Q & A STPM 2022 Physics

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STPM
A


. .
Semester 1 2 3 Q & A STPM Q

Physics [960] &



Q & A STPM fulfils the needs of students in mastering the
technique of answering questions effectively to excel in the
STPM exam. The questions and answers provided meet the
STPM exam standard and are useful as a revision tool.



FEATURES TITLES IN THIS SERIES: Physics
• Based on the latest syllabus ■ Pengajian Am Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
• Arranged according to subtopics ■ Bahasa Melayu
• Tips to guide students in answering ■ Sejarah
questions effectively ■ Geografi
• Common Errors to show common ■ Ekonomi
mistakes frequently made by students ■ Pengajian Perniagaan STPM
while answering questions ■ Mathematics (T)
• Model Paper which follows the latest ■ Physics
STPM assessment format to prepare ■ Chemistry
students for the actual exam ■ Biology . .
Pengajian Am Ekonomi Physics Semester 1 . 2 . 3 Semester 1 2 3
Bahasa Melayu Pengajian Perniagaan Chemistry
Sejarah Mathematics (T) Biology
Physics
Geograf


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[960]
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EC059242
ISBN: 978-967-0007-34-2



PELANGI Lin Poh Tin • Poh Liong Yong

Format: 146mm X 216mm TP Q&A STPM Phy pgi CRC
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STPM



. .
Semester 1 2 3


Physics



[960]


Lin Poh Tin • Poh Liong Yong





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ISBN: 978-967-0007-34-2
eISBN: 978-967-0007-35-9 (eBook)

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STPM SCHEME OF ASSESSMENT
STPM SCHEME OF ASSESSMENT

Paper
Semester Code and Theme / Title Type of Test Mark Duration Administration
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
of Study (Weighting)
Name
Written Test 60
(26.67%)
Section A 15
15 compulsory
multiple-choice
questions to be
answered.
Section B 15
2 compulsory
First 960/1 Mechanics and structured 1 Central
Semester Physics Thermodynamics questions to be 1— hours assessment
2
Paper 1
answered.
Section C 30
2 questions to
be answered
out of 3 essay
questions.
All questions
are based on
topics 1 to 11.
Written Test 60
(26.67%)
Section A 15
15 compulsory
multiple-choice
questions to be
answered.
Section B 15
2 compulsory
Second 960/2 Electricity and structured 1 Central
Semester Physics Magnetism questions to be 1— hours assessment
2
Paper 2
answered.
Section C 30
2 questions to
be answered
out of 3 essay
questions.
All questions
are based on
topics 12 to 18.

ii




SCHEME STPM Q&A PHY T1.indd 2 17/02/2022 2:50 PM

Paper
Semester Mark
of Study Code and Theme / Title Type of Test (Weighting) Duration Administration
Name
Written Test 60
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
(26.67%)
Section A 15
15 compulsory
multiple-choice
questions to be
answered.
Section B 15
Oscillations and 2 compulsory
960/3 structured
Central
1
Physics Waves, Optics questions to be 1— hours assessment
and Modern
Paper 3 2
Physics answered.
Third Section C 30
Semester 2 questions to
be answered
out of 3 essay
questions.
All questions
are based on
topics 19 to 25.


960/5 Written Physics Written 45
Physics Pratical Practical Test (20%)
Paper 5 3 compulsory 1 Central
structured 1— hours assessment
2
questions to be
answered.

School-based 225
Assessment of to be
First, Practical scaled Through-
Second 960/4 13 compulsory to 45 out the School-based
and Third Physics Physics Pratical experiments (20%) three assessment
Semester Paper 4 and one project semesters
to be carried
out.










iii




SCHEME STPM Q&A PHY T1.indd 3 17/02/2022 2:50 PM

CONTENTS
CONTENTS


STPM Scheme of Assessment ii
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
SEMESTER 1

CHAPTER Physical Quantities and Units
1 1.1 Base Quantities and SI Units   2
1
1.2 Dimensions of Physical Quantities
1.3 Scalars and Vectors 6
1.4 Uncertainties in Measurements 10


CHAPTER Kinematics

2 2.1 Linear Motion 12
2.2 Projectiles
17

CHAPTER Dynamics
3 3.1 Newton’s Laws of Motion 23
3.2 Linear Momentum and its Conservation
28
3.3 Elastic and Inelastic Collisions 30
3.4 Centre of Mass 33
3.5 Frictional Forces 35


CHAPTER Work, Energy and Power
4 4.1 Work 37
42
4.2 Potential Energy and Kinetic Energy
4.3 Power 48


CHAPTER Circular Motion
5 5.1 Angular Displacement and Angular Velocity 51
5.2 Centripetal Acceleration
54
5.3 Centripetal Force 56




iv




CONTENT STPM Q&A Physics.indd 4 16/02/2022 4:36 PM

CHAPTER Gravitation
6 6.1 Newton’s Law of Gravitation 63
6.2 Gravitational Field
65
6.3 Gravitational Potential 72
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6.4 Satellite Motion in Circular Orbit 78
6.5 Escape Velocity 83

CHAPTER Statics
7 7.1 Centre of Gravity 87
87
7.2 Equilibrium of Particles
7.3 Equilibrium of Rigid Bodies 90
CHAPTER Deformation of Solids

8 8.1 Stress and Strain 97
99
8.2 Force–extension Graph and Stress-strain Graph
8.3 Strain Energy 105

CHAPTER Kinetic Theory of Gases
9 9.1 Ideal Gas Equation 109
111
9.2 Pressure of a Gas
9.3 Molecular Kinetic Energy 113
9.4 The r.m.s Speed of Gas Molecules 116
9.5 Degrees of Freedom and Law of Equipartition of Energy 119
9.6 Internal Energy of an Ideal Gas 123

CHAPTER Thermodynamics of Gases
10 10.1 Heat Capacities 125
126
10.2 Work Done by a Gas
10.3 First Law of Thermodynamics 127
10.4 Isothermal and Adiabatic Changes 135

CHAPTER Heat Transfer

11 11.1 Conduction 146
11.2 Convection
157
11.3 Radiation 158
11.4 Global Warming 166
STPM Model Paper 960/1 168

v




CONTENT STPM Q&A Physics.indd 5 16/02/2022 4:36 PM

SEMESTER 2

CHAPTER Electrostatics
12 12.1 Coulomb’s Law 175
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180
12.2 Electric Field
12.3 Gauss’s Law 187
12.4 Electric Potential 192

CHAPTER Capacitors
13 13.1 Capacitors 202
202
13.2 Parallel – Plate Capacitors
13.3 Dielectrics 205
13.4 Capacitors in Series and in Parallel 207
13.5 Energy Stored in Charged Capacitor 210
13.6 Charging and Discharging of a Capacitor 215

CHAPTER Electric Current

14 14.1 Conduction of Electricity 220
222
14.2 Drift Velocity
14.3 Current Density 227
14.4 Electric Conductivity and Resistivity 232

CHAPTER Direct Current Circuits

15 15.1 Internal Resistance 238
243
15.2 Kirchhoff’s Law
15.3 Potential Divider 249
15.4 Potentiometer and Wheatstone Bridge 256


CHAPTER Magnetic Fields
16 16.1 Concept of a Magnetic Field 263
264
16.2 Force on a Moving Charge
16.3 Force on a Current-Carrying Conductor 269
16.4 Magnetic Fields due to Currents 275
16.5 Force between Two Current-Carrying Conductors 281
e
16.6 Determination of the Ratio m 286
16.7 Hall Effect 291

vi




CONTENT STPM Q&A Physics.indd 6 16/02/2022 4:36 PM

CHAPTER Electromagnetic Induction
17 17.1 Magnetic Flux 296
17.2 Faraday’s Law and Lenz’s Law
301
17.3 Self-Induction 307
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17.4 Energy Stored in an Inductor 311
17.5 Mutual Induction 312

