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SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1
PHYSICS 8867/01
Preliminary Examination 21th September 2018 Paper 1 Multiple Choice 1 hour Additional Materials: OMS.
READ THIS INSTRUCTIONS FIRST
Write your name, civics group and index number in the spaces at the top of this page. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. There are thirty questions in this section. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the OMS. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate.
This document consist of 15 printed pages and 1 blank page
For Examiners’ Use
MCQ / 30
NAME
CG INDEX NO.
2
SRJC 2018 8867/PRELIM/2018
DATA AND FORMULAE
Data
speed of light in free space c = 3.00 x 108 m s−1
elementary charge e = 1.60 x 10−19 C
unified atomic mass constant u = 1.66 x 10−27 kg
rest mass of electron me = 9.11 x 10−31 kg
rest mass of proton mp = 1.67 x 10−27 kg
the Avogadro constant NA = 6.02 x 1023 mol−1
gravitational constant G = 6.67 x 10−11N m2 kg−2
acceleration of free fall g = 9.81 m s−2
Formulae
uniformly accelerated motion s = ut + at2
v2 = u2 + 2as
resistors in series R = R1 + R2 + …
resistors in parallel
3
SRJC 2018 8867/PRELIM/2018
Answer all questions
1 The rate of heat flow, R, can be found using the equation
2 1(T - T )R = k A
L
where k is the thermal conductivity, A is the total cross sectional area of the conducting surface, and L is the thickness of conducting surface separating the 2 temperatures T1 and T2.
What is the SI base units for k?
A kg m s-3 K-1 B kg m s-2 K-1 C kg m s-3 D kg s-2 K-1
2 Which pair includes a vector quantity and a scalar quantity?
A magnetic flux density; moment B magnetic flux density; work done C work done; current D work done; potential difference
3 A student carries out an experiment to determine the acceleration of free fall g. The following
table shows the results.
g / m s-2
6.91
6.89
6.90
6.87
6.92
6.88 What can be said about this experiment? A It is accurate but not precise. B It is both precise and accurate. C It is neither precise nor accurate. D It is precise but not accurate.
4
SRJC 2018 8867/PRELIM/2018
4 The graph below represents the variation with time t of the acceleration a of a car starting from rest.
What is the total displacement of the car from the starting point at the end of 5 s?
A - 4 m B 4 m C 8 m D 12 m
5 Two solid spheres form an isolated system. Sphere 1 approached Sphere 2 along the same line
with velocities u1 and u2 respectively. They collide and after the collision, the velocities of Sphere 1 and 2 are v1 and v2 respectively.
Which set of values shows that collision is not elastic?
u1 / m s–1 u2 / m s–1 v1 /m s–1 v2 /m s–1 A 3 -3 -1 5 B 3 2 2 4 C 5 -3 -3 5 D 5 0 0 5
a / m s-2
t / s
2
- 4
2 1 3 4 5
5
SRJC 2018 8867/PRELIM/2018
6 A rigid cross-shaped structure having four arms PO, SO, QO and RO, each 1.80 m long, is pivoted at O. Forces act on the ends of the arms and on the midpoints of the arms as shown.
What is the net moment about O?
A 54 N m B 108 N m C 216 N m D 432 N m
7 Forces 5 N, 12 N and 13 N act at a point which is in equilibrium. What is the angle between the 5 N and 13 N force? A 23° B 67° C 90° D 113° 8 A mass of 400 g hangs in equilibrium from a system of 2 springs as shown below. The springs
are identical and have a spring constant of 15 N m-1. The angle subtended by springs A and B is 70°.
What is the extension of spring B? A 0.13 m B 0.16 m C 0.23 m D 0.26 m
rods of the clamp
70°
spring A spring B
400 g mass
6
SRJC 2018 8867/PRELIM/2018
9 The tension in a sample of wire varies with extension as shown in the diagram below.
The graph shows that the wire undergoes two types of deformation as it is extended to
15.6 mm. In the region where Hooke’s law applies, the deformation is elastic and the wire will lose this deformation when the tension is released. In the region where Hooke’s law does not apply, the deformation is plastic and the wire retains this deformation when the tension is released.
Which area represents the elastic potential energy that is stored in the wire when it is extended to
15.6 mm?
A Area Z B Area X + Area Y C Area Y + Area Z D Area X + Area Y + Area Z
10 Water from a reservoir is fed to the turbine of a hydroelectric system at a rate of 500 kg s-1. The
reservoir is 300 m above the level of the turbine. The electrical output from the generator driven by the turbine is 200 A at a potential difference of
6000 V. What is the efficiency of the system? A 8.0 % B 8.2 % C 80 % D 82 %
X
Y
Z
7
SRJC 2018 8867/PRELIM/2018
11 A constant force of 9.0 kN, parallel to an inclined plane, moves a body of weight 20 kN through a distance of 40 m along the plane at constant speed. The body gains 12 m in height as shown in the diagram.
How much work done is dissipated as heat? A 120 kJ B 240 kJ C 360 kJ D 600 kJ 12 A driving force of 200 N is needed for a car of mass 800 kg to travel along a level road at a
constant speed of 20 m s-1. What power is required to maintain the car at this speed up a gradient in which the car rises 1 m
for each 8 m travelled along the road? A 6.0 kW B 7.2 kW C 20 kW D 24 kW
12 m
40 m
8
SRJC 2018 8867/PRELIM/2018
13 A vehicle starts from rest and accelerates uniformly. Which graph shows how the power output of the vehicle varies with the distance travelled? A B C D 14 Sand is sprinkled on a turntable on points X and Y. The turntable is rotating with increasing speed. Which one of the following comparing sand at points X and Y is true?
angular speed linear speed A X = Y X > Y B X = Y X = Y C X > Y X = Y D X > Y X > Y
power
distance
power
distance
power
distance
power
distance
X
Y
0
0 0
0
9
SRJC 2018 8867/PRELIM/2018
15 A car travels over a curved bump on the road at a speed of 20 m s-1.
What is the minimum radius of curvature of the hump if the car is not to lose contact with the road? A 2.0 m B 41 m C 400 m D 410 m 16 What is the mean speed and mean angular speed of a geostationary satellite of Mars? Mean period of rotation of Mars about its axis = 24.6 hours Radius of Mars = 3390 km Radius of orbit of the satellite = 20400 km
speed / m s-1 angular speed / rad s-1
A 250 7.1 x 10-5
B 1500 7.1 x 10-5 C 250 4.3 x 10-3 D 1500 4.3 x 10-3
17 A 2.0 V battery is connected to a 5.0 Ω and 3.0 Ω resistor as shown in the circuit below.
What is the number of electrons that pass through the 5.0 Ω resistor in 3.0 s? A 1.6 x 1018 B 4.7 x 1018 C 7.5 x 1018 D 3.0 x 1020
5.0 Ω
3.0 Ω
2.0 V
10
SRJC 2018 8867/PRELIM/2018
18 A current I flows through a composite wire PQ made up of three different materials. The materials all have the same lengths and cross sectional areas but with resistivities ρ, 2ρ and 3ρ, respectively.
Distance x is measured from P. Which graph shows how the potential V along the wire varies with distance x between P and Q? A B
C D
P Q
ρ 2ρ 3ρ I
11
SRJC 2018 8867/PRELIM/2018
19 Six resistors are connected in a circuit as shown below.
What is the effective resistance of the circuit between terminals AB? A 2.0 Ω B 2.8 Ω C 3.4 Ω D 5.3 Ω
20 Parallel conductors WXYZ, carrying equal currents, pass vertically through the four corners of a
square. In two conductors, the current is flowing into the page, and in the other two out of the page.
What are the directions of the current in order to produce a resultant magnetic field in the direction at O, the centre of the square?
into the page out of the page A W and X Y and Z B W and Y X and Z C X and Z W and Y D Y and Z W and X
2.0 Ω
4.0 ΩA
3.0 Ω
B
8.0 Ω
7.0 Ω
3.0 Ω
O
resultant magnetic field
W
X
Y
Z
12
SRJC 2018 8867/PRELIM/2018
21 Which diagram shows the electric field between a positively charged metal sphere and an earthed metal plate?
A B
C D
22 A uniform wire AB of length 0.70 m and weight 0.40 N is suspended horizontally in a magnetic field. It is positioned at an angle to the magnetic field of flux density 1.5 T as shown in the diagram below. What is the magnitude and direction of the current in the wire? magnitude direction A 0.47 A to B B 0.66 A to B C 0.47 B to A D 0.66 B to A
B
A 35°
top view
13
SRJC 2018 8867/PRELIM/2018
23 The figure below shows the top view of a current balance where the rigid rectangular wire loop ABQR pivoted at PS is in equilibrium. It is connected in series with an ideal 2.0 V battery and a 0.025 Ω resistor of a total mass of 300 g. Part of the wire loop is placed inside a solenoid. The mass of the loop can be taken to be negligible and the wire has no resistance.
