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Anglo-Chinese Junior College Physics Preliminary Examination Higher 2

PHYSICS Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet

9646/01 18 September 2014

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 and Index number in the answer sheet provided.

There are forty questions in this section. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the 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 Question Paper.

2

2014 H2 9646 Prelim Exam P1

DATA AND FORMULAE Data speed of light in free space, c = 3.00 108 m s1

permeability of free space, o = 4 107 H m1

permittivity of free space, o = 8.85 1012 F m1

(1/(36)) 10-9 F m-1

elementary charge, e = 1.60 1019 C

the Planck constant, h = 6.63 1034 J s

unified atomic mass constant, u = 1.66 1027 kg

rest mass of electron, me = 9.11 1031 kg

rest mass of proton, mp = 1.67 1027 kg

molar gas constant, R = 8.31 J K1 mol1

the Avogadro constant, NA = 6.02 1023 mol1

the Boltzmann constant, k = 1.38 1023 J K1

gravitational constant, G = 6.67 1011 N m2 kg2

acceleration of free fall, g = 9.81 m s2

Formulae

uniformly accelerated motion, s = ut + 21 at 2

v 2 = u 2 + 2as

work done on/by a gas, W = p V 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 cos t

= 22 xxo

mean kinetic energy of a molecule of an ideal gas

E = 3

2kT

resistors in series, R = R1 + R2 +

resistors in parallel, 1/R = 1/R1 + 1/R2 +

electric potential, V = r

Q

o4

alternating current/voltage, x = xo sin t

transmission coefficient, T exp(2kd)

where k = 2

2 )(8

h

EUm

radioactive decay, x = xo exp(t)

decay constant, = 2

1

6930

t

.

3

2014 H2 9646 Prelim Exam P1 [Turn over

1 A student uses a digital ammeter to measure a current. The reading of the

ammeter is found to fluctuate between 1.98 A and 2.02 A. The manufacturer of the ammeter states that any reading has a systematic uncertainty of

1% . Which value of the current should be quoted by the student?

A

2.000.01 A

B

2.000.02 A

C

2.000.03 A

D

2.000.04 A

2 An athlete of mass 80 kg competes in a 100 m race.

What is the best estimate of his mean kinetic energy during the race?

A 4 x 102 J B 4 x 103 J C 4 x 104 J D 4 x 105 J

3 The figure below shows a circular path taken by a particle. At a certain instant,

the particle’s horizontal velocity, vx , is + 2 m s-1, and its vertical velocity,

vy, is – 2 m s-1.

Horizontally, right is taken to be positive x-direction.

Vertically, upward is taken to be positive y-direction.

Through which quadrant is the particle moving at this instant if it is travelling

anti-clockwise around the circle?

4 Two diamonds begin free fall from rest from the same height 1.0 s apart. How long after the first diamond begin to fall will the two diamonds be 10 m apart?

A 0.43 s B 1.43 s C 1.52 s D 2.43 s

A

B

D

C

O

4


5 A ball is released from rest on a smooth slope XY.

It moves down the slope, along a smooth surface YZ and rebounds inelastically at Z. Then it moves back to Y and comes to rest momentarily somewhere on XY. Which velocity-time graph represents the motion of the ball?

A

B

C

D

5


6 A varying force acts upon an object. The graph below shows how the force acting on the object varies with time.

The magnitude of the maximum change in momentum of the object is

A 3.0 Ns B 2.0 Ns C 1.5 Ns D 1.0 Ns

7 A stationary Thoron nucleus of mass 220u emits an alpha particle of mass 4u with

kinetic energy E. Which of the following gives the correct value of

the ratio speed of alpha particle

speed of the recoiling daughter nucleus

and kinetic energy of the daughter nucleus immediately after the emission?

speed of alpha particle

speed of the recoiling daughter nucleus

kinetic energy of the recoiling daughter nucleus

A 55 1

55 E

B 54 (1

54)2 E

C 54 1

54 E

D 1

54 54 E

0 0.1 0.2 0.3 time/s

Force/N

10

6


8 A cupboard is attached to a wall by a screw. Which force diagram shows the cupboard in equilibrium, with the weight W of the

cupboard, the force S that the screw exerts on the cupboard and the force R that the wall exerts on the cupboard?

The magnitude of the forces are not drawn to scale.

A B

C D

7


9 The hull of a ship may be assumed to have a uniform horizontal cross-sectional area in the region above and below the water line.

When in seawater of density 1.06 x 103 kg m-3, the ship floats with 3.00 m of its hull

below water. The ship then travels into a river estuary when the density of the water is 1.01 x 103 kg m-3.

What is the new submerged depth of the hull?

A 3.35 m B 3.15 m C 3.05 m D 2.86 m 10 A uniform diving board of length 5.0 m and mass 35 kg hangs horizontally in

equilibrium when it is hinged at one end and supported 2.0 m from this end by a spring of spring constant 10 kN m–1.

When a child of mass 40 kg stands at the far end of the board as shown in the diagram below, what is the extra compression of the spring caused by the child standing on the end of the board?

A 1.0 cm B 5.9 cm C 9.8 cm D 19.6 cm

8


11 A mass on a smooth horizontal table is attached by two light springs to two fixed supports. The mass executes linear simple harmonic motion.

Which of the graphs correctly show how the kinetic energy (KE), elastic potential

energy (EPE) and gravitational potential energy (GPE) would vary with time?

A

B

C

D

E

t

T

GPE

EPE

KE

E

t

T

KE

EPE

GPE

E

t

T

GPE

KE

EPE

KE

E

t

T

GPE

EPE

9


12 A car moving through air at velocity v experiences a resistive force F given by the expression F = kv2 where k is a constant.

Which of the following graphs show how the power supplied to the car P will vary at

various v to ensure that the car is moving without any acceleration?

A

B

C

D

13 A wind turbine has blades that sweep an area of 2000 m2. It converts the power

available in the wind to electrical power with an efficiency of 50%. What is the electrical power generated if the wind speed is 10 m s–1? (The density of air is 1.3 kg m–3.)