CHAPTER Alternating Current Circuits
18 18.1 Alternating Current Through a Resistor 314
321
18.2 Alternating Current Through an Inductor
18.3 Alternating Current Through a Capacitor 326
18.4 R-C and R-L Circuits in Series 331
STPM Model Paper 960/2 337


SEMESTER 3
CHAPTER Oscillations

19 19.1 Characteristics of Simple Harmonic Motion 344
19.2 Kinematics of Simple Harmonic Motion
347
19.3 Energy in Simple Harmonic Motion 355
19.4 Systems in Simple Harmonic Motion 358
19.5 Damped Oscillations 369
19.6 Forced Oscillations and Resonance 373

CHAPTER Wave Motion

20 20.1 Progressive Waves 377
383
20.2 Wave Intensity
20.3 Principle of Superposition 388
20.4 Standing Waves 391
20.5 Electromagnetic Waves 395

CHAPTER Sound Waves
21 21.1 Propagation of Sound Waves 399
401
21.2 Sources of Sound
21.3 Intensity Level of Sound 411
21.4 Beat 415
21.5 Doppler Effect 417

vii




CONTENT STPM Q&A Physics.indd 7 16/02/2022 4:36 PM

CHAPTER Geometrical Optics
22 22.1 Spherical Mirrors 423
428
22.2 Refraction at Spherical Surfaces
22.3 Thin Lenses 435
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CHAPTER Wave Optics
23 23.1 Huygens’s Principle 444
444
23.2 Interference
23.3 Two-slit Interference Pattern 449
23.4 Interference in a Thin Film 456
23.5 Diffraction by a Single Slit 460
23.6 Diffraction Gratings 465
23.7 Polarisation 469
23.8 Optical Waveguides 474

CHAPTER Quantum Physics
24 24.1 Photons 481
488
24.2 Wave-Particle Duality
24.3 Atomic Structure 493
24.4 X-Rays 499
24.5 Nanoscience 504

CHAPTER Nuclear Physics

25 25.1 Nucleus 505
509
25.2 Radioactivity
25.3 Nuclear Reactions 518

STPM Model Paper 960/3 529

ANSWERS 536













viii




CONTENT STPM Q&A Physics.indd 8 16/02/2022 4:36 PM

Chapter
1 Physical Quantities and

Units

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1.1 Base Quantities and SI Units


Section A Multiple-choice Questions Semester

Question 1 1
Which of the following is not true about the unit for each physical quantity?
A Momentum : kg m s –1
B Impulse : N s
C Energy : kg m s –2
–2
D Pressure : N m –2

Answer: C
A Momentum = mass × velocity
Exam Tips
Unit = kg m s –1 Exam Tips
B Impulse = force × time Base quantities and their SI
Unit = N s units:
C Energy (work) = force × displacement Mass (kg), length (m), time (s),
= mass × acceleration × displacement current (A), temperature (K),
2
–2
Unit = kg × m s × m = kg m s –2 quantity of matter (mol).
D Pressure = force ÷ area
2
Unit = N ÷ m = N m –2
Question 2

The gravitational force between two masses placed a distance r apart is
represented by the equation GMm . The unit for the gravitational constant G is
2
2
A N m kg –2 r
B N m kg 2
–1
C N m kg 2
–2
D N m kg –2
Answer: A 2
G = Fr
Mm
2
Unit G = N m 2 = N m kg –2
kg 2
1




01 STPM Q&A PHY T1.indd 1 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
Question 3

–1
–2
2
2
The permittivity of free space e has units A s N m and the permeability of
0
free space µ has units N A . Their relationship is as shown in the equation below.
–2
0 1
k =
2
e µ 0
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0
Determine the appropriate unit for k.
–1
2
A m s A C m s –2
2
–1
B m s A 2 D m s –1
Answer: D 1 1 1 1
Semester
2
1 Unit for = e µ 0 = A s N m –2 NA –2 = s m –2 = m s –2
2 2
–1
2
0
Unit for k = m s –1
1.2 Dimensions of Physical Quantities
Section A Multiple-choice Questions
Question 1
The drag force F is related to the density ρ and velocity v by the equation
2
F = kρv .
Where k is a constant. The dimension of the constant k is
A M L 2 C L 2
B L T 2 D L –2

Answer: C
–3
–2
[F] = M L T [ρ] = M L [v] = L T –1
k = F
ρv 2
M L T –2
[k] = = L 2
2
ML L T –2
–3

Exam Tips
Exam Tips
Base quantities and their dimensions:
Mass (M), length (L), time (T), current (A), temperature (θ), quantity of matter (N).




2




01 STPM Q&A PHY T1.indd 2 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
Question 2

Which of the following equations is homogeneous?
A T = 4π 2 l g where T = period, l = length, g = gravitational field strength
B E = ½ ρ v where E = energy, ρ = density, v = velocity
2
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C F = P/t where F = force, P = power, t = time
D E = PV where E = energy, P = pressure, V = volume

Answer: D Semester
–2
A [T] = T [l/g] = L/L T = T 2
B [E] = M L T –2 [ρv ] = M L L T = M L T –2
–3
2
–1
2
–2
2
2
C [F] = M L T –2 [P/t] = M L T / T = M L T –4 1
–3
2
2
2
3
D [E] = M L T –2 [PV] = M L T L = M L T –2
–1
–2
Exam Tips
Exam Tips
The equation is homogeneous when the dimensions in both sides of the
equation are equal.
Question 3
The force F between the two isolated point charges, Q separated with a distance
R is given by the equation
F = k Q R .
y
x
3
Where the unit for k is kg m A s . What are the values of x and y?
–2
–4
x y
A 1 1
B 1 –2
C –2 1
D 2 –2
Answer: D
–2
3
[F] = M L T –2 [Q]= A T [R] = L [k]= M L A T –4

–2
–2
3
M L T = (M L A T )(A T) L y
x
–4
= M L A –2+x T –4+x
3+y
Equating the coefficient
y : 1 = 3 + y y = –2
x : –2+ x = 0 x = 2 or –4 + x = –2 x = 2
3


01 STPM Q&A PHY T1.indd 3 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units

Section B Structured Questions
Question 4

(a) The force of attraction F between the Sun with mass M and a planet of mass
2Sdn Bhd. All Rights Reserved.
m is related by the expression
F = GMm .
r 2
Where r is the distance of separation between the Sun and the planet. Use
the expression to find units for the gravitational constant G in terms of base
units.

Semester
1 (b) The orbital period T of the planet about the Sun is related to the distance
r from the sun as shown in the equation below.
x
y
T = ( 4π 2 )M r z
G
Where M is the mass of the Sun and G is the gravitational constant.
Determine the values of x, y and z.

Answer:
(a) F = GMm
r 2
G = Fr 2
Mm
Unit of F = kg m s kg 2 –2
–2
Penerbitan Pelangi y z
kg m s m
Unit of G =

= m kg s
3


–1 –2
(b) [T] = T
–1
–2
3
[G] = L M T
[r] = L
[M] = M
x
–2 –1
T = (L M T ) M L
–1
3
1
–3
y z
= L M T M L
2
z–3
y+1
2
= L M T
Equating the coefficients:
x = 2
y + 1 = 0, y = –1
z – 3 = 0, z = 3
4
01 STPM Q&A PHY T1.indd 4 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
Question 5

(a) By reference to ideal gas equation, PV = nRT where
P = pressure of the gas
V = volume of the gas
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n = number of moles
R = molar gas constant
T = temperature of the gas
Express the unit of R in terms of J, mol and K.
(b) Some real gases obey the equation Semester
(p + a )(V – b) = RT.
V 2 m
m
Where V is the molar volume of the gas. Determine the units of a and b in 1
m
base units.

