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INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level
Higher 2
CANDIDATE NAME
CLASS INDEX NUMBER
PHYSICS Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet
9646/01
21 September 2015
1 hour 15 minutes
READ THESE INSTRUCTIONS FIRST
Write in soft pencil.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Write your name, civics group and index number on the Answer sheet in the spaces
provided unless this has been done for you.
There are forty questions on this paper. 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 separate
Answer Sheet.
Read the instructions on the Answer Sheet very carefully.
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 consists of 18 printed pages.
Innova Junior College [Turn over
© IJC 2015 9646/01/Prelim 2/15
2
Data speed of light in free space, c = 3.00 108 m s–1
permeability of free space, 0 = 4 10–7 H m–1
permittivity of free space, 0 = 8.85 10–12 F m–1
= (1/(36)) 10–9 F m–1
elementary charge, e = 1.60 10–19 C
the Planck constant, h = 6.63 10–34 J s
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
molar gas constant, R = 8.31 J K–1 mol–1
the Avogadro constant, NA = 6.02 1023 mol–1
the Boltzmann constant, k = 1.38 10–23 J K–1
gravitational constant, G = 6.67 10–11 N m2 kg–2
acceleration of free fall, g = 9.81 m s–2
Formulae
uniformly accelerated motion, s = ut + 2
1at2
v2 = u2 + 2as
work done on/by a gas, W = pV
hydrostatic pressure, p = g h
gravitational potential, =
r
Gm
displacement of particle in s.h.m. x = xo sin t
velocity of particle in s.h.m., v = vo cost
= )( 22 xxo
mean kinetic energy of a molecule of an ideal gas E = 2
3 kT
resistors in series, R = R1 + R2 + …
resistors in parallel,
R
1 = ...
11
21
RR
electric potential, V =
r
Q
04
alternating current/voltage, x = xo sin t
transmission coefficient T = exp(2kd)
where k = 2
2 )(8
h
EUm
radioactive decay, x = x0 exp(–t)
decay constant, =
2
1
6930
t
.
3
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
1 A student finds the density of a liquid by measuring its mass and its volume. The following is a summary of his measurements.
mass of empty beaker = (20 ± 1) g
mass of beaker + liquid = (70 ± 1) g
volume of liquid = (10.0 ± 0.6) cm-3 He correctly calculates the density of the liquid as 5.0 g cm-3. What is the uncertainty in this value?
A 0.4 g cm-3
B 0.5 g cm-3
C 0.6 g cm-3
D 2.6 g cm-3
2 Which statement using prefixes of the base unit metre (m) is not correct?
A 1 pm = 10-12 m
B 1 nm = 10-9 m
C 1 Mm = 106 m
D 1 Gm = 1012 m
3 An aeroplane flying in a straight line at a constant height of 500 m with a speed of 200 m s-1 drops an object. The object takes a time t to reach the ground and travels a horizontal distance d in doing so.
Ignoring air resistance, which one of the following gives the values of t and d?
t / s d / km
A 10 2
B 10 5
C 25 5
D 25 10
© IJC 2015 9646/01/Prelim 2/15
4
4 While watching the National Day Parade, David witnessed a parachutist jumped out of a plane. After some time, the parachutist pulled the cord to release his parachute. Applying what he had learnt from his recent Physics lesson, David tried to sketch the vertical acceleration – time graph of the parachutist. Which graph best represents the variation with time t, the vertical acceleration a during
the entire descent?
A B
C D
5 A ball falls vertically and bounces on the ground. The following statements are about the forces acting while the ball is in contact with the ground. Which statement is correct?
A The force that the ball exerts on the ground is always equal to the weight of the ball.
B The force that the ball exerts on the ground is always equal in magnitude and
opposite in direction to the force the ground exerts on the ball.
C The force that the ball exerts on the ground is always less than the weight of the
ball.
D The weight of the ball is always equal in magnitude and opposite in direction to the
force that the ground exerts on the ball.
5
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
6 A 4.0 kg block is pushed up a 36o incline by a force of magnitude F applied parallel to the incline. When F is 31 N, it is observed that the block moves up the incline with a
constant speed. Determine the magnitude of F that is required to lower the block down the incline at the same constant speed.
A 31 N B 17 N C 15 N D 13N
7 The diagrams show a metal cube suspended from a spring balance before and during immersion in water.
A reduction in the balance reading occurs as a consequence of the immersion. Which statement is correct?
A The balance reading will be further reduced if the cube is lowered further into the water.
B The balance reading during immersion corresponds to the upthrust of the water on the cube.
C The forces acting on the vertical sides of the cube contribute to the change in the balance reading.
D The gravitational pull on the cube is unchanged by the immersion.
© IJC 2015 9646/01/Prelim 2/15
6
8 A spanner is used to tighten a nut as shown.
A force F is applied at right-angles to the spanner at a distance of 0.25 m from the centre of the nut. When the nut is fully tightened, the applied force is 200 N. What is the resistive torque, in an anticlockwise direction, preventing further tightening?
A 8 N m B 42 N m C 50 N m D 1250 N m
9 A gas is enclosed inside a cylinder which is fitted with a frictionless piston.
Initially, the gas has a volume V1 and is in equilibrium with an external pressure p. The
gas is then heated slowly so that it expands, pushing the piston back until the volume of the gas has increased to V2.
How much work is done on the gas during this expansion?
A p(V2 – V1) B 2
1p(V2 – V1) C p(V1 – V2) D
2
1p(V1 – V2)
7
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
10 The diagram shows a wheel of circumference 0.30 m. A rope is fastened at one end to a force meter. The rope passes over the wheel and supports a freely hanging load of 100 N. The wheel is driven by an electric motor at a constant rate of 50 revolutions per second. When the wheel is turning at this rate, the force meter reads 20 N.
What is the output power of the motor?
A 0.3 kW B 1.2 kW C 1.8 kW D 3.8 kW
11 A street lamp is fixed to a wall by a metal rod and a cable.
Which vector triangle represents the forces acting at point P?
A B
C D
© IJC 2015 9646/01/Prelim 2/15
8
12 Alvis is sitting in a train carriage facing the direction in which the train is travelling. A pendulum hangs down in front of him from the carriage roof. The train travels along a circular bend, bending to his left. Which of the following diagrams shows the position of the pendulum as seen by Alvin and the direction of the forces acting on the pendulum bob?
A B C D
13 Two stars of mass M and 2M, a distance 3x apart, rotate in circles about their common centre of mass O.
The gravitational force acting on the stars can be written as 2
2
kGM
x.
What is the value of k?
A 0.22 B 0.50 C 0.67 D 2.0
14 A satellite orbits a planet at a distance r from its centre. Its gravitational potential energy
is –3.2 MJ. Another identical satellite orbits the planet at a distance 2r from its centre.
What is the sum of the kinetic energy and the gravitational potential energy of this second satellite?
A –0.40 MJ
B –0.80 MJ
C –1.6 MJ
D –6.4 MJ
9
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
15 The graph shows the variation of acceleration a with displacement x for an object undergoing simple harmonic motion.
The simple harmonic motion system is altered so that it has a period of oscillation twice that of before. Which line could be produced?
16 As the intensity of a single frequency sound wave travelling through the air is increased, how do the maximum speed of vibration of the air molecules and the speed of wave travel change?
maximum speed of vibration of air molecules
speed of wave travel
A increase increase
B increase no change
C no change increase
D no change no change
© IJC 2015 9646/01/Prelim 2/15
10
17 A car tyre, initially at o30 C , has been inflated to a pressure of 150 kPa as indicated by
the pressure gauge. This means that the pressure in the tyre is 150 kPa above atmospheric pressure of 100 kPa.
After driving on hot roads, the temperature of the air in the tyre is o60 C .
What is the percentage increase in the pressure gauge reading?
A 9.88% B 16.5% C 200% D 267%
18 An ideal gas is held in a constant volume container with a valve through which gas can escape. Starting at 250 K, the temperature of the gas is raised by 750 K, while its pressure in the container is allowed to triple. What fraction of the original number of gas molecules escape through the valve?
A 0 B 0.25 C 0.75 D 1.0
19 A light wave of amplitude A is incident normally on a surface of area S. The power per unit area reaching the surface is P.
The amplitude of the light wave is increased to 2A. The light is then focussed on to a
smaller area 3
S.
What is the power per unit area on this smaller area?
A 6 P
B 12 P
C 18 P
D 36 P
11
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
20 A stationary wave on a stretched string is set up between two points P and T.
Which statement about the wave is correct?
A Point R is a node.
B Point Q and S vibrate in phase.
C The distance between P and T is three wavelengths.
D The wave shown has the lowest possible frequency.
21 The basic principle of note production in a horn is to set up a stationary wave in an air column.
For any note produced by the horn, a node is formed at the mouthpiece and an antinode is formed at the bell. The frequency of the lowest note is 75 Hz. What are the frequencies of the next two higher notes for this air column?
first higher note / Hz second higher note / Hz
A 113 150
B 150 225
C 150 300
D 225 375
22 A parallel beam of white light passes through a diffraction grating. Orange light of wavelength 600 nm in the fourth order diffraction maximum coincides with blue light in the fifth order diffraction maximum. What is the wavelength of the blue light?
A 450 nm B 480 nm C 500 nm D 750 nm
© IJC 2015 9646/01/Prelim 2/15
12
23 The diagram shows two points P and Q which lie 90° apart on a circle of radius r. A positive point charge at the centre of the circle creates an electric field of magnitude E
at point P and Q.
Which expression gives the work done in moving a unit positive charge from P to Q?
A 0
B E × r
C E × (½πr)
D E × (πr)
24 Two oppositely-charged horizontal metal plates are placed in a vacuum. A positively-charged particle starts from rest and moves from one plate to the other plate, as shown
Which graph shows how the kinetic energy EK of the particle varies with the distance x moved from the positive plate?
A B C D
13
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
25 The belt in an electrostatic machine has width w and travels with velocity v. The amount
of charge per unit area on the surface of the belt is . As the belt passes a certain point, all the charge is removed and carried away as an electric current. Which of the following expressions gives the magnitude of this current?
A wv B wv2 C
w
v
D
v
w
26 A battery of e.m.f E and internal resistance r is connected to a variable resistor R as shown below. When R = 16 Ω, the current in the circuit is 0.50 A. It is found that the
battery supplies 4500 J of energy for a duration of 1.0 x 103 s.
What is the internal resistance r?
A 1.0 Ω B 2.0 Ω C 4.5 Ω D 9.0 Ω
27 When a battery is connected to a resistor, the battery gradually becomes warm. This causes the internal resistance of the battery to increase whilst its e.m.f. stays unchanged. As the internal resistance of the battery increases, how do the terminal potential difference and the output power change, if at all?
terminal potential
difference output power
A decrease decrease
B decrease unchanged
C unchanged decrease
D unchanged unchanged
r E
R
© IJC 2015 9646/01/Prelim 2/15
14
28 In the circuit shown, all the resistors are identical and all the ammeters have negligible resistance.
The reading on ammeter A1 is 0.6 A. What are the readings on the other ammeters?
reading on ammeter A2 / A
reading on ammeter A3 / A
reading on ammeter A4 / A
A 1.0 0.3 0.1
B 1.4 0.6 0.2
C 1.8 0.9 0.3
D 2.2 1.2 0.4
29 A beam of electrons is directed into an electric field and is deflected by it. Diagram 1 represents an electric field in the plane of the paper. Diagram 2 represents an electric field directed perpendicular to the plane of the paper. The lines A, B, C and D represent possible paths of the electron beam. All paths are in the plane of the paper. Which line best represents the path of the electrons inside the field?
15
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
30 A solenoid is connected in series with a battery and a switch as shown. A light, copper ring is held near to the solenoid and coaxial with it. What will happen to the copper ring immediately after the switch is closed?
A The ring remains stationary.
B The ring swings directly away from the solenoid.
C The ring swings directly towards the solenoid.
D The ring rotates about its vertical axis.
31 A soft-iron ring of variable cross-section has four coils wound round it at the positions shown. The coils have 2, 3, 3 and 4 turns. The 3-turn coil is connected to an a.c. supply. In which coil does the magnitude of the magnetic flux density have the largest variation?
copper ring solenoid
© IJC 2015 9646/01/Prelim 2/15
16
32 In the potentiometer setup below, the driver cell and the secondary cell are identical, with an e.m.f. of E and internal resistance r. The fixed resistor in the secondary circuit has a resistance R.
Which of the following graphs shows the variation of the balance length when fixed resistors of different values of resistance R is connected?
A B
C D
33 A 100 resistor conducts a current with changing direction and magnitude, as shown.
What is the mean power dissipated in the resistor?
A 100 W B 150 W C 250 W D 450 W
time / s
current / A
17
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
34 A generator produces a r.m.s current of 60 A at a r.m.s voltage of 120 V. The r.m.s voltage is stepped up to 4500 V by an ideal transformer and transmitted through a power line of total resistance 1.0 Ω. What is the percentage of power lost in the transmission line?
A 0.012 % B 0.036 % C 0.048 % D 0.060 %
35 An electron with kinetic energy E has a de Broglie wavelength of . Which of the
following graphs correctly represents the relationship between and E?
A B C D
36 Some of the energy levels of the hydrogen atom are shown below.
Electrons are excited to the 0.85 eV level. How many difference photon frequencies in the visible region will be observed in the emission spectrum of hydrogen?
