Download In the circuit shown below, a voltage of 10 V is applied across XY

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CCC Hoh Fuk Tong College
Second Term Examination 2011-2012
Physics Question-Answer Book
Name:
Class: 5E
Class No.:
Section B – Questions (84 marks)
Q. No.
1
2
3
4
5
6
7
8
9
10
Marks
8
7
8
11
8
9
8
7
12
6
1.
A uniform electric field is set up between two parallel charged plates with a
potential difference of 15 V. The separation between the plates is 0.02 m.
An electron is initially at rest and placed very close to the negative plate.
Under the effect of the electric field, the electron moves to the positive plate.
(a)
Find the electric field strength between the plates.
(2)
V
Electric field strength = d
15
= 0.02 = 750 V m–1
(b)
Find the magnitude of the electric force acting on the electron.
(2)
electric force = qE
= (1.6  10–19)(750) = 1.2  10–16 N
(c)
Take the mass of an electron as 9.1  10–31 kg. As the electron
arrives at the positive plate, find the
(i)
gain in kinetic energy
gain in KE = loss in PE = qV
= (1.6  10–19)(15) = 2.4  10–18 J
(2)
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(ii)
speed of the electron
(2)
1
By KE = 2 m v2
1
2.4  10–18 = 2  9.1  10–31  v2
v = 2.30  106 m s–1
2.
The tungsten filament of an electric lamp is a wire of radius 6  10–6 m.
Under normal operation temperature of the lamp, the resistivity of tungsten is
7.9  10–7  m. The lamp is operated at 240 V and there are 9.375  1019
electrons passing through the lamp in one minute.
(a)
If the lamp is operated at normal operation temperature, find
(i)
the current passing through the lamp,
(2)
Q
current = t
(9.3751019)(1.610-19)
=
= 0.25 A
60
(ii)
the resistance of the filament at this temperature.
(2)
V
resistance = I
240
= 0.25 = 960 
(b)
Determine the length of the filament.
(2)
l
By R =  A
960 = 7.9  10–7
l
(6 10-6)2
l = 0.137 m
(c)
Why is it difficult to quote a value for the resistance of a filament lamp?
(1)
This is because the resistance of the filament varies with its
temperature.
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3.
The figure below shows the circuit of an electric heater. It consists of two heating
elements R1 and R2 , rated at ΄220 V, 1 kW΄ and ΄220 V, 1.5 kW΄ respectively.
(a)
Which wire (live, neutral or earth) is switch K connected to?
your answer.
Explain
(2)
Switch K is connected to the live wire.
This ensures that no part of the appliance is still ‘live’ when
it is turned off.
(b)
When all switches are closed, what is the current passing through the
fuse?
(2)
Total power = 1000 + 1500 = 2500 W
By P = V  I
2500 = 220  I
I = 11.4 A
(c)
Which fuse is suitable for the heater, 10-A, 13-A or 20-A?
your answer.
Explain
(2)
A 13-A fuse is suitable.
Its fuse value is slightly larger than the rated current of the
circuit.
(d)
Find the resistance of R1 .
2
V
By P = R
2202
1000 = R
R = 48.4 
(2)
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4.
(a)
Figure 4(ai) shows a circuit of a current-carrying wire AB.
Figure
4(aii) shows the top view of a circular magnetic field line produced by
wire AB.
A
magnetic field line
wire AB
B
Figure 4(ai)
Figure 4(aii)
(b)
(i)
Draw an arrowhead on Figure 4(aii) to show its direction. (1)
(ii)
Sketch two additional magnetic field lines produced by the
wire AB to show the relationship between the magnetic field
strength and the distance from the current-carrying wire. (2)
A student used the set-up shown below to investigate the relationship
between the magnitude of the magnetic field (B) of a solenoid and its
number of turns per metre (n).
The student measured the values of B for solenoids of different n with
a constant current. The results are shown in the following table:
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Number of turns
50
100
150
200
Length of solenoid (m)
0.1
0.1
0.1
0.1
Number of turns per metre
n (m-1)
500
1000
1500
2000
Magnitude of the magnetic
310
600
980
1260
field B (  10 T)
-6
(i)
(ii)
Calculate the ‘Number of turns per metre n’ and complete the
above table.
(1)
Plot a graph of B against n.
(4)
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(iii)
What can be concluded from the graph plotted in (ii)?
(1)
The magnitude of the magnetic field is directly
proportional to the number of turns per unit length of
the solenoid (i.e. B  n)
(iv)
State and explain one precaution to improve the accuracy of the
measurements.
(2)
Any ONE:
 The relative positions between the solenoid and
the sensor must be fixed because the magnitude of
the magnetic field outside the solenoid is different
at different positions.
 The experiment should be conducted far away
from magnets/other electromagnetic devices so
that the magnetic field due to the solenoid is not
affected.
 Put the sensor inside the solenoid because the
magnetic field is uniform inside the solenoid.
5.(*)
A metal conductor has a rectangular cross-section and carries a current I in the
presence of a magnetic field of strength B. The magnetic field is
perpendicular to surface PWXQ (Fig 5(a)). The width and height of the
cross-section of the conductor are a and b respectively. The conductor has n
free electrons per unit volume. The charge of each electron is q. The Hall
voltage VH is measured by a micro-voltmeter. The current I is varied and the
corresponding values of VH are recorded. Figure 5(b) shows the graph of
VH against I.
Figure 5(a)
Figure 5(b)
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(a)
(i)
Across which two surfaces is the Hall voltage developed? (1)
Surfaces QRYX and PSZW
(ii)
Which surface is at a lower potential?
(1)
QRYX
(b)
Using the symbols given, write an equation for the Hall voltage.
(1)
BI
VH = nbq
(c)
Using (b), derive an equation for n and the slope of the graph of Figure
5(b).
BI
B
n = V bq = bq
H
(d)
I
B
1
=
VH
bq slope
Find n if B = 3.5 T, a = 3 mm and b = 2 mm.
n=
(e)
(1)
(2)
3.5
5-0