Answer:
(a) PV = nRT

PV
R =
nT
(N m )(m )
3
–2
unit of R =
mol. K
N m
–1
= = J mol K –1
mol. K
(b) unit of a = unit of pressure, P
V m 2
a = Pa
3
–1 2
(m mol )
∴ unit of a = Pa m mol –2
6
Unit of b = unit of V m
∴ = m mol –1
3

Exam Tips
Exam Tips
Unit of work done
N m = J








5




01 STPM Q&A PHY T1.indd 5 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
1.3 Scalar and Vectors


Section A Multiple-choice Questions
Question 1
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The diagram shows two forces acting on a box.
Exam Tips
Exam Tips
60 N
Resultant force is a single
60° force that is the sum of all the
80 N forces acting on an object.
Semester
1
What is the resultant force acting on the box?
A 41.1 N at 30° above the horizontal.
B 72.1 N at 60° above the horizontal.
C 100 N at 45° above the horizontal.
D 121.7 N at 25° above the horizontal.


Exam Tips
Answer: D Exam Tips
To find the resultant force,
(i) parallelogram method
60 N F
60° 120°

80 N F

Using cosine rule,
2
2
F = 60 + 80 – (2)(60)(80 cos 120°)
2
F = 121.7 N (ii) triangle method
sin θ = sin 120
60 121.7
θ = 25.3° above the horizontal. F




Question 2

A car travels with a velocity of 12 m s due North on a straight road. It then
–1
–1
travels due West with a velocity of 16 m s . Find the change in velocity of the car.
–1
–1
A 4 m s N 37° W C 20 m s S 37° W
–1
B 4 m s N 53° W D 20 m s S 53° W
–1
6


01 STPM Q&A PHY T1.indd 6 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
Answer: D
N
v
16 m s –1 16 m s –1

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–u u –1 12 m s –1
v – u 12 m s v
Common Error

Many students could not be Semester
2
2
(∆V) = 12 + 16 2 tan θ = 16 able to draw the direction of
–1
∆V = 20 m s 12 DV correctly.
θ = 53.1° 1
–1
∴ Change of velocity of the car is 20 m s S 53°W
Question 3

The diagram shows a horizontal force F pushing a block up an inclined plane.
F




If the weight of the block is W, what is the nett force acting on the block parallel
to the inclined plane?
A F cos θ C F cos θ – W sin θ
B W sin θ D F sin θ + W cos θ

Exam Tips
Answer: C Exam Tips
F cos 


F W sin 
F sin   F sin  F

W cos 
W

Nett force acting on the block parallel to F cos 
the inclined plane F can be resolved into two
= F cos θ – W sin θ components. F cos θ is
the horizontal component
and F sin θ is the vertical
component.


7




01 STPM Q&A PHY T1.indd 7 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units

Section B Structured Questions
Question 4

The diagram shows three coplanar forces acting on an object.
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y

50 N

40 N 40° x
27°
Semester
1 60 N



Calculate the
(a) total vertical force, (c) resultant force,
(b) total horizontal force,
acting on the object.


Answer:
Exam Tips
y Exam Tips
50 sin 40° Forces in many different
50 N
directions can all be
40 N 40° 50 cos 40° x resolved into two directions
27° 60 cos 27° perpendicular to each other.
60 N

60 sin 27° Common Error
(a) Total vertical force Forces along +x-axis and +y-axis
= 50 sin 40° + (–60 sin 27°) are positive.
= 4.9 N Forces along –x-axis and –y-axis are
negative.
(b) Total horizontal force
= 50 cos 40° + 60 cos 27° – 40
Exam Tips
= 51.8 N Exam Tips

2
(c) F = 4.9 + 51.8 2 The resultant force is
4.9 N F = 52.0 N calculated from the triangle
θ 4.9 of forces using Pythagoras
51.8 N tan θ = 51.8 theorem.

θ = 5.4° above the horizontal
∴ the resultant force is 52.0 N, 5.4° with the horizontal


8




01 STPM Q&A PHY T1.indd 8 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
Question 5

The diagram shows two tugs pulling a boat in a river along the direction due
North.

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15°
300 N 180 N
30° Semester

1




(a) Find the total vertical force acting on the boat.
(b) Find additional force needed to apply to the boat so that the boat continues
to move in a direction due North.


Answer:
(a)

180 cos 30°
300 cos 15°
15°
300 N
30° 180 N

300 sin 15° 180 sin 30°




Total vertical force = 180 cos 30° + 300 cos 15°
= 445.7 N
(b) Total horizontal force
= 180 sin 30° + (–300 sin 15°)
= 12.35 N
∴ Additional force of 12.35 N applied due West is needed.








9




01 STPM Q&A PHY T1.indd 9 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
1.4 Uncertainties in Measurements


Section A Multiple-choice Questions
Question 1
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The diagram shows a metre rule. x and x are two scale readings as shown.
1 2


25 26

Semester
1
x 1 x 2

Which of the following shows the correct reading of x and x ?
1
2
x x
1 2
A 25.40 ± 0.05 cm 26.00 ± 0.05 cm
B 25.45 ± 0.05 cm 25.95 ± 0.05 cm
C 25.45 ± 0.05 cm 26.00 ± 0.05 cm
D 25.40 ± 0.05 cm 25.95 ± 0.05 cm


Answer: A
When the reading is not at the middle, always take the reading which is nearer.


Question 2
The length and the width of a rectangle is 12.80 ± 0.05 cm and 23.7 ± 0.05 cm
respectively. Find the area of the rectangle.
A 304.5 ± 0.1 cm 2
B 305.8 ± 0.1 cm
2
C 304 ± 2 cm 2
D 305 ± 2 cm 2



Answer : C
A = L × W
A = (12.80 × 23.75) ± ∆A
= 304 ± ∆A




10




01 STPM Q&A PHY T1.indd 10 17/02/2022 10:03 AM

Physics Semester 1 STPM Chapter 1 Physical Quantities and Units
∆A = ∆L + ∆W = 0.05 + 0.05
A L W 12.8 23.75 Common Error

∆A = 0.006012(304) = 1.83 Uncertainty is in one significant
∴ A = 304 ± 2 cm 2 figure.
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Question 3

A student measured the length of a pendulum as 17.5 ± 0.05 cm. The Semester
fractional error in the measurement of the period T is 0.03. The student uses
the formula T = 2π l to calculate the value of g. Find the fractional error in
g 1
the calculated value of g.
A 0.03
B 0.04
C 0.05
D 0.06




Answer : D

T = 2π l
g
2
T = 4π 2 l
g
4π l
2
g =
T 2
∆g ∆l 2∆T
g = l + T


= 0.05 + 2(0.03)
17.5
= 0.063












11




01 STPM Q&A PHY T1.indd 11 17/02/2022 10:03 AM

Chapter
14 Electric Current




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14.1 Conduction of Electricity


Section A Multiple-choice Questions

Question 1
An electrical fan rotates when the equipment is switched on. What are the charge
carriers that convey the current in the fan?
A Electrons only
B Negative ions only
C Holes and electrons
D Lattice atoms and free electrons



Answer: A
The charge carriers in a metal conductor are purely free electrons.