A 1 B 2 C 3 D 6
37 Which of the following statements about lasers is incorrect?
A Excitation of the active medium in a laser can be achieved by absorption of light called optical pumping.
B Laser action cannot take place unless an inverted population of atoms is obtained in the active medium.
C Pumping results in electrons excited from so-called metastable states to higher energy states.
D Stimulated emission of radiation occurs when the frequency of the incident photon equals the resonant frequency of the transition between energy levels.
© IJC 2015 9646/01/Prelim 2/15
18
38 In the circuit involving a P-N junction below, holes migrate from
A X to Y through the cell.
B Y to X through the cell.
C P to N through the PN junction.
D N to P through the PN junction.
39 A 238
92Unucleus decays in two stages to a 234
91Pa nucleus.
What was emitted in these two stages?
A + B + C + D +
40 The tritium content of water from certain deep wells is only 25% of that in the water from recent rains. The half-life of tritium is 12.3 years. How long has it been since the water in the well came down as rain?
A 3.1 yrs B 6.2 yrs C 24.6 yrs D 36.9 yrs
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level
Higher 2
CANDIDATE NAME
CLASS INDEX NUMBER
PHYSICS Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet
9646/01
21 September 2015
1 hour 15 minutes
READ THESE INSTRUCTIONS FIRST
Write in soft pencil.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Write your name, civics group and index number on the Answer sheet in the spaces
provided unless this has been done for you.
There are forty questions on this paper. 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 separate
Answer Sheet.
Read the instructions on the Answer Sheet very carefully.
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 consists of 18 printed pages.
Innova Junior College [Turn over
© IJC 2015 9646/01/Prelim 2/15
2
Data speed of light in free space, c = 3.00 108 m s–1
permeability of free space, 0 = 4 10–7 H m–1
permittivity of free space, 0 = 8.85 10–12 F m–1
= (1/(36)) 10–9 F m–1
elementary charge, e = 1.60 10–19 C
the Planck constant, h = 6.63 10–34 J s
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
molar gas constant, R = 8.31 J K–1 mol–1
the Avogadro constant, NA = 6.02 1023 mol–1
the Boltzmann constant, k = 1.38 10–23 J K–1
gravitational constant, G = 6.67 10–11 N m2 kg–2
acceleration of free fall, g = 9.81 m s–2
Formulae
uniformly accelerated motion, s = ut + 2
1at2
v2 = u2 + 2as
work done on/by a gas, W = pV
hydrostatic pressure, p = g h
gravitational potential, =
r
Gm
displacement of particle in s.h.m. x = xo sin t
velocity of particle in s.h.m., v = vo cost
= )( 22 xxo
mean kinetic energy of a molecule of an ideal gas E = 2
3 kT
resistors in series, R = R1 + R2 + …
resistors in parallel,
R
1 = ...
11
21
RR
electric potential, V =
r
Q
04
alternating current/voltage, x = xo sin t
transmission coefficient T = exp(2kd)
where k = 2
2 )(8
h
EUm
radioactive decay, x = x0 exp(–t)
decay constant, =
2
1
6930
t
.
3
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
1 A student finds the density of a liquid by measuring its mass and its volume. The following is a summary of his measurements.
mass of empty beaker = (20 ± 1) g
mass of beaker + liquid = (70 ± 1) g
volume of liquid = (10.0 ± 0.6) cm-3 He correctly calculates the density of the liquid as 5.0 g cm-3. What is the uncertainty in this value?
A 0.4 g cm-3
B 0.5 g cm-3
C 0.6 g cm-3
D 2.6 g cm-3
2 Which statement using prefixes of the base unit metre (m) is not correct?
A 1 pm = 10-12 m
B 1 nm = 10-9 m
C 1 Mm = 106 m
D 1 Gm = 1012 m
3 An aeroplane flying in a straight line at a constant height of 500 m with a speed of 200 m s-1 drops an object. The object takes a time t to reach the ground and travels a horizontal distance d in doing so.
Ignoring air resistance, which one of the following gives the values of t and d?
t / s d / km
A 10 2
B 10 5
C 25 5
D 25 10
© IJC 2015 9646/01/Prelim 2/15
4
4 While watching the National Day Parade, David witnessed a parachutist jump out of a plane. After some time, the parachutist pulled the cord to release his parachute. Applying what he had learnt from his recent Physics lesson, David tried to sketch the vertical acceleration – time graph of the parachutist. Which graph best represents the variation with time t, the vertical acceleration a during
the entire descent?
A B
C D
5 A ball falls vertically and bounces on the ground. The following statements are about the forces acting while the ball is in contact with the ground. Which statement is correct?
A The force that the ball exerts on the ground is always equal to the weight of the ball.
B The force that the ball exerts on the ground is always equal in magnitude and
opposite in direction to the force the ground exerts on the ball.
C The force that the ball exerts on the ground is always less than the weight of the
ball.
D The weight of the ball is always equal in magnitude and opposite in direction to the
force that the ground exerts on the ball.
5
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
6 A 4.0 kg block is pushed up a 36o incline by a force of magnitude F applied parallel to the incline. When F is 31 N, it is observed that the block moves up the incline with a
constant speed. Determine the magnitude of F that is required to lower the block down the incline at the same constant speed.
A 31 N B 17 N C 15 N D 13N
7 The diagrams show a metal cube suspended from a spring balance before and during immersion in water.
A reduction in the balance reading occurs as a consequence of the immersion. Which statement is correct?
A The balance reading will be further reduced if the cube is lowered further into the water.
B The balance reading during immersion corresponds to the upthrust of the water on the cube.
C The forces acting on the vertical sides of the cube contribute to the change in the balance reading.
D The gravitational pull on the cube is unchanged by the immersion.
© IJC 2015 9646/01/Prelim 2/15
6
8 A spanner is used to tighten a nut as shown.
A force F is applied at right-angles to the spanner at a distance of 0.25 m from the centre of the nut. When the nut is fully tightened, the applied force is 200 N. What is the resistive torque, in an anticlockwise direction, preventing further tightening?
A 8 N m B 42 N m C 50 N m D 1250 N m
9 A gas is enclosed inside a cylinder which is fitted with a frictionless piston.
Initially, the gas has a volume V1 and is in equilibrium with an external pressure p. The
gas is then heated slowly so that it expands, pushing the piston back until the volume of the gas has increased to V2.
How much work is done on the gas during this expansion?
A p(V2 – V1) B 2
1p(V2 – V1) C p(V1 – V2) D
2
1p(V1 – V2)
7
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
10 The diagram shows a wheel of circumference 0.30 m. A rope is fastened at one end to a force meter. The rope passes over the wheel and supports a freely hanging load of 100 N. The wheel is driven by an electric motor at a constant rate of 50 revolutions per second. When the wheel is turning at this rate, the force meter reads 20 N.
What is the output power of the motor?
A 0.3 kW B 1.2 kW C 1.8 kW D 3.8 kW
11 A street lamp is fixed to a wall by a metal rod and a cable.
Which vector triangle represents the forces acting at point P?
A B
C D
© IJC 2015 9646/01/Prelim 2/15
8
12 Alvis is sitting in a train carriage facing the direction in which the train is travelling. A pendulum hangs down in front of him from the carriage roof. The train travels along a circular bend, bending to his left. Which of the following diagrams shows the position of the pendulum as seen by Alvin and the direction of the forces acting on the pendulum bob?
A B C D
13 Two stars of mass M and 2M, a distance 3x apart, rotate in circles about their common centre of mass O.
The gravitational force acting on the stars can be written as 2
2
kGM
x.
What is the value of k?
A 0.22 B 0.50 C 0.67 D 2.0
14 A satellite orbits a planet at a distance r from its centre. Its gravitational potential energy
is –3.2 MJ. Another identical satellite orbits the planet at a distance 2r from its centre.
What is the sum of the kinetic energy and the gravitational potential energy of this second satellite?
A –0.40 MJ
B –0.80 MJ
C –1.6 MJ
D –6.4 MJ
9
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
15 The graph shows the variation of acceleration a with displacement x for an object undergoing simple harmonic motion.
The simple harmonic motion system is altered so that it has a period of oscillation twice that of before. Which line could be produced?
16 As the intensity of a single frequency sound wave travelling through the air is increased, how do the maximum speed of vibration of the air molecules and the speed of wave travel change?
maximum speed of vibration of air molecules
speed of wave travel
A increase increase
B increase no change
C no change increase
D no change no change
© IJC 2015 9646/01/Prelim 2/15
10
17 A car tyre, initially at o30 C , has been inflated to a pressure of 150 kPa as indicated by
the pressure gauge. This means that the pressure in the tyre is 150 kPa above atmospheric pressure of 100 kPa.
After driving on hot roads, the temperature of the air in the tyre is o60 C .
What is the percentage increase in the pressure gauge reading?
A 9.88% B 16.5% C 200% D 267%
18 An ideal gas is held in a constant volume container with a valve through which gas can escape. Starting at 250 K, the temperature of the gas is raised by 750 K, while its pressure in the container is allowed to triple. What fraction of the original number of gas molecules escape through the valve?
A 0 B 0.25 C 0.75 D 1.0
19 A light wave of amplitude A is incident normally on a surface of area S. The power per unit area reaching the surface is P.
The amplitude of the light wave is increased to 2A. The light is then focussed on to a
smaller area 3
S.
What is the power per unit area on this smaller area?
A 6 P
B 12 P
C 18 P
D 36 P
11
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
20 A stationary wave on a stretched string is set up between two points P and T.
Which statement about the wave is correct?
A Point R is a node.
B Point Q and S vibrate in phase.
C The distance between P and T is three wavelengths.
D The wave shown has the lowest possible frequency.
21 The basic principle of note production in a horn is to set up a stationary wave in an air column.
For any note produced by the horn, a node is formed at the mouthpiece and an antinode is formed at the bell. The frequency of the lowest note is 75 Hz. What are the frequencies of the next two higher notes for this air column?
first higher note / Hz second higher note / Hz
A 113 150
B 150 225
C 150 300
D 225 375
22 A parallel beam of white light passes through a diffraction grating. Orange light of wavelength 600 nm in the fourth order diffraction maximum coincides with blue light in the fifth order diffraction maximum. What is the wavelength of the blue light?
A 450 nm B 480 nm C 500 nm D 750 nm
© IJC 2015 9646/01/Prelim 2/15
12
23 The diagram shows two points P and Q which lie 90° apart on a circle of radius r. A positive point charge at the centre of the circle creates an electric field of magnitude E
at point P and Q.
Which expression gives the work done in moving a unit positive charge from P to Q?
A 0
B E × r
C E × (½πr)
D E × (πr)
24 Two oppositely-charged horizontal metal plates are placed in a vacuum. A positively-charged particle starts from rest and moves from one plate to the other plate, as shown
Which graph shows how the kinetic energy EK of the particle varies with the distance x moved from the positive plate?
A B C D
13
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
25 The belt in an electrostatic machine has width w and travels with velocity v. The amount
of charge per unit area on the surface of the belt is . As the belt passes a certain point, all the charge is removed and carried away as an electric current. Which of the following expressions gives the magnitude of this current?
A wv B wv2 C
w
v
D
v
w
26 A battery of e.m.f E and internal resistance r is connected to a variable resistor R as shown below. When R = 16 Ω, the current in the circuit is 0.50 A. It is found that the battery supplies 4500 J of energy for a duration of 1.0 x 103 s.
What is the internal resistance r?
A 1.0 Ω B 2.0 Ω C 4.5 Ω D 9.0 Ω
27 When a battery is connected to a resistor, the battery gradually becomes warm. This causes the internal resistance of the battery to increase whilst its e.m.f. stays unchanged. As the internal resistance of the battery increases, how do the terminal potential difference and the output power change, if at all?
terminal potential
difference output power
A decrease decrease
B decrease unchanged
C unchanged decrease
D unchanged unchanged
r E
R
© IJC 2015 9646/01/Prelim 2/15
14
28 In the circuit shown, all the resistors are identical and all the ammeters have negligible resistance.
The reading on ammeter A1 is 0.6 A. What are the readings on the other ammeters?
reading on ammeter A2 / A
reading on ammeter A3 / A
reading on ammeter A4 / A
A 1.0 0.3 0.1
B 1.4 0.6 0.2
C 1.8 0.9 0.3
D 2.2 1.2 0.4
29 A beam of electrons is directed into an electric field and is deflected by it. Diagram 1 represents an electric field in the plane of the paper. Diagram 2 represents an electric field directed perpendicular to the plane of the paper. The lines A, B, C and D represent possible paths of the electron beam. All paths are in the plane of the paper. Which line best represents the path of the electrons inside the field?
15
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30 A solenoid is connected in series with a battery and a switch as shown. A light, copper ring is held near to the solenoid and coaxial with it. What will happen to the copper ring immediately after the switch is closed?
A The ring remains stationary.
B The ring swings directly away from the solenoid.
C The ring swings directly towards the solenoid.
D The ring rotates about its vertical axis.
31 A soft-iron ring of variable cross-section has four coils wound round it at the positions shown. The coils have 2, 3, 3 and 4 turns. The 3-turn coil is connected to an a.c. supply. In which coil does the magnitude of the magnetic flux density have the largest variation?
copper ring solenoid
© IJC 2015 9646/01/Prelim 2/15
16
32 In the potentiometer setup below, the driver cell and the secondary cell are identical, with an e.m.f. of E and internal resistance r. The fixed resistor in the secondary circuit has a resistance R.
