1202 Question Bank Physics Form 5

©PAN ASIA PUBLICATIONS

Contents












iii – x Chapter 5
Must Know Electronics 80 – 101


Chapter 1 Force and Motion II 1 – 20 NOTES 80
Paper 1
84
Paper 2 92
NOTES 1 Reinforcement and Assessment
Paper 1 3 of Science Process Skills for
Paper 2 11 Paper 3 (Practical Test) 100
Reinforcement and Assessment
of Science Process Skills for Chapter 6
Paper 3 (Practical Test) 19 Nuclear Physics 102 – 115

NOTES 102
Paper 1 104
Chapter 2 Pressure 21 – 42 Paper 2 108
Reinforcement and Assessment
NOTES 21 of Science Process Skills for
Paper 1 23 Paper 3 (Practical Test) 114
Paper 2 33
Reinforcement and Assessment
©PAN ASIA PUBLICATIONS
of Science Process Skills for Chapter 7
Paper 3 (Practical Test) 41 Quantum Physics 116 – 132

NOTES 116
119
Chapter 3 Electricity 43 – 61 Paper 1 124
Paper 2
Reinforcement and Assessment
NOTES 43 of Science Process Skills for
Paper 1 46 Paper 3 (Practical Test) 131
Paper 2 52
Reinforcement and Assessment
of Science Process Skills for
Answers 133 – 150
Paper 3 (Practical Test) 60


Chapter 4 Electromagnetism 62 – 79


NOTES 62
Paper 1 65
Paper 2 72
Reinforcement and Assessment
of Science Process Skills for
Paper 3 (Practical Test) 78




ii



Content_1202 Physics F5.indd 2 10/01/2022 12:27 PM

MUST


KNOW Mnemonics







Triangle of Forces Fleming’s Left-hand Rule

Triangle of forces:
• is a vector diagram whose sides represent three forces in The direction of the force can be determined by using Fleming’s
left-hand rule.
equilibrium.
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• is a closed circle with forces in clockwise or in anticlockwise
direction. Force, F Magnetic field, B

F
2 a
F F F
1 = 2 = 3 Current, I
c (q) sin a sin b sin c
F F = F + F + 2F F cos q
2
2
2
3
3 1 2 1 2
F
1
b Fleming’s left-hand rule → FBI
Mnemonics (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 4) 7 @ Pan Asia Publications Sdn. Bhd.

Factors Affecting Resistance of Wire Fleming’s Right-hand Rule
The direction of the induced current in a conductor can be
determined by using Fleming’s right-hand rule.
Mnemonic Factors
Last Length of wire, L Thumb
First finger
Rocket Resistivity of wire, r Movement
Magnetic field
Cross Cross-section of wire, A

Tower Temperature of wire, q
Current is
affected
Second finger



Mnemonics (Chapter 3) 3 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 4) 9 @ Pan Asia Publications Sdn. Bhd.

Current and Potential Difference npn Transistor and pnp Transistor


Electrons flow from the Negative to the Positive terminal ˜ ENP Collector Emitter Collector Emitter
C n P n E C P n P E
Current flow from the Positive to the Negative terminal ˜ CPN
B Base B Base
C C

B B


E E
npn transistor pnp transistor
Not Pointing IN Pointing IN


 Mnemonics (Chapter 5) 5 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 5) 11 @ Pan Asia Publications Sdn. Bhd.



Must know_1202 Physic F5.indd 1 07/01/2022 11:41 AM

MUST


KNOW Important Facts







Direction of Force on a Current-carrying Zero Resultant Force
Conductor in a Magnetic Field
1. Forces are balanced when the resultant force is zero.
2. When resultant force is nil, acceleration is nil and the body
• The direction of the magnetic field of the current-carrying wire
will be:
can be determined by using right-hand grip rule. When the wire
©PAN ASIA PUBLICATIONS
is placed in a magnetic field, a force acts on it, F. (a) Stationary or (b) Moves with uniform
velocity
I
F Steel
sphere
F F
N S F buoyant drag
friction
F = F Engine
friction mg oil
F
• The direction of the force can be determined by using Fleming’s mg = F + F
left-hand rule. buoyant drag
Important Facts (Chapter 4) 20 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 14 @ Pan Asia Publications Sdn. Bhd.


