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CCC HOH FUK TONG COLLEGE
Mock Examination, 2002 - 2003
Physics Paper 1
Secondary: Seven
Date: 03/03/2003
Time allowed: 3 hours (8:30 - 11:30 a.m.)
Marks: 122
Name: ___________________________
Class Number: ______
1.
Answer ALL questions.
2.
Write your answers in the spaces provided in the question paper. In calculations
you should show all the main steps in your working.
3.
Assume:
velocity of light in air = 3  108 m s-1
acceleration due to gravity = 10 m s-2
charge on an electron = 1.6  10-19 C
Planck’s constant = 6.6  10-34 J s
Avogardo’s constant = 6.02 x 1023 mol-1
Paper 1 : Page 2 of 17
1.
A student wants to use an ‘inertia table’ to determine the moment of inertia of a
cylinder about its central axis. The inertia table consists of a circular platform
suspended by a wire, and can be set into torsional oscillations.
A
O
(1)
(a)
Indicate on the figure above the directions of the torsional oscillations of
the table.
(1)
(b)
Briefly explain the meaning of moment of inertia of an object.
(2)
moment of inertia of an object about an axis
=  mi (ri)2,
where mi = mass of ith particle
ri = distance between the ith particle and the axis
(c)
If I = the moment of inertia of the table about the axis of oscillation, and
c = the torsional constant of the wire,
derive an expression for the period of oscillation of the table.
(4)
When displace through small angle ,
Restoring torque = -c  = I θ
I θ = -c 
c
θ = - 
I
Hence, period = 2
I
c
Paper 1 : Page 3 of 17
(d)
If the cylinder is placed on the platform with its axis lying along line AO,
the period of oscillations changes from 2.96 s to 3.21 s. Determine the
moment of inertia of the cylinder, assuming that the moment of inertia of
the table is 2.2  10-3 kg m2.
(4)
T1 = 2.96 = 2
2.2  10 -3
c
x  2.2  10 -3
c
2
x + 2.2 10-3
 3.21 

 = 2.2 10-3
 2.96 
x = 3.87  10-4 kg m2
T2 = 3.21 = 2
(e)
(a)
(1)
(1)
(1)
Mention two possible major sources of error in the experiment.
-
2.
(1)
(2)
difficulty in locating axis of cylinder along OA
cylinder may slip on the table during oscillations
oscillations too large amplitude
not purely torsional oscillations
(any 2)
When a monochromatic light of wavelength  is shone from above on a
converging lens of radius of curvature 0.8 m resting in contact with a
horizontal plane sheet of glass, Newton’s rings can be observed in
reflected light.
(i)
Draw the pattern observed from above.
(ii)
If the diameter of the 20th dark ring 5.2 mm, find the wavelength of
the light.
(2)
Since (dn)2 = 4 n R 
(5.2  10-3)2 = 4 (20) (0.8) ()
 = 4.225  10-7 m
(2)
Paper 1 : Page 4 of 17
(iii)
The student carefully raises the lens from the glass plate. He finds
that 50 extra fringes appear and move towards the center of the
fringe system. By what distance has the lens been raised? (2)
Let distance moved by lens = d
change in path difference = 2d
A shift of 50 fringes  the change in path is 50
2d = 50  4.225  10-7
d = 1.05  10-7 m
(b)
A tanker has accidentally leaked a large quantity of oil (refractive index
1.2) into the Ma Wan Channel, creating a large slick on the top of the
water (refractive index 1.3).
(i)
If you are looking straight down from an aeroplane onto a region of
the slick where its thickness is 460 nm, for which wavelength(s) of
visible light is the reflection greater?
(2)
For constructive interference of reflected light of
wavelength 
2t=m
2 (1.2) (460 nm) = m 
m  = 1104 nm
In order to be visible, m must be 2.
Then  = 552 nm
(ii)
If you are a diver directly under the same region of the slick, for
which wavelength (measured in vacuum) of visible light is the
transmitted light that has the strongest intensity?
(2)
For constructive interference of transmitted light of
wavelength 
2  t = (m + 1/2) 
(m + 1/2) = 1104 nm
In order to be visible, m must also be 2.