The length of the side AB is 6.0 cm and = 2SR ASR
5.
What is the magnitude of the magnetic flux density in the solenoid? A 0.37 T B 0.41 T C 3.7 T D 4.1 T
24 Two current-carrying wires A and B are placed a distance apart. Wire A carries a current of
magnitude I that is directed upwards, whereas wire B carries a current of magnitude ½ I that is directed downwards.
Which of the following is true about the electromagnetic force on wire A?
A It is greater in magnitude than electromagnetic force on wire B and acts away from B. B It is equal in magnitude as the electromagnetic force on wire B and acts away from B. C It is greater in magnitude than electromagnetic force on wire B and acts towards B. D It is equal in magnitude as the electromagnetic force on wire B and acts towards B.
I ½ I
wire A wire B
top view
resistor
battery
P
S
B Q
RA
14
SRJC 2018 8867/PRELIM/2018
25 Which of the following is the most complete description of the model of an atom based on Rutherford’s alpha scattering experiment only?
A Electrons orbit around a positive nucleus. B Atom is mostly empty space but has a positively charged nucleus. C Atom has a very small charged nucleus and is mainly made up of empty space. D Atom has a massive charged nucleus and is mainly made up of empty space.
26 Which of the following shows a pair of unknown nuclides X and Y that are isotopes and the
correct total number of nucleons of X and Y together?
A 108 10846 45 and X Y , total number of nucleons = 216
B 108 10846 45 and X Y , total number of nucleons = 125
C 105 10645 45 and X Y , total number of nucleons = 211
D 105 10645 45 and X Y , total number of nucleons = 121
27 Which of the following is a correct description of mass defect?
A The difference between the mass of the nucleus of the products and reactants in a nuclear reaction.
B It is the difference between the total mass of the neutrons and the mass of the nucleus. C It is equal to the energy gained when individual nucleons comes together to form a nucleus. D It is the binding energy of a nucleus divided by square of the speed of light.
28 Which of the following is a correct description of the path of the particles as they pass through a
magnetic field at a velocity perpendicular to the magnetic field strength?
alpha particle beta particle gamma radiation
A deflected deflected in the same direction as alpha
particle
straight path
B deflected deflected in the opposite direction with
respect to alpha particle
deflected
C straight path straight path deflected
D deflected deflected in the opposite direction with
respect to alpha particle
straight path
15
SRJC 2018 8867/PRELIM/2018
29 The graph below shows the variation of count rate from a particular radioactive sample with time.
What does the jagged feature of the graph indicate? A It indicates the presence of background radiation. B It indicates that the decay obeys radioactive decay law. C It indicates the spontaneous nature of the radioactive decay. D It indicates the random nature of the radioactive decay.
30 A radioactive sample has a half-life of 12 hours. A counter records an average count-rate of
20 s-1 in the absence of this sample. As soon as the sample is placed in a position near the counter, the average count-rates rises to 100 s-1.
With the sample remaining in the same position, what is the average count-rate 30 hours later?
A 14 s-1 B 17 s-1 C 34 s-1 D 200 s-1
END OF PAPER
count rate
time
16
SRJC 2018 8867/PRELIM/2018
BLANK PAGE
SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1
PHYSICS 8867/01
Preliminary Examination 21th September 2018 Paper 1 Multiple Choice 1 hour Additional Materials: OMS.
READ THIS INSTRUCTIONS FIRST
Write your name, civics group and index number in the spaces at the top of this page. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. There are thirty questions in this section. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the OMS. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate.
This document consist of 15 printed pages and 1 blank page
For Examiners’ Use
MCQ / 30
NAME
CG SOLUTIONSINDEX NO.
2
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use DATA AND FORMULAE
Data
speed of light in free space c = 3.00 x 108 m s−1
elementary charge e = 1.60 x 10−19 C
unified atomic mass constant u = 1.66 x 10−27 kg
rest mass of electron me = 9.11 x 10−31 kg
rest mass of proton mp = 1.67 x 10−27 kg
the Avogadro constant NA = 6.02 x 1023 mol−1
gravitational constant G = 6.67 x 10−11N m2 kg−2
acceleration of free fall g = 9.81 m s−2
Formulae
uniformly accelerated motion s = ut + at2
v2 = u2 + 2as
resistors in series R = R1 + R2 + …
resistors in parallel
3
SRJC 2018 8867/PRELIM/2018
Answer all questions
1 The rate of heat flow, R, can be found using the equation
2 1(T - T )R = k A
L
where k is the thermal conductivity, A is the total cross sectional area of the conducting surface, and L is the thickness of conducting surface separating the 2 temperatures T1 and T2.
What is the SI base units for k?
A kg m s-3 K-1 B kg m s-2 K-1 C kg m s-3 D kg s-2 K-1
2 Which pair includes a vector quantity and a scalar quantity?
A magnetic flux density; moment B magnetic flux density; work done C work done; current D work done; potential difference
3 A student carries out an experiment to determine the acceleration of free fall g. The following
table shows the results.
g / m s-2
6.91
6.89
6.90
6.87
6.92
6.88
What can be said about this experiment? A It is accurate but not precise. B It is both precise and accurate. C It is neither precise nor accurate. D It is precise but not accurate.
Ans: B
Ans: D
Ans: A
2 1
2
1 1 1 1 1 3 1
Lk =
(T - T )
W m =
m K
= W m K J s m K kg m s K
R
A
− − − − − − −
= =
4
SRJC 2018 8867/PRELIM/2018
4 The graph below represents the variation with time t of the acceleration a of a car starting from rest.
What is the total displacement of the car from the starting point at the end of 5 s?
A -4 m B 4 m C 8 m D 12 m
5 Two solid spheres form an isolated system. Sphere 1 approached Sphere 2 along the same line
with velocities u1 and u2 respectively. They collide and after the collision, the velocities of Sphere 1 and 2 are v1 and v2 respectively.
Which set of values shows that collision is not elastic?
u1 / m s–1 u2 / m s–1 v1 /m s–1 v2 /m s–1 A 3 -3 -1 5 B 3 2 2 4 C 5 -3 -3 5 D 5 0 0 5
Ans: B
RSA = 3 – 2 = 1, RSS = 4 - 2 = 2
a / m s-2
t / s
2
- 4
2 1 3 4 5
Ans: C change in displacement = Area under the v-t graph = ½ (1+4)(4) – ½ (1)(4) = 8 m
v / m s-1
4
1 2 3 4 5 t / s 0
5
SRJC 2018 8867/PRELIM/2018
6 A rigid cross-shaped structure having four arms PO, SO, QO and RO, each 1.80 m long, is pivoted at O. Forces act on the ends of the arms and on the midpoints of the arms as shown.
What is the net moment about O?
A 54 N m B 108 N m C 216 N m D 432 N m
7 Forces 5 N, 12 N and 13 N act at a point which is in equilibrium. What is the angle between the 5 N and 13 N force? A 23° B 67° C 90° D 113°
Ans: C sacw = (20+30+30) x 0.9 + (70+70) x1.8 = 324 N m scw = (20+50+50) x 0.9 = 108 N m Net moments = 216 N m
Ans: D Angle = 180 - tan-1 (12/5) = 180 – 67 = 113
5 N
12 N
13 N
6
SRJC 2018 8867/PRELIM/2018
8 A mass of 400 g hangs in equilibrium from a system of 2 springs as shown below. The springs are identical and have a spring constant of 15 N m-1. The angle subtended by springs A and B is 70°.
What is the extension of spring B? A 0.13 m B 0.16 m C 0.23 m D 0.26 m
Ans: B Vertically, 2Tcos 35° = mg 2 (kx) cos 35° = 0.4 x 9.81 x = 0.16 m
rods of the clamp
70°
spring A spring B
400 g mass
7
SRJC 2018 8867/PRELIM/2018
9 The tension in a sample of wire varies with extension as shown in the diagram below.
The graph shows that the wire undergoes two types of deformation as it is extended to
15.6 mm. In the region where Hooke’s law applies, the deformation is elastic and the wire will lose this deformation when the tension is released. In the region where Hooke’s law does not apply, the deformation is plastic and the wire retains this deformation when the tension is released.