Deflection of Cathode Rays in an Electric Field Buoyant Force
P
Cathode Anode Maltese Buoyant force = Weight of fluid displaced
cross
+
Shadow V 1 Air
– Air
Liquid
Filament
Liquid
Q with
density, V
If P and Q are not connected to the power supply, ρ V
2
cathode rays will move straight and shadow of the
maltase cross is at the center position of the screen. V object = V + V 2
1
F = rV g
+ B 2 F = rVg
B
If P and Q are connected to the power supply, where V is the volume of the
2
cathode rays is deflected upwards and shadow of part of the body submerged in where V is the volume of the
body
the maltase cross is deflected upwards. the liquid
–
Important Facts (Chapter 5) 22 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 16 @ Pan Asia Publications Sdn. Bhd.
4
• Equation for Alpha Decay (α or) He Internal Resistance
2
Parent nucleus has nucleon number, A and proton number, Z.
After a α-decay, the daughter nucleus has a nucleon number, The internal resistance of a dry cell is the obstruction or resistance
A – 4 and a proton number, Z – 2.
to the flow of charge by chemicals electrolyte in the dry cell.
4
A
X ˜ A–4 X + He + Energy
Z Z–2 2
0
• Equation for Beta Decay (β or) e
–1
Parent nucleus with nucleon number, A and proton number, Z.
After a β-decay, the daughter nucleus has a nucleon number, A
(no change) and a proton number, Z + 1. V Rheostat
0
A X ˜ A–0 X + e + Energy
Z Z+1 –1
• Equation for Gamma Decay (γ) A
There is no changes in nucleon number and proton number. The
nucleus is less energetic after gamma decay.
Ԑ = IR + Ir that is, V = IR
A
A X ˜ X + γ + Energy
Z Z Ԑ = V + Ir
Important Facts (Chapter 6) 24 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 3) 18 @ Pan Asia Publications Sdn. Bhd.
Must know_1202 Physic F5.indd 4 07/01/2022 11:41 AM

MUST


KNOW Common Mistakes







Resolution of Forces Current-carrying Conductor
• The diagram below shows a current-carrying conductor placed
Wrong Correct
in a magnetic field.
F = 100 sin 30°
x y I
F = 100 cos 30°
Vertical component 30° 100 N find the angle with the horizontal N S
labelled.©PAN ASIA PUBLICATIONS
or
line first:
100 N

Wrong
Correct
60° • State the direction of the force on the conductor.
F = 100 cos 30°
x Horizontal line Fleming’s right-hand rule The current-carrying conductor
F = 100 sin 30°
y is used to determine the is not an induced current, so need
F = 100 sin 30° or 100 cos 60°
x direction of force. to use Fleming’s left-hand rule to
F = 100 cos 30° or 100 sin 60°
y determine the direction of the force.
Common Mistakes (Chapter 1) 37 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 4) 43 @ Pan Asia Publications Sdn. Bhd.
Total Pressure Effect of Magnetic Field on Cathode Rays

What is the ratio of the pressure at X to the pressure at Y? • The diagram below shows a Maltese cross tube.
Cathode Anode Maltese
Wrong Correct cross
Air +
P 20 m H O P Shadow
Water x 2 x
20 m P = 30 m H O P –
30 m y 2 y Filament
= 2 : 3 (10 + 20) m H O
2
X =
(10 + 30) m H O
2
Y • What will happen to the velocity of the cathode ray if the
= 3 : 4
potential difference, V between the cathode and the anode is
Note: Total pressure = Atmospheric pressure + pressure due to water increased?
Wrong Correct
No change 2eV
Cathode rays velocity, v =  , thus
m
the velocity increases as V increases.
Common Mistakes (Chapter 2) 39 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 5) 45 @ Pan Asia Publications Sdn. Bhd.
Magnetic Field Working Functions for Photoelectric Effects

Wrong Correct If the threshold wavelength is 420 nm, calculate the metal working
function in units of eV.
• The electric • The electric
field lines – field lines Wrong Correct
incorrectly + are always +
directed away Students solve this question by Students need to simplify the
from positive using equation: photoelectric equation:
• Positively charges. hc 1
2
charged l = mv + W W = hf o
2
particles – • The electric
field lines are
• Negatively always directed –
charged toward negative
particles charges.


 Common Mistakes (Chapter 3) 41 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 7) 47 @ Pan Asia Publications Sdn. Bhd.



Must know_1202 Physic F5.indd 7 07/01/2022 11:41 AM

MUST


KNOW Important Diagrams







Magnetic Field Spring

P Q F P Q F
Q P
P Q
©PAN ASIA PUBLICATIONS
Field ines Short
x x
Copper Steel Long
Steel spring stiffer Short spring stiffer
Other fingers P Q
show the P Q F P F Q
direction of
the magnetic Right Q P
field hand
Current-carrying Small Large x Thin Thick x
conductor
Spring with smaller coil Spring with thicker wire stiffer
diameter stiffer
Important Diagrams (Chapter 4) 44 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 1) 38 @ Pan Asia Publications Sdn. Bhd.