Which of the following graphs shows the variation of the balance length when fixed resistors of different values of resistance R is connected?
A B
C D
33 A 100 resistor conducts a current with changing direction and magnitude, as shown.
What is the mean power dissipated in the resistor?
A 100 W B 150 W C 250 W D 450 W
time / s
current / A
17
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34 A generator produces a r.m.s current of 60 A at a r.m.s voltage of 120 V. The r.m.s voltage is stepped up to 4500 V by an ideal transformer and transmitted through a power line of total resistance 1.0 Ω. What is the percentage of power lost in the transmission line?
A 0.012 % B 0.036 % C 0.048 % D 0.060 %
35 An electron with kinetic energy E has a de Broglie wavelength of . Which of the
following graphs correctly represents the relationship between and E?
A B C D
36 Some of the energy levels of the hydrogen atom are shown below.
Electrons are excited to the 0.85 eV level. How many different photon frequencies in the visible region will be observed in the emission spectrum of hydrogen?
A 1 B 2 C 3 D 6
37 Which of the following statements about lasers is incorrect?
A Excitation of the active medium in a laser can be achieved by absorption of light called optical pumping.
B Laser action cannot take place unless an inverted population of atoms is obtained in the active medium.
C Pumping results in electrons excited from so-called metastable states to higher energy states.
D Stimulated emission of radiation occurs when the frequency of the incident photon equals the resonant frequency of the transition between energy levels.
© IJC 2015 9646/01/Prelim 2/15
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38 In the circuit involving a P-N junction below, holes migrate from
A X to Y through the cell.
B Y to X through the cell.
C P to N through the PN junction.
D N to P through the PN junction.
39 A 238
92Unucleus decays in two stages to a 234
91Pa nucleus.
What was emitted in these two stages?
A + B + C + D +
40 The tritium content of water from certain deep wells is only 25% of that in the water from recent rains. The half-life of tritium is 12.3 years. How long has it been since the water in the well came down as rain?
A 3.1 yrs B 6.2 yrs C 24.6 yrs D 36.9 yrs
19
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
Qn Ans Qn Ans Qn Ans Qn Ans 1 B 11 D 21 D 31 C 2 D 12 D 22 B 32 A 3 A 13 A 23 A 33 C 4 A 14 B 24 D 34 B 5 B 15 D 25 A 35 C 6 C 16 B 26 B 36 B 7 D 17 B 27 A 37 C 8 C 18 B 28 D 38 C 9 C 19 B 29 B 39 A
10 B 20 B 30 B 40 C
1 [Ans: B ] Answer: B Uncertainty for mass of liquid = 2 g ∆𝑝
𝑝=∆𝑚
𝑚+∆𝑉
𝑉
∆𝑝
5=
2
50+0.6
10
∆𝑝 = 0.5 g cm3
Comments
2 [Ans: D]
1 Gm should be 109 m.
Comments
3 [Ans: A ]
The object will have an initial speed of 200 m s-1 to the horizontal with initial height of 500 m. Using equation s = ut + 1/2at2 for the vertical component, -500 = 0 +1/2(-9.81)t2 t = 10 s d = 10 x 200 = 2 km
Comments
4 [Ans: A ]
After opening the parachute, the parachutist experience a net upwards force in order to slow down. So, there must be a negative acceleration after opening the parachute.
Comments
5 [Ans: B ]
Comments
© IJC 2015 9646/01/Prelim 2/15
20
By Newton’s 3rd law, the force the ball exerts on the ground is the action and reaction pair with the force the ground acts on the ball.
6 [Ans: C ] When pushing the object up, F - mg sin θ - resistive force = 0 31 – (4.0)(9.81)(sin 36°) + resistive force = 0 resistive force = 7.93 When pushing object down, F + mg sin θ - resistive force = 0 F = 7.93 - (4.0)(9.81)(sin 36°) = -15 N
Comments
7 [Ans: D] Option A is incorrect as there will not be any further change in the upthrust when the cube is lowered further as the entire volume of the cube is already immersed in water. Option B is incorrect as the balance reading corresponds to the tension in the spring. Option C is incorrect as upthrust and consequently the tension in the spring is caused by pressure differences on the top and bottom sides of the cube. Option D is correct.
Comments
8 [Ans: C] At equilibrium, Resistive torque = Tightening torque = 200(0.25) = 50 N m (ACW)
Comments
9 [Ans: C] Work done BY the gas = p x (change in volume during expansion) = p(V2 – V1) Work done ON the gas = - Work done BY the gas = p(V1 – V2)
Comments
10 [Ans: B] Work done by the motor in 1 rev = Friction force x circumference = (100 – 20) x 0.30 = 24 J Output power of motor = Work done in 50 rev / time = (24 x 50) / 1 = 1200 W
Comments
21
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
= 1.2 kW
11 [Ans: D] When in equilibrium, force vector diagram acting at point P must form a closed loop. Hence, option D is the only possible choice.
Comments
12 [Ans: D] The tension in the pendulum’s string provides for the centripetal force pointing to the left. Hence, the free body diagram of the pendulum consists of its weight and the tension force acting on it. Centripetal force is a resultant force and should not be shown in the free body diagram. Hence, option D is correct.
Comments
13 [Ans:A]
1 2G 2
2
2
2
2
2
( )(2 )
(3 )
2
9
0.22
Gm mF
r
G M M
X
GM
X
GM
X
Comments
14 [Ans: B]
Comments
Recall:
GPE (1)
KE (2)2
Total E (3)2
GPE of new satellite = 3.2 MJ
From eqn (3), Total Energy of new satellite 0.8 MJ2(2 )
GMm
r
GMm
r
GMm
r
GMm
r
GMm
r
© IJC 2015 9646/01/Prelim 2/15
22
15 [Ans: D ]
2
2
2
2
2
4
when period is doubled,
gradient of straight line graph of
1 against will be multplied by
4
i.e. less steep, reduced from 10 to 2.5
a x
xT
xT
T
a x
Comments
16 [Ans: B ]
2
2 2
o
max
Since intensity increases, , amplitude increases
from data sheet,
Hence,
Thus maximum velocity of oscillating air molecules
will increase.
However, there is no change to frequency
o
I A
v x x
v x
or wavelength,
hence speed of wave travel, , no changev f
Comments
23
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17 [Ans:B ]
Process is isovolumetric i.e. , , are constant:
constant
150 100 250 kPa
[Don't forget to add atm. pressure]
273.15 30 303.15 K
273.15 60 333.15 K
250
303.15 333.15
i
i
f
i f
i f
f
f
PV nRT
n R V
P nR
T V
P
T
T
P P
T T
P
P
274.74 kPa
New reading of Pressure Gauge
274.7 100 174.7 kPa
174.7 150%tage increase 100% 16.5%
150x
Comments
18 [Ans: B]
o
new
o
is tripled, while is increased
from 250 K to 1000 K (increase of 750 K)
3
4
3
4
1Hence, of original no. of
4
molecules must have escaped
PV nRT
PVn
RT
P T
P Vn
R T
n
Comments
© IJC 2015 9646/01/Prelim 2/15
24
19 [Ans: B]
2Intensity Amplitude
Since amplitude is doubled, the intensity will be multiplied by 4.
PowerIntensity
Area
1Since Area is now of original, intensity will be multiplied by 3
3
In the intensity will be m
ultiplied by 4 3=12 times original
Comments
20 [Ans: B] A Point R is moving with maximum amplitude, thus it is an
anti-node. B Point R is in the adjacent loop to both point Q and S. As
such, both point Q and S are antiphase to point R. As such, point Q and S are in phase.
C There are three loops between point P and T, thus the distance between P and T is 3 × ½ wavelength, i.e. 1.5 wavelength.
D The wave shown is in 3rd harmonic, thus it does not have the lowest possible frequency.
Comments
21 [Ans: D] For the fundamental frequency, the distance between the mouthpiece (node) and the bell (antinode) is 0.25λ1. As for the frequencies of the next two higher notes, the distance between the mouthpiece and the bell is 0.75λ2 and 1.25λ3. i.e. L = 0.25λ1 = 0.75λ2 = 1.25λ3 λ1 = 3λ2 = 5λ3 Since v = fλ with the speed of the note constant, v = 75 × λ1 = f2 × 1/3 λ1 = f3 × 1/5 λ1 f2 = 225 Hz f3 = 375 Hz
Comments
22 [Ans: B] dsinθn = nλ Since the orange and the blue light passes through the same diffraction grating, d is the same. Since the orange and blue light coincides at the same spot, θ is the same. dsinθn = nλ = 4 × 600 = 5 λ λ = 480 nm
Comments
25
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23 [Ans: A] Since P and Q are at the same distance away from the positive point charge, P and Q have the same electric potential, i.e. ∆VP
to Q = 0 W = q∆V = 1 × 0 = 0 J
Comments
24 [Ans: D] Since the positively-charge particle is inside the parallel plates, it experiences uniform electric field and thus constant electric force. ∆Ek = Work done by the electric force = F × x Hence, the kinetic energy EK of the particle vary directly proportionally to the distance x moved from the positive plate.
Comments
25 [Ans: A] In 1 second, the area of belt that passes by the point = wv
The amount of charge on this area is wv
Current = charge collected in 1 second = wv
Comments
26 [Ans:B]
2 2
2 2
3
P=I R+I r
4500(0.50) (16) (0.50) r
1.0 x10
r = 2
Comments
27 [Ans: A]
2
Total
Total
Terminal P.d.
As internal resistance increases, terminal p.d. decreases.
Power output by battery
increases, Power output decreases.
RE
R r
r
Emf
R
Since R
Comments
© IJC 2015 9646/01/Prelim 2/15
26
28 [Ans: D]
Based on the circuit diagram, there are three branches connected in parallel to the e.m.f. Thus these three branches would have the same p.d. across it. i.e. E = I12R = I3R = I43R Since I1 = 0.6 A, I3 = 1.2 A and I4 = 0.4 A I2 = I1 + I3 + I4 = 0.6 + 1.2 + 0.4 = 2.2 A
Comments
29 [Ans: B] The electric force acting on the electrons is acting in the opposite direction to that of the electric field. With the electric force acting to the right, the electron will deflect to the right.
Comments
30 [Ans: B] When the switch is closed, there is a change in magnetic flux through the ring. By Faraday’s law, an e.m.f. is induced in the ring. By Lenz’s law, the direction of induced current is such that it opposes the change in magnetic flux causing it. Hence, the induced current in the ring causes the ring to move away from the solenoid.
Comments
31 [Ans: C] Recall that magnetic flux density, B = Φ/A Hence, the smallest cross-sectional area will give the largest variation in magnetic flux density for the same amount of magnetic flux (magnetic flux in concentrated within the soft-iron ring). Points to note: At any time the magnetic flux is the same for all the coils. The flux density is largest where the area is
Comments
27
© IJC 2015 9646/01/Prelim 2/15 [Turn Over
smallest, so the largest variation of flux density is for coil C.
32 [Ans: A ] At balance length,
( )
( )( )
( )
,
ALAB
AB
AB
ABAL AB
AB
AL
lR
l R
R r R r
R R rl l
R r R
Hence
Rl
R r
Comments
33 [Ans: C]
A58.121
22 22
RMS
T
TT
I
W25010058.1 22RMS RP I
Comments
34 [Ans: B]
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
2 2
( )
120 60 7200
4500 7200
1.6
1.6 1
generated mean rms generated rms generated
generated mean rms transmission rms transmission rms transmission
rms transmission
lost rms transmission
P V I W
P V I I W
I A
P I R
2.6 W
Percentage power loss 2.6100% 0.036 %
7200
Comments
35 [C ]
Momentum p = 2mE
De Broglie wavelength = h/p = h/2mE Thus to obtain a straight line graph through the origin, one
needs to plot a graph of against 1/E.
Comments
36 [B ] Photons in the visible region have energies ranging from 1.8 eV to 3.1 eV.
When the electrons are excited to the 0.85 eV, they emit photons when they move to a lower energy level.
From 4 2, E = 2.54 eV
From 4 3, E = 0.66 eV
From 3 2, E = 1.88 eV Excitation to ground state will give rise to ultraviolet rays. Number of emission lines in the visible region = 2
Comments
© IJC 2015 9646/01/Prelim 2/15
28
37 [C ] Pumping results in electrons excited to metastable state is an incorrect statement.
Comments
38 [C ] In the forward-bias, a current will be flowing from the p-type semiconductor to the n-type semiconductor. This results in the holes moving towards the depletion region to the N-type semiconductor.
Comments
39 [A ] 238
92U undergoes decay to 234
90Th , which then undergoes
decay to 234
91Pa
Comments
40 [C ] 25% = ¼ of the initial amount which means two half life’s have
passed. Therefore time = 12.3 2 = 34.6 years
Comments
© IJC 2015 9646/02/Prelim 2/15 [Turn over
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level
Higher 2
CANDIDATE NAME
CLASS INDEX NUMBER
PHYSICS Paper 2 Structured Questions Candidates answer on the Question Paper No Additional Materials are required.
9646/02
15 September 2015
1 hour 45 minutes
READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions.
At the end of the examination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question.
This document consists of 19 printed pages and 1 blank page.