Which area represents the elastic potential energy that is stored in the wire when it is extended to
15.6 mm?
A Area Z B Area X + Area Y C Area Y + Area Z D Area X + Area Y + Area Z
10 Water from a reservoir is fed to the turbine of a hydroelectric system at a rate of 500 kg s-1. The
reservoir is 300 m above the level of the turbine. The electrical output from the generator driven by the turbine is 200 A at a potential difference of
6000 V. What is the efficiency of the system? A 8.0 % B 8.2 % C 80 % D 82 %
Ans : D Power input = rate of change of GPE = 500 x 9.81 x 300 = 1.47 x 106 W Power output = IV = 200 x 6000 = 1.2 x 106 W Therefore, efficiency = 1.2 / 1.47 = 82%
X
Y
Z
Ans : A
8
SRJC 2018 8867/PRELIM/2018
11 A constant force of 9.0 kN, parallel to an inclined plane, moves a body of weight 20 kN through a distance of 40 m along the plane at constant speed. The body gains 12 m in height as shown in the diagram.
How much work done is dissipated as heat? A 120 kJ B 240 kJ C 360 kJ D 600 kJ 12 A driving force of 200 N is needed for a car of mass 800 kg to travel along a level road at a
constant speed of 20 m s-1. What power is required to maintain the car at this speed up a gradient in which the car rises 1 m
for each 8 m travelled along the road? A 6.0 kW B 7.2 kW C 20 kW D 24 kW
Ans : A Work done by force = 9 x 40 = 360 kJ Gain in GPE = mgh = 20 x 12 = 240 kJ Therefore work done dissipated as heat = 360 – 240 = 120 kJ.
12 m
40 m
Ans : D Resistive force, Fr = 200 N Driving force required up a slope = mgsin θ + Fr = (800)(9.81)(1/8) + 200 = 1,181 N Therefore power required = Force x velocity = 1181 x 20 = 24 kW.
9
SRJC 2018 8867/PRELIM/2018
13 A vehicle starts from rest and accelerates uniformly. Which graph shows how the power output of the vehicle varies with the distance travelled? A B C D
power
distance
power
distance
power
distance
power
distance
Ans : D Power = Force x velocity Since v2 = u2 + 2as & u = 0, v = √ (2as) Therefore Power = Force x √ (2as) Since a constant, therefore force constant, P varies with distance s according to P α √s shape.
0
0 0
0
10
SRJC 2018 8867/PRELIM/2018
14 Sand is sprinkled on a turntable on points X and Y. The turntable is rotating with increasing speed. Which one of the following comparing sand at points X and Y is true?
angular speed linear speed A X = Y X > Y B X = Y X = Y C X > Y X = Y D X > Y X > Y
15 A car travels over a curved bump on the road at a speed of 20 m s-1.
What is the minimum radius of curvature of the hump if the car is not to lose contact with the road? A 2.0 m B 41 m C 400 m D 410 m
Ans: B By Newton’s 2nd Law, Fnet = mv2/r W – N = mv2/r N = W - mv2/r For car to not lose contact, N >= 0, Therefore, W - mv2/r >= 0 r >= v2/g = 202 / 9.81 = 41 m
X
Y
Ans: A ω on the same turntable is the same for both points. v = rω, since rx > ry, therefore vx > vy
11
SRJC 2018 8867/PRELIM/2018
16 What is the mean speed and mean angular speed of a geostationary satellite of Mars? Mean period of rotation of Mars about its axis = 24.6 hours Radius of Mars = 3390 km Radius of orbit of the satellite = 20400 km
speed / m s-1 angular speed / rad s-1
A 250 7.1 x 10-5
B 1500 7.1 x 10-5 C 250 4.3 x 10-3 D 1500 4.3 x 10-3
17 A 2.0 V battery is connected to a 5.0 Ω and 3.0 Ω resistor as shown in the circuit below.
What is the number of electrons that pass through the 5.0 Ω resistor in 3.0 s? A 1.6 x 1018 B 4.7 x 1018 C 7.5 x 1018 D 3.0 x 1020
5.0 Ω
3.0 Ω
2.0 V
Ans: B
1819
2.0 3.04.7 10
8.0 1.6 10
Q It Vtn
e e Re −
×= = = = = ×× ×
A: Use V/Re (1.6 x 1018) C: Never consider 3.0 Ω resistor (7.5 x 1018) D: Use VRt/e (3.0 x 1020)
Ans: B ω = 2π/T = 2π/(24.6 x 60 x 60) = 7.1 x 10-5
v = rω = 20400 x 103 x 7.1 x 10-5
= 1500 m s-1
12
SRJC 2018 8867/PRELIM/2018
18 A current I flows through a composite wire PQ made up of three different materials. The materials all have the same lengths and cross sectional areas but with resistivities ρ, 2ρ and 3ρ, respectively.
Distance x is measured from P. Which graph shows how the potential V along the wire varies with distance x between P and Q? A B
C D
Ans: A I x
V IR V xA
ρ ρ= = ∝ since I and A are the same.
Therefore the gradient of the graph gives the resistivity, where the higher the resistivity, the greater the gradient of the graph is expected to be.
P Q
ρ 2ρ 3ρ I
13
SRJC 2018 8867/PRELIM/2018
19 Six resistors are connected in a circuit as shown below.
What is the effective resistance of the circuit between terminals AB? A 2.0 Ω B 2.8 Ω C 3.4 Ω D 5.3 Ω
20 Parallel conductors WXYZ, carrying equal currents, pass vertically through the four corners of a
square. In two conductors, the current is flowing into the page, and in the other two out of the page.
What are the directions of the current in order to produce a resultant magnetic field in the direction at O, the centre of the square?
into the page out of the page A W and X Y and Z B W and Y X and Z C X and Z W and Y D Y and Z W and X
2.0 Ω
4.0 ΩA
3.0 Ω
B
8.0 Ω
7.0 Ω
3.0 Ω
Ans: C (2+7)//[3 + (3// (8+4))] = 3.4 Ω Distractors: A: (2+7)//3//(3+4+8) = 2.0 Ω B: (2+7)//(3+3)//(8+4) = 2.8 Ω D: (2+7)//[3+3//4+8] = 5.3 Ω
O
resultant magnetic field
W
X
Y
Z
Ans: D
14
SRJC 2018 8867/PRELIM/2018
21 Which diagram shows the electric field between a positively charged metal sphere and an earthed metal plate?
A B
C D
Ans: D Field lines should radiate out of positive charge, hence B and C are wrong. Field lines should be normal to any conducting surface, hence A and B are wrong. Since positive charge is at a higher potential than the earthed plate, electric field lines are directed from the charge to the plate.
15
SRJC 2018 8867/PRELIM/2018
22 A uniform wire AB of length 0.70 m and weight 0.40 N is suspended horizontally in a magnetic field. It is positioned at an angle to the magnetic field of flux density 1.5 T as shown in the diagram below. What is the magnitude and direction of the current in the wire? magnitude direction A 0.47 A to B B 0.66 A to B C 0.47 B to A D 0.66 B to A
Ans: A 0.40
sin 0.47sin 1.5 0.70sin55
FF BIL I
BLθ
θ= = = =
×A
Using FLHR, direction of current must be from A to B.
B
A 35°
top view
16
SRJC 2018 8867/PRELIM/2018
23 The figure below shows the top view of a current balance where the rigid rectangular wire loop ABQR pivoted at PS is in equilibrium. It is connected in series with an ideal 2.0 V battery and a 0.025 Ω resistor of a total mass of 300 g. Part of the wire loop is placed inside a solenoid. The mass of the loop can be taken to be negligible and the wire has no resistance.
The length of the side AB is 6.0 cm and = 2SR ASR
5.
What is the magnitude of the magnetic flux density in the solenoid? A 0.37 T B 0.41 T C 3.7 T D 4.1 T
Ans: B By principle of moments,
20.060 3 0.300 9.81 2
0.0250.41T
× = ×
× × × = × ×
=
BF AS mg SR
B L L
B Distractors: Never convert units – 4.1 T Use wrong distance – AS = 5L and SR = 3L – 0.37 T Use wrong distance and never convert units = 3.7 T
top view
resistor
battery
P
S
B Q
RA
17
SRJC 2018 8867/PRELIM/2018
24 Two current-carrying wires A and B are placed a distance apart. Wire A carries a current of magnitude I that is directed upwards, whereas wire B carries a current of magnitude ½ I that is directed downwards.