Then  = 441.6 nm
(iii)
The oil is cleaned as much as possible until a very thin film
remains. What will be observed in looking at the region from
above?
(2)
It appears to be white.
The phase changes of reflections at the two surfaces
of the kerosene are the same (  radians).
If the film is very thin, the optical path difference of
the two rays reflected by the two surfaces of the film
is effectively zero. Then constructive interference
occurs for all wavelengths of light to give white
light there.
Paper 1 : Page 5 of 17
3.
Doppler effect provides a convenient means of tracking an artificial satellite
which is emitting a radio signal of constant frequency f. The signal received at the
tracking station is combined with a constant signal of frequency F generated in the
receiver to give beats.
tracking station (F)
Earth
Figure below shows the change in beat frequency as the satellite passes over the
station, with F set at 40 MHz.
(a)
(b)
If v and c represent the speed of the satellite and the radio signal
respectively, write down an expression for the frequency of the received
signal at station when
(1)
(i)
the satellite is distant and moving towards the station (as f1)
(ii)
the satellite is distant and moving away from the station (as f2)
(i)
f1 = (1 + v/c)f or f1 = f (c/c-v) (0.5)
(ii)
f2 = (1 – v/c)f or f2 = f (c/c+v) (0.5)
Hence, write down an expression for the beat frequency when
(i)
the satellite is distant and moving towards the station
(ii)
the satellite is distant and moving away from the station
(Hint: f1 and f2  F)
(i)
beat frequency = (1 + v/c)f – F
(0.5)
(ii)
beat frequency = (1 – v/c)f – F
(0.5)
(1)
Paper 1 : Page 6 of 17
(c)
Using (b) and data from the graph, find the speed of the satellite.
(4)
(1 + v/c)f – F = 4000
 (1 + v/c)f = 40004000 ----(1)
(1 - v/c)f – F = 2000
 (1 - v/c)f = 40002000 ----(2)
c+v 40004
(1)/(2) gives c-v = 40002 where c = 3  108
Solving gives v = 7499 m s-1
(d)
Find the value of f.
(2)
Put v = 7499 into (1)
7499
(1 +
) f = 40004000
3  108
f = 40003000 Hz
(e)
From the graph, identify the time at which the satellite is directly above the
station. Give reason(s) to your answer.
(3)
When the satellite is directly above the station, its velocity
does not have a component towards the station.
The frequency of the received signal is the same as the
emitted frequency, i.e. 40003000 Hz.
Beat frequency = 40003000 – 40000000 = 3000 Hz.
This occurs at 3:30 p.m.
(f)
If the frequency F were adjusted to be 5 kHz higher, on the graph provided
below, draw how the beat frequency varies with time.
(2)
The frequency of the received signal would be lower than F.
When the satellite is distant and approaching, beat freq.= 5 – 4
= 1 kHz
When the satellite is distant and moving away, beat frequency
= 5 – 2 = 3 kHz
3:30 p.m.
Paper 1 : Page 7 of 17
4.
A particle of mass 4.0  10-6 kg carrying a charge 5  10-3 C is injected with a
certain speed at point C into the region between two parallel plates such that it just
reaches point O.
10 cm
+75 V
O
X
10 cm
C
0V
(a)
What type of charge (positive / negative) does the particle carry?
(1)
positive
(b)
Calculate the electric field strength between the two plates.
V 75
= d = 0.1
= 750 V m-1
E
(c)
(2)
Calculate the minimum speed of injection at C.
(2)
½ m v2 = q V
v
(d)
2qV
=
m
= 433 m s-1
=
2 (5  10 3 ) (75)
4  10 -6
Find the time taken by the particle to reach O.
(3)
ma = qE
qE
(510-3)(750)
a= m =
= 9.375  105
-6
410
s = ½ a t2
t=
(e)
2s
=
a
2 (0.1)
= 4.62  10-4 s
5
9.375  10
If a vertical magnetic field is applied such that the particle will hit the edge
at X, find the direction and magnitude of the field.
(3)
Direction : out of the paper
Arc CX is part of a circle of radius 10 cm
mv (4  10-6)(433)
Since B = qr =
= 3.464 T
(5  10-3) (0.1)
Paper 1 : Page 8 of 17
5.