A 130 kW B 650 kW C 1300 kW D 2600 kW

14 A car is making a turn at speed v. The radius of the turn is r and the centripetal force

on the car is F. If the car rounds the same curve at speed 2v, the required centripetal force is

A 1

2F B F C 2F D 4F

P

v

P

v

P

v

P

v

10


15 The Earth has approximately 81 times the mass of the Moon. There is a point between the Earth and the Moon where the resultant gravitational force on a mass m is zero. If the distance to this point from the centre of the Earth is y and from the centre of the Moon it is x, the ratio y / x

A 4

1

81 B 4

1

81 2

1

81 C 81 D 281

16 Fig. 16 shows the gravitational equipotentials near a certain non-spherical body.

Fig. 16

What is the work that must be done on 100 g mass to move it from P to Q

A 50 J B - 50 J C 5000 J D -5000 J

17 Simple harmonic motion is defined as the motion of a particle such that

A its displacement x from the equilibrium position is always given by the expression x = xosinωt

B its displacement x from the equilibrium position is related to its velocity by the expression v = ωx

C its acceleration is proportional to, and in the opposite direction to, the displacement from the equilibrium position.

D Its acceleration is always ω2xo and is directed at right angles to its motion.

11


18 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 half of that before. Which line shows the new variation of acceleration a with displacement x for the object?

19 A particle is in simple harmonic motion. What is the magnitude of the phase difference between the velocity and

displacement of particle?

A 4

rad

B 2

rad

C rad D 3

2rad

12


20 The variation with frequency f of the amplitude xo of the forced oscillations of a machine is as shown.

Which of the following statement is false?

A Resonance frequency of the above system is 12 Hz

B For no damping, the maximum amplitude is 1.6 cm.

C Increased damping reduces the sharpness of the resonance peak

D Damping decreases the maximum amplitude of the oscillations

21 Argon and chlorine are gases. One mole of argon has mass 40 g and one mole of

chlorine has mass 71 g.

What is the ratio of the

A 0.56 B 1 C 1.8 D 2

22 An ideal gas undergoes a cycle of processes as shown in the PV diagram below.

Which statement correctly describes the situation?

A The internal energy of the gas increases over one complete cycle. B The gas gives out more heat than it absorbs over the whole cycle. C Over the entire cycle, the gas absorbs heat and does network on its

environment. D The two curved portions of the graph represent adiabatic process.

xo / cm

13


23 Equal masses of four different liquids (A, B, C and D) at 20°C are separately heated at

the same rate. Their boiling points and specific heat capacities are as shown in the table below.

Liquid Boiling Point / °C Specific Heat Capacity / J kg-1

K-1

A 50 1000

B 60 530

C 80 850

D 360 140

Which liquid will be the last to boil?

A Liquid A B Liquid B C Liquid C D Liquid D 24 A progressive wave in a stretched string has a speed of 2 m s-1 and a frequency of

100 Hz. What is the phase difference between two points 25 mm apart?

A zero B 4

C

2

D

25 Two coherent monochromatic waves of equal amplitude are brought together to form

an interference pattern on a screen. Which of the following graphs could represent the variation of intensity with position x

across the pattern of fringes.

A B

C D

14


26 A double-slit interference experiment is set up as shown.

Fringes are formed on the screen which is 40.0 cm away. The distance between the

slits is 9.0 cm apart. The microwave source has a wavelength of 2.0 cm and intensity of I.

What is the resultant intensity at X?

A Zero B I C 2I D 4I 27 Fig 27 shows some equipotential lines in an electric field.

Fig 27

The magnitude of the electric field strength at X is EX and at Y is EY. Which of one of the following correctly compares EX and EY and gives the correct

directions of the electric field?

Magnitude of electric field strength Direction of electric field

A EX >EY X Y

B EX >EY Y X

C EX <EY X Y

D EX <EY Y X

X Microwave Source

15


28 Two charged conducting spheres each of radius 1.0 cm are placed with their centres 10.0 cm apart, as shown in Fig 28.1.

Fig 28.1

Sphere A carries a charge of + 2.0 x 10-10 C. The graph in Fig 28.2 shows how the resultant electric field strength E, between the

two spheres varies with distance x.

Fig 28.2

What is the magnitude of the electric field strength due to the charge on sphere B at

the 5.0 cm mark. Identify the nature of the charge on sphere B.

Magnitude of EB / Vm-1 Nature of charge on sphere B

A 1.08 x 103 positive

B 1.08 x 103 negative

C 1.76 x 103 positive

D 1.76 x 103 negative

16


29 A battery of internal resistance r and e.m.f. E can supply a current of 6.0 A to a resistor R as shown in Fig 29.1.

The I V characteristics of the resistors R and R1 respectively is shown in Fig 29.2.

R

Fig 29.1. Fig 29.2

When the resistor R is replaced by R1, the current becomes 5.0 A. What are the values of the e.m.f. E and the internal resistance r?

E / V r /

A 7.6 0.073

B 12 2.0

C 12 0.80

D 15 8.0

30 Four identical resistors are connected as shown in Fig 30.

Fig 30

How will the powers to the resistors change when the resistor W is removed?

A The powers to X,Y and Z will all increase. B The powers X will decrease and the powers to Y and Z will increase. C The powers to X will increase and the powers to Y and Z will decrease. D The powers to X will increase and the powers Y and Z will remain unaltered.

17


31 Fig 31 shows the top view of a current balance. The wire frame ABCD is supported by the pivots P and Q. PBCQ lies within the magnetic field, whose flux density is to be measured. The sides AD and BC are equidistant from the pivots. Electrical connections are made to the frame through the pivots.

Fig 31

Given that the current I flowing in the current balance is 1.2 A, and the total mass of small weights m is 30 g, calculate the flux density of the magnetic field.

A 0.625 T B 3.1 T C 6.1 T D 625 T

32 A soft-iron ring of variable cross-section has four coils wound round it at the positions

shown. The coils have 2, 3, 4 and 5 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 least variation?