Which of the following is true about the electromagnetic force on wire A?
A It is greater in magnitude than electromagnetic force on wire B and acts away from B. B It is equal in magnitude as the electromagnetic force on wire B and acts away from B. C It is greater in magnitude than electromagnetic force on wire B and acts towards B. D It is equal in magnitude as the electromagnetic force on wire B and acts towards B.
25 Which of the following is the most complete description of the model of an atom based on
Rutherford’s alpha scattering experiment only?
A Electrons orbit around a positive nucleus. B Atom is mostly empty space but has a positively charged nucleus. C Atom has a very small charged nucleus and is mainly made up of empty space. D Atom has a massive charged nucleus and is mainly made up of empty space.
26 Which of the following shows a pair of unknown nuclides X and Y that are isotopes and the
correct total number of nucleons of X and Y together?
A 108 10846 45 and X Y , total number of nucleons = 216
B 108 10846 45 and X Y , total number of nucleons = 125
C 105 10645 45 and X Y , total number of nucleons = 211
D 105 10645 45 and X Y , total number of nucleons = 121
Ans: C
Ans: C A and B is wrong because they are not isotopes. D is wrong because it shows total number of neutrons.
I
Ans: B Force that wires exert on each other are repulsive, and they are of the same magnitude (action-reaction pairs), hence magnetic force is equal in magnitude and away from B.
½ I
wire A wire B
18
SRJC 2018 8867/PRELIM/2018
27 Which of the following is a correct description of mass defect?
A The difference between the mass of the nucleus of the products and reactants in a nuclear reaction.
B It is the difference between the total mass of the neutrons and the mass of the nucleus. C It is equal to the energy gained when individual nucleons comes together to form a nucleus. D It is the binding energy of a nucleus divided by square of the speed of light.
28 Which of the following is a correct description of the path of the particles as they pass through a
magnetic field at a velocity perpendicular to the magnetic field strength?
alpha particle beta particle gamma radiation
A deflected deflected in the same direction as alpha
particle
straight path
B deflected deflected in the opposite direction with
respect to alpha particle
deflected
C straight path straight path deflected
D deflected deflected in the opposite direction with
respect to alpha particle
straight path
Ans: D Since gamma radiation is chargeless, it will go pass undeflected. For alpha and beta particle, they are opposite charged and hence they will be deflected in opposite directions.
Ans: D
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SRJC 2018 8867/PRELIM/2018
29 The graph below shows the variation of count rate from a particular radioactive sample with time.
What does the jagged feature of the graph indicate? A It indicates the presence of background radiation. B It indicates that the decay obeys radioactive decay law. C It indicates the spontaneous nature of the radioactive decay. D It indicates the random nature of the radioactive decay.
30 A radioactive sample has a half-life of 12 hours. A counter records an average count-rate of
20 s-1 in the absence of this sample. As soon as the sample is placed in a position near the counter, the average count-rates rises to 100 s-1.
With the sample remaining in the same position, what is the average count-rate 30 hours later?
A 14 s-1 B 17 s-1 C 34 s-1 D 200 s-1
count rate
time
Ans: D
Ans: C 1/2
3012
1
12
1100 20 2
1414 20 34
tt
o
A
A
A
A
C s−
=
= − == + =
SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1
PHYSICS 8867/02
Preliminary Examination 14 September 2018 Paper 2 Structured Questions 2 hours Candidates answer on the Question Paper. No Additional Materials are required.
READ THIS INSTRUCTIONS FIRST
Write your name, civics group and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all questions. Section B Answer one question only. At the end of the examination, fasten all your work securely together. The number of marks is given in bracket [ ] at the end of each question or part question.
This document consists of 23 printed pages and 1 blank page.
For Examiners’ Use
Q1 / 7
Q2 / 12
Q3 / 10
Q4 / 15
Q5 / 16
Q6 / 20
Q7 / 20
Total marks
/ 80
NAME
CG INDEX NO.
2
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use DATA AND FORMULAE
Data
speed of light in free space, c = 3.00 × 108 m s−1
elementary charge, e = 1.60 × 10−19 C
unified atomic mass constant, u = 1.66 × 10−27 kg
rest mass of electron, me = 9.11 × 10−31 kg
rest mass of proton, mp = 1.67 × 10−27 kg
the Avogadro constant NA = 6.02 x 1023 mol−1
gravitational constant G = 6.67 x 10−11N m2 kg−2
acceleration of free fall, g = 9.81 m s−2
Formulae uniformly accelerated motion, s = ut + ½ at2 v2 = u2 + 2as resistors in series, R = R1 + R2 + … resistors in parallel, 1/R = 1/R1 + 1/R2 + …
3
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use Section A
Answer all the questions in this section in the spaces provided.
1 Information related to the Earth and the Moon is given below:
Radius of Earth3.7
Radius of Moon=
Mass of Earth
= 81Mass of Moon
The center-to-center distance of the Moon from the Earth is 3.84 x 108 m and the gravitational field strength due to the Earth at its surface is 9.8 N kg-1.
(a) Using these data, calculate the gravitational field strength due to the Moon at its surface.
gravitational field strength = …………………. N kg-1 [2]
(b) There is a point on the line between the Earth and the Moon at which their combined gravitational field strength is zero.
Calculate the distance between this point and the centre of the Earth.
distance = ……………………………… m [2]
4
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (c) The Moon orbits around the Earth with a period of 27.3 days.
(i) Calculate the angular speed of the Moon.
angular speed = ……………….. rad s-1 [1]
(ii) Calculate the mass of the Earth.
mass = ………………………………… kg [2]
5
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use 2 (a) By reference to energy transfers, distinguish between electromotive force (e.m.f.) and
potential difference (p.d.).
e.m.f. …………………………………………………………………………………………………
…...…………………………………………………………………………………………………… p.d. …...……………………………………………………………………………………………… ……………………………………………………………………………..……………………... [2]
(b) A circuit is set up as shown in Fig. 2.1.
Fig. 2.1
The source is found to provide 2.4 x 105 J of electrical energy to the 2000 Ω resistor and thermistor when a charge of 2.2 x 104 C passes through the ammeter. At room temperature, the thermistor has a resistance of 1800 Ω. (i) Sketch on Fig. 2.2 the variation with temperature θ of resistance R in a thermistor.
[1]
Fig. 2.2
(ii) For the thermistor at room temperature,
1. show that the e.m.f. of the source is 11 V. [1]
A
2000 Ω
E
0
R
θ0
6
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 2. determine the fraction of power dissipated in the thermistor.
fraction = ……………… [2]
(c) A uniform resistance wire PQ of length 1.2 m is subsequently connected across the resistor and thermistor, as shown in Fig. 2.3. A sensitive voltmeter is connected between point Y and a moveable contact M on the wire.
Fig. 2.3
(i) At room temperature, the contact M is moved along PQ until the voltmeter shows
zero reading. Calculate the length of wire between M and Q.
length of wire = ……………………. m [2]
P
V
E
2000 Ω
M
X Z
Q
Y
1.2 m
7
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use (ii) State and explain the effect, if any, on the length of the wire between M and Q for
the voltmeter to remain at zero deflection if each of the following changes takes place independently.
1. The thermistor is warmed slightly.
………………………………………………………………………………….………..
………………………………………………………………………………….………..
………………………………………………………………………………….………..
……………………………………………………………………………………….. [2]
2. A uniform wire with a bigger cross sectional area is used.
…………………………………………………………………………………………….
…………………………………………………………………………………………….
…………………………………………………………………………………………….
..……………………………………………………………………………………….. [2]
8
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 3 (a) Define the tesla.
……..…………………………………………………………………………………………………..
……..…………………………………………………………………………………………………..
……..…..………………………………………………………………………………………….. [2]
(b) A particle of mass m and charge q is moving with speed v normal to a magnetic field of flux density B. (i) Show that the particle will move in a circular path of radius r given by the expression
mvr
Bq=
[2]
(ii) An electron with a momentum of 1.8 x 10-23 kg m s-1 moves in a magnetic field of field strength 8.9 mT.
Calculate the radius of its path.
radius = …………………. m [1]
9
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use
(c) A uniform magnetic field is produced in the region ABCD, as shown in Fig. 3.1. Two particles of the same mass at point X undergo two different paths.