Semester
2 Question 2
When a conductor carries a current, the number of conduction electrons that
flow in the conductor for 10 s is 3.75 × 10 . What is the current that flows in
20
the conductor?
A 3.8 A C 6.0 A
B 4.2 A D 10 A



Answer: C Exam Tips
Exam Tips
Q
I =
t Electric current is defined
as the rate of flow of charge
= Ne in the circuit in a definite
t direction.
–19
20
(3.75 × 10 )(1.6 × 10 ) dQ
= I =
10 dt
= 6.0 A

220




014 STPM Q&A Physics T2.indd 220 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
Question 3

A neon lamp is connected to a battery. If a constant current of 1.25 A flows to
the lamp, how long will it take for a charge of 15 C to flow through the lamp?
A 8.3 s C 16 s
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B 12 s D 19 s

Answer: B
Q = It
t = Q
I
= 15
1.25
= 12 s




Section B Structured Questions
Question 4
(a) Free electrons are charge carries in metals. What is meant by free electrons?
(b) Explain why the random motion of free electrons does not produce a
current. Semester
(c) When a metal conductor is connected to a battery, a current flows in the
circuit. Explain the mechanism of conduction of electricity in metals.
2
Answer:
(a) Free electrons are electrons that are not used in bonding. They move
randomly throughout the metal. There is no net flow in any specific
direction.
(b) When there is no electric field applied to a conductor, the free electrons
move randomly. There is no net flow of electrons in any specific direction.
Hence the average velocity of the free electrons is zero. Thus there is no
net flow of charge and the current is zero.
(c) When a battery is connected across the ends of a metal conductor, an
electric field is set up. Electrons are negatively charged, so they drift in the
opposite direction to the field. The free electrons are now accelerated by
the field. They drift in a particular direction from the low potential end to
the high potential end. This produces the electric current.






221




014 STPM Q&A Physics T2.indd 221 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current

14.2 Drift Velocity

Section A Multiple-choice Questions

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Question 1
A conductor of cross-sectional area 0.10 mm carries a current of 2.5 A. If the
2
number of conduction electrons per cubic metre is 1.0 × 10 m . What is the
–3
28
value of drift velocity of the free electrons in the conductor?
–2
A 1.6 × 10 m s C 3.0 × 10 m s –1
–3
–1
B 3.9 × 10 m s –1 D 6.2 × 10 m s –1
–3
–2
Answer: A
I = nAve
v = I
nAe
= 2.5
(1 × 10 )(0.1 × 10 )(1.6 × 10 )
28
–6
–19
= 1.56 × 10 m s –1
–2
Question 2
Which statement is true about the drift velocity of the electrons in a current-
carrying metallic conductor?
Semester
A It is directly proportional to the diameter of the conductor.
2 B It is inversely proportional to the current.
C It is inversely proportional to the number of conduction electrons per unit
volume.
D It is directly proportional to the resistance of the conductor if the potential
difference across the conductor is kept constant.



Answer: C
v = I ,
Exam Tips
nAe Exam Tips
1
v ∝ I , v ∝ R Formulae used
v ∝ 1 • I = nAve
n • I = V
v ∝ 1 ∝ 1 R
A d 2




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Physics Semester 2 STPM Chapter 14 Electric Current
Question 3

A conductor of length 2.0 m and cross-sectional area 5.0 × 10 m carries
2
–6
a current of 6.0 A. If the drift velocity of the conduction electrons is
1.5 × 10 m s , what is the number of electrons in the conductor?
–4
–1
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A 2.5 × 10 C 4.2 × 10 23
23
B 2.9 × 10 23 D 5.0 × 10 23
Answer: D
I = nAve
= N Ave = Nve
Al l
N = Il
ve
= 6(2)
(1.5 × 10 )(1.6 × 10 )
–4
–19
= 5.0 × 10
23

Question 4
A tungsten wire of length 1.00 m and cross-sectional area 2.01 × 10 m carries
2
–6
a current of 5.00 A.
If the number of free electrons per unit volume is 4.3 × 10 , what is the time Semester
28
taken by an electron to travel from one end of the wire to the other end?
A 1.81 × 10 s C 4.51 × 10 s
3
3
B 2.77 × 10 s D 7.25 × 10 s 2
3
3

Answer: B
I = nAve
v = I
nAe
= 5
(2.01 × 10 )(4.3 × 10 ) (1.6 × 10 )
28
–19
–6
= 3.616 × 10 m s –1
–4
t = l
v
= 1
3.616 × 10 –4
= 2.77 × 10 s
3


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Physics Semester 2 STPM Chapter 14 Electric Current

Section B Structured Questions
Question 5
A current flows in a wire of length L and radius r with the free electrons travelling
–1
at a drift velocity of 6.25 × 10 m s for 1.0 minute.
–3
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(a) If the number of free electrons in 1.0 m of wire is 5.0 × 10 , calculate the
19
3
current flow in the circuit.
(b) Calculate the drift velocity for the same current in a wire of the same
1
materials but with radius r.
2
(c) The wire is then replaced with a wire of the same length that tapers
uniformly from end P to end Q as shown in the diagram.
P Q

The diameter at P is twice that at Q. Compare the drift velocity of the
conduction electrons at P and at Q.

Answer:

(a) N = 5.0 × 10 N = 5.0 × 10 19
19
A
Total charge = Ne
= (5 × 10 )(1.6 × 10 )
19
–19
= 8.0 C
Q
Current, I =
Semester
2 t
= 8 = 0.13 A
1 × 60
(b) v = I = I
nAe nπr e
2
v ∝ 1
r 2
= 2
v 2 r 1 2
v 1 1 r 2
v 2 = r 2
1
6.25 × 10 –3 1 2
r
2
= 4
v 2 = 4(6.25 × 10 )
–3
= 0.025 m s –1


224




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Physics Semester 2 STPM Chapter 14 Electric Current

(c) v = I = I
nAe πd 2
n 1 4 2 e
\ v ∝ 1
d 2
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v P ∝ 1 .......................a
d P 2
v Q ∝ 1 .......................b

d Q 2
a ÷ b ,
v P = 1 d Q 2 2
v Q d P
d
= 1 2d 2 2

= 1
4
\ v Q = 4v P
The drift velocity at Q is four times that at P.



Section C Essay Question
Question 6 Semester
(a) What do you understand by drift velocity of the charge carries in a metal
wire? 2
(b) When a battery is connected to the ends of a metal conductor of cross-
sectional area A, the charge carriers in the conductor move with drift
velocity v. Thus a current I flows in the conductor. Derive an expression for
the current in terms of drift velocity and the cross-sectional area.
(c) The current in the metal conductor increases when the potential difference
between its ends is increased. Explain what will happen to the number of
charge carriers per unit volume in the conductor and the drift velocity of
the charge carriers.
(d) Copper has a density of 8.93 × 10 kg m and its relative atomic mass is
–3
3
63.5. The density of free electrons in the copper is 8.34 × 10 .
28
(i) Find the number of atoms in 1.0 m of copper.
3
(ii) Find the effective number of free electrons contributed by each copper
atom.