18


33 The primary of an ideal transformer has 200 turns and is connected to a 15 V root-mean-square (r.m.s.) supply. The secondary 1600 turns and is connected to a resistor of resistance 80 Ω, as shown in the diagram.

What are possible values of the secondary voltage, the secondary current and the mean power dissipated in the resistor?

secondary voltage / V r.m.s.

secondary current / A r.m.s.

mean power in resistor / W

A 1.9 0.02 0.38

B 1.9 42 80

C 120 0.67 80

D 120 1.5 180

34 A laser point creates a spot on a screen as it reflects 70% of the light striking

it. This light exerts radiation pressure on the screen. The laser point is now moved twice as far away from the screen.

i. The radiation pressure remains the same because the intensity of the laser light remains constant.

ii. The radiation pressure decreases because the beam diverges and area of illumination increases.

iii. The radiation pressure decreases because energy of the light is

lost due to scattering from air molecules and dust particles light travels a longer distance to the screen.

Which of the statements above is/are correct?

A i only

B ii only

C iii only

D ii and iii only

15 V a.c. r.m.s. 80 Ω

19


35 Which of the following statement on photoelectric effect is not an evidence

for particulate nature of light?

A Emission of electrons happens as soon as light shines on metal.

B Increasing intensity of light increases rate at which electrons leave metal.

C Maximum speed of emitted electrons is dependent on the frequency of

incident light.

D A minimum threshold frequency of light is needed.

36 Which of the following about doped semiconductors is correct?

Charge of semiconductor Majority charge carriers

A

p – type n – type

positive

negative

holes

electrons

B


neutral

neutral

holes

electrons

C


positive

neutral

protons

neutrons

D


neutral

neutral

protons

electrons

37 Suppose Fuzzy, a quantum-mechanical duck of mass 2.00 kg, lives in a world in which

h, the Planck constant, is 2 J s. Fuzzy is initially known to be within a pond 1.00 m wide. What is the minimum uncertainty in the component of his velocity parallel to the width of the pond?

A 0.250 m s−1 B 1.00 m s−1 C 2.50 m s−1 D 3.14 m s−1

20


38 The figure below shows the wave function ψ(x) of an electron.

Which of the following statements is correct?

A The probability of locating the electron between positions P and Q is

Q

Px dx .

B 2

x is the probability of locating the electron within a given region.

C There is greater probability of locating the electron on the left of the vertical axis.

D

The probability of locating the electron at x = 0 is the highest.

39 An electron is incident on a rectangular potential barrier with a kinetic energy of 2.0 eV.

The barrier height is 6.0 eV and its width is d = 10100.1 m. If the width of the barrier is

reduced to 'd and the transmission coefficient is doubled, the ratio d

d ' is

A 0.50 B 0.66 C 0.72 D 2.0

40 The activity of a radioactive sample decreases to one third of its original activity Ao in a

period of 3 years. After 3 more years, its activity would be

A 0

9

1A B 0

6

1A C 0

3

1A D 0

3

2A

P 0 Q

ψ(x)

x

H2 Physics P1

Qn Ans Key

Discussions

1 D random errors is ±0.02 + systematic error is ±0.02

2 B estimate average speed. apply ½ mv2.

3 C

4 C

Solving 101

2

1

2

1 22 tggtgives t = 1.52 s

5 A The 2 slopes have the same gradient since same magnitude of acceleration. The time taken to travel back to slope is longer after collision since speed after rebound is lower.

6 C Area under F-t graph gives change in momentum

7 C Use Conservation of momentum to get ratio of speed, hence KE

8 A No resultant force 3 forces in equilibrium must be concurrent

9 B When floating, U = W

Hence 1h1Ag = 2h2Ag

10 C The additional clockwise moment of spring on board = additional anti-clockwise moments when man steps on board.

11 C Recall the nature of KE and EPE graphs in SHM. GPE depends on reference point.

12 C Power = Fv = kv3

13 B 9702/June 2013 P1 Q18

Power i/p = ½ Av3

14 D Since

, therefore the force will be 4 times.

15 B At the point where gravitational force on mass m is zero the gravitational force on the mass m exerted by earth is equal and opposite to that exerted by moon on mass m . Applying Newton’s law of gravitation to determine magnitude of these two forces

16 A Work done by external agent on mass = change in GPE of the 100g mass

U=m(f - i) = 0.100( -2000-(-2500)) = 50 J

17 C Definition of SHM

18 C Use amax = ω2xo

19 B

20 B For no damping, the maximum amplitude is infinity.

21 B The Avogadro number is a constant.

22 B As can be seen from the area under the graph, there is more work done on the gas than by the gas when expanding, hence, there is net heat loss by the gas each cycle.

23 C By considering Q = mcΔθ

24 C 1λ=2 cm, hence 2.5 cm is 1.25 λ; 0.25 x 2π = 0.5 π

25 A The standard double slit interference pattern is shown in A

26 A Path difference of 1 cm, destructive interference results.

27 A Electric field strength is the –ve of the electric potential gradient which is the change in electric potential per unit distance hence EX > EY Direction of field always in direction of decreasing potentials.

28 B Resultant field strength E = EA + EB

Calculate EA by applying E = 2

0r 4

Q

= 0.72 x103 Vm-1

From the graph at x = 5 cm = 5.0 x 10-2 m the magnitude of resultant field is 1.8 x103 Vm-1 hence EB = 1.08 x103 Vm-1; and QB is a negative charge as its direction is towards +ve x direction; so the resultant field is greater than that due to QA.

if candidate gave EB = 1.76 x103 Vm-1, they must have use V = r 4

Q

0

instead

29 C From the IV graphs determine R when current is 6.0 A; R=1.2

When current is 5.0 A; R= R1 =1.6 Using equation E=IR+Ir Substituting corresponding values of I and R Determine E and r OR E=IR+Ir = E=V+Ir Substituting corresponding values of V and I Determine E and r

30 C Let the potential difference across the supply be V Before W is removed the potential drop across each resistor is the same Hence power dissipated in each is the same and is the current through each

resistor is equal to R 2

V

as the total resistance the circuit is R

When W is removed, total resistance in circuit increases 1.5 R

Hence the total current in the circuit is 3R

2V

The current through X is 3R

2V

and that through each of Y and Z is 3R

V

Thus the potential drop across X increases and that across each of Y and Z decreases, hence the power dissipated in X will increase and that in each of Y and Z will decrease.

