```Tung Wah Group of Hospitals Kap Yan Directors' College
2012-2013
84
MOCK EXAMINATION
19 Feb 2013
LPK
Name: ____________________________
Class no.: _____________
PHYSICS PAPER 1
This paper must be answered in English
INSTRUCTIONS
(1) This section carries 84 marks. Answer ALL questions.
(2) Write your answers in the spaces provided in this Question-Answer Book. Do not write in the margins.
Answers written in the margins will not be marked.
(3) Graphs and supplementary answer sheets will be provided on request. Write your name, mark the question
number box on each sheet and fasten them with a string INSIDE this Question-Answer Book.
P.1
1. An electrical heater is placed in an insulated container holding 100 g of ice at a temperature of –14oC. The
heater supplies energy at a rate of 98 J per second.
(a) After an interval of 30 s, all the ice has reached a temperature of 0 oC. Calculate the specific heat capacity
of ice.
(2 marks)
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(b) Calculate the final temperature of the water formed when the heater is left on for a further 500 s. Given that
specific heat capacity of water = 4200 J kg-1 oC-1
specific latent heat of fusion of water = 3.3 x 105 J kg –1
(3 marks)
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(c) The whole procedure is repeated in an uninsulated container in a room at a temperature of 25 oC. State and
explain whether the final temperature of the water formed would be higher or lower than that calculated in
part (b).
(2 marks)
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P.2
2. A fixed mass of ideal gas at a low temperature is trapped in a container at constant pressure. The gas is then
heated and the volume of the container changes so that the pressure stays at 1.00 x 105 Pa.
When the gas reaches a temperature of 0 oC, the volume is 2.20 x 10–3 m3.
(a) Draw a graph on the axes below to show how the volume of the gas varies with temperature in oC.
(2 marks)
(b) Calculate the number of moles of gas present in the container.
(2 marks)
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(c) Calculate the average kinetic energy of a molecule when this gas is at a temperature of 50 oC.
(2 marks)
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P.3
(d) By considering the motion of the molecules, explain why the volume of the container must change if the
gas pressure is to remain constant as the temperature increases.
(3 marks)
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3. The figure below shows a uniform plank, of weight 30 N and length 3 m, resting on two supports. The supports
are 0.5 m and 2.0 m from the left hand end of the plank. A weight of 18 N is suspended from the left hand end
of the plank.
(a) Find the reactions, X and Y, at the two supports.
(3 marks)
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(b) By how much should the weight on the left hand end be increased so that the reaction Y becomes zero?
(2 marks)
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P.4
4. The graph shows the variation in the horizontal force acting on a tennis ball with time while the ball is being
served.
(a) (i) Using the information from the graph, what is the impulse that acts on the tennis ball? (2 marks)
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(ii) Assume that the ball had no horizontal speed before the impulse was applied. After given the impulse
from part (a)(i), the horizontal speed of the ball is 20 m s -1 at the moment the ball leaves the racquet .
What is the mass of the tennis ball?
(3 marks)
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(b) During flight the ball accelerates due to gravity. When it reaches the ground the vertical component of the
velocity is 6.1 m s-1. Assume that air resistance is negligible.
(i) Calculate the speed of the ball as it reaches the ground.
(2 marks)
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(ii) Calculate the angle between the direction of travel of the ball and the horizontal as it reaches the
ground.
(2 marks)
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P.5
5.
You are given a convex lens (in lens holder), a translucent screen, an illuminated letter “ F ” and a plane
mirror as shown. With the aid of a diagram, describe an experiment to determine the focal length of the
convex lens using the above apparatus. State briefly the physics principle of such an experiment.
(6 marks)
translucent screen
plane mirror
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P.6
6. Just over two hundred years ago Thomas Young demonstrated the interference of light by illuminating two
closely spaced narrow slits with light from a single light source.
(a) What did this suggest to Young about the nature of light?
(1 mark)
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(b) The demonstration can be carried out more conveniently with a laser. A laser produces coherent,
monochromatic light.
(i) State what is meant by monochromatic.
(1 mark)
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(ii) State what is meant by coherent.