Fig. 3.1 (i) Suggest why the paths of the particles are different. ……………………………………………………………………………………………... [1] (ii) Suggest, with a reason, why both of their paths are a spiral, rather than an arc of a
circle.
…..….………………………………………………………………………………………….. ……....…………………………………………………………………………………………. ….………………………………………………………………………………………….. [2]
(iii) State and explain what can be deduced from the paths about the initial speeds of the two particles.
….………………………………………………………………..…………………………….. ….……………………………………………………………………..……………………….. ….…………………………………………………………………………..………………. [2]
uniform magnetic field
A B
C D
10
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 4 (a) An induced nuclear fission reaction maybe represented by the equation
235 1 141 92 192 0 56 36 03U n Ba kr n+ → + +
(i) Sketch the variation with nucleon number of the binding energy per nucleon. [2]
Fig. 4.1
(ii) Hence, explain why Uranium – 235 is more likely to undergo fission than fusion.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
………………………………………………………………………………………………… …………………………………………………………………………………………………
………………………………………………………………………………………………… ………………………………………………………………………………………………… …………………………………………………….………………………………………. [3]
nucleon number
binding energy per nucleon/ MeV
11
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use (iii) The mass of the nucleus of the various nuclides are as listed below:
nuclide mass / u
uranium - 235 235.04393 barium – 141 140.91441 krypton - 92 91.92616
neutron 1.00867
Determine the energy released in the reaction.
energy released = ……………………………. J [2]
(b) A stationary nucleus of a radioactive nuclide, 21884 Po , underwent a decay by the emission of
an α–particle. The decay is represented by the equation
21884 Po D α→ +
where D is the nuclide formed after the decay.
(i) State the nuclear notation for D. [1]
................D
12
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use
(ii) Determine the ratio of the kinetic energy of the α–particle to the total kinetic energy of D and α–particle.
ratio = ……………………… [3]
(iii) The number of particles in a 4.05 g of 21884 Po sample dropped by 59.4% after 243 s.
1. Determine the number of particle 21884 Po in the initial sample.
number of particles = ………………….. [1]
2. Determine the half life of 21884 Po .
half life = …………………. s [2]
13
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use
(iv) The count rate at t = 0 and after 1 half life 21884 Po of were measured. Suggest a
possible reason why the count rate is not exactly halved.
…………………………………………………………………….……………………………
……………………………………………………..……………………………………….. [1]
14
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 5 A wire-wound resistor is manufactured by winding resistance wire on an insulating former. A
commonly used material for the wire is an alloy of nickel and chromium called nichrome. The wire is produced by pulling the nichrome through a suitable sized hole. Nichrome is sufficiently ductile to be drawn into a wire without danger of it cracking or breaking after winding. It resists corrosion and has a fairly high resistivity. The wire itself must be uniform and thin, and is covered with an insulating material.
A manufacturer of resistors of this type supplies information concerning them in the form of a family of lines shown in the graph of Fig. 5.1. Resistors of different resistance R1, R2, ….. R5 are shown by the separate lines.
0.001
0.002
0.003
0.01
0.02
0.03
0.10
0.20
0.30
current I / A
1
1.00
3 30 10 300 100 1000
Fig. 5.1
potential difference V / V
1R
2R
3R
4R
5R
15
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use (a) Explain the meaning of resistivity.
……………………...…………………………………………………………………………………
………………………..…………………………………………………………………………… [1]
(b) Using your answer to (a), explain why the wire must be thin and the material must have a fairly high resistivity. ………………………………………………………………………………………………………... …………………………………………………………………………………………………….. [2]
(c) By choosing some values of potential difference and current from Fig. 5.1, complete the
table in Fig 5.2 showing the resistances of R1, R2, ….. R5. [2]
Resistor
resistance / Ω
R1
R2
R3
R4
R5
Fig. 5.2
(d) Draw two additional lines on Fig. 5.1: (i) One line for a resistance of 2000 Ω. (ii) One line for a resistance of 47 Ω. [3]
16
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (e) This particular set of resistors is manufactured so that the resistors can safely be used
with power dissipation of up to 1 W. Complete the following table in Fig. 5.3 to show the maximum safe current in the resistors for the potential difference given. [2]
potential difference / V maximum current / A
1000
100
10
1
Fig. 5.3
(f) Plot the points in (e) on Fig. 5.1. On the graph, shade the region of safe use for all these
resistors. [4]
(g) The lines on Fig. 5.1 represent ideal behaviour. Suggest, with a reason, how the line for a real resistor might differ from the ideal.
……………………………………………………………………………………………………….. ……………………………………………………………………………………………………….. ……………………………………………………………………………………………………. [2]
17
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use Section B
Answer one question from this section in the spaces provided.
6 (a) (i) Define acceleration.
……………….………………………………………………..…………………………… [1]
(ii) Hence, derive s =ut + ½ at2 for a motion of uniform acceleration a, displacement s
and initial velocity u in the time taken t. [2] (iii) A student conducts an experiment to determine the acceleration of free fall g. He
takes measurements of a steel ball as it falls from rest alongside a vertical scale marked in metres. In addition, he measured the time for the fall of the steel ball.
The following measurements are recorded: Time taken for ball to fall from start point to end point, t = (0.60 ± 0.02) s Reading at start point R1 = (0.040 ± 0.005) m Reading at end point R2 = (1.605 ± 0.005) m
Assuming negligible air resistance, determine g together with its associated uncertainty, ∆g.
g ± ∆g = ……………..………. m s-2 [4]
18
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (b) A raindrop falls vertically from rest. (i) Assume that air resistance is negligible.
On Fig. 6.1, sketch a graph to show the variation with time t of the velocity v of the raindrop for the first 1.0 s of the motion. [1]
Fig. 6.1
(ii) In practice, air resistance D on the raindrop is not negligible.
D is given by the expression
D = k v2
where k is a constant and v is the speed. 1. The raindrop has mass 1.38 × 10–5 kg and k is 2.76 × 10–6 N m-2 s2.
Calculate the terminal velocity of the raindrop.
terminal velocity = …………… m s-1 [2] 2. The raindrop reaches terminal velocity at t = 3.0 s.
On Fig. 6.1, sketch the variation with time t of velocity v for the raindrop. The sketch should include the first 5.0 seconds of the motion. [2]
v /m s-1
t / s
10.0
8.0
6.0
4.0
2.0
0
0 1.0 2.0 3.0 4.0 5.0
19
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use 3. Describe and explain the acceleration of the raindrop as it starts from rest and
reaches terminal velocity. …………………………………………………………………………………………... …………………………………………………………………………………………... …………………………………………………………………………………………... ……….…………………………………………………………………………………. ……………………………………………………………………………………..… [3]
(c) A raindrop falls on a roof and rebounds off with a velocity of 5.5 m s–1 at an angle 60o with respect to the horizontal as shown in Fig. 6.2.
Fig. 6.2
Assume air resistance is negligible. The maximum horizontal distance travelled by the raindrop is 3.8 m.
(i) Calculate the time taken for the raindrop to hit the ground.
time = …………………. s [1]
(ii) Determine the speed of the raindrop as it hits the ground.
speed = …………………...m s-1 [2]
ground
60o
5.5 m s–1
roof
20
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (iii) Discuss quantitatively whether the assumption that air resistance is negligible is
justified by considering the vertical component of the initial velocity of the raindrop. The raindrop has the same mass and dimension as in (b)(ii). ………………………………………………………………………………………………….. …………………………………………………………………………………………………. …………………………………………………………………………………………….…[2]
21
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use 7 (a) In a car test, a car with a dummy driver and passenger, moving at a speed of 6.9 m s-1,
collides head-on into a wall. The mass of the car is 1250 kg, the mass of the driver is 85 kg and the mass of the front passenger is 65 kg. The average deceleration of the car as it comes to a stop is 48 m s-2. Both passenger and driver have their seat belts tightly fastened.
(i) Define impulse.
…………………………………………………………………………………………………..
……………………………………………………………………………………….……... [1]
(ii) Calculate the magnitude of the average force exerted on the car and its occupants.
average force = ……………..………. N [1] (iii) Determine the magnitude of impulse of the car and its occupants.
impulse = ……………..………. N s [2]
(iv) Hence, calculate the time taken for the car to come to a stop.
time = ……………..………. s [2]
(v) Assuming that the average deceleration remains the same, state and explain how your answer in (iv) will change (if any) when the total mass of car and occupants has doubled.