31 C

0.030 9.81

1.2 0.04

6.1

Bweight F

mg BIL

mgB

IL

T

32 A

,

Since magnetic flux

magnetic flux density

BA

BA

Magnetic flux is the same throughout the core assuming no leakage.

Since section (A) has the biggest cross-sectional area, it has the smallest magnetic flux density B. Note that as N is the smallest for A, it also has the smallest flux linkage, hence A is the region with the least variation of magnetic flux density.

33 D Vs/Vp = Ns/Np V = IR Pmean = IrmsVrms

34 D

35 B

36 B P and N-type semiconductors are neutral. You simply doped with valency V or III elements.

37 A Apply ∆p∆x ≥ h/4

38 D Know the idea behind finding the probability of locating a particle with its wave function

39 B Find k and apply T α e-2kd

k d dTe

T

Tk d d

T

2 ( ' )'

'ln( ) 2 ( ' )

40 A Apply I = Ioe-t when t = 3 and t = 6

This paper consists of 22 printed pages.


CANDIDATE NAME

CLASS

CENTRE NUMBER s

INDEX NUMBER

PHYSICS Paper 2 Structured Questions Candidates answer on the Question Paper. No Additional Materials are required

9646/02 27 Aug 2014

1 hour 45 minutes


Write your Name and Index number in the spaces on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiners’ use only

1 / 8

2 / 7

3 / 7

4 / 8

5 / 7

6 / 7

7 / 16

8 / 12

Total / 72

2


For Examiner’s

Use DATA AND FORMULAE

Data speed of light in free space, c = 3.00 108 m s1



(1/(36)) 109 F m1











Formulae


v 2 = u 2 + 2as

work done on/by a gas, W = p V

hydrostatic pressure, p = g h


Gm



= 22 xxo


E = 3

2kT




Q

o4



where k = 2

2 )(8

h

EUm


decay constant, = 2

1

6930

t

.

3


For Examiner’s

Use

Answer all the questions in the spaces provided. 1 (a) Define work. [1]

(b) Energy can be defined as the ability to do work.

In each of the following cases, state a possible object on which work may be done and explain how the law of conservation of energy can be applied. An example of how the question should be answered is provided below.

Situation On what object is the work done?

Application of conservation of energy

Car stopping without skidding from a speed of 40 m s-1.

Work is done on the brakes. Almost all of the kinetic energy becomes heat energy.

Steam engine where a constant mass of steam at constant pressure increases its volume.

[1]

[1]

Photocell / Light Dependent Resistor (LDR) in which light incident on a metal surface results in decrease in resistance.

[1]

[1]

Battery pushing charge through an electric chainsaw cutting a thick slab of wood.

[1]

[2]

4


For Examiner’s

Use 2 (a) (i) Given that the mass of an oxygen molecule is 5.3 × 10–26 kg, show that the root

mean square (rms) speed for oxygen molecules in the atmosphere when the temperature is 23 °C is 480 m s-1.

[1] (ii) Explain why the rms speed of argon atoms in the atmosphere at 23 °C will be

different from that of the oxygen molecules given in (i)

[1]

(b) The rms speed of hydrogen molecules at 23 °C is 1920 m s–1. The escape velocity

from the Earth is 11 000 m s–1. Explain why almost all the molecules of hydrogen that have ever been in the Earth’s atmosphere have escaped into space but many oxygen molecules have remained in the atmosphere.

[2]

(c) Using the Kinetic model of gases, explain how gases exert a pressure on the sides

of its container. [3]

5


For Examiner’s

Use

3 (a) An electron is travelling at right angles to a uniform magnetic field of flux density

1.5 mT, as illustrated in Fig 3.1.

Fig 3.1

The magnetic field is acting out of the plane of the paper. When the electron is at P, its velocity is 1.8 x 107 m s-1 in the direction shown. This is normal to the magnetic field.

(i) On Fig 3.1, sketch the path of the electron, assuming that it does not

leave the region of the magnetic field. [1]

(ii) 1. Show that the force on the electron due to the magnetic field is

4.32 x 10-15 N.

[1]

2. Hence calculate the radius of the electron’s path.

radius = ……………………….. m [3]

(b) A uniform electric field is now produced in the same region and in the opposite direction to the magnetic field. Suggest the shape of the resultant path of the electron and draw a sketch to illustrate the path.

……………………………………………………………………………………………………… …………………………………………………………………………………………………. [2]

P

Region of uniform magnetic field out of plane

of paper

6


For Examiner’s

Use 4 (a) Explain what is meant by the term interference.

[2]

(b)

A Kundt’s tube is an experimental acoustical instrument that serves to measure the speed of sound in different medium. It comprises of a long horizontal tube, containing a fine powder, which is closed at one end. A loudspeaker connected to a signal generator is positioned at the other end. The device is shown in Fig 4.1 and the signal generator is set to a frequency of 400 Hz. From the interference of waves resulting in stationary waves being formed, an interesting pattern can be observed as seen in Fig 4.1.

Fig 4.1

(i) Label the positions of nodes (N) and antinodes (A) on Fig. 4.1 [1] (ii) Explain the formation of the piles of fine powder at the positions

shown on Fig 4.1.

[3]

(iii) Hence or otherwise, determine the speed of the sound waves in that medium.

Speed of Sound = m s-1 [2]

123 cm

Signal Generator

400 Hz

Loudspeaker

Piles of powder

7


For Examiner’s

Use

5 (a)

Stimulated emission and spontaneous emission are two processes in which photon emissions can take place. Explain the main difference between how these processes can happen.

[1]

(b) State and explain the importance of stimulated emission in the production of lasers.