Two parallel horizontal conducting rails separated by a distance 50 cm are
connected by a resistor 50  as shown. A metal rod PQ of length 70 cm is placed
on the rails and is given an initial speed of 1.5 m s-1. A magnetic field of 4 T is
applied perpendicularly out of the paper. The rails and the metal rod are assumed
to be frictionless with no resistance.
magnetic field out
P
50 
1.5 m s-1
50 cm
70 cm
Q
(a)
Calculate the induced e.m.f. in the circuit at the beginning.
(2)
=Blv
= 4  0.5  1.5 = 3 V
(b)
What is the direction and magnitude of the induced current in the metal
rod at the beginning?
(2)
I = /R = 3/50 = 0.06 A
Direction : P to Q along the rod
(c)
What is the direction and magnitude of the magnetic force acting on the
metal rod at the beginning?
(3)
F = B I l = 4  0.06  0.5 = 0.12 N
Direction : to the left
(d)
What effect (increased / decreased / no change) will be produced on the
magnitude of the induced current and magnetic force on the rod a short
time later? Explain briefly.
(3)
The magnetic force will decelerate the rod and makes it
move slower. (1)
The induced emf and hence the induced current will
decrease. (1)
The magnetic force will decrease subsequently. (1)
Paper 1 : Page 9 of 17
(e)
If the rod, initially at rest, is pulled towards the right by a constant force 2
N, find the final speed of the rod.
(3)
Let v be the final speed
Blv
Induced current = R
Blv
B2 l2 v
Induced magnetic force = B I l = B R l = R
=2N
B2 l2 v (4)2 (0.5)2 v
2= R =
50
-1
v = 25 m s
6.
V1
V2
100 
12 V a.c. 200 Hz
(a)
The circuit shown has a sinusoidal a.c. supply set at 12 V r.m.s. and 200
Hz. A 100  resistor is connected in series with the supply and with an
inductor which has some resistance (a few tens ohms) and a removable
iron core.
The readings on the voltmeters with the core in position are as follows:
V1 = 6.1 V r.m.s.
V2 = 8.3 V r.m.s.
(i)
What is the current in the circuit?
(1)
I = V1/R = 6.1/100 = 0.061 A
(ii)
Draw a phasor diagram relating V1, V2, the supply voltage and the
voltage(s) across the inductor. Hence, show that the phase angle
between the current and the supply voltage is about 40o.
(4)
VL
V = 12 V
(2)

V2 = 8.3 V
V1 = 6.1 V
(VR)L
Paper 1 : Page 10 of 17
122 + 6.12 - 8.32
cos  =
= 0.7673
2  12  61
 = 39.89o
(iii)
(2)
Hence calculate the inductance L of the inductor and the resistance
R of the inductor.
(4)
VL = 12 sin = 12 sin39.89o = 7.696 V
Since VL = I XL = I  (2 fL)
7.696 = 0.061  (2 100L)
L = 0.1 H
[(VR)L]2 = (V2)2 – (VL)2 = 8.32 – 7.6962
(VR)L = 3.108 V
(VR)L 3.108
R = I = 0.061 = 50.96 
(b)
The core of the inductor in the circuit is now removed.
(i)
What effect does this have on the inductance of the inductor?
(1)
Inductance will be decreased
(ii)
Explain any change in the reading on voltmeter V1.
Since Z = R T2  ( L) 2
As L is decreased, Z will decrease
I will increase
V1 = I  100 will increase
7.
An experiment is carried out to study the characteristics of a transistor.
6V
2.0 k
15 k
B
V VB
C
E
V VA
(2)
Paper 1 : Page 11 of 17
The input voltage VA and the voltage across the load resistor, VB, are measured by
voltmeters of very high resistance. The results are plotted as below:
(a)
From the plot, state
(2)
(i)
the minimum voltage across the collector and the emitter, VCE.
0.1 V
(ii)
the maximum voltage across the base and the emitter, VBE.