………………………………………………………………………………………….……..
……………………………………………………………………………………….……. [2]
22
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use
(vi) For the collision between the car and the wall, state how the following laws apply:
1. Newton’s third law
……………….…………………………………………………………………………..
………………………………………………………………………………….……. [1] 2. The law of conservation of linear momentum
………………………………………………………………….………………………..
………………………………………………………………….………………………..
………………………………………………………………………………….……. [2]
(vii) Explain if the collision between the car and the wall is elastic.
…………………………………………………………………………………………………..
……………………………………………………………………………………….……... [2]
(b) A uniform rectangular card is suspended from a wooden rod. The card is held at one of its
end as shown in Fig. 7.1. The force by the hand on the card acts horizontally to the right.
Fig. 7.1 (i) On Fig. 7.1,
1. mark with an ‘X’ the position of the centre of gravity of the card. [1]
2. draw an arrow labelled with W to represent the weight of the card. [1]
wooden rod card
force by hand on card
23
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use (ii) State the conditions for the card to be in equilibrium.
………………………………………………………………………….……………………… ………………………………………………………………………….……………………… ………………………………………………………………………….………………………
……..…………………………………………………………………….………………… [2]
(iii) Draw an arrow labelled with R on Fig. 7.1 to represent the force exerted by the
wooden rod on the card. Show your construction clearly. [2]
(iv) The card is now released. It swings on the wooden rod and eventually comes to a rest.
By reference to the completed diagram in Fig. 7.1, describe the final position in
which the card comes to a rest.
……………………………………………………………………….…………………….….. ……………………………………………………………………….…………………….…..
…….……………………………………………………………………….…………….… [1]
END OF PAPER
24
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use BLANK PAGE
SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1
PHYSICS 8867/02
Preliminary Examination 14 September 2018 Paper 2 Structured Questions 2 hours Candidates answer on the Question Paper. No Additional Materials are required.
READ THIS INSTRUCTIONS FIRST
Write your name, civics group and index number in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all questions. Section B Answer one question only. At the end of the examination, fasten all your work securely together. The number of marks is given in bracket [ ] at the end of each question or part question.
This document consist of 22 printed pages and 2 blank pages.
For Examiners’ Use
Q1 / 7
Q2 / 12
Q3 / 10
Q4 / 15
Q5 / 16
Q6 / 20
Q7 / 20
Total marks
/ 80
NAME
CG INDEX NO. SOLUTIONS
2
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use DATA AND FORMULAE
Data
speed of light in free space, c = 3.00 × 108 m s−1
elementary charge, e = 1.60 × 10−19 C
unified atomic mass constant, u = 1.66 × 10−27 kg
rest mass of electron, me = 9.11 × 10−31 kg
rest mass of proton, mp = 1.67 × 10−27 kg
the Avogadro constant NA = 6.02 x 1023 mol−1
gravitational constant G = 6.67 x 10−11N m2 kg−2
acceleration of free fall, g = 9.81 m s−2
Formulae uniformly accelerated motion, s = ut + ½ at2 v2 = u2 + 2as resistors in series, R = R1 + R2 + … resistors in parallel, 1/R = 1/R1 + 1/R2 + …
3
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use Section A
Answer all the questions in this section in the spaces provided.
1 Information related to the Earth and the Moon is given below :
Radius of Earth3.7
Radius of Moon=
Mass of Earth
= 81Mass of Moon
The center-to-center distance of the Moon from the Earth is 3.84 x 108 m and the gravitational field strength due to the Earth at its surface is 9.8 N kg-1.
(a) Using these data, calculate the gravitational field strength due to the Moon at its surface.
gravitational field strength = …………………. N kg-1 [2]
(b) There is a point on the line between the Earth and the Moon at which their combined gravitational field strength is zero.
Calculate the distance between this point and the centre of the Earth.
distance = ……………………………… m [2]
Since g α M/r, = ( ) ( )2 [M1]
= (1/81)(3.7)2 = 0.169 Therefore, g due to Moon = 0.169 X 9.81 = 1.66 N kg-1 [A1]
=
81 = [M1]
r = 3.46 x 108 m [A1]
4
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (c) The Moon orbits around the Earth with a period of 27.3 days.
(i) Calculate the angular speed of the Moon.
angular speed = ……………….. rad s-1 [1]
(ii) Calculate the mass of the Earth.
mass = ………………………………… kg [2]
ω = 2π/T = 2π/(27.3 x 24 x 60 x 60) = 2.66 x 10-6 rad s-1 [1]
Gravitational acceleration provides centripetal acceleration. GM/r2 = rω2 M = r3ω2/G [M1] = (3.84 x 108)3 x (2.66 x 10-6)2 / 6.67 x 10−11 = 6.01 x 1024 [A1]
5
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use 2 (a) By reference to energy transfers, distinguish between electromotive force (e.m.f.) and
potential difference (p.d.).
e.m.f. …………………………………………………………………………………………………
………………………………………………………………………………………………………... p.d. …………………………………………………………………………………………………...
………………………………………………………………………………………………..…... [2]
(b) A circuit is set up as shown in Fig. 2.1.
Fig. 2.1
The source is found to provide 2.4 x 105 J of electrical energy to the 2000 Ω resistor and thermistor when a charge of 2.2 x 104 C passes through the ammeter. At room temperature, the thermistor has a resistance of 1800 Ω. (i) Sketch on Fig. 2.2 the variation with temperature θ of resistance R in a thermistor.
[1]
Fig. 2.2
(ii) For the thermistor at room temperature,
1. show that the e.m.f. of the source is 11 V. [1]
A
2000 Ω
5
4
2.4 1011
2.2 10
WE
Q
×= = =×
V [M1]
E
0
R
θ0
Emf: electrical energy converted from other forms of energy per unit charge delivered round a complete circuit. Pd: The potential difference between two points in a circuit is the amount of energy converted from electrical to other forms of energy per unit charge passing from one point to the other
6
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 2. determine the fraction of power dissipated in the thermistor.
fraction = ……………… [2]
(c) A uniform resistance wire PQ of length 1.2 m is subsequently connected across the resistor and thermistor, as shown in Fig. 2.3. A sensitive voltmeter is connected between point Y and a moveable contact M on the wire.
Fig. 2.3 (i) At room temperature, the contact M is moved along PQ until the voltmeter shows
zero reading. Calculate the length of wire between M and Q.
length of wire = ……………………. m [2]
fraction = 2
2 2
1800
1800 2000T T
T T
I R R
I R I R R R= =
+ + + [M1]
0.474= [A1]
Potential difference between MQ and YZ has to be the same for voltmeter to register zero reading.
1800
1.2 3800
L = [M1]
L = 0.568 m [A1]
P
V
E
2000 Ω
M
X Z
Q
Y
1.2 m
7
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use (ii) State and explain the effect, if any, on the length of the wire between M and Q for
the voltmeter to remain at zero deflection if each of the following changes takes place independently.
1. The thermistor is warmed slightly.
…………………………………………………………………………………………... ……………………………………………………………………………………….….
……………………………………………………………………………………….….
……………………………………………………………………………………….. [2]
2. A uniform wire with a bigger cross sectional area is used.
……………………………………………………………………………………………. …………………………………………………………………………………………….
…………………………………………………………………………………………….
……………………………………………………………………………………….... [2]
While the bigger cross-sectional area will result in a lower resistance in the wire PQ, as the potential difference across PQ did not change, the potential difference per unit length remains the same. Since potential difference across thermistor (and resistor) did not change, the length of wire will not need to change.
(As temperature rises), resistance of thermistor decreases and by Potential Divider Principle, the potential difference across the thermistor drops. For zero deflection, the potential difference across the thermistor and across MQ have to be the same. Since the pd across MQ is proportional to its length, MQ is shorter.
8
SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 3 (a) Define the tesla.
………………………………………………………………………………………………………... ……..…………………………………………………………………………………………………..
……..…………………………………………………………………………………………………..
……..…..………………………………………………………………………………………….. [2]
(b) A particle of mass m and charge q is moving with speed v normal to a magnetic field of flux density B. (i) Show that the particle will move in a circular path of radius r given by the expression
mvr
Bq=
[2]
(ii) An electron with a momentum of 1.8 x 10-23 kg m s-1 moves in a magnetic field of field strength 8.9 mT.