[2]

(c) Explain what is meant by population inversion and why is it an essential condition in

laser production.

[2]

8


For Examiner’s

Use 5 (d) Fig 5(a) shows the energy level diagram of a three-level laser. Lasing takes place

between E2 and E1 while Fig 5(b) shows the energy level diagram of a four-level laser. Lasing takes place between E3 and E2.

Fig 5(a) Fig 5(b)

State the advantage of the four-level laser over the three-level laser.

[2]

E3

Ground state

E4

E2

E1 E1

E3

Ground state

E2

9


For Examiner’s

Use

6 (a) Thoron is a radioactive gas. The variation with time t of the detected count rate C from a sample of the gas is shown in Fig. 6.1

Fig 6.1

(i) Radioactive decay is said to be a random and spontaneous process.

State the feature of Fig 6.1 which indicates that the process is random.

[1]

(ii) A second similar sample of thoron is prepared but it is at a much higher temperature. The variation with time of the count rate for this second sample is determined.

State the expected feature of the decay curve for the second sample, with

reference to Fig 6.1 that suggests that radioactive decay is a spontaneous process.

[1]

10


For Examiner’s

Use 6 (b) In order to identify the radioactive particles emitted by a given sample of radioactive

isotope, a student set up the apparatus as illustrated in Fig 6.2 and vary the thickness of Al sheets used. The separation from the window of the detector from the shielding is about 6 cm.

Fig 6.2 Fig 6.3

Fig 6.3 is the best fit line showing the variation of the count rate with the thickness x of Al sheets used. The count rate plotted in Fig 6.3 is after accounting for background count.

(i) Explain why this experiment cannot provide evidence to show the presence of

alpha particles.

[1]

(ii) Indicate the evidence from Fig 6.3 that indicates the presence of 1. beta particles

[1]

2. gamma particles

[1]

2000

4000

6000

11


For Examiner’s

Use 6 (c) A point source of alpha particles 241

95Am

with decay constant 4.80 x 10-11 s-1 is

mounted 7.0 cm in front of a Geiger Muller(GM) tube whose mica window has a receiving area of 3.0 cm2, as shown in Fig 4. The whole setup is enclosed in a vacuum enclosure.

Fig 4 The counter linked to the GM tube records 5.4 x 104 counts per minute. (i) Show that the activity of the source is 18.5 x 104 s-1

[1]

(ii) Hence determine the number of 241

95Am atoms in the source.

Number of atoms = …………… [1]

7.0 cm

3.0 cm2

GM tube connected to ratemeter

source

12


For Examiner’s

Use 7 (a) Niels Bohr presented a new model of the hydrogen atom in 1913. Bohr combined

ideas from Planck’s original quantum theory, Einstein’s concept of the photon, Rutherford’s planetary model of the atom and Newtown mechanics to arrive at a model of the hydrogen atom. He postulated that the electron moves in circular orbit around the proton under the influence of the electric force of attraction as shown in Fig. 7.1. The atom emits radiation when the electron makes a transition from a more energetic initial state to a lower-energy stationary state. The approximation is made such that the proton is so much massive than the electron that it can be regarded as stationary. The smallest radius of the orbit r is 5.29 x 10-11 m.

Bohr postulated that the angular momentum of the electron was quantised and predicted that the allowable energies of the hydrogen atom are given by

2

13.6nE

n eV where n = 1, 2, 3… ------------- equation (1)

(i) Considering the forces acting on an electron,

1. show that the momentum p, of an electron in an orbit of radius r is given

by 2

4 o

mep

r

[2] 2. hence, determine the magnitude of the momentum of the electron when in

orbit of radius r = 5.29 x 10-11 m. momentum = kg m s-1 [1]

proton

r

+e

electron

v

-e

Fig. 7.1

13


For Examiner’s

Use

7 (a) (ii) Hence explain, how this model of the hydrogen, where the electron orbits in a circular path of a fixed radius, violates the Heisenberg uncertainty principle.

[1]

(b) When high-energy electrons or any other charged particles bombard a metal

target, x-rays are emitted. The x-ray spectrum typically consists of a broad continuous band containing a series of sharp lines which is due to bremsstrahlung radiation. Characteristics x-rays occurs when a bombarding electron collides with a target atom. The electrons have sufficient energy to remove an inner-shell electron from the atom. The vacancy created in the shell is filled when an electron in a higher level drops down into the level containing the vacancy.

A diagram showing the origin of the K , K and other peaks are as shown in

Fig. 7.2 below.

Fig. 7.2 For a multi-electron atom, because of the presence of one k-shell electron, the

other electron will ‘see’ an effective nuclear charge of approximately (Z - 1)e, where e is the elementary charge and Z is the atomic number of the element.

As such equation 1 in (a) can be approximated as

2

2

(13.6 )( 1)n

eV ZE

n

.

n=1

n=2

n=3

n=4

n=5

K K

K

L L

L

M M M

M

N

O

L

K

14


For Examiner’s

Use For an electron that makes a transition from the L shell (with n = 2) to the K shell (with n = 1),

(i) express the energy change in terms of Z, [2]

(ii) hence, derive the relationship between the frequency f of the K radiation and

Z. [2]

(c) It is thought that the wavelength of K x-ray radiation, varies with atomic

number of element Z according to the expression below as suggested by Mosley in 1914.

1

A Z B where A and B are constants

The frequency f of K x-ray radiation, of a few elements are given in Fig. 7.3.

Element Atomic number Z

Frequency f / 1018 Hz

1

/ 104 m

12

Titanium 22 1.08 6.016 6.01 Chromium 24 1.30 6.59 6.59

Iron 26 1.54 7.16 7.16 Nickel 28 1.79

Zinc 30 2.07 8.30 8.30 Gallium 31 2.21 8.59 8.59

Fig. 7.3

(i) Complete Fig. 7.3 for Nickel. [1]

15


For Examiner’s

Use

7 (c) (ii) The data from Fig. 7.3 are used to plot the graph of Fig. 7.4.

