed from the body by biological processes are decayed.
A.
B.
C.
D.
A
(1) only
(2) only
(1) and (3) only
(2) and (3) only
C
D
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PP-DSE-PHY 2−17
B
17
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PP-DSE-PHY 2−20
20
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PP-DSE-PHY 2−21
21
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PP-DSE-PHY 2−22
22
List of data, formulae and relationships
Data
R = 8.31 J mol 1 K 1
NA = 6.02 × 1023 mol 1
g = 9.81 m s 2 (close to the Earth)
G = 6.67 × 10 11 N m2 kg 2
c = 3.00 × 108 m s 1
e = 1.60 × 10 19 C
me = 9.11 × 10 31 kg
0 = 8.85 × 10 12 C2 N 1 m 2
0 = 4 × 10 7 H m 1
u = 1.661 10−27 kg
(1 u is equivalent to 931 MeV)
AU = 1.50 1011 m
ly = 9.46 1015 m
pc = 3.09 1016 m = 3.26 ly = 206265 AU
 = 5.67 10−8 W m−2 K−4
h = 6.63 × 10 34 J s
molar
gas
constant
Avogadro
constant
acceleration due to gravity
universal gravitational constant
speed of light in vacuum
charge of electron
electron rest mass
permittivity of free space
permeability of free space
atomic mass unit
astronomical unit
light year
parsec
Stefan constant
Planck constant
Rectilinear motion
Mathematics
For uniformly accelerated motion :
Equation of a straight line
y = mx + c
Arc length
=rθ
Surface area of cylinder
= 2rh + 2r2
Volume of cylinder
= r2h
Surface area of sphere
= 4r2
Volume of sphere
=
v = u + at
1
s = ut + at 2
2
v2 = u2 + 2as
4 3
r
3
For small angles, sin  ≈ tan  ≈  (in radians)
U=
GMm
r
a itati
te a
al
te tial e e
k
t
ate
k
la
e e
ta
e
c
t e mal t a mitta ce
cti
al e
le e ect
0
0
vma
= f
ma im m
Ei tei

t elect ic e ati
Ra lei
d
e
e e
=
h
h
=
p mv

eB
le el e
lie
ati
e at m
= c
m la
c ite i
e l i
e
I
Z
Z
=
I
Z +Z
I = I 0e
d
ac
x
i
c ite i
t
i e
e l i
e
a le
i te it le el
=
Ra lei
e
e
B
tic im e a ce
i te it
e lecti
c e icie t
t a mitte i te it t
a me i m
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PP-DSE-PHY 2–23
23
Q1Q2
A1. E = mc ∆T
energy transfer during heating
and cooling
D1.
F=
A2. E = l ∆m
energy transfer during change
of state
D2.
E=
pV = nRT
equation of state for an ideal gas
D3.
V=
pV =
kinetic theory equation
D4.
E=
molecular kinetic energy
D5.
I = nAvQ
D6.
R=
A3.
Nmc 2
3
3RT
A5. EK =
2N A
A4.
v
=
t t
B1. F = m
4π 0 r 2
Q
4π 0 r
electric field strength due to
a point charge
4π 0 r
electric potential due to
a point charge
d
electric field between parallel plates
(numerically)
general current flow equation
l
A
resistance and resistivity
D7. R = R1 + R2
Force
Coulomb’s law
B2. moment = F d moment of a force
D8.
1
R
B3. EP = mgh
gravitational potential energy
D9.
P = IV = I 2 R
kinetic energy
D10. F = BQv sin 
force on a moving charge in a
magnetic field
mechanical power
D11. F = BIl sin 
force on a current-carrying
conductor in a magnetic field
centripetal acceleration
D12. V =
Hall voltage
Newton’s law of gravitation
D13. B =
mv 2
B4.
EK =
B5.
P = Fv =
2
t
2
B6.
B7.
v
a=
=  2r
r
Gm1 m2
F=
r2
D
C1.
y =
C2.
d sin  = n
1
R1
nQt
0
2πr
 0 NI
D14. B =
l
Φ
D15.  = N
t
V
N
D16. s  s
fringe width in
double-slit interference
diffraction grating equation
p
C3.
1
1
u
v
1
f
1
R2
resistors in series
p
resistors in parallel
power in a circuit
magnetic field due to a long
straight wire
magnetic field inside a long
solenoid
induced e.m.f.
ratio of secondary voltage to
primary voltage in a transformer
equation for a single lens
E1.
E2.
N = N0 e
t1 =
2
kt
ln 2
k
law of radioactive decay
half-life and decay constant
E3.
A = kN
activity and the number of
undecayed nuclei
E4.
E = mc 2
mass-energy relationship
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PP-DSE-PHY 2–24
24
鳴謝
Acknowledgements
本專輯的試題曾引用下列刊物的資料：
Material from the following publications has been used in question papers in this volume:
Leisure and Cultural Services http://www.lcsd.gov.hk/beach/b5/swim-address-s.php#pao
Department, The Government of yuekong
HKSAR
Electrical and Mechanical http://www.energyland.emsd.gov.hk/en/appAndEquip/appli
Services
Department,
The ances/meels.html
Government of HKSAR
University of Szeged, Hungary
http://www.szote.u-szeged.hu/radio/potlap1/amel2 9c.htm
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