Calculation of Potential Difference Pressure
Through Resistors
• Liquid pressure is due to weight of liquid above.
Surface
of liquid
R h W Container
1
filled with
liquid
A
V
in
• Atmospheric pressure is due to weight of atmosphere above.
0 kPa
R V
2 out
31 kPa
1 ( R 1 ) 2 ( R 2 )
V = × V or V = × V in 101 kPa
in
R + R R + R
1 2 1 2
Important Diagrams (Chapter 5) 46 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 2) 40 @ Pan Asia Publications Sdn. Bhd.

Electron Interference Experiment Electric Field

Electric field pattern between two charged spheres and a charged
sphere with a plate:
Double-
slit Screen
Electron + –
+
Electron
beam gun
Interference
pattern
– –
+ –




Important Diagrams (Chapter 7) 48 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 3) 42 @ Pan Asia Publications Sdn. Bhd.


Must know_1202 Physic F5.indd 8 07/01/2022 11:41 AM

1
Chapter Force and Motion II





NOTES


1.1 Resultant Force 3. By drawing a scaled diagram:
A C
©PAN ASIA PUBLICATIONS
1. A resultant force is a single force that produces the
same effect when it replaces two or more forces F
R
acting on the object.
F
1 θ
a = 3 m s –2 a = 3 m s –2 O B
=
F Use scale = 1 cm : p N
m R m
(a) Draw OA and OB in the same direction as F
1
F and F respectively.
2 2
F = ma (b) Draw a parallelogram OACB.
R
(c) Measure the length of OC = q cm.
2. Simple rules:
F = p µ q N
(a) Forces in same direction R
F = 3 N 4. By vice versa, if F = 0, object is stationary / moving
1
F = 7 N
R with uniform velocity. If F ≠ 0, object moves with
=
uniform acceleration.
F = 4 N
2
u = v = 0
F = F + F Stationary a = 0
R 1 2 123 F = ma
v – u u = v ≠ 0
a = –––– = 0
(b) Forces in opposite direction t Uniform velocity a = 0
F = 3 N F = 4 N F = 1 N u v u ≠ v F = ma
1 2
R
= a ≠ 0 ≠ 0
Uniform acceleration
F = F – F
R 1 2 1.2 Resolution of Forces
(c) Two perpendicular forces
1. A force can be resolved into two components, usually
vertical component, F and horizontal component, F
y x.
F sin θ
F
F R F
2 2
F
θ
F 1 θ

2
F = F + F 2 Angle with horizontal line F cos θ
R 1 2
2. On a smooth plane, where mass of object = m and R
(d) Two non-perpendicular forces
= normal reaction force.
R = mg cos θ
F
1
a = g sin θ
θ
Mass, m
F
2
F = F + F + 2F F cos q θ
2
2
2
R 1 2 1 2
1

Chp 1_1202 QB Physics F5.indd 1 10/01/2022 12:33 PM

1.3 Forces in Equilibrium
1.4 Elasticity
1. Forces are in equilibrium when the resultant force is
equal to zero. 1. Hooke’s law states that the extension of spring is
proportional to the applied force.
2. Methods of solving numerical problems for three
forces in equilibrium. (a) F ∝ x
(a) By resolution of forces F = kx
where k = the force constant of the spring
Gradient of graph = k
F sin α
F F F sin β 1 Vertically
2 1 2
β α F sin α + F sin β = F
Vertically 2 3
1
= ©PAN ASIA PUBLICATIONS
Horizontally
F cos β F cos α F sin a + F sin b = F 4
2 1 F cos α = F cos β 3
1
2
1
2
F F Horizontally 3
3 3 F cos a = F cos b
1 2 Force/ N 2
(b) By drawing a scaled triangle of force 1
F 0
2 0.1 0.2 0.3 0.4
α Extension/ m
F
3 (b) Gradient of X > Gradient of Y
Steeper slope ˜ Greater value of k ˜ Stiffer
F
1
spring.
β

F/ N
(i) Use suitable scale.
(ii) Measure the corresponding sides.
(iii) Calculate the required force using the scale
used. X
(c) If the triangle is right-angled, use simple
Y
trigonometry.
Example:
F F F F Example:
2 2 1 1 F F β β x/ cm
F = F sin α
1 1 F = F sin α
β β α α 1 1 3 3
F = F cos α
F = F cos α
2 2 3 3
2. Area under force-extension graph:
F F
3 3
F F
3 3
F F Force
2 2 α α
β β
F
3. Alternative method: 1
Area = – Fx
2
(a) Using sine rule (b) Using cosine rule
= Elastic potential
energy
sin a sin b sin c
2
2
2
F a –––– = –––– = –––– F = F + F – 2F F cos θ
F
2 F F 2 F 3 1 2 1 2
1 2 3
O x
c θ Extension
F F
3 3
Elastic potential energy = Area under the graph
F b F
1 1
1
E = Fx
P 2
sin a sin b sin c F = F + F – 2F F cos q 1
2
2
2
= 3 1 2 1 2 = (kx)(x)
F F F 2
1 2 3
1
= kx 2
2
2
Chp 1_1202 QB Physics F5.indd 2 10/01/2022 12:33 PM

PAPER 1

Answer all questions.