Fig. 7.4

On Fig. 7.4,

1 Plot the point corresponding to Nickel. [1] 2 Draw the best-fit line for all the points. [1]

(iii) Use the line drawn in (ii) to determine the magnitude of the constants A and B

in the expression in (c).

A = m1

2

B = [3]

6.00

6.50

7.00

7.50

8.00

8.50

9.00

22 23 24 25 26 27 28 29 30 31 32

1

/ 104 m

12

Atomic number Z

16


For Examiner’s

Use 7 (d) One of the possible use for characteristic spectrum is to allow impurity in

specimens to be detected by scientists. A cobalt target is bombarded with electrons, and the wavelengths of its characteristic x-ray spectrum are measured. There is also a second fainter characteristic spectrum, which is due to an impurity

in the cobalt. The wavelengths of the K lines are 178.9 pm (cobalt) and 143.5 pm

(impurity), and the atomic number Z for cobalt is 27.

Using Fig. 7.3, deduce the impurity in the cobalt. [2]

17


For Examiner’s

Use

8 The exact relationship between the temperature of an object and the amount of radiation emitted has been explored by a number of famous physicists. In 1879, the law that describes this relationship was first experimentally discovered Josef Stefan. Shortly later, Ludwig Boltzmann derived it theoretically. The relationship between the power radiated Prad by an object of temperature T (in kelvin), independently formulated by Stefan and Boltzmann states that

Prad = A T4

where is the emissivity of the object

is the Stefan-Boltzmann constant and A is the surface area of the object The emissivity constant depends on the material of the object. For practical purposes, the net power being radiated is more useful than the absolute radiated power. The net power radiated by an object at temperature T in an environment T0

is given by

Pnet = Prad – Pabsorb

This leads to

Pnet = A T4 - A T04

For a given radiating body , and A are constant and if the room temperature T0 (295K) is

much lower than the object’s temperature T ( ranged from 1500 to 2000K), then Prad is

much greater than Pabsorb. We can then assume that for a given radiating body power

radiated is given by

P T4

Design a laboratory experiment to investigate how the power P radiated from a tungsten

lamp at high temperature varies with its thermodynamic temperature T.

The equipment available includes the following: Leads/ connecting wires 12 V 400 W Tungsten filament bulb A variable power supply 13 V MAX Radiation detector Voltmeter Ammeter Rheostat Digital multimeters Metre rule Retort stand, boss and clamp Thermocouple Platinum Resistance thermometer

Data showing the temperature dependence of relative resistance for tungsten and resistivity of tungsten as a function of temperature are provided and are shown in Fig 8.1 and Fig 8.2 respectively.

18


For Examiner’s

Use Fig 8.1: Temperature dependence of relative resistance for tungsten

temperature / kelvin

R/Rref

19


For Examiner’s

Use

Fig 8.2 Resistivity of tungsten as a function of temperature

R/Rref Temperature / K Resistivity / cm 1.00 300 5.65

1.43 400 8.06

1.87 500 10.56

2.34 600 13.23

2.85 700 16.09

3.36 800 19.00

3.88 900 21.94

4.41 1000 24.93

4.95 1100 27.94

5.48 1200 30.98

6.03 1300 34.08

6.58 1400 37.19

7.14 1500 40.36

7.71 1600 43.55

8.28 1700 46.78

8.86 1800 50.05

9.44 1900 53.35

10.03 2000 56.67

10.63 2100 60.06

11.24 2200 63.48

11.84 2300 66.91

12.46 2400 70.39

13.08 2500 73.91

13.72 2600 77.49

14.34 2700 81.04

14.99 2800 84.70

15.63 2900 88.33

16.29 3000 92.04

16.95 3100 95.76

17.62 3200 99.54

18.28 3300 103.3

18.97 3400 107.2

19.66 3500 111.1

26.35 3600 115.0

20


For Examiner’s

Use You may also use any of the other equipment usually found in a Physics laboratory. You should draw a labelled diagram to show the arrangement of your apparatus. In your account you should pay attention to

(a)

(b)

The identification and control of variables,

the equipment you would use,

(c) the procedure to be followed,

(d) how the power radiated from a tungsten lamp could be determined ,

(e) any precautions that you would take to improve the accuracy and safety of the

experiment.

Diagram .…………………...…….…………..……………………………………………………...……………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

21


For Examiner’s

Use

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

22


For Examiner’s

Use ……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

.…………………...…….…………..……………………………………………………...……………

……………………………………........…………………………………………..……………………

……………………………………........…………………………………………..……………………

End of paper

[12]

Qn Suggested Guide

1

(a) Product of force and displacement moved in the direction of the force.

1 (b)

Situation On what object is the work done?

Application of conservation of energy

Steam engine

Work is done on the Piston of the engine.

Internal energy of steam is converted to to KE of piston (to drive engine).

Photocell / Light Dependent Resistor (LDR)

Work is done on the Electrons on metal surface

Photon energy to EPE of electrons (to jump energy gap from VB to CB) (and KE of electrons).

Battery

Word is done on the Motor in chainsaw.

Electrical energy of charges to rotational KE of motor to KE of chainsaw and subsequently heat due to work done against friction(and sound) on drilling of wood.

Qn Suggested Guide

2(a)(i) 2

232

26

1 3

2 2

3(1.38 10 )(296)

(5.3 10 )

m c kT

xc

x

= 481 m s-1

(ii) According to Kinetic Theory, the mean KE is the same since at the same temperature. However as mass is different they have different rms speed.

(b) Reference made to distribution of speeds. (ie idea on molecules having speed above and below rms speeds)

For hydrogen the probability of molecules having speed greater than escaped speed is higher than oxygen as rms speed is higher.

(c) When molecule collide with the wall, the wall exerts a force on the molecule causing the molecule to experience a change in momentum, hence it rebounds.

By N3L, it exerts a force of the same magnitude back on the wall.

As there are many molecules moving about randomly, colliding with the wall, an average constant force and hence pressure is exerted on the wall.