225




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Physics Semester 2 STPM Chapter 14 Electric Current
Answer:
(a) Drift velocity is the average velocity of the electrons when they move
and collide with the positive ions in the opposite direction to the electric
field.
(b)
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E
+ e –
Z A
Q P
v
Let A = cross-sectional area
n = density of free electrons / number of electrons per unit volume
v = drift velocity
In one second, the electrons move from P to Q by a distance of v.
The volume from P to Q = Av
\ The total number of free electrons flowing from P to Q in 1 second
= nAv
The total charge flow from P to Q = nAve
\ Current flow = rate of flow of charge
I = nAve
Exam Tips
Exam Tips
(c) The number of charge carriers per unit
volume remains unchanged. Formulae used:
The drifts velocity increases as the increase V = IR ρl
of potential difference will increase the R = A
Semester
2 current flow in the conductor. v = I
v = I , V = IR nAe
nAe
The drift velocity is directly proportional to the current.


(d) (i) N = m
N A M
N = mN A
M
N = mN A
V VM

= eN A
M
(8.93 × 10 )(6.02 × 10 )
3
23
=
63.5 × 10 –3
= 8.466 × 10 atoms per m 3
28
226




014 STPM Q&A Physics T2.indd 226 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
(ii) In 1.0 m of copper, 8.466 × 10 atoms of copper contribute
3
28
8.34 × 10 free electrons.
28
8.466 × 10 atoms ⎯→ 8.34 × 10 free electrons
28
28
8.34 × 10 28
\ 1 atom ⎯→
8.466 × 10 28
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= 0.985 electrons
\ Each copper atom contributes 0.985 free electron.

14.3 Current Density

Section A Multiple-choice Questions

Question 1
A current of 3.0 A flows in an aluminium wire of cross-sectional area
2.5 × 10 m . What is the current density in the wire?
2
–6
A 1.2 × 10 A m C 7.5 × 10 A m –2
–2
6
6
–2
6
6
B 1.5 × 10 A m D 8.3 × 10 A m –2
Answer: A
Current density, J = I = 3 Semester
A 2.5 × 10 –6
6
= 1.2 × 10 A m
–2
2
Question 2
There are 6.23 × 10 free electrons per cm in a wire. The average drift velocity
3
24
of the electrons is 1.1 × 10 m s . Calculate the current density in the wire.
–1
–3
A 1.0 × 10 A m C 2.1 × 10 A m –2
–2
9
3
B 6.9 × 10 A m –2 D 1.1 × 10 A m –2
9
6
Answer: D
J = nve
6.23 × 10
= 1 1 × 10 m 24 3 2 (1.1 × 10 m s ) (1.6 × 10 C)
–3
–19
–1
–6
Exam Tips
= 1.1 × 10 C s m –2 Exam Tips
–1
9
I
= 1.1 × 10 A m –2 J = = nAve = nve
9
A A
n = density of electrons
v = drift velocity
227
014 STPM Q&A Physics T2.indd 227 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
Question 3

A charge of 60 Coulomb flows in a wire for 45 minutes. If the diameter of the
wire is 1.0 mm, what is the current density in the wire?
A 2.2 × 10 A m C 7.6 × 10 A m –2
–2
7
4
4
–2
B 2.8 × 10 A m D 5.6 × 10 A m –2
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8
Answer: B
Exam Tips
Q 60 C Exam Tips
I = =
t 45 × 60 s
= 2.222 × 10 C s –1 Unit of current is Ampere (A) or
–2
Coulomb per second (C s ).
–1
Area = πd 2

4
–3 2
= π(1 × 10 )
4
= 7.8540 × 10 m 2
–7
Current density, J = I
A
2.222 × 10 –2
=
7.854 × 10 –7
= 2.83 × 10 A m –2
4
Semester
2 Question 4
A current flows in a copper wire. The current density in the wire is 1.4 × 10
6
28
A m . If the number of free electrons in 1 m of wire is 8.4 × 10 , what is the
3
–2
drift velocity of the electrons?
–4
A 1.04 × 10 m s
–1
B 3.10 × 10 m s –1
–4
C 1.67 × 10 m s
–3
–1
–3
D 1.90 × 10 m s –1
Answer: A
J = nve
1.4 × 10 = (8.4 × 10 ) v (1.6 × 10 )
6
28
–19
v = 1.04 × 10 m s –1
–4
228




014 STPM Q&A Physics T2.indd 228 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current

Section B Structured Questions
Question 5
A current of 1.1 A flows in a copper wire of diameter 1.2 mm. The density of
copper is 8.93 × 10 kg m and the relative atomic mass of copper is 63.5.
3
–3
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Assuming that each copper atom contributes one free electron, calculate
3
(a) the number of free electrons in 1.0 m of wire,
(b) the current density in the wire,
(c) the drift velocity of the electrons in the wire.
Answer:
(a) Density = mass Exam Tips
volume Exam Tips
N × molar mass Number of
ρ = N A atoms = Mass
V Avogdro Molar mass
N = ρN A number
V M N = m
M
23
3
= (8.93 × 10 )(6.02 × 10 ) N A
63.5 × 10 –3
= 8.466 × 10 28
\ Number of the electrons per m = 8.466 × 10 28 Semester
3
(b) Cross-sectional area, A = πd 2
4
–3 2
= π(1.2 × 10 ) 2
4
= 1.131 × 10 m 2
–6
Current density, J = I
A
= 1.1
1.131 × 10 –6
= 9.726 × 10 A m –2
5
(c) J = nve
v = J
ne
= 9.726 × 10 5
(8.466 × 10 )(1.6 × 10 )
28
–19
= 7.18 × 10 m s –1
–5


229




014 STPM Q&A Physics T2.indd 229 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
Question 6

(a) Define electric current density and conductivity.
(b) (i) When a potential difference, V is applied across the ends of a metallic
wire of length l, cross-sectional area A and resistance R, a current I
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flows through the wire. Deduce an expression to show the relationship
between current density and conductivity.
(ii) There are 5.0 × 10 free electrons in one cubic metre of the wire. If
28
–1
the potential gradient applied across the wire is 2.0 V m and the drift
velocity of the electrons is 3.91 × 10 m s , calculate the conductivity
–3
–1
of the wire.
Answer:
(a) Electric current density, J = electric current, I
cross-sectional area, A

Conductivity, σ = 1
resistivity, ρ
Exam Tips
Exam Tips
(b) (i) Electric field strength E at any point dV V
in the wire, E = – dx or l
E = V ⇒ V = El E = electric field
l strength
dV
Current density, J = I ⇒ I = JA dx = potential
A gradient
Semester
2 Resistance of wire, R = ρ l
A Exam Tips
= l ⇐ σ = 1 Exam Tips
σA ρ To find unit of σ:
Ohm’s Law: V = IR ρ = RA
l
l
El = JA 1 2 ⇒ (Ω)(m ) = Ωm
2
σA
\ J = σE 1 m
σ =
ρ
–1
(ii) J = nve = Ω m –1
σE = nve
σ(2) = (5 × 10 )(3.91 × 10 )(1.6 × 10 )
28
–3
–19
7
σ = 1.56 × 10 Ω m –1
–1


230




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Physics Semester 2 STPM Chapter 14 Electric Current

Section C Essay Question
Question 7
A constant current of 2.0 A flows through a metal wire. Its diameter at end P is
smaller than the diameter at end Q as shown in the diagram.
(a) V = IRPelangi Sdn Bhd. All Rights Reserved.