(1 mark)
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(c) The figure below shows the maxima of a two slit interference pattern produced on a screen when a laser
was used as a monochromatic light source.
Given:
The slit spacing = 0.30mm.
The distance from the slits to the screen = 10.0 m.
Use the figure above to calculate the wavelength of the light that produced the pattern.
(2 marks)
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(d) The laser is replaced by another laser emitting visible light with a shorter wavelength. State and explain
how this will affect the spacing of the maxima on the screen.
(2 marks)
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P.7
(e) Explain the formation of the interference pattern seen on the screen.
(4 marks)
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7. A cell of emf, ε, and internal resistance, r, is connected to a variable resistor R. The current through the cell
and the terminal potential difference of the cell are measured as R is decreased. The circuit is shown in the
Figure.
The graph below shows the results from the experiment.
P.8
(a) Explain why the terminal potential difference decreases as the current increases.
(2 marks)
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(b) (i) Use the graph to find the emf, ε, of the cell.
(1 mark)
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(ii) Use the graph to find the internal resistance, r, of the cell. Explain briefly.
(3 marks)
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8. A loop of wire is placed in the magnetic field produced by an electromagnet (not shown in the figure below).
The loop of wire has a resistance of 2.6  and an area of 4.0 × 10–3 m2. The electromagnet, when switched on,
takes 0.8 s to reach its maximum flux density of 600 μT.
(a) Assuming all the field is directed out of paper and links with the loop, and the loop’s area is perpendicular
to the field, calculate the average current that flows in the wire in the 0.8 s after the electromagnet is
switched on.
(3 marks)
magnetic field directed
out of paper produced
by electromagnet
wire loop
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(b) What is the direction of the induced current?
(1 mark)
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P.9
9. The figure shows a horizontal wire, held in tension
between fixed points at P and Q. A short section of the
wire is positioned between the pole pieces of a
permanent magnet, which applies a uniform horizontal
magnetic field at right angles to the wire. Wires
connected to a circuit at P and Q allow an electric
current to be passed through the wire.
(a) (i) State the direction of the force on the wire
when there is a direct current from P to Q, as
shown in the figure above.
(1 mark)
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(ii) In a second experiment, an alternating current is passed through the wire. Explain why the wire will
vibrate vertically.
(3 marks)
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(b) The permanent magnet produces a uniform magnetic field of flux density 220 mT over a 55 mm length of
the wire. What is the maximum magnitude of force on the wire when there is an alternating current of rms
value 2.4 A in it?
(3 marks)
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(c) The length of PQ is 0.40 m. When the wire is vibrating, transverse waves are propagated along the wire at
a speed of 64 m s–1. Explain why the wire is set into large amplitude vibration when the frequency of the
a.c. supply is 80 Hz.
(3 marks)
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P.10
10. (a) Long cables are used to send electrical power from a supply point to a factory some distance away, as
shown in the figure below. An input power of 500 kW at 25 kV is supplied to the cables.
(i) Calculate the current in the cables.
(2 marks)
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(ii) The total resistance of the cables is 30 Ω. Calculate the power supplied to the factory by the cables.
(2 marks)
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(b) Describe briefly how a transformer works.
(3 marks)
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(c) In Great Britain, the electrical generators at power stations provide an output at 25 kV. Most homes,
offices and shops are supplied with electricity at 230V. Power is transmitted from the power stations to the
consumers by the grid system, the main principles of which are shown in the figure below. In this network,
T1, T2, T3, etc, are transformers.
(i) Explain how a step-down transformer differs in construction from a step-up transformer.
(1 mark)
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P.11
(ii) Explain why the secondary windings of a step-down transformer should be made from thicker copper
wire than the primary windings.
(2 marks)
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11. Radioisotopes are often used for medical applications. 131I is a β-emitter, and can be used to treat an overactive
thyroid gland. When a small dose of 131I is swallowed, it is absorbed into the bloodstream. It is then
concentrated in the thyroid gland, where it begins destroying the gland’s cells.
up to 3 months after the treatment. Explain why it is possible that the activity of the 131I can be detected
outside the patient’s body.
(2 marks)
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(b) (i) The half-life of 131I is 8 days. What fraction of the original number of iodine atoms will have decayed
after a period of 24 days?