Qn Suggested Guide

3 (a)(i)

(ii) 1. FB = Bqv = 1.5 x 10-3 x 1.6 x 10-19 x 1.8 x 107

= 4.32 x 10-15 N (Shown)

2. Magnetic force provides centripetal force for circular motion

215

31 7 2

15

4.32 10

9.11 10 (1.8 10 )

4.32 10

B cF F

mv

r

r

0.068 r m

(b)

The helical path has progressively larger pitch (since FE acts out of the plane of the paper Electrons accelerates out)

P

Region of uniform

magnetic

Qn Suggested Guide

4

(a)(i) Interference is observation in the variation of the intensity of the resultant wave when two of more waves meet due to the resultant displacement being given by the vector sum of the displacement of the individual wave.

(b) (i)

(b)(ii) When the sound wave that are produced by the source is reflected at the glass boundary with a phase

difference of . The incident and reflected wave overlaps and as they have the speed and frequency, resulting in the formation of nodes and antinodes of a standing wave if the length

of the tube is given by odd multiples of /4. The piles of power gather at the nodes as at the nodes no oscillation occurs.

(b)(iii) Wavelength of stationary wave =123 X 2/3 = 82 cm Hence speed of wave = 400 Hz X 0.82 m = 328 = 330 m

123 cm Signal Generator

400 Hz

Loudspeaker

Piles of powder

N A N A N A

Qn Suggested Guide

5(a) Stimulated emission is triggered by an external photon whereas spontaneous emission happens on its own accord.

(b)

In stimulated emission, a photon of energy equal to the energy difference between the metastable state and lower excited state will trigger the transition of the atom to the lower energy state.

This results in 2 emitted photons that have the same frequency and hence coherent. In addition, the photons emitted are in phase, have the same polarization and are moving in the same direction

These photons further stimulate more photons that are in phase, creating a coherent beam of laser light of high intensity.

(c) Population inversion that is, there being more excited atoms in the higher energy state than a lower one is one of the conditions required

This ensures that the probability that an incident photon will stimulated emission exceeds the probability that the photon will be absorbed as Incident photons can cause either stimulated absorption or stimulated emission. (If there is no population inversion, incoming photons will be more likely absorbed to cause excitation rather than to result in stimulated emission as upper energy level is unoccupied. It is required so that the stimulated emission dominates spontaneous emission in the excited atoms.)

(d) The 4 level system ensures there is always population inversion between E3 and E2 as E2 undergoes rapid spontaneous emission to E1 This reduces the loss of photons by stimulated absorption between the upper (E3) and lower (E2) laser level. (Note Level 4 is practically empty due to fast spontaneous emission, hence any appreciable population accumulating in level 3, the upper laser level, will form a population inversion with respect to level 2. That is, as long as population in level 3 > 0, then a population inversion is achieved between E3 and E2. Since only a few atoms must be excited into the upper laser level to form a population inversion, a four-level laser is much more efficient than a three-level one, and most practical lasers are of this type.)

6(a)(i)

Count rate flunctuates / curve not smooth

(ii) The new graph should give the same half-life.

(b)(i) Alpha particle is stopped in air, hence cannot be detected.

(ii)1. As thickness of Al sheets increases, the count rate decreases rapidly as beta particles are stopped by mm of Al.

(ii)2. The count rate became a constant for thickness beyond 3 mm. This indicates presence of gamma particles as they are not stopped by mm of Al.

(c)(i) Total disintegrations =

42 5 4 10 60

4 73

x. /( ) [ ]

= 184725.6 s-1

=18.5 x 104 s

-1

(ii) A =N

184725.6 = 4.80 x 10-11

N

Hence N = 3.85 x 1015

Qn Suggested Guide

7(a)(i) 1

Electrostatic force provides the centripetal force

E CF F

2 2

24 o

e mv

r r

22 2

4 o

mem v

r

2

4 o

mep

r

2 = 1.99 x 10-24

kg m s-1

(ii) If momentum of electron in radial direction is fixed, then its uncertainty is zero.

Using 4

hp

, would be infinite. However the

above model has a fixed radius.

(b)(i) 1 2 E E E

2 2

2 2

(13.60)( 1) (13.60)( 1)

1 2

Z Z

2(10.2)( 1) Z eV

(ii) 2 1

E E

fh

19 2

34

(10.2 1.6 10 )( 1)

(6.63 10 )

Z

15 2(2.46 10 )( 1)Z

(c)(i) 7.72

(ii)1. Plot point at coordinates (28, 7.72) to half smallest square

2. Points evenly scattered on both sides of line

(iii) Knowing A = gradient

Correct value of A (= 0.286 x 104)

-AB = vertical intercept = -0.281 x 104

Correct value of B (= 0.98)

(d) Calculate

1

To get the impurity as Zinc.

Q8 – Suggested Mark Scheme and Proposed answers

Q8 www.physics.nus.edu.sg/~L1000/PC1142/StefansLaw.pdf ;

personal.tcu.edu/zerda/manual/lab22.htm (adapted)

(Annotate all marks awarded with the corresponding alphabet)

Basic procedure (2 marks)

Radiation from lamp directed to radiation sensor B1

Vary Power/voltage supplied to lamp, find power radiated

B1

(max B2)

Diagram (2 marks)

OR

Sensor connect to DMM (voltmeter) / data logger / Mentioned radiation meter gives direct reading

D1

Electrical circuit for lamp given (allow use of ohmmeter to measure resistance.) D1

(max D2)

Measurements (2 marks)

Determination the resistance R and temperature T of the filament

1. For each voltage setting, using the corresponding values of I and V from the ammeter and voltmeter respectively, calculate the

corresponding resistance RT of the filament using RT = I

V

Or mention use of ohmmeter to give readings for each value of PS.

2. Calculate the ratio of

ref

T

R

R ; the corresponding temperature T can

be determined from the graph of

ref

T

R

R against T.

M1

Determination of Power Radiated from Bulb

For each T, record readings of Power radiated from the voltmeter which is a measure of the power radiated.