For Examiner’s Use
1
9
2
6
3
12
4
6
5
6
6
6
7
15
8
12
Penalty
Total
72
Innova Junior College [Turn over
2
© IJC 2015 9646/02/Prelim 2/15 [Turn over
For Examiner’s Use
Data
speed of light in free space, c = 3.00 x 108 m s-1
permeability of free space, o = 4 x 10−7 H m−1
permittivity of free space,
o
= 8.85 x 10−12 F m−1
(1/(36)) x 10−9 F m−1 elementary charge, e = 1.60 x 10−19 C
the Planck constant, h = 6.63 x 10−34 J s
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
molar gas constant, R = 8.31 J K−1 mol−1
the Avogadro constant, NA = 6.02 x 1023 mol−1
the Boltzmann constant, k = 1.38 x 10−23 J K−1
gravitational constant, G = 6.67 x 10−11 N m2 kg−2
acceleration of free fall, g = 9.81 m s−2
Formulae
uniformly accelerated motion, s = ut + ½at2
v2 = u2+ 2as
work done on/by a gas, W = pV
mean kinetic energy of a molecule of an ideal gas E = kT2
3
hydrostatic pressure, p = gh
gravitational potential, = GM
r
displacement of particle in s.h.m. x = xosin ωt
velocity of particle in s.h.m. v = vocos ωt
2 2
ox x
resistors in series, R = R1 + R2 + …
resistors in parallel, 1/R = 1/R1 + 1/R2 + …
electric potential V = Q/4or
alternating current/voltage, x = xo sint
transmission coefficient T = exp (−2kd)
where k = 2
2
8 m U E
h
radioactive decay, x = xo exp(−t)
decay constant, = ½
0.693
t
3
© IJC 2015 9646/02/Prelim 2/15 [Turn over
For Examiner’s Use
Answer all the questions.
1 A simple pendulum may be used to determine a value for the acceleration of free fall g.
Measurements are made of the length L of the pendulum, and the period T of the
oscillation. The values obtained, with their uncertainties, are as shown.
T = (1.93 ± 0.03) s
L = (92 ± 1) cm
(a) Calculate the percentage uncertainty, to two significant figures, in the measurement
of
(i) the period T,
percentage uncertainty = ……………………. % [1]
(ii) the length L.
percentage uncertainty = ……………………. % [1]
(b) The relationship between T, L and g is given by
2
2
4
gTL
Use your answers in (a) to determine the absolute uncertainty in g.
Hence state the value of g, with its uncertainty, to an appropriate number of
significant figures.
uncertainty = ……………………. m s-2 [1]
g = ……………………. ± ……………………. m s-2
[1]
4
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For Examiner’s Use
(c) A ball is thrown horizontally from the top of a building, as shown in Fig. 1.1.
Fig. 1.1
The ball is thrown with a horizontal speed of 8.2 m s–1. At point P, the path of the ball is moving at an angle of 60° to the horizontal. Air resistance is negligible. For the ball at point P,
(i) show that the vertical component of its velocity is 14.2 m s–1, [1]
(ii) determine the vertical distance through which the ball has fallen.
distance = ……………………. m [2]
(d) On Fig. 1.1, sketch the new path of the ball, for the ball having an initial horizontal speed
(i) greater than 8.2 m s–1 and with negligible air resistance (label this path G),
[1]
(ii) equal to 8.2 m s–1 but with air resistance (label this path A). [1]
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For Examiner’s Use
2 The mean value of an alternating current is zero. Explain
(a) (i) why an alternating current gives rise to a heating effect in a resistor,
……………………………………………………………………...…………………..
……………………………………………………………………...…………………..
……………………………………………………………………...…………………..
……..…………………………………………………………………………..
[2]
(ii) by reference to the heating effect, what is meant by the root-mean-square
(r.m.s.) value of an alternating current.
………………………………………………………………………...………………..
………………………………………………………………………...………………..
…………………………………………………………………………….…...
[1]
(b) Use Faraday’s law to explain why the current in the primary coil is not in phase with
the e.m.f. induced in the secondary coil.
…………………………………………………………………………………..…………….
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…………………………………………………………………………………..……
[3]
6
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For Examiner’s Use
3 (a) Define electric field strength.
.....................................................................................................................................
…………………………………………………………………………………..…… [2]
(b) A small charged sphere is suspended from a fine spring (made of an insulating material) between horizontal parallel metal plates as shown in Fig. 3.1.
Fig. 3.1
Initially, the plates are uncharged. Switch S is then set to position X. This causes the sphere to move vertically downwards so that eventually it comes to a rest 18 mm below than its original position.
(i) State and explain the nature of the charge on the sphere.
............................................................................................................................
............................................................................................................................
............................................................................................................................
................................................................................................................ [2]
(ii) Given that the magnitude of the charge on the sphere is 41 nC, determine the
spring constant of the spring.
spring constant = ……………………. N m-1
[3]
Charged sphere
Fine spring
plates
Small hole
in top plate
10.5 cm
4.8 cm
7
© IJC 2015 9646/02/Prelim 2/15 [Turn over
For Examiner’s Use
(c) Switch S is now moved to position Y.
(i) State and explain the effect of this on the electric field between the plates.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
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............................................................................................................... [2]
(ii) Hence by considering the forces acting on the sphere, explain the
subsequent motion of the sphere.
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...........................................................................................................................
............................................................................................................... [3]
8
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For Examiner’s Use
4 The volume of some air, assumed to be an ideal gas, in the cylinder of a car engine is 500 cm3 at a pressure of 1.1 × 105 Pa and a temperature of 27 °C. The air is suddenly compressed to a volume of 50 cm3. No energy enters or leaves the gas by heating during the compression. The pressure rises to 6.6 × 106 Pa.
(a) Determine the temperature of the gas after the compression.
temperature = ……………………. K [2]
(b) (i) State the first law of thermodynamics.
.........................................................................................................................
.........................................................................................................................
................................................................................................................
[1]
(ii) Use the law to explain why the temperature of the air changed during the
compression.
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[3]
9
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For Examiner’s Use
5 (a) Explain how a line emission spectrum leads to an understanding of the existence of discrete electron energy levels in atoms.
.....................................................................................................................................
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……………………………………………………………………………………..… [3]
(b) Some of the lines of the emission spectrum of atomic hydrogen are shown Fig. 5.1.
Fig. 5.1
The photon energies associated with some of these lines are shown in Fig. 5.2.
wavelength / nm photon energy / 10-19 J
434 4.58
486 ……………
656 3.03
Fig. 5.2
(i) Complete Fig. 5.2 by calculating the photon energy for a wavelength of
486 nm.
[1]
10
© IJC 2015 9646/02/Prelim 2/15 [Turn over
For Examiner’s Use
(ii) Energy levels of a single electron in a hydrogen atom are shown in Fig. 5.3.
Fig. 5.3
(not to scale)
Use data from (b)(i) to show, on Fig. 5.3, the transitions associated with each
of the three spectral lines shown in Fig. 5.1. Show each transition with an arrow.
[2]
11
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For Examiner’s Use
6 (a) Describe and explain how n-type doping changes the conduction properties of semiconductors.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
……………………………………………………………………………………..… [2]
(b) A junction is formed between slices of p-type and n-type semiconductor material. Describe the origin of the depletion region at the junction.
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…………………………………………………………………………………..…… [4]
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For Examiner’s Use
7 The variation with temperature / C of the resistance R of a small device X is as shown in Fig. 7.1. Device X is made from intrinsic semiconductor.
Fig. 7.1
(a) A student hypothesizes that the resistance R may be inversely proportional to
over the range of temperature from 5.0 C to 30 C. Show, without drawing a graph, that his hypothesis in not correct.
[2]
R /
/ C
13
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For Examiner’s Use
(b) Describe how band theory is used to explain the trend of the graph shown in
Fig. 7.1.
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.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
……………………………………………..…………………………………………
[3]
(c) Another student proposes that R decreases exponentially with thermodynamic
temperature, T. Fig. 7.2 shows some of the data for R, , T-1 and ln(R /
R / / C T / K T-1 / 10-3 K-1 ln(R / )
100 0.0 273.2 3.660 4.61
90 2.0 275.2 3.634 4.50
80 4.3 277.5 3.604 4.38
70 7.0 280.2 3.569 4.25
60
50 14.5 287.7 3.476 3.91
Fig. 7.2
Complete Fig. 7.2 for the resistance of 60 .
[1]
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(d) Fig. 7.3 is a graph of some of the data of Fig. 7.2.
Fig. 7.3
Complete Fig. 7.3 using your data for the resistance of 60 . [1]
(e) The resistance and temperature of device X is found to be related by the expression
B
TR Ae
where A and B are constants and T represents the thermodynamic temperature. Use Fig. 7.3 to determine the values for A and B. Express the answers in the
appropriate units.
A = …………………….
B = ……………………. [4]
3.70 3.45 3.50 3.55 3.60 3.65
3.80
4.00
4.20
4.40
4.60
4.80
ln(R / )
1 3 1 10 KT
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(f) (i) Device X is now connected to a fixed resistor of resistance 40 as shown in Fig. 7.4.
Fig. 7.4
Determine the value for the voltmeter reading when device X is immersed in
water at a temperature of 40.0 C.
V = ……………………. V [2]
(ii) If the temperature of the water is raised, would the voltmeter reading increase
or decrease? Explain.
............................................................................................................................
............................................................................................................................
............................................................................................................................
................................................................................................................ [2]
ideal
voltmeter
6.0 V
V
40 X
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8 Two identical coils are connected together and arranged as shown in Fig. 8.1
Fig. 8.1 The coils are in the vertical plane and are parallel to each other. When the coils are connected to a power supply, there is a magnetic field between them. It is suggested that the magnetic flux density B of the field at the point X is related to the radius r of the coils by the relationship
00.72μ NB
r
I
where N is the number of turns on each coil, I is the current in the coils and μ0 is the
permeability of free space. You are provided with two flat circular coils and a hall probe.
Design a laboratory experiment to test the relationship between B and I and determine a
value for μ0 .
You should draw a labelled diagram to show the arrangement of your apparatus. In your account you should pay particular attention to: (a) the identification and control of variables, (b) the equipment you would use, (c) the procedure to be followed, (d) how the relationship between B and r is determined from your readings,
(e) any precaution that would be taken to improve the accuracy and safety of the experiment.
[12]
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Diagram
…....…………………………………………………………………………………………………….... …....……………………………………………………………………………………………………....
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BLANK PAGE
© IJC 2015 9646/02/Prelim 2/15 [Turn over
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level
Higher 2
CANDIDATE NAME
CLASS INDEX NUMBER
PHYSICS Paper 2 Structured Questions Candidates answer on the Question Paper No Additional Materials are required.
9646/02
15 September 2015
1 hour 45 minutes
READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions.
At the end of the examination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question.
This document consists of 19 printed pages and 1 blank page.
For Examiner’s Use
1
9
2
6
3
12
4
6
5
6
6
6
7
15
8
12
Penalty
Total
72
Innova Junior College [Turn over
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Data
speed of light in free space, c = 3.00 x 108 m s-1
permeability of free space, o = 4 x 10−7 H m−1
permittivity of free space,
o
= 8.85 x 10−12 F m−1
(1/(36)) x 10−9 F m−1 elementary charge, e = 1.60 x 10−19 C
the Planck constant, h = 6.63 x 10−34 J s
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
molar gas constant, R = 8.31 J K−1 mol−1
the Avogadro constant, NA = 6.02 x 1023 mol−1
the Boltzmann constant, k = 1.38 x 10−23 J K−1
gravitational constant, G = 6.67 x 10−11 N m2 kg−2
acceleration of free fall, g = 9.81 m s−2
Formulae
uniformly accelerated motion, s = ut + ½at2
v2 = u2+ 2as
work done on/by a gas, W = pV
mean kinetic energy of a molecule of an ideal gas E = kT2
3
hydrostatic pressure, p = gh
gravitational potential, = GM
r
displacement of particle in s.h.m. x = xosin ωt
velocity of particle in s.h.m. v = vocos ωt
2 2
ox x
resistors in series, R = R1 + R2 + …
resistors in parallel, 1/R = 1/R1 + 1/R2 + …
electric potential V = Q/4or
alternating current/voltage, x = xo sint
transmission coefficient T = exp (−2kd)
where k = 2
2
8 m U E
h
radioactive decay, x = xo exp(−t)
decay constant, = ½
0.693
t
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Answer all the questions.
1 A simple pendulum may be used to determine a value for the acceleration of free fall g.
Measurements are made of the length L of the pendulum, and the period T of the
oscillation. The values obtained, with their uncertainties, are as shown.
T = (1.93 ± 0.03) s
L = (92 ± 1) cm
(a) Calculate the percentage uncertainty, to two significant figures, in the measurement
of
(i) the period T,
(2.s.f)
percentage uncertainty = ……………………. % [1]
(ii) the length L.
(2.s.f)
percentage uncertainty = ……………………. % [1]
(b) The relationship between T, L and g is given by
2
2
4
gTL
Use your answers in (a) to determine the absolute uncertainty in g.
Hence state the value of g, with its uncertainty, to an appropriate number of significant figures.
Rearranging equation,
= 1.1 + 2(1.6) = 4.3 %
= ± 0.4 m s-2 (1 s.f.)