(2 marks)
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(ii) Doctors wish to prescribe a sample of 131I of activity 1.5 MBq. The sample is prepared exactly 24
hours before it is due to be swallowed by the patient. Calculate the activity that the sample should have
when it is prepared.
(3 marks)
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END OF SECTION B
P.12
F6 Physics Mock Examination Suggested Answers (12-13)
1. (a) E = mc T
30 x 98 = 0.1 c (14)
-1 o
c = 2100 J kg
(1M)
-1
C
(1A)
(b) E = ml + mcT
500 x 98 = 0.1 x 3.3 x 105 + 0.1 x 4200 x T
(1M for ml, 1M for mc T)
T = 38oC. Thus required temperature is 38oC.
(1A)
(c) The temperature would be higher
(1A)
as the ice/water spends more time below 25oC
or (1M)
heat travels in the direction of hot to cold
or (1M)
2. (a) Graph passes through given point 2.2 x 10-3 m3 at 0oC straight line with positive slope.
(1A)
Straight line passing through –273oC at zero volume.(1A)
(b) n = PV/RT
(1M)
5
-3
= 1 x 10 x 2.2 x 10 / (8.31 x 273) = 0.097 mole (1A)
(c) Average KE = 3/2 (R/NA)T = 3/2 x (8.31 / 6.02 x 1023) x (273+50)
= 6.69 x 10-21 J
(1M)
(1A)
(d) When the temperature of a gas rises, the kinetic energy and hence the speeds of the molecules increase. If
the volume remains constant, the pressure will increase as the molecules will hit the walls more frequently
and more violently. If the pressure is to remain unchanged, the volume must increase to spread the increased
force over a larger area. (1A+1A+1A)
3. (a) X + Y = 48 N
(1M)
18 x 0.5 + 1.5 Y = 30 x 1.0
(1M)
Y = 14 N and X = 34 N
(1A)
(b) Takes moments about X with Y = 0
30 x 1 = W x 0.5
(1M)
Increase in weight = 60 – 18 = 42 N
(1A)
4. (a) (i) Impulse = area under graph = 24 x 10-3 x 106 / 2
(1M)
= 1.27 Ns
(1A)
(ii) Impulse = change in momentum = m (v-0) = 1.27
(1M)
m (20 – 0) = 1.27
m = 63.5 g (0.0635 kg)
(b) (i) Required speed =
202  6.12  20.9m / s
(ii)  = tan-1 vy / vx
= tan-1 6.1/20 = 17.0o
(1A)
(1M + 1A)
(1M)
(1A)
5.
Diagram 1M
The apparatus is set up as shown. The lens-mirror combination is moved until a sharp image of the object (in
this case an illuminated ' F ') is also formed on the screen beside the object. The distance from the screen to the
lens is measured. This distance is the focal length of the lens. (1M+1M+1M)
The principle is as follows:
When an object is placed on the focal plane of a convex lens, light rays from it become parallel after passing
through the lens. On hitting a plane mirror behind the lens, the rays will be reflected back along their original
paths. An image is so formed at the same position as the object.
(1M+1M)
6.
(a)
Light is a wave.
(1A)
(b)
(i) Monochromatic means single frequency or wavelength.
(1A)
(ii) Coherent means constant phase difference (and same frequency).
(c)
(d)
s = 0.16/8 = 0.02 m
(1A)
(1A)
 = s a / D = 0.02 x 0.3 x 10-3 / 10 = 6 x 10-7 m
(1A)
The maxima become closer together.
(1A)
Since s = D/a, now  is decreased, so s decreases. (1A)
(e) Light waves from the two slits are two coherent sources because they come from a single source and
are in phase and of the same frequency. At places where the light waves arrive in phase (corresponding to
path difference = m  where m is an integer), constructive interference takes place and a bright fringe is
formed. At places where the light waves arrive exactly out of phase, destructive interference takes place
(corresponding to path difference = (m+1/2)), a dark fringe is formed. Since the path difference changes
continuously, alternate bright and dark fringes are formed.
(1A+1A+1A+1A)
7.