0.6 V
(b)
Find the base current Ib and the collector current Ic when
(i)
VA = 1.0 V
(ii)
VA = 1.2 V
VA - VBE 1 - 0.6
-5
15 k = 15000 = 2.66710 A(26.7 A)
VB 1.93(or 2)
Ic = 2 k = 2000
= 9.6510-4 A(110-3 A)
(i) Ib =
1.2 - 0.6
(ii) Ib = 15000 = 4.010-5 A(40 A)
2.90(or 3)
Ic = 2000 = 1.4510-3 A(1.510-3 A)
(4)
Paper 1 : Page 12 of 17
(c)
Hence, find the current gain .
(1)
Ic (1.45-0.965)10-3
=
=
= 36.38 (or 37.5)
Ib
(4-2.667)10-5
(d)
The load resistor is replaced by a buzzer which has a resistance of 2.0 k
and operates with a minimum current of 1.0 mA. What is the input
voltage that will just cause the buzzer to sound?
(2)
When Ic = 1.0 mA, VB = (110-3)  (2000) = 2.0 V
From graph, VA = 1.0 V
(f)
Complete the table below to show the voltage VCE for different values of
VA.
(2)
VA / V
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
VB / V
2.0
2.5
3.0
3.5
3.9
4.4
4.9
5.4
5.9
5.9
5.9
VCE / V
4.0
3.5
3.0
2.5
2.1
1.6
1.1
0.6
0.1
0.1
0.1
If the input voltage VA varies with time as shown, draw how the voltage
VCE across the collector and the emitter varies with time.
(2)
6
5
vo ltag e / V
(e)
4
3
2
1
0
0
50
100
150
200
250
time / s
300
350
400
450
500
Paper 1 : Page 13 of 17
8.
(a)
One mole of water at 100 oC is heated in open air until all the water has
turned into steam.
(i)
Assuming that water vapour behaves as ideal gas with a molar
volume of 2.24  10-2 m3 at s.t.p. (0 oC and 1 atmosphere), estimate
the increase in the volume from water to steam.
(2)
V2
2.2410-2
By Charles Law : 373 = 273
V2 = 3.06  10-2 m3
Increase in volume  3.06  10-2 m3
(ii)
If the atmospheric pressure is found to be 1.01  105 Pa, how much
work is done on the atmosphere during the boiling process? (2)
Work done on the atmosphere = P  V
= (1.01  105 )  (3.06  10-2)
= 3.09  103 J
(iii)
Given that the specific latent heat of vaporization of water is 2.26 
106 J kg-1 and that the molar mass of water is 0.018 kg mol-1,
calculate the fraction of energy used in breaking the bonds between
water molecules as water turns to become steam.
(2)
Energy needed to boil away 1 mole of water
= 2.26106  0.018 = 4.07104 J
Fraction to break up bonds
= (4.07104 - 3.09  103)/(4.07104)
= 0.924
(b)
The cylinder in figure 8(b)(i) holds a volume V1 = 1000 cm3 of air at an
initial pressure P1 = 1.10  105 Pa and temperature T1 = 300 K. Assume
that air behaves like an ideal gas.
P2
C
piston
P1
T1
Figure 8(b)(i)
P1
B
A
V1
Figure 8(b)(ii)
V2
Paper 1 : Page 14 of 17
Figure 8(b)(ii) shows a sequence of changes imposed on the air in the
cylinder.
(i)
AB - the air is heated to 375 K at constant pressure.
(ii)
BC - the air is compressed isothermally to volume V1.
(iii) CA - the air cools at constant volume to pressure P1.
(i)
Calculate the new volume V2 and hence the work done on the air in
stage (i).
(4)
V1 300
V2 = 375
375
V2 = 300  1000 = 1250 cm3
Work done on air = - P  V
= - (1.10 105)  [(1250 – 1000) 10-6]
= -27.5 J
(ii)
Calculate the new pressure P2.
(2)
P1 V1 = P2 V2
(1.10 105)  1250 = P2 (1000)
P2 = 1.375  105 Pa
(iii)
State how a value for the work done on the air during the full
sequence of changes may be found from the graph in Figure
8(b)(ii).
(1)
The workdone can be found from the area enclosed
by the curves.
9.