– 1
–3
–3
Take g = 10 N kg , density of Hg, r = 13 600 kg m and density of air, r = 1 000 kg m unless stated
otherwise.
2.1 Pressure in Liquid 4. Diagram 3 shows a tube filled with liquid P.
The density of the liquid is 1.2 g cm . The angle
-3
1. Diagram 1 shows the water supply system to a between the tube and the horizontal line is 50 .
o
SPM house.
©PAN ASIA PUBLICATIONS
CLONE
Liquid P
h
50° X

Diagram 3
X
Diagram 1 If the pressure at X due to liquid P is 7 200 Pa,
what is the length of the column of liquid in the
If the pressure meter reading at X is 6.2 × 10 5 tube?
Pa, what is the value of h? A 78 cm C 100 cm
A 6.2 m C 620 m B 90 cm D 110 cm
B 62 m D 6 200 m
5. Diagram 4 shows three containers with mercury,
2. Diagram 2 shows that the speed of water
SPM draining out can be increased by tilting the water and cooking oil. Liquid pressures at the
CLONE bases are P , P dan P respectively.
water container. X Y Z



Water
container Mercury Water Cooking
oil
Wooden
block
X Y Z
Diagram 4
Which comparison is correct?
Diagram 2
A P = P  P C P  P  P
Y
X
Z
X
Z
Which factor produces such an effect? B P  P  P D P  P  P Y
A The density of water X Y Z Z Y X
B Water depth 6. Diagram 5 shows a metal cylinder immersed in
C Gravity a liquid. The upper surface of the cylinder, X is
D Volume of water h m from the surface of the liquid.
3. Liquid pressure depends on
I the depth of the liquid Liquid
II the density of the liquid h
III volume of liquid Y surface
IV the acceleration due to gravity there Cylinder
A I and II C I, II and III
B II and IV D I, II and IV
Diagram 5
23
Question 4:
P = hrg
h = depth from the liquid surface SOS TIP







Chp 2_1202 QB Physics F5.indd 23 07/01/2022 12:08 PM

The pressure on surface X depends on the Which comparison is true?
following except A x < x ’
A the density of a liquid B x = x ’
B the acceleration due to gravity C x > x ’
C value of h
D area of X 10. Diagram 8 shows water ejected out at same
velocity at points normally perforated to the
7. The pressure due to a liquid at the bottom of surface of a can filled with water.
the container containing the liquid is 3 600 Pa.
What is the pressure when the container is on a
©PAN ASIA PUBLICATIONS
planet where the gravitational field strength is Hole at the
– I
5.6 N kg ? HOTS Applying same position
A 640 Pa C 2 000 Pa
B 1 800 Pa D 3 600 Pa

8. Diagram 6 shows two liquids, P and Q with
densities of 1 200 kg m and 1 600 kg m –3
–3
respectively. Diagram 8
The following statements are conclusions from
Liquid P 20 cm
observations except
A liquid pressure acts normally to the surface
X B liquid pressure at same level is equal
C liquid pressure acts in all directions
Liquid Q
D liquid pressure depends on shape of container

11. Diagram 9 shows a container with water.
Diagram 6
What is the depth in liquid Q where the pressure
is equal to the pressure at 20 cm below the Water
surface of liquid P?
A 12.0 cm C 18.0 cm
B 15.0 cm D 26.7 cm

9. Diagram 7 shows two holes, P and Q (same Diagram 9
level) of different sizes on the wall of a tall
cylinder. The water from the holes P and Q Which of the following is a correct vector
crosses a horizon distance of x (in the diagram) diagram of water pressure acting on the walls?
and x’ respectively. A C



Small hole Big hole
P Q

B D





x
Diagram 7
24 Question 7: Question 9:
SOS TIP P = hrg Liquid pressure does not depend on area.
P ∝ g







Chp 2_1202 QB Physics F5.indd 24 07/01/2022 12:08 PM

12. Diagram 10 shows a container with water. 17. The equipment commonly used to read altitude
above sea level is basically a
A Aneroid barometer
B Fortin barometer
C Bourdon gauge
D mercury manometer