Hence, g ± Δg = (9.8 ± 0.4) m s-2
uncertainty = ……………………. m s-2 [1]
g = ……………………. ± ……………………. m s-2
[1]
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(c) A ball is thrown horizontally from the top of a building, as shown in Fig. 1.1.
Fig. 1.1
The ball is thrown with a horizontal speed of 8.2 m s–1. At point P, the path of the ball is moving at an angle of 60° to the horizontal. Air resistance is negligible. For the ball at point P,
(i) show that the vertical component of its velocity is 14.2 m s–1, [1]
vertical velocity at P = 8.2 tan 60° = 14.2 m s-1 (shown)
(ii) determine the vertical distance through which the ball has fallen. Using, vy
2 = uy2 + 2ays
(14.2)2 = 0 + 2(9.81)(s) s = 10.3 m
distance = ……………………. m [2]
(d) On Fig. 1.1, sketch the new path of the ball, for the ball having an initial horizontal speed
(i) greater than 8.2 m s–1 and with negligible air resistance (label this path G),
[1]
(ii) equal to 8.2 m s–1 but with air resistance (label this path A). [1] *Initial paths of A and G are horizontal
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2 The mean value of an alternating current is zero. Explain
(a) (i) why an alternating current gives rise to a heating effect in a resistor,
either The heating effect in a resistor is directly proportional to I2 The square of value of an alternating current is always positive Hence there will be a heating effect OR The current moves in opposite directions through the resistor during half-cycles. However, the heating effect in the resistor is independent of the direction of the current passing through it.
[2]
(ii) by reference to the heating effect, what is meant by the root-mean-square
(r.m.s.) value of an alternating current.
The root-mean-square (rms) value, Irms, of an alternating current is equivalent to that value of a steady direct current which would dissipate energy in a given resistance at the same rate (i.e. produce the same heating effect) as that provided by the alternating current.
[1]
(b) Use Faraday’s law to explain why the current in the primary coil is not in phase with
the e.m.f. induced in the secondary coil.
The varying magnetic flux in the iron core, due to the alternating current (AC) in the primary coil, is in phase with the AC.
On the other hand, An emf is induced in the secondary coil due the changing magnetic flux linkage with the secondary coil. Faraday’s law states that the induced emf is directly proportional to the rate of change of magnetic flux linkage.
Since the The rate of change of magnetic flux linkage is not in phase with the magnetic flux and we can conclude both the induced emf and alternating
current are not in phase. For instance, if the AC in the primary coil is sinusoidal, the induced emf will also be sinusoidal, but π/2 out of phase.
[3]
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3 (a) Define electric field strength.
.....................................................................................................................................
…………………………………………………………………………………..…… [2]
(b) A small charged sphere is suspended from a fine spring (made of an insulating material) between horizontal parallel metal plates as shown in Fig. 3.1.
Fig. 3.1
Initially, the plates are uncharged. Switch S is then set to position X. This causes the sphere to move vertically downwards so that eventually it comes to a rest 18 mm below than its original position.
(i) State and explain the nature of the charge on the sphere.
............................................................................................................................
............................................................................................................................
............................................................................................................................
................................................................................................................ [2]
(ii) Given that the magnitude of the charge on the sphere is 41 nC, determine the
spring constant of the spring.
9
1
3
0
41 10 50000.24 N m
18 10 0.048
F
kx qE
qE qVk
x xd
Charged sphere
Fine spring
plates
Small hole
in top plate
10.5 cm
4.8 cm
Electric field strength at a point is the electric force per unit positive charge acting on a small test charge placed at that point.
The bottom plate is connected to the terminal at lower potential and the charged sphere is moved further down to equilibrium, it implies that there is an attractive force between the sphere and the bottom plate [M1]. Thus sphere is positively charged [A1]. OR
Electric field points from the higher to lower potential plate (downwards), since the sphere moves downwards in the direction of electric field [M1], it is positively charged [A1].
[M1]
[M1]
= 0.24 N m-1 [A1]
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spring constant = ……………………. N m-1
[3]
(c) Switch S is now moved to position Y.
(i) State and explain the effect of this on the electric field between the plates.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
............................................................................................................... [2]
(ii) Hence by considering the forces acting on the sphere, explain the
subsequent motion of the sphere.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
............................................................................................................... [3]
The electrons/charges flow from one plate to the other until there is no potential difference between the plates. (discharge) [M1] OR
Since both plates are now at the same potential, there is no potential difference [M1] between the plates.
Electric field is zero [A1].
Electric force that was initially acted downwards is now removed, hence tension acting on the sphere is now larger than weight. [M1]
The net force is proportional to extension and directed opposite to displacement [M1].
Sphere moves in simple harmonic motion [A1], it will oscillate about the equilibrium position.
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4 The volume of some air, assumed to be an ideal gas, in the cylinder of a car engine is
500 cm3 at a pressure of 1.1 × 105 Pa and a temperature of 27 °C. The air is suddenly compressed to a volume of 50 cm3. No energy enters or leaves the gas by heating during the compression. The pressure rises to 6.6 × 106 Pa.
(a) Determine the temperature of the gas after the compression.
5 6
6 6
1 1 10 500 10 8 31 273 15 27
0 02205
6 6 10 50 10 0 02205 8 31
1800 9
1800
( . )( ) ( . )( . )
. moles
( . )( ) ( . )( . )( )
.
K (3 sf)
i i i
f f f
f
f
PV nRT
n
n
P V nRT
T
T
temperature = ……………………. K [2]
(b) (i) State the first law of thermodynamics.
The increase in internal energy of a closed system ∆U, is given by the sum of the heat added to the system Q, and the work done on the system, W i.e. ∆U = Q + W. Symbols must be explained.
[1]
(ii) Use the law to explain why the temperature of the air changed during the
compression.
Since no energy entered or left the system by heating, Q = 0. Hence, by the 1st law of thermodynamics, any increase in the internal energy of the system must be due to work done one the system ∆U = W. Since gas is compressed, W is positive and internal energy U increases. For an ideal gas, the internal energy is equal to the sum of the microscopic kinetic energies of the gas molecules, which in turn is directly proportional to the temperature of the gas. Hence as internal energy increases, temperature increases.
[3]
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5 (a) Explain how a line emission spectrum leads to an understanding of the existence of discrete electron energy levels in atoms.
When the excited electrons in the atoms changes from high energy level to low energy level, they emit photons as seen in the line emission spectrum [B1]. Each line corresponds to a specific photon energy [B1]. Such specific photon energy must be due to discrete energy changes when the electrons de-excite. Thus it leads to an understanding of the existence of discrete electron energy levels in atoms. [B1]. [3]
(b) Some of the lines of the emission spectrum of atomic hydrogen are shown Fig. 5.1.
Fig. 5.1
The photon energies associated with some of these lines are shown in Fig. 5.2.
wavelength / nm photon energy / 10-19 J
434 4.58
486 ……………
656 3.03
Fig. 5.2
(i) Complete Fig. 5.2 by calculating the photon energy for a wavelength of
486 nm.
34 8
9
hc 6.63 10 3.0 10E
486 10
= 4.09 × 10-19 J [B1] [B1] [1]
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(ii) Energy levels of a single electron in a hydrogen atom are shown in Fig. 5.3.
Fig. 5.3
(not to scale)
Use data from (b)(i) to show, on Fig. 5.3, the transitions associated with each
of the three spectral lines shown in Fig. 5.1. Show each transition with an arrow.
Ep =(– 0.87 × 10-19) – (– 5.45 × 10-19) = 4.58 × 10-19 J Ep =(– 1.36 × 10-19) – (– 5.45 × 10-19) = 4.09 × 10-19 J Ep =(– 2.42 × 10-19) – (– 5.45 × 10-19) = 3.03 × 10-19 J Three transitions to/from – 5.45 × 10-19 J level [M1] Since the photons are from line emission spectrum, the electron transitions should be from higher to lower energy (level) [A1] [2]
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6 (a) Describe and explain how n-type doping changes the conduction properties of semiconductors.
A n-type extrinsic semiconductor is doped with Group IV dopants, which are known as donor dopant. In a n-type semiconductor, a donor energy level is created just below the conduction band [B1], thus reducing the energy difference between the donor level and the conduction band. An electron in the donor energy level can be easily thermally excited to conduction band, creating more mobile electrons in the conduction band [B1]. Thus the number of charge carriers is increased and the conductivity is increased as well. [2]
(b) A junction is formed between slices of p-type and n-type semiconductor material. Describe the origin of the depletion region at the junction.
Due to concentration difference, electrons and holes in n-type and p-type material respectively diffuse into the other type of material [B1]. The electrons then recombine with holes on both sides [B1]. These diffusions also expose positively charged ions (donors) in the n-type material and negatively charged ions (acceptors) in the p-type material [B1]. It set up an internal electric field that opposes any further diffusion of charge carriers across the junction [B1]. Hence there is a depletion of mobile charge carriers in this region. [4]
12
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7 The variation with temperature / C of the resistance R of a small device X is as shown in Fig. 7.1. Device X is made from intrinsic semiconductor.
Fig. 7.1
(a) A student hypothesizes that the resistance R may be inversely proportional to
over the range of temperature from 5.0 C to 30 C. Show, without drawing a graph, that his hypothesis in not correct.
If (constant)
@(50, 14.5): 725
@(90, 2): 180
Since hypothesis is not correct
[2]
R /
/ C
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(b) Describe how band theory is used to explain the trend of the graph shown in
Fig. 7.1.
Since Device X is an intrinsic semiconductor, the band gap in this device is small and usually of value about 1 eV. When temperature increases, electrons in the valence band could easily gain enough energy to be excited to the conduction band, leaving behind holes. Holes are positive charge carriers and conduct electricity. With an increase in the charge carriers (electrons in the conduction band and holes in the valence band), current increases for the same potential difference and resistance decreases, as shown in the figure. As temperature increases further, the lattice vibrations of the atoms become more significant and the rate of change of the resistance decreases with temperature.
[3]
(c) Another student proposes that R decreases exponentially with thermodynamic
temperature, T. Fig. 7.2 shows some of the data for R, , T-1 and ln(R /
R / / C T / K T-1 / 10-3 K-1 ln(R / )
100 0.0 273.2 3.660 4.61
90 2.0 275.2 3.634 4.50
80 4.3 277.5 3.604 4.38
70 7.0 280.2 3.569 4.25
60 10.5 283.7 3.525 4.09
50 14.5 287.7 3.476 3.91
Fig. 7.2
Complete Fig. 7.2 for the resistance of 60 .
[1]
(d) Fig. 7.3 is a graph of some of the data of Fig. 7.2.
ln(R / )
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Fig. 7.3
Complete Fig. 7.3 using your data for the resistance of 60 . [1]
(e) The resistance and temperature of device X is found to be related by the expression
B
TR Ae
where A and B are constants and T represents the thermodynamic temperature.
Use Fig. 7.3 to determine the values for A and B. Express the answers in the
appropriate units.
Linearising:
Since A is a constant and k is also a constant, plotting lnR vs. graph will yield a
straight line graph with gradient B and y-intercept lnA
Gradient of graph = 3
4.56 3.80
(3.65 3.45)10
= 3.80 103
B = 3.80 103 K Using (3.65 × 10-3, 4.56):
4.56= (3800)(3.65 × 10-3
) + ln A
A = 9.05 10-5
3.70 3.45 3.50 3.55 3.60 3.65
3.80
4.00
4.20
4.40
4.60
4.80
1 3 1 10 KT
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A = …………………….
B = ……………………. [4]
(f) (i) Device X is now connected to a fixed resistor of resistance 40 as shown in Fig. 7.4.
Fig. 7.4
Determine the value for the voltmeter reading when device X is immersed in
water at a temperature of 40.0 C.
For a temperature of 40 C, T = 313.2 K,
1/T = 3.193 10-3
lnR = ln(9.05 10-5
) + 3800/313.2 = 2.823
R= 16.8
Voltmeter reading = p.d across 40 resistor = 40/(40+16.8) 6.0 = 4.2 V
V = ……………………. V [2]
(ii) If the temperature of the water is raised, would the voltmeter reading increase
or decrease? Explain.
If the temperature is raised, the resistance of the thermistor will decrease. This
results in a smaller potential difference across the thermistor and an increase in
the potential difference across the 40 resistor. Thus, the voltmeter reading,
which measures the p.d. across the 40 resistor, will increase. ............................................................................................................................
............................................................................................................................
............................................................................................................................
................................................................................................................ [2]
ideal
voltmeter
6.0 V
V
40 X
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For Examiner’s Use
8 Two identical coils are connected together and arranged as shown in Fig. 8.1
Fig. 8.1
The coils are in the vertical plane and are parallel to each other. When the coils are connected to a power supply, there is a magnetic field between them. It is suggested that the magnetic flux density B of the field at the point X is related to the radius r of the coils by the relationship
00.72μ NB
r
I
where N is the number of turns on each coil, I is the current in the coils and μ0 is the
permeability of free space. You are provided with two flat circular coils and a hall probe.
Design a laboratory experiment to test the relationship between B and I and determine a
value for μ0 .
You should draw a labelled diagram to show the arrangement of your apparatus. In your account you should pay particular attention to:
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(a) the identification and control of variables, (b) the equipment you would use, (c) the procedure to be followed, (d) how the relationship between B and I is determined from your readings,
(e) any precaution that would be taken to improve the accuracy and safety of the experiment.