210-3  1.610-19
0.210-6 - 0
= 2.73  1029 m-3
Explain why the sensor used in a Hall probe is usually a semiconductor
but not a metal.
(2)
The number of charge carriers per unit volume in a
semiconductor is much smaller than that in a metal.
BI
By VH = nbq, the Hall voltage is much higher and can be
measured more easily.
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6.
Figure 6a shows the structure of an a.c. generator.
Figure 6a
(a)
What is the energy change in the generator?
(1)
Mechanical energy is changed to electrical energy.
(b)
What are the names of X and Y?
(2)
X: carbon brush.
Y: slip ring.
(c)
Determine the direction of the current passing through R at the moment
shown in Figure 6a.
(1)
Current flows from B to A via R.
(d)
Figure 6b below shows the variation of the voltage across AB against
time.
Figure 6b
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(e)
(i)
At which points (P, Q, R or S) in Figure 6b does the plane of
the coil lie perpendicularly to the magnetic field?
(2)
Q and S
(ii)
On Figure 6b, draw the variation of the output voltage against
time if the rotation speed of the coil is doubled.
(2)
How can the a.c. generator be converted into a d.c. generator.
(1)
Replace the slip rings with a commutator.
7.
In a power station, the output power is 15 MW and the output voltage is 4 kV.
The electricity is stepped up by an ideal transformer and transmitted to a town
160 km away through transmission cables.
It is known that the
cross-sectional area of the transmission cables is 3 × 10-4 m2 and its resistivity
is 1.68  10-8  m. The power loss in the cable is 5% of the output power of
the step-up transformer.
15 MW
4 kV
(a)
What is the power loss in the cables?
(2)
power loss = 15  106  5%
= 7.5  105 W
(b)
What is the total resistance of the cables?
(2)
l
Total resistance = A
1.68  10-8  (160 103  2)
=
= 17.92 
3 × 10-4
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(c)
What is the current flowing in the cables?
(2)
By P = I2 R
7.5  105 = I2  17.92
I = 204.6 A
(d)
What is the output voltage of the step-up transformer?
(2)
By P = V I
15  106 = V  204.6
V = 7.33  104 V
8.
The following figure shows part of a radioactive series.
mass number
234