(a)
Terminal p.d. V = E –Ir
(1A)
If current increases, terminal p.d. V decreases.
(1A)
Or when current increases, the p.d. across internal resistance increases /
energy loss in internal resistance increases.
(b)
(i) Emf = y-intercept = 1.52 V (1.51-1.53V)
(ii) Since V = E – Ir,
(1M)
r = - slope = - 0.45  (0.43-0.47)
8.
(a)
(1A)
(1M for substitution, 1A)
Induced emf = change in flux linkage / time
E
= 600 x 10-6 x 4.0 x 10-3 / 0.8
=
(1M)
3 x 10-6 V
(1M)
Induced current I = E/R = 3 x 106 / 2.6 = 1.15 x 10-6 A
9.
(1A)
(b)
Direction: clockwise
(a)
(i) Downward
(1A)
(ii) When the current flows from P to Q, the force acts vertically downward.
When the current flows in opposite direction, the force acts vertically upward.
The process repeats and so the wire vibrates up and down.
(b)
(1A)
Maximum current = peak current = 2.4 x
2 = 3.39 A
(1A)
(1A)
(1M for peak current)
Maximum force = BIL = 220 x 10-3 x 3.39 x 55 x 10-3 = 0.041N (1M+1A)
(c)
Wavelength of waves = v/f = 64/80 = 0.8m (1A)
length of wire = 0.4 m = /2
So stationary waves are formed and the wire resonates at frequency of ac supply or fundamental
frequency of the wire is the same as applied frequency.
10.
(a)
(b) • A.c.voltage at primary generates alternating B field
• Magnetic field links with secondary coil through the core
• Changing magnetic field induces emf in secondary
(1A)
(1A)
(1A)
(c)
11.
MC
1-5 A D D B B
21-25 D B D B D
6-10
11-15 A A B A C
16-20 B C A B B
26-30 D C C C D
31-35 B B A B C
36 B
Explanations to selected MC
1.
mP = 5 mQ, CP = CQ / 3
E = Pt = CP TP = CQ TQ => TP > TQ => TP > TQ
3.
Let the rate be R.
For cooling of gas, R (10) = mc (80-40) = 2500 m (40)
For condensation, R (35-10) = m l
l = 250 kJ kg-1
4.
crmsO2
H 2
2 1



crms H 2
O 2
32 4
5. Since the trucks move off together, this is a completely inelastic collision, KE decreases.
6. At t = 0.4 s, v = u + (F/m) t = 0 + (5/2) (0.5) = 1.25 m/s
At t = 1 s, v = 1.25 + (15/2) (0.5) = 5 m/s
7. Work done = Fs cos  = 20 x 8000 = 160000 J
8. R – mg = ma
906 – 600 = (600/9.81) a
=> a = 5 m s-2
Since R > mg, so the acceleration is upward, so the possible motions are (1) moving upward and accelerating
at 5 m s-2 or moving downward and decelerating at 5 m s-2.
9. v = r = (2/T) r
10. Acceleration is negative and decreases with time, so the slope becomes less steep.
11. Maximum friction = 200 = 20 (3)2 rmax
rmax = 1.11 m
12. g is inversely proportional r2.
15. n = sin 65o / sin 35o = 1.58
c = sin-1 (1/n) = 39.3o
16. P is trough meeting trough, so a bigger trough is formed, the amplitude increases, so constructive interference
takes place. Trough appears dark.
17. Statement 3, sound wave travels faster in liquid such as water than in air.
23. Heat therapy makes use of infrared, not microwave.
25. a = F/m = eE/m = e (V/d) / m = 1.6 x 10-19 x (150/0.01) / 9.1 x 10-31 = 2.64 x 1015 m s-2
26. 45 =
50 =
Q
4π 0 r
Q
4π 0 ( r  1.5)
50 = 45 r / (r-1.5) => r = 15 m
27. R = V/I increases when the temperature increases as current increases.
31. s = (1/2)(F/m) t2 = (1/2)(qE/m) t2  q/m
Now q  2q, m  4m, so s  0.5 s
35. Rate of decay of X decreases with time and so the formation of product Y also decreases with time.
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