[12]
18
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Diagram
Control:
1. The separation of the 2 coils should be kept constant by using a half meter ruler and measure the distance between the 2 coils each time before taking measurement.
2. The position of the hall probe between the 2 coils should be kept constant by clamping the hall probe with clamp on a retort stand and measuring the distance of point X from the two coils each time before taking a measurement.
Procedure:
1. Set up the apparatus as shown above. 2. Switch on the power source to pass a current through both coils to produce a
magnetic field. 3. Record the value of I from the ammeter.
4. Place the hall probe between the coils at point X. Record the value of the magnetic flux density B from the C.R.O.
5. Repeat steps 3 to 5 to obtain further values of B for different I by changing the
resistance of variable resistor. 6. Record N the number of turns on the coil and r radius of the coil using a ruler.
7. From , plot a graph of B against I. Value of can be obtained from the
gradient from the following relationship: gradient =
Safety:
1. Switch off the power source when not in use to prevent overheating of the coil. Reliability:
1. The plane of the hall probe should be perpendicular to the magnetic field within the 2 coils.
2. The axes of the 2 coils must coincide to ensure that the magnetic field in the region between the 2 coils is uniform.
A
X C.R.O
coil
Hall Probe
Retort Stand
19
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3. The coils are to be close to one another to ensure that the magnetic field between the coils is uniform.
4. The position of the hall probe is along the centre axis of the 2 coils and it is equidistant from the coils.
20
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BLANK PAGE
© IJC 2011 9646/MCE/11 [Turn over
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level
Higher 2
CANDIDATE
NAME
CLASS INDEX NUMBER
PHYSICS
Paper 3 Longer Structured Questions
Candidates are to answer on the Question Paper
No Additional Materials are required
9646/03
17 September 2015
2 hours
READ THESE INSTRUCTIONS FIRST
Write your name and class on all the work you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams, graphs or rough
working.
Do not use staples, paper clips, highlighters, glue or correction
fluid.
The use of an approved scientific calculator is expected where
appropriate.
Section A Answer all questions.
Section B Answer any two questions. You are advised to spend about one hour on each section. At the end of the examination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question. Marks will be deducted for using inappropriate number of significant figures or wrong value of g.
This document consists of 24 printed pages.
Innova Junior College [Turn over
For Examiner’s Use
Section A
1
8
2
6
3
8
4
9
5
9
Section B
6
20
7
20
8 20
Penalty
Total
80
Percentage
2
© IJC 2015 9646/03/Prelim 2/15 [Turn over
Data
speed of light in free space, c = 3.00 × 108 m s-1
permeability of free space, o = 4 × 10-7 H m-1
permittivity of free space, o = 8.85 × 10-12 F m-1
(1/(36)) × 10-9 F m-1
elementary charge, e = 1.60 × 10-19 C
the Planck constant, h = 6.63 × 10-34 J s
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
molar gas constant, R = 8.31 J K-1 mol-1
the Avogadro constant, NA = 6.02 × 1023 mol-1
the Boltzmann constant, k = 1.38 × 10-23 J K-1
gravitational constant, G = 6.67 × 10-11 N m2 kg-2
acceleration of free fall, g = 9.81 m s-2
Formulae
uniformly accelerated motion, s = ut + ½at2
v2 = u2+ 2as
work done on/by a gas, W = pV
hydrostatic pressure, p = gh
gravitational potential, = GM
r
displacement of particle in s.h.m., x = xosin ωt
velocity of particle in s.h.m., v = vocos ωt
= 2 2
0x x
mean kinetic energy of a molecule of an ideal gas E = kT2
3
resistors in series, R = R1 + R2 + …
resistors in parallel, 1/R = 1/R1 + 1/R2 + …
electric potential V = Q/4or
alternating current/voltage, x = xo sint
transmission coefficient T = exp (-2kd)
where k = m U - E
h
2
2
8
radioactive decay, x = xo exp(-t)
decay constant, = ½t
0.693
3
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Section A
Answer all the questions in the spaces provided.
1 A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g. The block
is suspended from strings so that it is free to move in a vertical plane. The length of the string is 1.50 m. The bullet buries itself in the block. The block and bullet rise together through a vertical distance of 8.6 cm, as shown in Fig. 1.1
Fig. 1.1
(a) Determine the tension of the string after the block and bullet rise to a vertical distance of 8.6 cm.
tension = ……………………. N [2]
(b) (i) Calculate the change in gravitational potential energy of the block and bullet.
energy = ……………………. J [1]
4
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(ii) Show that the initial speed of the block and the bullet, after they began to move off together, was 1.3 m s-1.
[1]
(iii) Using the information in (b)(ii) and the principle of conservation of
momentum, determine the speed of the bullet before the impact with the block.
speed = ……………………. m s-1 [1] (d) (i) Calculate the kinetic energy of the bullet just before impact.
kinetic energy = ……………………. J [1] (ii) State and explain what can be deduced from your answers to (b)(i) and
(d)(i) about the type of collision between the bullet and the block.
………………………………………………………………………………………..
………………………………………………………………………………………..
……………………………………………………………………………….. [2]
5
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2
A cell of e.m.f. of 6.0 V with an internal resistance of 0.5 Ω is connected in an electrical circuit comprising four resistors and a resistance wire W as shown in Fig. 2.1.
Fig. 2.1
(a) The resistivity of W is 1.25 x 10-5 Ω m. Its length and diameter is 1.20 m and 3.0 x 10-3 m respectively. Calculate the effective resistance across AB due to the four resistors and wire W.
resistance = ……………………. Ω [3]
6.0 Ω
20.0 Ω
A
5.0 Ω
B
6.0 V 0.5 Ω
5.0 Ω
W
6
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(b) The four resistors were removed, leaving the cell with wire W. In addition, a 8.0 Ω resistor, a cell of e.m.f of 1.4 V with internal resistance 0.3 Ω, and a sensitive ammeter are connected to the circuit as shown in Fig 2.2.
Fig 2.2
State and explain if a balance point can be obtained.
……………………………………………………………………………………………….
……………………………………………………………………………………………….
……………………………………………………………………………………………….
………………………………………………………………………………………
[3]
0.5 Ω 6.0 V
W
1.4 V 0.3 Ω
8.0 Ω
7
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3 Some water becomes contaminated with radioactive iodine-131 ( I131
53 ). A graph of the
natural logarithm of the activity A of a radioactive iodine plotted against time is
illustrated in Fig. 3.1.
Fig. 3.1
(a) Define decay constant.
……………………………………………………………………………………………….
………………………………………………………………………………………
[2] (b) Using Fig. 3.1, show that the decay constant of the radioactive iodine-131 is
1.01 × 10-6 s-1. [2]
(c) The initial activity of the iodine-131 in 1.0 kg of this water is 1810 Bq. (i) Calculate the initial number of iodine-131 atoms in 1.0 kg of this water.
initial number = ……………………. [2]
8
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(ii) An amount of 1.0 mol of water has a mass of 18 g. Calculate the ratio
number of molecules of water in 1.0 kg of water
number of atoms of iodine-131 in 1.0 kg of contaminated water
ratio = ……………………. [2]
9
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4 A laser is placed in front of a double slit, as shown in Fig. 4.1.
Fig. 4.1
The laser emits light of frequency 6.7 x 105 GHz. Interference fringes are observed on the screen.
(a) Explain how the interference fringes on the screen are formed.
……………………………………………………………………………………………….
……………………………………………………………………………………………….
……………………………………………………………………………………………….
………………………………………………………………………………………
[4]
(b) Show that the wavelength of the light is 450 nm. [1]
10
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(c) The separation of the maxima P and Q observed on the screen is 12 mm. The distance between the double slit and the screen is 2.8 m. Calculate the separation of the two slits.
separation = ……………………. m [2] (d) The laser is replaced by a laser emitting red light. State and explain the effect on
the interference fringes seen on the screen.
…………………………………………………………………………………………........
…………………………………………………………………………………………........
………………………………………………………………………………………
[2]
11
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5 A particle has mass m and charge +q, and is travelling with speed v through a vacuum.
The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 5.1.
Fig. 5.1
The uniform electric field between the plates has magnitude 2.8 × 104 V m-1 and is zero outside the plates. The particle passes between the plates and emerges beyond them, as illustrated in Fig. 5.1.
(a) Explain why the path of the particle in the electric field is not an arc of a circle.
…………………………………………………………………………………………........
………………………………………………………………………………………
[1] (b)
A uniform magnetic field is now formed in the region between the metal plates. The magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 5.2.
Fig. 5.2
(i) State and explain the direction of the magnetic field.
………………………………………………………………………………...
………………………………………………………………………………...
…………………………………………………………………………
[2]
12
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(ii) The particle has speed 4.7 × 105 m s-1.
Calculate the magnitude of the magnetic flux density.
magnetic flux density = ……………………. T [2] (c) The particle in (b) has mass m, charge +q and speed v.
Without any further calculation, state and explain the effect, if any, on the path of a particle that has
(i) mass m, charge –q and speed v,
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[2] (ii) mass m, charge +q and speed 2v,
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[2]
13
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Section B
Answer two questions from this Section in the spaces provided.
6 (a) (i) State Newton’s law of gravitation.
………………………………………………………………………………………..
………………………………………………………………………………………..
……………………………………………………………………………….. [2] (ii) Newton, without knowledge of the numerical value of the gravitational
constant G, was nevertheless able to calculate the ratio of the mass of the
Sun to the mass of any planet in the solar system, provided that the planet had at least a moon. One such situation is shown in Fig. 6.1 where planet A is in circular orbit around the Sun. A moon is in circular orbit around planet A.
Fig. 6.1 1. By considering the relation of the period of orbit to the orbital radius
show that,
23
p
m
m
p
p
s
T
T
R
R
M
M
where Ms = mass of Sun; Mp = mass of planet A; Rp = distance of planet A from Sun; Rm = distance of moon from planet A; Tm = period of moon orbiting planet A; and Tp = period of planet A orbiting Sun.
You may assume that the gravitational force between the Sun and the moon is negligible.
[3]
Sun
Planet A
Moon
14
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2. If planet A is the Earth, where
Rp = 1.50 x 108 km,
Rm = 3.84 x 105 km, Tm = 27.3 days, Tp = 365.2 days,
calculate the ratio of Earthofmass
Sunofmass.
ratio = ……………………. [2]
(b) (i) Define gravitational potential, .
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[2]
15
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(ii) Fig. 6.2 shows a line joining the centre of the Earth to the centre of the moon. Q is a point on this line between the Earth and the Moon where the magnitude of the gravitational potential is a maximum. The ratio of the distances of Q from the centre of the Earth, and to the centre of the Moon,
is given by 9E Mr r .
Fig. 6.2 Given that the Earth’s mass is 81 times that of the moon,
1. Explain why point Q must lie nearer to the centre of the Moon.
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………
[2] 2. Determine the gravitational potential,
Q at point Q.
You may take the mass of the Earth to be 245.97 10 kg .
Q = ……………………. J kg-1 [3]
3.84 x 105 km
Earth
Q
rM Moon
rE
16
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(iii) The gravitational potential at the Moon’s and Earth’s surface are – 3.89 x 106 J kg-1 and – 62.8 x 106 J kg-1 respectively. Sketch in Fig. 6.3, the variation of gravitational potential along the line joining the centre of the Earth and Moon. You do not need to sketch the gravitational potential within the Earth and the Moon. Label the sketch with appropriate numerical values.
Fig. 6.3
[2] (iv) In order to reach the moon, a satellite projected from the Earth’s surface
must at least be able to reach point Q in space. Explain why this is so.
………………………………………………………………………………………..
.……………………………………………………………………………………….
………………………………………………………………………………
[1] (v) Determine the minimum speed at which a satellite must be projected from
the Earth’s surface for it to be able to reach the Moon. Neglect air resistance from the Earth’s atmosphere.
minimum speed = ……………………. m s-1 [3]
Moon
r / 105 km
/ 106 J kg-1
Earth Q
17
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7 (a) State what is meant by simple harmonic motion.
……………………………………………………………………………………………….
……………………………………………………………………………………………….
………………………………………………………………………………………
[2] (b) A small ball rests at point P on a curved track of radius r, as shown in Fig. 7.1.
Fig. 7.1
The ball is moved a small distance to one side and is then released. The horizontal displacement x of the ball is related to its acceleration a towards P by
the expression
g
a xr
where g is the acceleration of free fall.
(i) Show that the ball undergoes simple harmonic motion.
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[2] (ii) In deriving the above expression, it was assumed that the curved track is
smooth. Suggest one other assumption.
………………………………………………………………………………………..
………………………………………………………………………………..
[1]
18
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(iii) The radius of curvature of the track, r is 28 cm.
Determine the time interval, τ between the ball passing point P and then
returning to point P.
τ = ……………………. s [3]
(iv) Suppose that the ball, of mass 5.0 g, is moved up the curved track through
an angle of 1.0 before being released. By assuming that the amplitude is
approximately equal to the arc length or otherwise,
1. show that the amplitude is 34.89 10 m ,
[1]
2. calculate the maximum potential energy of the ball.
maximum potential energy = ……………………. J [2]
19
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(v) On the axes below, show qualitatively the variation with time of the ball’s potential energy, kinetic energy and total energy.
[3] (vi) State and explain how the period of the oscillation would change if a ball
with a larger mass was used instead.