I
P  Q


(a) Calculate the potential difference across PQ if the resistance of the wire is
0.5 Ω.
(b) Compare the
(i) mean drift speeds of the electrons flowing through the cross section
at P and Q,
(ii) current density at P and Q,
(iii) on the axes below, sketch a graph of current density against the distance
along the wire from P.
Current
density



P Q Semester
(c) The current is increased by increasing the potential difference across the
wire. State the effects on the concentration and the drift velocity of the 2
charge carriers. Exam Tips


Answer:
Exam Tips
Penerbitan at P is greater than that at Q. The v = nAe
I
= 2(0.5) = 1.0 V

(b) (i) The mean drift speeds of electrons
1
v ↑∝
A↓
drift velocity increases as the cross-
sectional area decreases.
A P , A Q
\ v P . v Q
(ii) The current density at P is greater

I
than at Q. The current density
increases as the cross-sectional area
J ∝
decreases. J = A I
A
A P , A Q
J P . J Q
231



014 STPM Q&A Physics T2.indd 231 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
(iii) Current
density



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P Q
(c) The increase in potential difference would increase the drift velocity of the
charge carriers.
v = I v ∝ I
nAe
The concentration of charge carriers remains unchanged because the
number of free electrons in the wire is the same.



14.4 Electric Conductivity and Resistivity

Section A Multiple-choice Questions

Question 1
–7
2
A metal wire has length 15 m, cross-sectional area 6.0 × 10 m and resistance
5.0 Ω. What is the conductivity of the wire?
A 2.0 × 10 Ω m –1 C 5.0 × 10 Ω m –1
–1
–1
–7
6
–7
–1
6
B 3.0 × 10 Ω m –1 D 7.5 × 10 Ω m –1
–1
Semester
2 Answer: C –7
ρ = RA = (5)(6 × 10 )
l 15
= 2.0 × 10 Ω m
–7
σ = 1 = 1
ρ 2 × 10 –7
= 5.0 × 10 Ω m –1
6
–1
Question 2
A copper wire of length 2.0 m has resistance 20 Ω. The wire is stretched to an
extension of 0.001 m. If the volume of the wire remains unchanged, what is the
new resistance of the wire?
A 20.01 Ω C 20.03 Ω
B 20.02 Ω D 20.04 Ω





232




014 STPM Q&A Physics T2.indd 232 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
Answer: B
Volume is constant,
A 1 l 1 = A 2 l 2
A 1 = l 2
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A 2 l 1
ρl
R = ⇒ R ∝ l
A A
⇒ R 1 ∝ l 1
A 1
R 2 ∝ l 2
A 2
\ R 2 = l 2 × A 1
R 1 A 2 l 1
= l 2 × A 1
l 1 A 2
= 1 2 2
l 2
l 1
R 2 = 1 2 + 0.001 2 2
20 2
R 2 = 20.02 Ω

Question 3 Semester

A hollow copper rod has an inner radius of 0.65 cm and an outer radius of
1.0 cm as shown in the diagram.
P 2
Q
O
0.65 cm
20 cm
1.0 cm
A current flows from P to Q. If the length of the rod is 20 cm and its resistivity
is 1.7 × 10 Ω m, what is the resistance of the copper rod?
–8
A 3.4 × 10 Ω C 1.8 × 10 Ω
–4
–9
B 1.9 × 10 Ω D 2.9 × 10 Ω
–5
–3

Answer: B
ρ l
R =
A
2 2
Cross-sectional area, A = πr 2 – πr 1
= π(1 × 10 ) – π(0.65 × 10 )
–2 2
–2 2
= 1.8143 × 10 m 2
–4
233


014 STPM Q&A Physics T2.indd 233 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
–8
–2
R = (1.7 × 10 )(20 × 10 )

1.8143 × 10 –4
= 1.87 × 10 Ω
–5
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Question 4
The conductivity of a copper wire decrease as the temperature increases because
A the number of free electrons decreases.
B the lattice atoms vibrate with smaller amplitude.
C the drift velocity of the electrons decreases.
D the mean time between collision decreases.


Answer: D

Exam Tips
Exam Tips
Copper wire has a constant number of free electrons.
When the temperature increases, the lattice atoms vibrate with greater amplitude.
Hence the mean time between collision, τ decreases because the mean free path
of the free electrons decreases.
2
ne τ↓
↓σ =
m
Section B Structured Questions
Question 5
Semester
2 A metal rod consists of a copper sheath with an iron core as shown in the
diagram.
Sheath



Core

The diameter of the sheath is 4 mm and the diameter of the core is 3 mm. The
length of the rod is 50 cm. If the potential difference applied across the ends of
the rod is 12 mV, calculate the
(a) resistance of the copper shealth,
(b) resistance of the iron core,
(c) total resistance of the metal rod,
(d) current flow in the metal rod.
[Resistivity of copper = 1.7 × 10 Ω m, resistivity of iron = 1.0 × 10 Ω m]
–7
–8


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Physics Semester 2 STPM Chapter 14 Electric Current
Answer: Exam Tips
(a) R c = ρ c l = ρ c l Exam Tips
2
2
A c π(r 2 – r 1 ) Cross-sectional area of
(1.7 × 10 )(0.5) copper sheath:
–8
= 2 2
2
2
π(2 – 1.5 ) × 10 –6 A = π(r 2 – r 1 )
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= 1.546 × 10 Ω
–3
–7
(b) R i = ρ i l = (1 × 10 )(0.5)
–3 2
A i π(1.5 × 10 )
= 7.074 × 10 Ω
–3
(c) The sheath and the core are arranged in parallel.
1 = + 1
1
R R c R i
1 1
= +
1.546 × 10 –3 7.074 × 10 –3
\ R = 1.269 × 10 Ω
–3
(d) V = IR
0.012 = I(1.269 × 10 )
–3
I = 9.46 A
Question 6 Semester
A thin square slab is cut from a sheet of metal of uniform thickness t as shown
in the diagram.
2
L
Y t
L
x

t
Z
(a) If the resistivity of the metal is ρ, find the resistance of the metal slab when
a current flows in the direction of
(i) x (ii) y (iii) z
Explain whether the resistance depends on the value of L and t in each case.
(b) Another square slab of sides 2L is cut from the same metal. If the current
flows in the direction of y, find the ratio of the resistance of the second slab
to the resistance of the first slab.




235




014 STPM Q&A Physics T2.indd 235 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
Answer:
Exam Tips
ρL ρ Exam Tips
(a) (i) R x = Lt = t
The resistance is independent of L Formula used:
ρl
but dependent on t. R = A
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ρt
(ii) R y =
L 2
The resistance is dependent on L and t.
ρL ρ
(iii) R z = Lt = t
The resistance is dependent on t but independent of L.
ρt ρt
(b) R 1 = L 2 , R 2 = (2L) 2
ρt
= 4L 2
ρt L 2
R 2
\ = 2 ×
R 1 4L ρt
1
=
4

Section C Essay Question
Question 7

(a) What is meant by resistivity and conductivity of a substance? [2]
Semester
2 (b) Show that the conductivity of a substance σ = ne τ . [5]
2
m
(c) The conductivity of a copper wire decreases with increasing temperature
but the conductivity of a sample of pure silicon increases as temperature
rises. Explain the dependence of conductivity on temperature for copper
and silicon. [5]
(d) State the effects of temperature change on the resistivity of conductors,
semiconductors and superconductors. [3]



Answer
(a) Resistivity of material, ρ = RA , R = resistance of a wire of that material
l
A = cross-sectional area of the wire
l = length of the wire
1
Conductivity, σ = resistivity, ρ of materials