18. Diagram 11 shows a tube with mercury at sea
X Y Z
level.
Diagram 10
Vacuum
©PAN ASIA PUBLICATIONS
Which comparison of water pressure at points
X, Y and Z is true?
A P  P  P
X Y Z
B P = P  P h mm
X Y Z
C P = P = P P atm
X Y Z
D P  P = P
X Y Z
Mercury
13. Dams are built with the bottom being thicker
than the top. What is the reason for doing so?
Diagram 11
A Water pressure increases with depth
B The deeper the water, the colder it is What is the value of h?
C The density of water is directly proportional A 760 mm
to the depth of water B 800 mm
D The size of water molecules increases with C 100 cm
the depth of water D 120 cm
14. What is the force due to water pressure on a fish 19. Diagram 12 shows two turtles.

with a body surface area of 360 cm at a depth
2
of 25 m below the surface of sea water where Big turtle
the density of seawater is 1 025 kg m .
–3
A 900 N C 9 000 N
B 8 600 N D 9 200 N
Small turtle
15. A cylinder with diameter D contains mercury.
The pressure due to mercury is P. If the mercury Diagram 12
is poured into another cylinder with diameter
2D, what is the new pressure P’ due to mercury? The small turtle is 10 m below the surface of the
HOTS Applying water. What is the ratio of the total pressure on
1 the large turtle to the small turtle?
A P C P A 1:1 C 2:1
4
1 1 B 1:2 D 4:1
B P D P
2 8
20. Consider the density of air is 1.29 kg m
–3
2.2 Atmospheric Pressure and does not change with height. What is the
difference in pressure between the ground and
16. Mercury is used in Fortin barometers because the top of a hill with height 300 m?
A it is opaque A 890 Pa
B it is shiny B 1 927 Pa
C it does not stick to the glass tube C 3 870 Pa
D it has very high density D 5 600 Pa

25
Question 12: Question 20:
The length of the arrow shows the magnitude of the pressure. ∆ = hr∆h
pressure
Question 15: SOS TIP
Base area ∝ diameter . 2





Chp 2_1202 QB Physics F5.indd 25 07/01/2022 12:08 PM

PAPER 2

Section A

Answer all questions.

1. Diagram 1 shows the thermal emission and production of cathode ray generation.
SPM
CLONE X Y


©PAN ASIA PUBLICATIONS
e
e

Discharge tube


– +
E.H.T.
Diagram 1

(a) What is meant by the term thermionic emission? [1 mark]




(b) State the name of X and Y. [2 marks]






(c) Why vacuum is created in discharge tube? [1 mark]



(d) State two factors which affects the rate of thermionic emission. [2 marks]







(e) Potential difference, V = 2 500 V is supplied between X and Y, HOTS Applying
(i) calculate the velocity of electron. [2 marks]










(ii) state any change in speed of electron if the potential difference between X and Y increased?
[1 mark]



92 Question 1:
SOS TIP (b) Using the concept of conservation of energy, the work done (W = eV) is equal to the increase in kinetic energy mv . )
(
1
2
2





Chp 5_1202 QB Physics F5.indd 92 07/01/2022 2:09 PM

Section B

Answer all questions.

6. (a) Diagram 6.1 shows the transistor symbol.


C

B


©PAN ASIA PUBLICATIONS
E
Diagram 6.1

(i) State the type of transistor above. [1 mark]
(ii) Write an equation to relate I , I and I . [1 mark]
B C E
(iii) Explain the principle of operation of transistor above using a simple circuit. [4 marks]
(b) Diagram 6.2 shows a transistor circuit for activating the fire alarm.




Diode Switch
Thermistor
Bell
100 kΩ 5V

Extra high
tension
1 kΩ R



Diagram 6.2
Explain the operation of the transistor circuit as an auto switch for a fire alarm system. [4 marks]

(c) Diagram 6.3 shows a circuit with transistor used as a sound amplifier.


R


S
Q
P




Diagram 6.3









97
Question 6:
(c) On all four electronic devices, the loudspeaker is an important electronic device for the sound amplifier. SOS TIP








Chp 5_1202 QB Physics F5.indd 97 07/01/2022 2:09 PM

Section C

Answer all questions.