………………………………………………………………………………………..
………………………………………………………………………………..
[2] (c) The variation with time t of the displacement x of the ball in (b) is shown in
Fig. 7.2. T is the period of the oscillation.
Fig. 7.2
Some moisture now forms on the track, causing the ball to come to rest after approximately 15 oscillations.
E / J
t / s
T
20
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(i) State and explain the nature of the damping caused by the moisture on the track.
……………………………………………………………………………………..…
………………………………………………………………………………..
[2]
(ii) On the axes of Fig. 7.2, sketch the variation with time t of the displacement x
of the ball, for the first two periods after the moisture has formed. Assume the moisture forms at time t = 0.
[2]
21
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8 (a) Fig. 8.1 shows the playback head of an audiotape player. The tape is played by passing it at constant speed beneath the gap in the playback head.
Fig. 8.1
(i) Explain, using the principles of electromagnetic induction, how an output current is generated.
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[3]
(ii) Suggest how the magnitude of the output current would be affected by the
speed of the tape.
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[2]
22
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(b) A small coil is positioned so that its axis lies along the axis of a large bar magnet, as shown in Fig. 8.2.
Fig. 8.2
The coil has a cross-sectional area of 0.40 cm2 and contains 150 turns of wire. The average magnetic flux density B through the coil varies with the distance x
between the face of the magnet and the plane of the coil, as shown in Fig. 8.3.
Fig. 8.3
(i) The coil is 5.0 cm from the face of the magnet. Use Fig. 8.3 to determine the
magnetic flux density in the coil.
magnetic flux density = ……………………. T [1]
leads to
coil
axis of coil
and magnet
coil
pole of
magnet
23
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(ii) Calculate the magnetic flux linkage of the coil.
magnetic flux linkage = ……………………. Wb [2] (iii) The coil is moved along the axis of the magnet so that the distance x
changes from x = 5.0 cm to x = 15.0 cm in a time of 0.30 s. Calculate
1. the change in magnetic flux linkage of the coil,
change in magnetic flux linkage = ……………………. Wb [2] 2. the average e.m.f. induced in the coil.
induced e.m.f. = ……………………. V [2] (iv) State and explain the variation, if any, of the speed of the coil so that the
induced e.m.f. remains constant during the movement in (iii).
……………………………………………………………………………………......
……………………………………………………………………………………......
……………………………………………………………………………………......
……………………………………………………………………………………......
………………………………………………………………………………..
[3]
24
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(c) A rectangular loop of mass M, resistance R, and dimensions w by l falls into a
magnetic field as shown in Fig. 8.4. During the time interval before the top edge of the loop reaches the field, the loop approaches a terminal velocity vT.
Fig. 8.4
(i) Sketch the variation of the induced e.m.f. E in the loop with time t.
[2]
(ii) Show that T 2 2
MgRv
B w .
[3]
l
w
E / V
t / s
© IJC 2011 9646/MCE/11 [Turn over
INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level
Higher 2
CANDIDATE
NAME
CLASS INDEX NUMBER
PHYSICS
Paper 3 Longer Structured Questions
Candidates are to answer on the Question Paper
No Additional Materials are required
9646/03
17 September 2015
2 hours
READ THESE INSTRUCTIONS FIRST
Write your name and class on all the work you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams, graphs or rough
working.
Do not use staples, paper clips, highlighters, glue or correction
fluid.
The use of an approved scientific calculator is expected where
appropriate.
Section A Answer all questions.
Section B Answer any two questions. You are advised to spend about one hour on each section. At the end of the examination, fasten all your work securely together. The number of marks is given in the brackets [ ] at the end of each question or part question. Marks will be deducted for using inappropriate number of significant figures or wrong value of g.
This document consists of 24 printed pages.
Innova Junior College [Turn over
For Examiner’s Use
Section A
1
8
2
6
3
8
4
9
5
9
Section B
6
20
7
20
8 20
Penalty
Total
80
Percentage
2
© IJC 2015 9646/03/Prelim 2/15 [Turn over
Data
speed of light in free space, c = 3.00 × 108 m s-1
permeability of free space, o = 4 × 10-7 H m-1
permittivity of free space, o = 8.85 × 10-12 F m-1
(1/(36)) × 10-9 F m-1
elementary charge, e = 1.60 × 10-19 C
the Planck constant, h = 6.63 × 10-34 J s
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
molar gas constant, R = 8.31 J K-1 mol-1
the Avogadro constant, NA = 6.02 × 1023 mol-1
the Boltzmann constant, k = 1.38 × 10-23 J K-1
gravitational constant, G = 6.67 × 10-11 N m2 kg-2
acceleration of free fall, g = 9.81 m s-2
Formulae
uniformly accelerated motion, s = ut + ½at2
v2 = u2+ 2as
work done on/by a gas, W = pV
hydrostatic pressure, p = gh
gravitational potential, = GM
r
displacement of particle in s.h.m., x = xosin ωt
velocity of particle in s.h.m., v = vocos ωt
= 2 2
0x x
mean kinetic energy of a molecule of an ideal gas E = kT2
3
resistors in series, R = R1 + R2 + …
resistors in parallel, 1/R = 1/R1 + 1/R2 + …
electric potential V = Q/4or
alternating current/voltage, x = xo sint
transmission coefficient T = exp (-2kd)
where k = m U - E
h
2
2
8
radioactive decay, x = xo exp(-t)
decay constant, = ½t
0.693
3
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Section A
Answer all the questions in the spaces provided.
1 A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g. The block
is suspended from strings so that it is free to move in a vertical plane. The length of the string is 1.50 m. The bullet buries itself in the block. The block and bullet rise together through a vertical distance of 8.6 cm, as shown in Fig. 1.1
Fig. 1.1
(a) Determine the tension of the string after the block and bullet rise to a vertical distance of 8.6 cm.
cos = (1.50 – 0.086)/1.50
= 19.5 At the highest point, v = 0 and the net force along the string is zero.
T = mg cos T = (0.602)(9.81) cos(19.5) = 5.6 N
tension = ……………………. N [2]
(b) (i) Calculate the change in gravitational potential energy of the block and bullet.
GPE = mgh = 0.602 (9.81)(0.086)
= 0.51 J
energy = ……………………. J [1]
(ii) Show that the initial speed of the block and the bullet, after they began to move off together, was 1.3 m s-1.
[1]
4
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By conservation of energy, Loss in KE = Gain in GPE
½ mv2 – 0 = GPE v = 0.51(2)/(0.602) = 1.3 m s-1
(iii) Using the information in (b)(ii) and the principle of conservation of
momentum, determine the speed of the bullet before the impact with the block. By conservation of linear momentum, Momentum, p of bullet + p of block = total final p of bullet and block mbvb + mblockublock= Mv 0.002vb + 0 = 0.602(1.3) vb = 390 m s-1
speed = ……………………. m s-1 [1] (d) (i) Calculate the kinetic energy of the bullet just before impact.
KE = ½ mv2 = ½ (0.002)(390)2 = 152 J
kinetic energy = ……………………. J [1] (ii) State and explain what can be deduced from your answers to (b)(i) and
(d)(i) about the type of collision between the bullet and the block.
The kinetic energy before impact (152J) is more than the kinetic energy after impact (0.51 J), total kinetic energy of the system decreases after the interaction and thus this is an inelastic collision. .
……………………………………………………………………………….. [2]
2
A cell of e.m.f. of 6.0 V with an internal resistance of 0.5 Ω is connected in an electrical circuit comprising four resistors and a resistance wire W as shown in Fig. 2.1.
6.0 Ω
20.0 Ω
A
5.0 Ω
B
6.0 V 0.5 Ω
5.0 Ω
W
5
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Fig. 2.1
(a) The resistivity of W is 1.25 x 10-5 Ω m. Its length and diameter is 1.20 m and 3.0 x 10-3 m respectively. Calculate the effective resistance across AB due to the four resistors and wire W.
1
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒=
1
6.0+4.0+
1
5.0+
1
2.1
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 1.3 Ω
resistance = ……………………. Ω [3] (b) The four resistors were removed, leaving the cell with wire W. In addition, a 8.0 Ω
resistor, a cell of e.m.f of 1.4 V with internal resistance 0.3 Ω, and a sensitive ammeter are connected to the circuit as shown in Fig 2.2.
Fig 2.2
State and explain if a balance point can be obtained.
Considering the primary circuit.
0.5 Ω 6.0 V
W
1.4 V 0.3 Ω
8.0 Ω
resistance of W = 𝜌𝑙
𝐴
= (1.25 x 10−5)(1.20)
𝜋(1.5×10−3)2
= 2.1 Ω
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The p.d across the wire = 2.1
2.1+8.0+0.5× 6.0 = 1.2 V
In order to have a balance point pd across the wire > 1.4 V, thus a balance point
cannot be obtained.
[3]
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3 Some water becomes contaminated with radioactive iodine-131 ( I131
53 ). A graph of the
natural logarithm of the activity A of a radioactive iodine plotted against time is
illustrated in Fig. 3.1.
Fig. 3.1
(a) Define decay constant.
Decay constant is the probability that a radioactive nucleus in a sample would decay [M1] per unit time [A1]. [2]
(b) Using Fig. 3.1, show that the decay constant of the radioactive iodine-131 is
1.01 × 10-6 s-1.
0tA A e
0ln lnA A t [B1]
Gradient = - λ
7.5 4.0
0 40 24 60 60
[M1]
λ = 1.01 × 10-6 s-1 [A0]
[2]
(c) The initial activity of the iodine-131 in 1.0 kg of this water is 1810 Bq. (i) Calculate the initial number of iodine-131 atoms in 1.0 kg of this water.
A = λN 1810 = 1.01 × 10-6 × N [C1] N = 1.79 × 109 atoms [A1]
initial number = ……………………. [2] (ii) An amount of 1.0 mol of water has a mass of 18 g.
Calculate the ratio
number of molecules of water in 1.0 kg of water
number of atoms of iodine-131 in 1.0 kg of contaminated water
No of water molecules in 1.0 kg
8
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2310006.02 10
18 [C1]
= 3.34 × 1025 molecules
Ratio 25
9
3.34 10
1.79 10
= 1.87 × 1016 [A1]
ratio = ……………………. [2] 4 A laser is placed in front of a double slit, as shown in Fig. 4.1.
Fig. 4.1
The laser emits light of frequency 6.7 x 105 GHz. Interference fringes are observed on the screen.
(a) Explain how the interference fringes on the screen are formed.
……………………………………………………………………………………………….
……………………………………………………………………………………………….
……………………………………………………………………………………………….
………………………………………………………………………………………
[4]
Light waves emerging from each of the double slits are coherent (or have constant phase difference) as they come from the same laser source.
When the light waves meet at a point on the screen they superpose to form interference fringes.
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Bright fringes are formed at the point when the path difference between the waves are an integer multiple of wavelength (n λ) (or have an integer multiple phase difference of 2 π) as they interfere constructively.
Dark fringes are formed at the point when the path difference between the waves are (n + ½) λ (or have a phase difference of (2n + 1) π as they interfere destructively.
(b) Show that the wavelength of the light is 450 nm. [1]
Speed of laser light = 3.0 x 108 m s-1
λ = c / f
= 3.0 x 108 / 6.7 x 105 x 109 = 448 x 10-9 m = 450 nm (shown)
(c) The separation of the maxima P and Q observed on the screen is 12 mm. The
distance between the double slit and the screen is 2.8 m. Calculate the separation of the two slits. Fringe separation, x = 0.012 / 9 = 0.00133 m Slit separation, a = λ D / x = (450 x 10-9)(2.8) / 1.33 x 10-3 = 9.5 x 10-4 m
separation = ……………………. m [2] (d) The laser is replaced by a laser emitting red light. State and explain the effect on
the interference fringes seen on the screen.
…………………………………………………………………………………………........
…………………………………………………………………………………………........
………………………………………………………………………………………
[2]
Red laser light has a wavelength in the region of 700 nm which is higher than the wavelength of 450 nm. Hence, with the same set of double slits and similar slit to screen distance, the fringe separation will be larger.
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5 A particle has mass m and charge +q, and is travelling with speed v through a vacuum.
The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 5.1.
Fig. 5.1
The uniform electric field between the plates has magnitude 2.8 × 104 V m-1 and is zero outside the plates. The particle passes between the plates and emerges beyond them, as illustrated in Fig. 5.1.
(a) Explain why the path of the particle in the electric field is not an arc of a circle.
The resultant force (i.e. the electric force) acting on the particle is always downwards and not acting towards a fixed point. So it is not a circular motion. [1]
(b)
A uniform magnetic field is now formed in the region between the metal plates. The magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 5.2.
Fig. 5.2
(i) State and explain the direction of the magnetic field.
Since the particle passes undeviated between the plates, the direction of the magnetic force on the charged particle should be upwards to balance the downwards electric force [B1]. Using Fleming’s left hand rule, the magnetic field should be acting into the plane of the page [B1]. [2]
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(ii) The particle has speed 4.7 × 105 m s-1. Calculate the magnitude of the magnetic flux density.
FB = FE Bqv sin θ = qE Bq × 4.7 × 105 × sin 90° = q × 2.8 × 104 [C1] B = 5.96 × 10-2 T [A1]
magnetic flux density = ……………………. T [2] (c) The particle in (b) has mass m, charge +q and speed v.