236




014 STPM Q&A Physics T2.indd 236 17/02/2022 10:12 AM

Physics Semester 2 STPM Chapter 14 Electric Current
(b) When a potential difference is applied across a conductor, free electron
esperience electric force and move with acceleration, a,
F eE
\ a = =
m m
let t = mean time interval between collisions of free electrons and the
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lattice atoms
mean initial velocity of electrons = 0
eE
Mean final velocity of electron, v = 0 + m t
J = nve
2
2 1
= n 1 eE t e = ne t 2 E
m
m
J = σE
2
\σ = ne t
m
(c) In copper wire, when the temperature increases the lattice ions vibrate
with bigger amplitude. The mean free path of the free electrons decreases
and the rate of collision between free electrons and ions increases.
ne t
2
Hence the mean time interval between the collisions decreases. σ = m ,
conductivity decreases as t decreases.
Silicon is a semiconductor. When the temperature increases, more holes Semester
and free electrons are produced. Holes and electrons are the charge
carriers in semiconductor. Thus the number of charge carriers increases. 2
The mean free time interval between collisions also decreases but the
effect of the number of charge carriers is greater than that of mean free
time interval.
ne τ
2
Hence, σ = , n increases, σ increases.
m
(d) At high temperature, the resistance of metal conductors increases. Hence
the resistivity of metal conductor increases as temperature rises.
At high temperature, the resistance of semiconductor decreases. So, the
resistivity of semiconductor decreases with increasing temperature.
At very low temperature of near-absolute zero, the resistivity of certain
materials becomes zero. Superconductors have zero resistivity at very low
temperature.





237




014 STPM Q&A Physics T2.indd 237 17/02/2022 10:12 AM

Chapter
21 Sound Waves




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21.1 Propagation of Sound Waves


Section A Multiple-choice Questions

Question 1
The figure below is the displacement-distance graph of a sound wave travelling
in air in the +x-direction. The points P, Q and R have the same displacement.
Displacement, y

P Q R
0
Distance, x


The instantaneous directions of motion of the air molecules at P, Q and R are
correctly shown in
P Q R
A
B
C

D



Answer: A
Sound wave is a longitudinal wave. Displacement, y
Vibration of the air molecules is parallel
to the direction of wave propagation. P Q R
The figure shows the y-x graphs at time
= 0, and at time = ∆t. 0 Distance, x Semester
The arrows at P, Q and R indicate
the directions of the instantaneous
velocities, P (–), Q (+) and R (–). 3



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Physics Semester 3 STPM Chapter 21 Sound Waves
Question 2

The figure below is the displacement-distance graph of sound wave travelling
in air in the +x-direction. At which of the points A, B, C and D is the greatest
instantaneous pressure?
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Displacement, y



0
A B C D Distance, x






Answer: B
From the y-x graph, the instantaneous displacements (+ or – ) of air molecules
to the left and right of each of the points are shown in the figure below.

A B C D

Pressure at point B is the greatest because the air molecules on either sides
move inwards.

Question 3
The variation of displacement y from the equilibrium position with time t for
air molecules at a distance x from a point O when sound ware travels in air is
represented by the equation
y = y 0 sin (ωt – kx)
What is the expression for the variation of pressure p with time t for points
distance from O?
[ p 0 is the amplitude of pressure change]
A p = p 0 sin (ωt – kx)
B p = p 0 cos (ωt – kx)
C p = –p 0 sin (ωt – kx)
D p = –p 0 cos (ωt – kx)


Common Error
Semester
1
3 The peak for the curve for p occurs – cycle after and not before the peak for
4
the curve of y.

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Physics Semester 3 STPM Chapter 21 Sound Waves
Answer: D
Refer to the answer in Question 2. On the same axes, sketch the y-x graph and
p-x graph (see figure below). From the shape of the p in the graph, it can be
deduced that the expression for p is p = –p 0 cos (ωt – kx).

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y = y sin (ωt – kx)
0
p
0
Distance, x





21.2 Sources of Sound


Section A Multiple-choice Questions

Question 1
Which of the following is the equation for a standing wave?
A y = A sin 5t C y = A cos (5t + 2x)
B y = A sin (5t – 2x) D y = A cos 2x sin 5t


Answer: D
Exam Tips
Exam Tips
Amplitude of standing wave varies with distance x. In A, B and C the amplitude
is constant, A. In D, amplitude = A cos 2x, varies with x.


Question 2
A standing wave set up in a string fixed at both ends has three nodes. If the
frequency of the wave is doubled, how many nodes are there?
A 4 C 6
B 5 D 7


Answer: B
The wavelength is halved when the frequency is doubled. Semester

N N N
3
N N N N N


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Physics Semester 3 STPM Chapter 21 Sound Waves
Question 3

A string of length L is stretched between two fixed points. Two successive
resonant frequencies emitted by the string are 315 Hz and 420 Hz. What is the
fundamental resonant frequency of the string?
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A 105 Hz C 210 Hz
B 140 Hz D 280 Hz

Answer: A
Resonant frequencies of a stretched string are: f 0 , 2f 0 , 3f 0 , 4f 0 , 5f 0 , 6f 0 , …
Differences between successive resonant frequencies = 2f 0 - f 0 = f 0 , 3f 0 - 2f 0 = f 0
Therefore fundamental frequency, f 0 = (420 – 315) Hz = 105 Hz

Question 4
Suppose all six equal-length strings of an acoustic guitar are played without
fingering, that is, without being pressed down at any frets. What quantities are
the same for all six strings?
A Fundamental frequency and speed of wave along the strings
B The speed of wave along the strings and the fundamental wavelength
C Fundamental wavelength and speed of sound emitted
D The speed of sound emitted and the fundamental frequency


Answer C
Fundamental wavelength = length of string, which is the same for all the
strings.
Sound travels with the same speed in air irrespective of frequency.
Speed of wave in string, v = T , fundamental frequency, f 0 = 1 T ,
μ 2L μ
mass per unit length of string, μ is different.



Question 5
Which of the following statements about the standing waves in a stretched
string of fixed length is true?
A The mid-point of the string is always a node.
B The mid-point of the string is always an antinode.
C There is always even number of antinodes.
D The total number of nodes and antinodes is always odd.
Semester
3



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Physics Semester 3 STPM Chapter 21 Sound Waves
Answer: D
Refer to the figure below.

N A N

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N N N




N N N N N




Question 6
A flute has a length of 33.0 cm. The speed of sound in air is 330 m s . What is
–1
the fundamental frequency of the flute, assuming it is a tube closed at one end
and open at the other?
A 100 Hz C 500 Hz
B 250 Hz D 1000 Hz


Answer: C



A

I = 0.33 m
Length of flute, l = 0.33 m = λ
2
v
Fundamental frequency, f 0 = = 330 Hz
λ 2(0.33)
= 500 Hz

Question 7
The lengths of the air column in a tube that resonant with a tuning fork of
frequency 512 Hz are 16.0 cm and 47.0 cm respectively. No resonance occurs
between 16.0 cm and 47.0 cm. What is the speed of sound in air? Semester
A 300 m s –1 C 330 m s
–1
B 317 m s –1 D 334 m s –1
3



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Physics Semester 3 STPM Chapter 21 Sound Waves
Answer: B
A A
c
I
1
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I 2










From the above figure, l 2 – l 1 = (0.470 – 0.160) m = λ
2
Speed of sound, v = f λ = (512)(2 × 0.310) m s
–1
= 317 m s –1
Question 8
A stationary wave is formed inside a tube which is open at both ends. Which
diagram shows the locations of the node N and antinode A in the tube for the
first overtone?

A A N A C A N A N A

B N A N D N A N A N




Answer: C
Exam Tips
Exam Tips
Draw the stationary wave in the tube. Antinodes are found at the open ends.

A A A A A
N N N


Common Error

There is an antinode and not a node (N) at either of the open ends of the tube.
Semester
3 Then draw figures of the fundamental mode and the first overtone to get the
right answer.