7. Diagram 7.1(a) and 7.1(b) show the effects on cathode rays using a deflection tube.

Vacuum Vacuum
Cathode Anode Cathode Anode
Power Power
supply supply


©PAN ASIA PUBLICATIONS
0.5 kV 2.5 kV
Potential difference Potential difference
Spotted Spotted
light light

Diagram 7.1(a) Diagram 7.1(b)

(a) What is meant by thermionic emission? [1 mark]

(b) (i) Observe Diagram 7.1(a) and 7.1(b). Compare the size of the light spot seen on the screen with
the given potential difference. [2 marks]
(ii) State the relationship between size of the light spot and the potential difference to make a
deduction about the relationship the size of light spot and kinetic energy of cathode ray.
[2 marks]
(iii) State the physical quantity to be conserved in 7(b)(ii). [1 mark]

(c) Diagram 7.2 shows the cathode rays entering the region of the electric and magnetic field in the
deflection tube.
Electron
beam
+


N N
O
1 cm

–

Diagram 7.2
State and explain in which direction the cathode ray will be deflected. [4 marks]

(d) You are given a task to modify the design of cathode ray deflection tube so that it can increase the
rate of thermionic emission and brightness of light spot on the screen.
State and explain the modification based on the following aspects: [10 marks]
(i) Type of metal used as cathode
(ii) Type of the wire as a filament
(iii) Surface area of cathode
(iv) Potential difference charged in cathode gun
(v) Type of the screen


99
Question 7:
(b) (ii) Direction of magnetic force exerted on cathode rays can be determined by using Fleming left-hand rule. SOS TIP








Chp 5_1202 QB Physics F5.indd 99 07/01/2022 2:10 PM

Reinforcement and Assessment of Science Process Skills
for Paper 3 (Practical Test)


1. In this experiment, you will investigate the relationship between the base current, I with the collector
B
current, I for npn transistor.
C
(a) Set up the apparatus as shown in Diagram 1.

R = 1 kΩ
1
©PAN ASIA PUBLICATIONS
R = 1 kΩ A
3
2
I
R = 50 kΩ C
2
A 6 V
1
I
B
R = 0 – 1 kΩ
4

Diagram 1
(i) Close the switch and adjust the rheostat until the reading on ammeter A is 10 mA.
1
(ii) Record the reading on ammeter A in the table.
2
(iii) Repeat the above procedure by adjusting the rheostat until the readings on ammeter A are
2
20 mA, 30 mA 40 mA and 50 mA. [6 marks]

I / mA I / mA
B C













(b) State the variables for this experiment;
(i) Manipulated variables [1 mark]
(ii) Responding variable [1 mark]

(c) (i) Plot a graph of collector current, I (y-axis) against base current, I (x-axis).
C B
(ii) Calculate the gradient of the graph. [3 marks]
(d) (i) The equation relating the collector current, I to the base current, I is
C B
I = kI
C B
where k is a constant
Using answer from (c)(ii), calculate the value of the constant k.
(ii) Calculate the magnitude of the collector current, I when the base current, I is 85 mA.
C B
[3 marks]
100 Question 1:
SOS TIP (d) Linearise the equation and establish a relationship with y = mx + c.









Chp 5_1202 QB Physics F5.indd 100 07/01/2022 2:10 PM

(e) State one precaution that needs to be taken to improve the accuracy of the readings in this
experiment. [1 mark]
If you are unable to carry out the experiment described, you can answer this question by using the data
obtained as shown in Table 1.
Table 1
I /A I /A
B C