Without any further calculation, state and explain the effect, if any, on the path of a particle that has
(i) mass m, charge –q and speed v,
With the polarity of the charge changed, the electric force will have the same magnitude but now upwards. On the other hand, the magnetic force also has the same magnitude but downwards [M1]. Therefore, the particle will pass undeviated [A1]. [2]
(ii) mass m, charge +q and speed 2v,
When the speed of the particle is increased to 2v, the magnetic force acting upwards is increased to two times while the magnitude of the electric force is unchanged. With the resultant force acting upwards, the particle will deviate upwards [A1]. [2]
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Section B Answer two questions from this Section in the spaces provided.
6 (a) (i) State Newton’s law of gravitation. ………………………………………………………………………………………..
………………………………………………………………………………………..
……………………………………………………………………………….. [2]
Newton’s law of gravitation states that the gravitational force of attraction F between two point masses M and m is directly proportional to the product of their masses and inversely proportional to the square of their separation distance r.
(ii) Newton, without knowledge of the numerical value of the gravitational
constant G, was nevertheless able to calculate the ratio of the mass of the
Sun to the mass of any planet in the solar system, provided that the planet had at least a moon. One such situation is shown in Fig. 6.1 where planet A is in circular orbit around the Sun. A moon is in circular orbit around planet A.
Fig. 6.1 1. By considering the relation of the period of orbit to the orbital radius
show that, 23
p
m
m
p
p
s
T
T
R
R
M
M
where Ms = mass of Sun; Mp = mass of planet A; Rp = distance of planet A from Sun; Rm = distance of moon from planet A; Tm = period of moon orbiting planet A; and Tp = period of planet A orbiting Sun. You may assume that the gravitational force between the Sun and the moon is negligible.
[3]
Sun
Planet A
Moon
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The gravitational force acting on a planet provides for the centripetal force acting on it
to keep it in circular orbit.
For planet A orbiting the sun,
ppp
p
spRM
R
MGM
2
3
2
2p
p
s RT
GM
For the moon orbiting planet A,
mmm
m
mpRM
R
MGM
2
3
22
m
m
p RT
GM
Solving,
23
p
m
m
p
p
s
T
T
R
R
M
M (shown)
2. If planet A is the Earth, where
Rp = 1.50 x 108 km,
Rm = 3.84 x 105 km, Tm = 27.3 days, Tp = 365.2 days,
calculate the ratio of Earthofmass
Sunofmass.
23
p
m
m
p
p
s
T
T
R
R
M
M
23
5
8
365.2
27.4
10 x 3.84
10 x 1.50
p
s
M
M
= 3.36 x 105
ratio = ……………………. [2]
Sun (Ms)
Planet A (Mp)
Moon
Rp Rm
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(b) (i) Define gravitational potential, .
………………………………………………………………………………………..
………………………………………………………………………………………..
……………………………………………………………………………….. The gravitational potential at a point in a gravitational field is the work done per unit mass by an external agent in bringing a point mass from infinity to that point.
[2]
(ii) Fig. 6.2 shows a line joining the centre of the Earth to the centre of the moon. Q is a point on this line between the Earth and the Moon where the magnitude of the gravitational potential is a maximum. The ratio of the distances of Q from the centre of the Earth, and to the centre of the Moon,
is given by 9E Mr r .
Fig. 6.2
Given that the Earth’s mass is 81 times that of the moon,
1. Explain why point Q must lie nearer to the centre of the Moon.
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………
[2]
Since point Q is a point with maximum gravitational potential, the
resultant gravitational field strength which is numerically equal to the
potential gradient at that point is zero. Hence, Q must lie further away
from the centre of the Earth and nearer to the centre of the Moon as
the gravitational field strength due to the Earth is much greater than
that due the Moon at that point because of its much greater mass.
3.84 x 105 km
Earth
Q
rM Moon
rE
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2. Determine the gravitational potential, Q at point Q.
You may take the mass of the Earth to be 245.97 10 kg .
810x843. ME rr m
ME rr 9
Solving, 710x843.Mr m and 810x463.Er m
M
M
E
EQ
r
GM
r
GM
7
24
8
2411
10x843
10x98581
1
10x463
10x98510x676
.
.
.
..
610x281. J kg-1
Q = ……………………. J kg-1 [3]
(iii) The gravitational potential at the Moon’s and Earth’s surface are
– 3.89 x 106 J kg-1 and – 62.8 x 106 J kg-1 respectively. Sketch in Fig. 6.3, the variation of gravitational potential along the line joining the centre of the Earth and Moon. You do not need to sketch the gravitational potential within the Earth and the Moon. Label the sketch with appropriate numerical values.
Fig. 6.3
- for correct graph shape (smooth curve) ; – for correct numerical values
[2]
Moon
r / x 105 km
Φ / x 106 J kg-1
Earth
- 1.28
- 3.89
- 62.8
Q
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(iv) In order to reach the moon, a satellite projected from the Earth’s surface
must at least be able to reach point Q in space. Explain why this is so.
………………………………………………………………………………………..
.……………………………………………………………………………………….
………………………………………………………………………………
[1]
The gravitational field strength due to the Earth is directed to the left at point Q and that due to the Moon is directed to the right. Hence, a satellite projected from the Earth’s surface must have sufficient kinetic energy to at least overcome the gravitational potential barrier between the Earth’s surface and Point Q for it to be subsequently pulled towards the moon’s surface by the moon’s gravitational field.
(v) Determine the minimum speed at which a satellite must be projected from the Earth’s surface for it to be able to reach the Moon. Neglect air resistance from the Earth’s atmosphere.
Let the minimum projection speed be vmin and the mass of the satellite be ms.
By applying conservation of energy, KE lost ≥ GPE gain
sESurfQs mvm 02
1 2
min
ESurfQv 22
min
6210x8622812 ..min v
410x111.min v m s-1
Therefore, the minimum projection speed = 1.11 x 104 m s-1
minimum speed = ……………………. m s-1 [3]
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7 (a) State what is meant by simple harmonic motion.
Simple harmonic motion is defined as the periodic motion of a body in which its acceleration is directly proportional to its displacement from a fixed point, and acceleration is in the opposite direction to the displacement.
[2]
(b) A small ball rests at point P on a curved track of radius r, as shown in Fig. 7.1.
Fig. 7.1 The ball is moved a small distance to one side and is then released. The horizontal displacement x of the ball is related to its acceleration a towards P by the expression
g
a xr
where g is the acceleration of free fall.
(i) Show that the ball undergoes simple harmonic motion.
Since g and r are both constants, the equation is of the form y= - kx, hence
the acceleration of the ball is directly proportional to its displacement. Also, the negative sign indicates that the ball’s acceleration is in the opposite direction to its displacement. Hence ball undergoes SHM.
[2] (ii) In deriving the above expression, it was assumed that the curved track is
smooth. Suggest one other assumption.
Ball is displaced by a small angle θ, or The radius of curvature is large
[1]
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(iii) The radius of curvature of the track, r is 28 cm.
Determine the time interval, τ between the ball passing point P and then
returning to point P.
2
1
[M1 with correct substitution]
9.815.919 rad s
0.28
25.919
1.0615 s [M1]
The time interval to pass and return to it is 2
1.06150.531 s [A1]
2
g
r
T
T
TP
τ = ……………………. s [3]
(iv) Suppose that the ball, of mass 5.0 g, is moved up the curved track through
an angle of 1.0 before being released. By assuming that the amplitude is
approximately equal to the arc length or otherwise,
1. show that the amplitude is 34.89 10 m ,
[1]
3
3
o
1(0.28)( 2 )
360
4.8869 10 m [M1]
Assuming 4.8869 10 m
s r
x s
2. calculate the maximum potential energy of the ball.
2 2
o
3 2
6
6
1Max PE
2
1(0.005)(35.03)(4.8869 10 )
2
2.09 10 J
Alternatively,
Max PE mgh
(0.005)(9.81)(0.28 0.28cos1 )
2.09 10 J
m x
maximum potential energy = ……………………. J [2]
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(v) On the axes below, show qualitatively the variation with time of the ball’s
potential energy, kinetic energy and total energy.
[3] (vi) State and explain how the period of the oscillation would change if a ball
with a larger mass was used instead.
No change, since the angular frequency g
r , is independent of the ball’s
mass.
[2] (c) The variation with time t of the displacement x of the ball in (b) is shown in
Fig. 7.2. T is the period of the oscillation.
Fig. 7.2
T
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Some moisture now forms on the track, causing the ball to come to rest after approximately 15 oscillations.
(i) State and explain the nature of the damping caused by the moisture on the track.
Light damping since the amplitude of the oscillations die gradually i.e. after 15 oscillations.
[2]
(ii) On the axes of Fig. 7.2, sketch the variation with time t of the displacement x
of the ball, for the first two periods after the moisture has formed. Assume the moisture forms at time t = 0.
[2]
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8 (a) Fig. 8.1 shows the playback head of an audiotape player. The tape is played by passing it at constant speed beneath the gap in the playback head.
Fig. 8.1
(i) Explain, using the principles of electromagnetic induction, how an output current is generated.
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[3]
(ii) Suggest how the magnitude of the output current would be affected by the
speed of the tape.
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………………..
………………………………………………………………………………..
[2]
Magnetic field lines from the tape are channelled via the iron, through the coil
[M1].
As the magnetisation of the tape surface changes, the magnetic flux linkage
through the coil changes [M1].
By Faraday’s Law, there is an induced emf/current in the coil due to rate of change of
magnetic flux linkage [A1], which is amplified and then sent to a loudspeaker.
As the speed of tape increases, the rate at which the magnetic flux linkage through
the coil increases [B1]. Hence, by Faraday’s Law, induced emf, and therefore current,
in the coil increases. [B1]
OR
As the speed of tape decreases, the rate at which the magnetic flux linkage
through the coil decreases [B1]. Hence, by Faraday’s Law, induced emf, and
therefore current, in the coil decreases. [B1]
22
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(b) A small coil is positioned so that its axis lies along the axis of a large bar magnet, as shown in Fig. 8.2.
Fig. 8.2
The coil has a cross-sectional area of 0.40 cm2 and contains 150 turns of wire. The average magnetic flux density B through the coil varies with the distance x
between the face of the magnet and the plane of the coil, as shown in Fig. 8.3.
Fig. 8.3
(i) The coil is 5.0 cm from the face of the magnet. Use Fig. 8.3 to determine the
magnetic flux density in the coil.
magnetic flux density = ……………………. T [1]
leads to
coil
axis of coil
and magnet
coil
pole of
magnet
50 mT (allow 50 ± 1 mT for full credit) [A1]
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(ii) Calculate the magnetic flux linkage of the coil.
magnetic flux linkage = ……………………. Wb [2] (iii) The coil is moved along the axis of the magnet so that the distance x
changes from x = 5.0 cm to x = 15.0 cm in a time of 0.30 s. Calculate
1. the change in magnetic flux linkage of the coil,
change in magnetic flux linkage = ……………………. Wb [2] 2. the average e.m.f. induced in the coil.
induced e.m.f. = ……………………. V [2] (iv) State and explain the variation, if any, of the speed of the coil so that the
induced e.m.f. remains constant during the movement in (iii).
……………………………………………………………………………………......
……………………………………………………………………………………......
……………………………………………………………………………………......
……………………………………………………………………………………......
………………………………………………………………………………..
[3]
flux linkage = BAN [C1]
= 50 × 10−3 × 0.4 × 10−4 × 150 [M1] = 3.0 × 10−4 Wb [A1]
new flux linkage = 8.0 × 10−3 × 0.4 × 10−4 × 150 = 4.8 × 10−5 Wb [M1] change = |(4.8 × 10−5) – (3.0 × 10−4)| = 2.52 × 10−4 Wb [A1]
Ave e.m.f. = |∆∅
∆𝑡|
= 2.52 × 10−4
0.30[M1]
= 8.4 × 10−4 V [A1]
The magnetic flux density of the magnet, and hence magnetic flux linkage with coil, decreases as the separation increases. Hence, the change in flux linkage over a given distance is now smaller when the coil is further. [B1] For the induced emf to remain the same, the rate of change of flux linkage with respect to time must be constant. [B1]
24
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(c) A rectangular loop of mass M, resistance R, and dimensions w by l falls into a
magnetic field as shown in Fig. 8.4. During the time interval before the top edge of the loop reaches the field, the loop approaches a terminal velocity vT.
Fig. 8.4
(i) Sketch the variation of the induced e.m.f. E in the loop with time t.
[2]
(ii) Show that T 2 2
MgRv
B w .
[3]
l
w
E / V
t / s
BwVT
𝐸𝑖 = 𝑑Ф
𝑑𝑡=
𝑑(BA)
𝑑𝑡=
𝐵𝑑(A)
𝑑𝑡
=𝐵𝑑(𝑙 ∙ w)
𝑑𝑡
=𝐵𝑤𝑑(𝑙)
𝑑𝑡
𝐼 = 𝐸𝑖
𝑅=
𝐵𝑤𝑣𝑇
𝑅 [M1]
𝑀𝑔 = 𝐵(𝐵𝑤𝑣𝑇
𝑅)𝑤
𝑀𝑔 = 𝐵𝐼𝑤 [M1]
= 𝐵𝑤𝑣𝑇 [M1]
T 2 2
MgRv
B w [A0]