In an experiment with a vacuum photocell to verify the Einstein’s equation for
emission of photoelectrons from a metal surface, the maximum kinetic energy of
the emitted photoelectrons was determined for different wavelength of the
incident radiation. The following results were obtained:
Wavelength / nm
300
350
400
450
500
Maximum KE / eV
2.03
1.43
0.99
0.64
0.36
(a)
Use these results to plot a linear graph and derive a value for the Planck’s
constant, in J s.
(6)
Paper 1 : Page 15 of 17
2.5
maxi. KE / eV
2
(c(c) same slope (1)
smaller intercept (1)
1.5
1
1
/ 106 m-1

0.5
0
1
-0.5
2
3
4
-1
-1.5
-2
-2.15
-2.5
1.9 - 0.4
3.25106 - 2.05106
= 1.25 10-6 eV m = 2.0 10-25 J m
slope of graph =
h=
(b)
slope 2.0 10-25
-34
c = 3 108 = 6.7 10 J s
(2)
Determine the value of work function of the metal, in eV.
(1)
Work function = - y-intercept = 2.15 eV
(c)
Another metal which emits photoelectrons easier were used for the surface
of the photocathode. Draw on graph (a) the graph for that metal. (2)
(d)
A source emits monochromatic light of wavelength 400 nm at a rate of 0.1
W. Of the photons given out, 0.8% fall on the cathode of the photocell in
part (a) which gives a current of 6 A in an external circuit. You may
assume that this current consists of all the photoelectrons emitted.
Calculate
(4)
(i)
the energy of a photon,
Energy of a photon
h c (6.710-34)(3108)
=
=
= 5.02510-19 J (1)

40010-9
(ii)
the number of photons leaving the source per second,
number of photons =
0.1
17
-19 = 1.9910
5.02510
(1)
Paper 1 : Page 16 of 17
(iii)
the percentage of the photons falling on the cathode which produce
photoelectrons.
number of photons emitted per s
610-6
=
= 3.75 1013
1.610-19
3.751013
percentage =
= 2.36% (2)
1.991017 0.8%
10.
Figure below shows an ionization chamber linked to a 0 – 500 V variable voltage
supply and a very sensitive electrometer with a range of 0 – 100 pA (1 pA = 10-12
A). The apparatus can be used to measure the activity of the alpha source.
source
insulator
0 – 500 V
pA
(a)
When the voltage supply was increased from 300 V to 400 V, the current
recorded on the meter increased from 60 pA to 72 pA, but when the
voltage supply was further increased to 500 V the current remained at 72
pA. Suggest a reason in each case why
(2)
(i)
the first increase in voltage supply raised the current;
When the voltage is increased, ions travel faster to
the electrodes with less recombination. More ions
are received per second. (1)
(ii)
but the second increase in voltage supply did not
All ions generated per second are received. (1)
(b)
The supply voltage is set at 500 V. Find the number of air molecules
ionized in the chamber in one second.
(1)
7210-12
number =
= 4.5108
1.610-19
Paper 1 : Page 17 of 17
(c)
The energy of each emitted alpha particle is 5.4 MeV and the average
ionization energy of the molecules of air is 30 eV.
(3)
(i)
Estimate the activity of the source.
number of molecules ionised by each -particle
5.4106
= 30
= 1.8105
activity = number of -particles emitted per second
4.5108
=
= 2500 Bq
1.8105
(ii)
It is known that the range of the alpha particles in air is
approximately 30 mm. Estimate the average number of ionization
produced per mm of path of an alpha particle.
average number of ionisation per mm of path
1.8105
= 30
= 6103 mm-1
(d)
If the source has a half-life of 5000 years, find the number of radioactive
nuclei inside the source.
(2)
decay constant, k
ln2
ln2
= half-life =
= 4.39610-12
5000365246060
activity = kN
2500 = 4.39610-12N  N = 5.691014
(e)
If the volume of the chamber were reduced significantly, what would be
the effect on the saturated ionization current? Explain briefly.
(2)
The saturated ionisation current would be decreased,
because the -particles will reach the wall of the chamber
before losing all of their energies. Less ion-pairs would be
produced by each particle.
End of Paper 1