20 30 10 15
©PAN ASIA PUBLICATIONS
10 40 5 20
0 50 0 25
µA µA








20 30 10 15
10 40 5 20
0 50 0 25
µA µA







20 30 10 15
10 40 5 20
0 50 0 25
µA µA








20 30 10 15
10 40 5 20
0 50 0 25
µA µA







20 30 10 15
10 40 5 20
0 50 0 25
µA µA






101
Question 1:
(e) Avoid parallax errors when taking readings from measuring apparatus. SOS TIP








Chp 5_1202 QB Physics F5.indd 101 07/01/2022 2:10 PM

Answers







CHAPTER 1 50. A The addition of a load of 150 g causes a compression of
6 cm.
Paper 1 51. A
52. A Note the half -length spring can still only stand 10 kg (even
1. D
it is stiffer now).
2. C
53. C
3. C
4. B
5. C Paper 2
6. A
7. A Section A
8. A When the resultant force is zero, the acceleration is zero. 1. (a)
Velocity is zero or uniform.
9. D Reading = Weight + ma
10. C Upward resultant force = Balancing weight 6 cm R
120 o
11. A F = ma
R
500 – 120 = (60 + 10)a 8 cm
a = 5.4 m s –1
12. D (i) Use a scale of 1 cm: 10 N. Draw a parallelogram
13. D Diagonal length = 7.0 cm
Magnitude of resultant force = 7 µ 10 N
14. C Friction always opposes motion
15. B Apply formula: o = 70 N o
F = F + F + 2F + 2F F cos q Direction = 39 counter clock wise from 80 N or N51 E
2
2
2
p 1 2 1 1 2 (by measurement using protractor)
16. C F = ma
R (ii) F = 60 + 80 + 2(60)(80) cos 120°
2
2
2
F – 20 = 5(2) p
F = 30 N F = 72 N
p
(b) F-friction = ma
17. A
70 − 30 = 20a
18. B a = 2.0 m s –2 Chapter 1
C
19. D©PAN ASIA PUBLICATIONS
(c) Magnitude of opposite force = 40 N direction
20. D F = 30 – 12 sin 30°
o
g = S51 W
= 24 N
40 + 30 = 70 N
21. B F is the frictional force while R is the normal reaction force.
–2
f (d) The new acceleration exceeds 4.0 m s because the resultant
22. D force = 2(70) − 30
23. C
= 110 > 2 times the previous resultant force (40 N)
24. B
(e) Fʹ = 70 + 30
25. B Magnitude of resultant force is doubled and toward left. = 100 N
26. A Weight down the track = mg sin q because now the frictional force acts in the opposite direction.
27. D Moving along OX means the vertical forces are balanced.
2. (a) Weight = mg
28. A = 5 µ 10
29. B = 50 N
30. D (b) 50 – 30 = 20 N
31. A (c) (i) F = m µ a
system
32. A 20 = (50 + 30)a
33. B a = 0.25 m s –2
34. B Resolve the forces vertically. (ii) Consider 5 kg mass only
35. D What is the meaning of balanced forces? F = m µ a
36. A 50 – T = 5(0.25)
T
37. A T = 48.75 N
38. B 5 kg
39. B 50 N a = 0.25 m s -2
40.
41. C Weight down the track is balanced by friction.
42. B (d) (i) a’  a
43. B (ii) Because now, the tension in the rope is 50 N, exceeding
44. D the tension before.
45. B Hooke’s Law is true for compression and stretching. 3. (a) (i) When the resultant is nil.
46. D (ii) 10 N
47. A (b) F = ma
net
48. B 30 – 10 = 5a
49. B The elastic energy is directly proportional to the square of a = 4 m s –2
the elongation.
133
Answer_1202 Physics F5.indd 133 10/01/2022 12:50 PM

(c) F = ma 1
net (d) W = kʹk 2
30 – 10 – friction = ma 2
1
20 – 10 = 5a = (9 600)(0.15) 2
a = 2 m s 2
–2
= 108 J
(d) F = ma
net
q
2(30) cos – 10 = 0 Section B
2
q = 80.4° 6. (a) • The length of the board should be long so that the component
2 of weight down the board is less. The workers need less force
q = 160.8° to do their work.
(e) (i) Trolley will decelerate • The board should be strong and rigid so that it won’t be
(ii) Trolley will accelerate broken by the heavy load.
(f) Because the component to the right for both 30 N will • The surface of the board should be smooth in order to reduce
q the frictional force resisting the movement of the load.
decrease when angle increases (F cos will decrease when q
increase). 2 • Angle q = 0° so that the rope is parallel to the board to
ensure pulling force needed is minimal.
4. (a)
• The arrangement for Q is chosen because the board is long,
R rigid, smooth and the pulling force is parallel to the board.
(b) Let frictional force = F , F = ma
g net
5 – F = 2(1)
g
mg F = 3 N
g
Apply F = ma again
net
20 – 10 – 3 = 2a
(b) W = mg sin q a = 3.5 m s –2
x
(c) Resultant force
(c) (i)
(d) a = g sin q
Bouyant
a / m s -2 Dragging force, F
force,F V
B
9.8
W = mg
Chapter 1 (e) ©PAN ASIA PUBLICATIONS
(ii) Initially, the weight of the sphere exceeds the buoyant force
and dragging force. The sphere accelerates. However, as
q/°
the velocity of the sphere increases, the dragging force
90°
also increases until finally, the sum of the buoyant force
and dragging force is equal in magnitude to the weight of
u = 0
sphere moves with a uniform downward terminal velocity.
t = 1.2 s the sphere. Then and after, the forces are balanced and the
S h Section C
7. (a) (i) Stiffness is a measure of the resistance offered by an elastic
30 o body to deformation.
(ii) • The thickness of the coil winding for Y is more.
a = g sin q = 9.8 sin 30° • The elongation for Y is less.
= 4.9 m s –2 • The time for 20 oscillations for Y is less, thus the
1
s = ut + at 2 oscillation period of the load is less.
2 • The stiffer the spring the less the swing period.
1
h = 0 + (4.9)(1.2) 2
sin 30° 2 (b) • The spring constant should be high so that it is hard and can
h = 1.8 m support and withstand the weight of the child including a fat
one.
5. (a) Vector: force • The material for the spring and seat should be durable and
Scalar: extention, energy, work, length able to resist wear due to the weather so that it can last longer.
Direction is the one that differentiates vector from scalar. • The diameter of the spring should be large to add stability
(b) (i) Directly proportional and prevent the child from slipping down.
(ii) • The center of gravity of the spring rider is on the axis of the
x / m
spring to add stability while the boy uses it.
(a) • The seat base should be rough so that the child does not slip
while using it.
• The rider’s spring should be mounted on a coarse fabric
0.5
material to prevent injury in case the child falls.
(b)
(c)
F / N
1600
(c) F = kx
1 600 l o
k =
0.5
= 3 200 N m –2
kʹ = 3k x
= 3 µ 3 200
= 9 600 N m –2
134
Answer_1202 Physics F5.indd 134 10/01/2022 12:50 PM