232
A
230
228
B 

C
D
226
224
222
87
E 
88
89
90
91
atomic number
(a)
Write equations for
(i)
nuclide A decaying to nuclide B,
(1)
(ii)
nuclide B decaying to nuclide C.
(1)
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(b)
State nuclides in the series that are isotopes of each other.
(2)
A and D (or B and E)
(c)
The final stable nuclide of the above series is
208
82
X.
In decaying
from A to X, how many  particles and  particles are emitted?
(3)
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9.
Technetium-99m
99
43
Tc is a radioisotope widely used as a medical tracer for
taking images of the kidney. It emits  radiation when it decays.
(a)
Write an equation for the decay of technetium-99m.
99
43
(b)
Tc * 
99
43
(1)
Tc + 
The figure below shows the decay curve of a sample of
technetium-99m.
(i)
Why do the data points not fall exactly on the decay curve? (1)
Decay is a random process.
(ii)
Estimate the half-life of technetium-99m from the decay curve.
6 hours (from graph)
(iii)
(2)
The initial activity of technetium-99m injected into a patient is
370 MBq. Find the activity of the technetium-99m in the
patient 24 hours later.
(2)
24
No. of half-lives = 6 = 4
1
Activity 24 hours later = 370  (2)4 = 23.1 MBq
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(c)
The characteristics of two radioisotopes, X and Y, are as shown.
Radioisotope
Radiation emitted
Half-life
X


40.3 days
Y
5.8 hours
Explain why neither X nor Y is suitable for replacing technetium-99m
as medical tracer.
(2)
X is not suitable because it has a long half-life. It will
accumulate in the body for a long time, rendering it
hazardous to health.
Y is not suitable because the radiation emitted has a strong
ionization power, which causes damage to body cells.
(d)*
Two radiologists P and Q are discussing how frequently images of the
kidneys of a patient with mild kidney problems should be taken.
P:
We should take an image of the patient’s kidney once every
month.
Q:
It is sufficient for diagnostic purposes to take an image of the
patient’s kidney once every 6 months.
In each injection, the radiation dose is about 0.0015 Sv.
With reference to the effects of radiation dose on the human body,
comment on the two radiologists’ different approaches to using
technetium-99m.
(4)
Annual dose < 0.08 Sv
0.08 – 2.00 Sv
2 – 8 Sv
Effects on
human body
Some increase in
cancer rates,
mutation induction
Various
symptoms
observed
No health effects
observed
Approach by P: the annual dose = 0.0015  12 = 0.018 Sv
Approach by Q: the annual dose = 0.0015  2 = 0.003 Sv
For either approach, the annual dose is within the ‘no health
effects observed’ zone.
Either one comment:
For P, it gives a closer monitoring of the patient’s
condition.
For Q, the use of radioisotopes is kept to a minimum
as long as diagnosis can be made.
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10.
When a uranium-235 nucleus is bombarded by a slow moving neutron, it may
undergo fission. The equations below show two possible reactions:
(a)
reaction 1:
1
0
139
n  235
92 U  54 Xe 
reaction 2:
1
0
116
1
n  235
92 U 2 46 Pd  x 0 n  energy
95
38
Sr  2 01 n  energy
Find the value of x in reaction 2.
(2)
Consider the mass numbers.
1 + 235 = 2  116 + x
x=4
(b)
Given the following data:
mass of
139
54
mass of
95
38
mass of
116
46
Xe
Sr
Pd
mass of a neutron
= 138.918793 u
= 94.919359 u
= 115.914159 u
= 1.008665 u
Which reaction releases more energy?
Explain briefly.
(4)
Total mass after reaction 1
= 138.918 793 + 94.919 359 + 2  1.008 665
= 235.855 482 u
Total mass after reaction 2
= 2  115.914 159 + 4  1.008 665
= 235.862 978 u
The total mass after reaction 1 is smaller, meaning that the
loss of mass in reaction 1 is greater.
According to the mass-energy relationship, reaction 1
releases more energy.
END OF SECTION B