Calculate the radius of its path.
radius = …………………. m [1]
Electromagnetic force provides the centripetal force. [1] 2
Bqv=mv
r [1]
mvr
Bq =
23
3 19
1.8 100.0126
8.9 10 1.6 10
mv pr r
Bq Bq
−
− −
×= = = =× × ×
m [1]
If a straight conductor carrying a current of 1 ampere is placed at right angle to a uniform magnetic field of flux density 1 tesla, then the force per unit length on the conductor is 1 newton per metre.
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SRJC 2018 8867/PRELIM/2018 [Turn Over
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Use (c) A uniform magnetic field is produced in the region ABCD, as shown in Fig. 3.1. Two
particles of the same mass at point X undergo two different paths.
Fig. 3.1 (i) Suggest why the paths of the particles are different. ……………………………………………………………………………………………… [1] (ii) Suggest, with a reason, why both of their paths are a spiral, rather than an arc of a
circle.
….………………………………………………………………………………………………. ….………………………………………………………………………………………………. ….…………………………………………………………………………………………... [2]
(iii) State and explain what can be deduced from the paths about the initial speeds of the two particles.
….…………………………………………………………………………………………….... ….…………………………………………………………………………………………….... ….…………………………………………………………………………………………... [2]
uniform magnetic field
A B
C D
Both their momenta/velocities are decreasing [1]. Since r = mv/Bq, radius of both paths are decreasing [1], hence their paths are spiral.
The two paths have same initial radius suggesting that particles have the same initial speed [1] assuming that the magnitudes of their charge are the same. [1]
They have opposite charge. (Different charge not accepted)
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SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 4 (a) An induced nuclear fission reaction maybe represented by the equation
235 1 141 92 192 0 56 36 03U n Ba Kr n+ → + +
(i) Sketch the variation with nucleon number of the binding energy per nucleon. [2]
(ii) Hence, explain why Uranium – 235 is more likely to undergo fission than fusion.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………….………………………………………..[3]
1 mark for shape and label, 1 mark for values (8.8 MeV per nucleon)
Uranium-235 is on the right hand side of graph, beyond Fe-56. BE is the BE per nucleon multiply by the number of nucleons. [B1] Fission involve breaking a larger nucleus into smaller nuclei [B1] This will involve products with higher BE per nucleon, hence more stable products and more likely to occur. [B1] OR Fission implies increase of total BE and hence release of energy, hence more likely to occur. [B1]
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Use
(iii) The mass of the nucleus of the various nuclides are as listed below:
nuclide mass / u uranium - 235 235.04393 barium – 141 140.91441 krypton - 92 91.92616 neutron 1.00867
Determine the energy released in the reaction.
energy released = ……………………………. J [2]
(b) A stationary nucleus of a radioactive nuclide, 21884 Po , underwent a decay by the emission of
an α–particle. The decay is represented by the equation
21884 Po D α→ +
where D is the nuclide formed after the decay.
(i) State the nuclear notation for D. [1]
................D
2
2
11 ]
Energy released = mass of reactants - mass of products=(235.04393 1.0086 140.91441 91.92616 3(1.00867)] [M1]
0.186022.78 10 J
7) [
[A1
ucuc−
+ − + +== ×
21482 D
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SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (ii) Determine the ratio of the kinetic energy of the α–particle to the total kinetic energy
of D and α–particle.
ratio = ……………………… [3]
(iii) The number of particles in a 4.05 g of 21884 Po sample dropped by 59.4% after 243 s.
1. Determine the number of particle 21884 Po in the initial sample.
number of particles = ………………….. [1]
2. Determine the half life of 21884 Po .
half life = …………………. s [2]
By conservation of linear momentum,
2
2 2
0 [M1]
KE of -particletotal KE
[M1]
1
1 1
14
1 14 2140.982 [A1]
D
D
D
D
D
p pp p
pm
p pm m
m
m m
u
u u
α
α
α
α
α
α
α
α
α
= += −
=+
=+
=+
=
218 2284 218
4.05No. of particles N = 1.119 10 [A1]218
Pou
= ×
12
12
0
243
12
1( )2
10.406 ( ) [M1]2
187 s [A1]
tt
t
N N
t
=
=
=
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SRJC 2018 8867/PRELIM/2018 [Turn Over
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Use
(iv) The count rate at t = 0 and after 1 half life 21884 Po of were measured. Suggest a
possible reason why the count rate is not exactly halved.
………………………………………………………………………………….………………
…………………………………………………………………………………………….…[1]
1. Existence of background count. [B1] 2. Product Y is also giving off radiation that adds to count rate. 3. Random nature of radioactive decay.
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SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use 5 A wire-wound resistor is manufactured by winding resistance wire on an insulating former. A
commonly used material for the wire is an alloy of nickel and chromium called nichrome. The wire is produced by pulling the nichrome through a suitable sized hole. Nichrome is sufficiently ductile to be drawn into a wire without danger of it cracking or breaking after winding. It resists corrosion and has a fairly high resistivity. The wire itself must be uniform and thin, and is covered with an insulating material.
A manufacturer of resistors of this type supplies information concerning them in the form of a family of lines shown in the graph of Fig. 5.1. Resistors of different resistance R1, R2, ….. R5 are shown by the separate lines.
0.001
0.002
0.003
0.01
0.02
0.03
0.10
0.20
0.30
current I / A
1
1.00
3 30 10 300 100 1000
Fig. 5.1
potential difference V / V
1R
2R
3R
4R
5R
2000 Ω
47 Ω
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SRJC 2018 8867/PRELIM/2018 [Turn Over
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Use
(a) Explain the meaning of resistivity. ……………………………………………………………………………………………………………… ………………………………………………………………………………………………………….….[1]
(b) Using your answer to (a), explain why the wire must be thin and the material must have
a fairly high resistivity.
……………………………………………………………………………………………………………… ………………………………………………………………………………………………………….….[2]
(c) By choosing some values of potential difference and current from Fig. 5.1, complete the
table in Fig 5.2 showing the resistances of R1, R2, ….. R5. [2]
resistor resistance / Ω R1 R2 R3 R4 R5
Fig. 5.2
(d) Draw two additional lines on Fig. 5.1: (i) One line for a resistance of 2000 Ω. (ii) One line for a resistance of 47 Ω. [3]
A relationship between the dimensions of a specimen of a material (and its resistance that is constant at constant temperature.) Can define using equation as well but must define all the terms used and area must be defined as cross-sectional area.
With a fairly high resistivity and small cross-sectional area, there can be high resistance within a short length [M1] and thus save on the material used to make the resistor. [A1]
1
2
3
4
5
10 101100 100
11000 1000
11000 100000.1
10000 1000000.01
R
R
R
R
R
= = Ω
= = Ω
= = Ω
= = Ω
= = Ω
1 mark for 2000 Ω line 2 marks for 47 Ω line
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SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (e) This particular set of resistors is manufactured so that the resistors can safely be used
with power dissipation of up to 1 W. Complete the following table in Fig. 5.3 to show the maximum safe current in the resistors for the potential difference given. [2]
potential difference / V maximum current / A
1000 0.001
100 0.01
10 0.1
1 1
Fig. 5.3
(f) Plot the points in (e) on Fig. 5.1. On the graph, shade the region of safe use for all these resistors. [4]
(g) The lines on Fig. 5.1 represent ideal behaviour. Suggest, with a reason, how the line for a real resistor might differ from the ideal.
………………………...………………………………………………………………………………
…………………………………………………………..……………………………………………
………………………………………………………………………………………………….….[2]
All 4 plotted points correct, 3 marks. Every wrong plotted points minus 1 mark. 1 mark for correct shading.
In a real resistor, there is heating effect which causes the resistance to increase as V increases. [M1] So, I increases less than proportionately to V and therefore, the line should be a curve with decreasing gradient. [A1]
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SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use Section B
Answer one question from this section in the spaces provided.
6 (a) (i) Define acceleration. ……………………………………………………………………………………………….[1] (ii) Hence, derive s =ut + ½ at2 for a motion of uniform acceleration a, displacement s
and initial velocity u in the time taken t. [2] (iii) A student conducts an experiment to determine the acceleration of free fall g. He
takes measurements of a steel ball as it falls from rest alongside a vertical scale marked in metres. In addition, he measured the time for the fall of the steel ball.