(e) h r = h r • Mass is equal to the product of density and volume.
1 1 2 2
900 µ 1.2 = r µ 0.85 • Since the density of the oil is less, the volume to be
Y
r = 1.27 µ 10 kg m –3 displaced must be more. So, the wooden cube sinks more
3
Y
6. (a) (i) Vacuum (Torricelli) to do so.
(ii) Same
(b) h r = h r Paper 3
1 1 2 2
0.760 µ 13.6 = h µ 1
2
h = 10.3 m 1. (a) (i) Manipulated variable: Height of turpentine column, h
water 1
(c) h r = h Vr (ii) Responding variable: Height of water column, h 2
1 1 2 2
0.760 µ 13 600 = h µ 1.2
2 (b) h / cm h / cm
2
1
h = 8 613 m
air 3.1 2.6
(d) The mercury column on the right will drop less because the
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larger space there will produce less trapped air pressure than 6.0 5.2
the space on the left.
9.2 7.6
(e) 2 µ 13.6 cm = 27.2 cm
11.8 10.2
Section B
15.0 12.8

7. (a) • The cross-sectional area of the output piston should be large (c)
to produce a large output force.
• The boiling point of the liquid should be high so that it h / cm Graph of h against h 1
2
2
does not boil or evaporate easily. If bubbles are produced
in such a way in a liquid, the efficiency of the system will
be reduced. 12
• The compressibility of the fluid should be low so that no
work is used to compress it.
10
• The material for the transmission pipe should be steel so
that the corrosion rate is slow and the transmission pipe is
durable. 8
• System S was chosen because it has a large output piston,
high boiling point and hydraulic fluid of low compressibility,
6
and its transmission pipe is made of steel.
F A
(b) 1 = 1
F A 4
2 2
p 2
80 + W 4 (3.5d)
= p 2
6 + 2 d 2 Chapter 2 – Chapter 3
4
W = 18 N
0 h / cm
1
(c) (i) Same 2 4 6 8 10 12 14 16
(ii) The mercury column of the barometer Q rises higher. 11.0 – 3.4
The area for the container in Q is less, so the height of (d) k = 13.0 – 4.0
the water in the container is more for the same volume = 0.844
of water poured in. As pressure of liquid increases with
depth, the pressure exerted by the water on mercury in (e) The gradient remains the same
container Q is more and pushes the mercury column up
higher. CHAPTER 3
Section C
Paper 1
8. (a) (i) The pressure of a liquid acts in all directions, acts
normally to the surface and increases with the depth of the 1. C Based of meaning of electric field.
liquid. 2. D Direction of electric field must be outward.
(ii) h is more for Diagram 8.1. 3. D Magnitude of the electric field at P and Q is same. Since
The horizontal distance traversed by the water before it electric field is vector quantity, net electric field at X
touches the basin is further away for Diagram 8.1. become zero.
(iii) The deeper the water, the faster the water gushes out. F
Deduction: The water pressure increases with the depth of 4. B Based on definition, E = .
water. Q q
(b) • The shape of the boat should be streamlined to reduce the 5. C Based on definition, I = .
t
dragging force by water. 6. C Heavier positive ions are pulled towards the negative plate
• The overall density of the boat should be less and enable it slowly.
to accommodate more tourists.
• The bottom of the boat should be of glass material to 7. B Q = It
facilitate viewing of the coral reef. It is also better because = 0.85(8 × 60)
it can eliminate while viewing directly by the side of the = 408 C
boat reflection of sunlight. 8. C Work, W done to move one coulomb charge, Q between two
• The glass used must be of high quality so that it does not points in electric field.
break and is easily scratched. W
• The glass should be slightly convex to magnify the image 9. B Based on definition, V = .
Q
seen. 10. B W = QV
(c) • The weight of a liquid is equal to the weight of a floating V = 85 000 × 10 –3
cube. Thus, the weight or mass of water displaced is equal to 18
the weight or mass of oil displaced. = 4.72 V
137
Answer_1202 Physics F5.indd 137 10/01/2022 12:50 PM

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