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Physics Semester 3 STPM Chapter 21 Sound Waves
Question 9

A tube P is closed at one end and open at the other end. Another tube Q is
open at both ends. The tubes P and Q are of the same length. What is the ratio
of the frequencies of the first overtone of tube P to tube Q?
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A 1 C 2
4 3
B 1 D 3
2 4


Answer: D

N A A
N
I
A A A
N N

Tube P: l = 3   = 3v
λ 1
4
4f 1
v
Tube Q: l = 4   = f 2
λ 2
4
f 1 = 3
f 2 4
Question 10
Three successive resonance frequencies of a tube are 450 Hz, 750 Hz and
1050 Hz. Which of the following is the correction deduction about the tube and
its fundamental frequency?
Tube Fundamental frequency
A Closed at one end and open at the other 150 Hz
B Closed at one end and open at the other 300 Hz
C Open at both ends 150 Hz
D Open at both ends 300 Hz


Answer: A
Resonance frequencies of a tube closed at one end and open at the other are Semester
f 0 , 3f 0 , 5f 0 , 7f 0 , …
Difference of frequencies between successive overtones = 2f 0 = (750 – 450) Hz
= 300 Hz 3




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Physics Semester 3 STPM Chapter 21 Sound Waves
Hence fundamental frequency f 0 = 150 Hz
Resonance frequencies of a tube open at both ends are f 0 , 2f 0 , 3f 0 , 5f 0 …
Difference of frequencies between successive overtones = f 0 = (750 – 450) Hz
= 300 Hz
Then the resonance frequencies would be 300 Hz, 600 Hz, 900 Hz, …
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Hence the tube is not open at both ends.


Section B Structured Questions


Question 11
A 30.0 cm long guitar string with a mass per unit length of 9.00 × 10 kg m is
–1
–3
stretched to a tension of 28.0 N.
Find
(a) the fundamental frequency, and
(b) the frequencies of the next three harmonics.


Answer
(a) Fundamental frequency, f 0 = 1 T
μ
2l
1 28.0
= –3 Hz
2(0.30) 9.00 × 10
= 93.0 Hz
(b) The next three harmonics are of frequencies: 2(93.0 Hz), 3(93.0 Hz) and
4(93.0 Hz)
= 186 Hz, 279 Hz, 372 Hz

Question 10

A length of a string is L and its mass per unit length is μ. The string is
stretched and the tension in the string is T .The string vibrates at its fundamental
frequency, f.
(i) Draw a diagram to show the vibration of the string.
(ii) Write the expression for the speed of transverse wave along the string.
(iii) Deduce the expression for the fundamental frequency f in terms of L, m and
T.
(iv) Find in terms of f, the new fundamental frequency when (I) the tension in
the string is doubled, and (II) when the length of the string is halved.
Semester
3



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Physics Semester 3 STPM Chapter 21 Sound Waves
Answer
(i)


L
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T
(ii) Speed of transverse wave, v =
μ
λ
(iii) Length of string L = , λ = 2L
2
λ
Fundamental frequency, f = = 1 T
2 2L μ
(iv) f = 1 T
2L μ
1 2T
(I) When tension is 2T, fundamental frequency = = 2f
2L μ
1 1 2T
(II) When the length is L, fundamental frequency = = 2f
2 L μ
2  
2
Question 13
Two strings P and Q on a piano have fundamental frequencies of 262 Hz and
440 Hz respectively.
(a) The strings have the same mass per unit length m but the length of string
P is only 64% of the string Q. Determine the ratio of tensions in the two
strings.
(b) Another string R is of the same material as strings P and Q, but its diameter
is 150% of string Q, and its length is 120% of Q. Strings R and Q have the
same tension. What is the fundamental frequency of string R?


Answer
(a) The fundamental frequencies of the two strings are
f P = 1 T P and f Q = 1 T Q (L P = 0.64L Q )
2L P m 2L Q m

f P = 1 T P
f Q 0.64 T Q

T P = (0.64) 2  262 2 Semester
T Q 440
= 0.145
3




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Physics Semester 3 STPM Chapter 21 Sound Waves

(b) Mass per unit length of string Q, m =  πd 2  ρ ∝ d 2
4
Mass per unit length of string R, m’ = (1.50) m
2
1
f Q = T Q = 440 Hz
2L m
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T Q
Fundamental frequency of string R, f R = 1 1.50 m
2
2(1.20)L
= 1 (440) Hz
(1.20)(1.50)
= 244 Hz


Section C Essay Question

Question 14
(a) A small loudspeaker which is connected to an audio oscillator of adjustable
frequency is held at the open end of a burette which is filled with water. The
water is then slowing run off from the burette. First resonance occurs when
the length of the air column in the burette is 7.2 cm, and again at 20.2 cm.
(i) What is the frequency of sound from the loudspeaker?
(ii) Determine the end correction.
(Speed of sound in air = 330 m s )
–1
(b) The frequency of the oscillation can be varied and read from a meter.
Describe how would you use the arrangement described in (a) above to
determine experimentally the speed of sound in air. You should state the
measurements to be taken, and how the speed of sound is deduced from
a suitable graph.


Answer
(a)
c c
7.2 cm



20.2 cm




Semester
3




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Physics Semester 3 STPM Chapter 21 Sound Waves

(i) In Figure (a), 7.2 + c = 1 λ (1)
4
In Figure (b), 20.2 + c = 3 λ (2)
4
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1
(2) – (1): λ = 13.0 cm
2
λ = 26.0 cm
v
Frequency of tuning fork, f =
λ
= 330 Hz
0.26
= 1270 Hz
(ii) From equation (1),
1
End correction, c = (7.2 – λ) cm
4
1


= 7.2 – (26.0) cm
4
= 0.70 cm
(b) With the frequency oscillation set at a known frequency f, the first
resonance length l of the air column in the burette is measured. The
procedure is repeated for different values of f, and the corresponding
value of l obtained.
1
For the first resonance, l + c = λ = v
4 4f
A graph of l against 1/f is plotted (see figure below).

I (m)



Gradient = v/4

0 1/f (s)



Gradient of graph = v/4
Speed of sound, v = 4(gradient of graph) Semester



3




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Physics Semester 3 STPM Chapter 21 Sound Waves
Question 15

(a) A string is stretched between two fixed points. The mid point of the string
is plucked and then released. Explain how a standing wave is set up in the
string.
Draw the fundamental mode of vibration of the standing wave.
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(b) A string fixed at both ends and having a mass of 4.80 g, a length of 2.00 m,
and a tension of 48.0 N vibrates in its second mode (n = 2).
(i) What is the ratio of the wavelength in air of the sound emitted by this
vibrating string and the wavelength of the wave in the string?
(ii) Find the three lowest frequencies of standing waves in the string?
(Speed of sound in air = 330 m s )
–1

Answer
(a) After the string is plucked and released, pulses of wave travel along the
string toward the two fixed ends of the string where the wave pulses are
repeatedly reflected. Superposition of the reflected waves travelling in
opposite directions along the string produces the standing wave.




Fundamental mode

(b)

L

(i) Wavelength in string, λ = L = 2.00 m
2 T
Frequency of sound, f =
2L μ
v
Wavelength of sound wave, λ sound = (v = speed of sound)
f
μ
= vL T
(4.80 × 10 /2.00)
–3
= (330)(2.00) m
48.0
= 4.67 m
Wavelength of sound 4.67
= = 2.34
Wavelength in string 2.00
Semester
3




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