The following measurements are recorded: Time taken for ball to fall from start point to end point, t = (0.60± 0.02) s Reading at start point R1 = (0.040 ± 0.005) m Reading at end point R2 = (1.605 ± 0.005) m
Assuming negligible air resistance, determine g together with its associated uncertainty, ∆g.
g ± ∆g = ……………..………. m s-2 [4]
s = R2 - R1 = 1.605 – 0.04 = 1.565 m s =1/2 gt2 g = 2s/ t2 = 8.6944 m s-2 [1] Δs = 0.005+0.005 = 0.01 [1]
0.02 0.01 [1]0.60 1.565
0.68.7 0.6 [1] for 1sf & place value
∆g ∆t ∆s= 2 +
g t s
∆g= 2
8.6944∆g
g ∆g
+
=± = ±
s = <v>t
s = 2+
v u t … (A)
a = −v ut
v = u + at … (B) [1] Substitute (B) into (A)
s = ( )
2 + +
u at ut [1]
s = 212
+ut at
Acceleration is the rate of change of velocity. [1]
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SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (b) A raindrop falls vertically from rest. (i) Assume that air resistance is negligible.
On Fig. 6.1, sketch a graph to show the variation with time t of the velocity v of the raindrop for the first 1.0 s of the motion. [1]
Fig. 6.1
(ii) In practice, air resistance D on the raindrop is not negligible.
D is given by the expression
D = k v2
where k is a constant and v is the speed. 1. The raindrop has mass 1.38 × 10–5 kg and k is 2.76 × 10–6 N m-2 s2.
Calculate the terminal velocity of the raindrop.
terminal velocity = …………… m s-1 [2] 2. The raindrop reaches terminal velocity at t = 3.0 s.
On Fig. 6.1, sketch the variation with time t of velocity v for the raindrop. The sketch should include the first 5.0 seconds of the motion. [2]
v /m s-1
t / s
10.0
8.0
6.0
4.0
2.0
0
0 1.0 2.0 3.0 4.0 5.0
Straight line passing through origin and (1, 9.8)
At terminal velocity, acceleration is zero. By Newton’s Law, Fnet = 0 Weight = D (+ upthrust) 1.38 × 10–5 x 9.81 = 2.76 × 10–6v2 [1] v = 7.0 m s–1 [1]
Same initial gradient [1] Curve passing through origin and tends towards a horizontal line and should reach terminal velocity at 3 seconds [1]
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Use 3. Describe and explain the acceleration of the raindrop as it starts from rest and
reaches terminal velocity. …………………………………………………………………………………………... …………………………………………………………………………………………... …………………………………………………………………………………………... ……….…………………………………………………………………………………. ……………………………………………………………………………………..…[3]
(c) A raindrop falls on a roof and rebounds off with a velocity of 5.5 m s–1 at an angle 60o with respect to the horizontal as shown in Fig. 6.2.
Fig. 6.2
Assume air resistance is negligible. The maximum horizontal distance travelled by the raindrop is 3.8 m.
(i) Calculate the time taken for the raindrop to hit the ground.
time = …………………. s [1]
(ii) Determine the speed of the raindrop as it hits the ground.
speed = …………………...m s-1 [2]
Sx = ux t 3.8 = (5.5 cos 60o) t t = 1.38s
vx = ux = 5.5 cos 60o = 2.75 m s-1 vy = (-5.5 sin 60o) + (9.81) (1.38) = 8.77 m s-1 [M1]
v = 2 2 12 9. .75 8 19.77 m s−=+ [A1]
ground
60o
5.5 m s–1
roof
initial acceleration is g /9.81 (m s )/acceleration of free fall as speed is zero since no drag force, only force acting on it is its weight [1] acceleration decreases as D increases, Fnet decreases [1] acceleration decreases to zero as D = W (do not accept Fnet =0) [1]
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SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (iii) Discuss quantitatively whether the assumption that air resistance is negligible is
justified by considering the vertical component of the initial velocity of the raindrop. The raindrop has the same mass and dimension as in (b)(ii). ………………………………………………………………………………………………….. …………………………………………………………………………………………………. …………………………………………………………………………………………….…[2]
Drag force = 2.76 × 10–6 (5.5 sin 60) 2 = 6.26 x10–5 N [1] this is 0.4 times of the weight of raindrop [1] so assumption is not justified
21
SRJC 2018 8867/PRELIM/2018 [Turn Over
For Examiner’s
Use 7 (a) In a car test, a car with a dummy driver and passenger, moving at a speed of 6.9 m s-1,
collides head-on into a wall. The mass of the car is 1250 kg, the mass of the driver is 85 kg and the mass of the front passenger is 65 kg. The average deceleration of the car as it comes to a stop is 48 m s-2. Both passenger and driver have their seat belts tightly fastened.
(i) Define impulse.
…………………………………………………………………………………………………..
……………………………………………………………………………………….……... [1]
(ii) Calculate the magnitude of the average force exerted on the car and its occupants.
average force = ……………..………. N [1] (iii) Determine the magnitude of impulse of the car and its occupants.
impulse = ……………..………. N s [2] (iv) Hence, calculate the time taken for the car to come to a stop.
time = ……………..………. s [2]
(v) Assuming that the average deceleration remains the same, state and explain how your answer in (iv) will change (if any) when the total mass of car and occupants has doubled.
………………………………………………………………………………………….……..
……………………………………………………………………………………….……. [2]
Impulse = change in momentum = mv – mu = 0 – (1250+85+65) (6.9) [1] = - 9660 Ns Ans: 9660 [1]
By Impulse - momentum theorem, Ft = mv – mu t = 9660/67200 [1] = 0.144 s [1] If use kinematics, zero marks.
No change [A1]. Since the change in velocity and the deceleration remains unchanged (a = ∆v/∆t) [M1] Or since the impulse (the change in momentum) and the average force both doubled, the time taken remains unchanged [M1]
Product of (average) force and the time that the (average) force acts on the body [1] (or time of impact)
average force = ma = (1250+85+65) (48) = 67200 N
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SRJC 2018 8867/PRELIM/2018
For Examiner’s
Use (vi) For the collision between the car and the wall, state how the following laws apply:
1. Newton’s third law
…………………………………………………………………………………….……..
……………………………………………………………………………….………. [1] 2. The law of conservation of linear momentum
…………………………………………………………………………………………..
…………………………………………………………………………………………..
………………………………………………………………………………….……. [2]
(vii) Explain if the collision between the car and the wall is elastic. ………………………………………………………………………………………………..
……………………………………………………………………………………….……. [2]
(b) A uniform rectangular card is suspended from a wooden rod. The card is held at one of its
end as shown in Fig. 7.1. The force by the hand on the card acts horizontally to the right.
Fig. 7.1 (i) On Fig. 7.1,
1. mark with an ‘X’ the position of the centre of gravity of the card. [1]
2. draw an arrow labelled with W to represent the weight of the card. [1]
Accept answers that off-centre (below and towards the right) [1]
arrow points downwards starting from X [1]
Inelastic [A1] as there is loss in KE of the system after the collision [M1]
Force acting on the car by wall is equal in magnitude but opposite direction as the force on the wall by the car
System consisting of the car and wall only. There are forces by the ground acting on the system. Since there are external forces acting on the system, momentum is not conversed. or System consisting of the car, wall and ground such that there is no net external force acting on the system, Change in momentum of the car = change in momentum of the wall and ground
wooden rod card
force by hand on card
W
R
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SRJC 2018 8867/PRELIM/2018 [Turn Over
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Use (ii) State the conditions for the card to be in equilibrium.
…………………………………………………………………………………………………. …………………………………………………………………………………………………. …………………………………………………………………………………………………. ………………………………………………………………………………………………[2]
(iii) Draw an arrow labelled with R on Fig. 7.1 to represent the force exerted by the
wooden rod on the card. Show your construction clearly. [2] (iv) The card is now released. It swings on the wooden rod and eventually comes to a
rest. By reference to the completed diagram in Fig. 7.1, describe the final position in
which the card comes to a rest.
……………………………………………………………………….…………………….….. ……………………………………………………………………….……………………..….
…….……………………………………………………………………….……………… [1]
END OF PAPER
Concurrent forces, 3 forces passes through same point. [1] Correct direction of arrow [1]
Position of X directly below point of contact of card with rod/ line of action of W passing through the point of contact of card with rod [1] Such that there is no resultant moment Correct diagrams are accepted as part of the working.
Net force is zero [B1] Net torque is zero [B1] or the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about that same point or the lines of action of the 3 forces (weight, force by hand on card and force by rod on card) passes through a common point
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For Examiner’s
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