M1

Max 2

Analysis (1 mark)

Assuming that the power P radiated by the filament is proportional to

Tn (or nT

ref

R

R( )

)where n is a constant, plot a graph of log P against log T

(or log T

ref

R

R( ))to investigate how P depends on T. The gradient will

give n if a straight line graph is obtained.

(Accept any other suitable analysis including )

(Do not accept if student only mentioned plotting of graph without equation involved)

A1

(Max A1)

Control of variables (1 marks)

Radiation sensor should be kept at a fixed distance from the filament. C1

Radiation sensor should be set at the same height as the filament. C1

Alignment of filament with respect to sensor window that is angle of inclination of filament to the vertical should be kept fixed

C1

(Max C1)

Further details (3 marks)

mV range used to measuring o/p from sensor F1

For each voltage setting of the power supply, be brisk when recording the data points and the sensor readings as lamp will begin to heat up the sensor.

F1

In between readings cover shield the sensor with aluminized foam-core board, with aluminized side facing the bulb.

F1

Remove all other objects in the vicinity of the radiation sensor to ensure that its output is not influenced by extraneous radiation sources.

F1

(Max F3)

Satety (1 mark)

Bulb used in this experiment will cause burns if you touch it; do not move the bulb while it is on, but if necessary move it by the base only.

S1

Place bulb on good footing as it is light and could tip over easily. S1

Do not exceed voltages of 12 V or currents of 3 A on the bulb filament.

S1

When maximum power voltage is reached gradually turn down the voltage to 0 V and turn off the supply.

S1

(Max S1)

This paper consists of 20 printed pages


CANDIDATE NAME

CLASS

CENTRE NUMBER

S 3 0 0 4 INDEX

NUMBER

PHYSICS Paper 3 Longer Structured Questions Candidates answer on the Question Paper. No Additional Materials are required

9646/03 17 Sept 2014

2 hours


Write your Name and Index number in the spaces on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. 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 brackets [ ] at the end of each question or part question.

For Examiners’ use only

Section A

1 / 7

2 / 7

3 / 7

4 / 7

5 / 7

6 / 5

Section B

7 / 20

8 / 20

9 / 20

Total / 80

2

2014 ACJC H2 9646 Prelim Exam P3

For Examiner’s

Use DATA AND FORMULAE

Data speed of light in free space, c = 3.00 108 m s1



(1/(36)) 10 F m











Formulae


v 2 = u 2 + 2as

work done on/by a gas, W = p V

hydrostatic pressure, p = g h


Gm



= 22 xxo


E = 3

2kT




Q

o4



where k = 2

2 )(8

h

EUm


decay constant, = 2

1

6930

t

.

3

2014 ACJC H2 9646 Prelim Exam P3 [Turn over

For Examiner’s

Use

Section A

Answer all the questions in the section.

1 An electron having a horizontal velocity of 1.5 x 105 m s-1 enters a region of uniform

electric field as shown in Fig. 1.1. The electric field E

of magnitude 10 V m-1 is applied perpendicularly to the horizontal velocity. The electron travels a horizontal displacement of 5.0 cm before leaving the plate.

Fig. 1.1

(a) Show that the vertical velocity of the electrons as it exits the plates is 5.9 x 105 m s-1 .

[3]

(b) Determine the speed of the electron as it exits the parallel plate.

Speed of electron = ………………….. m s-1

[1]

(c) For motion within the plates, on Fig. 1.2, sketch the variation with horizontal

displacement the

(i) kinetic energy , and the [2]

(ii) electric potential energy of the electron. [1]

Label the initial and final values of the energies.

Fig. 1.2

E

O 5.0 cm

O Horizontal displacement

energy

5.0 cm from O

4


For Examiner’s

Use 2 (a) Explain what is meant by the radian.

[1]

(b) Fig. 2.1 shows a stone of mass 150 g at point P which is at the top of a circular path. The stone is suspended on one end of a string. The radius of

the circle is 70 cm and the speed of the stone at point P is 3.5 m s.

Fig. 2.1

When the stone is at point P, (i) Calculate the resultant force acting on the stone,

resultant force = …………… N [2]

(ii) Hence determine the tension in the string.

tension = …………….. N [1] (iii) Draw on Fig. 2.1 the forces acting on the stone. [1]

3.5 m s1

P

70 cm

5


For Examiner’s

Use

(c) If the speed of the stone remains constant at 3.5 m sduring the circular motion, sketch on Fig. 2.2 a force-time graph showing the variation of tension in the string as the stone travels a complete circle.

[Take t = 0 s when stone is at point P]

Label M on your sketch the point of maximum tension in the string.

[2]

Fig. 2.2

force / N

time / s 0

6


For Examiner’s

Use 3 The circuit in Fig. 3.1 is use to detect changes in temperature. The variation of the

resistance of the thermistor with temperature is shown in Fig. 3.2

Fig 3.1 Fig 3.2

(a) (i) State the potential difference Vout across R in terms of R, RT and V Vout = ………….. V [1]

(ii) The constant resistance R is 3.0 k and the potential +V is +12 V.

A fire alarm will be activated when the potential difference applied across it is 8.0 V. Explain how the above circuit could be used to activate the fire alarm.

[3]

(iii) Hence determine the temperature at which the fire alarm will be activated.

Temperature = …………….oC [2]

(b) State the purpose of the resistor of constant resistance in the circuit.

[1]

7


For Examiner’s

Use

4 Fig 4.1 shows the cross-section of two long straight wires X and Y perpendicular to the page. There is an electric current in both wires out of the page. The current in wire X is twice the current in wire Y.

(a) Sketch the resultant magnetic field pattern around the wires within the box.

Include direction arrows on the field lines. [3] (b) Each wire exerts a force on the other wire. Draw arrows on the wires to show

the direction of these forces. [1] (c) Explain what Newton’s third law implies about the magnitude of these forces.

[1]

(d) Hence, deduce the expression for the force per unit length for each of these

two forces, given that the wires X and Y are of infinite length, a distance d

apart, and carrying current I in wire Y. You are given that the magnetic flux density B at a perpendicular distance r

from a long straight wire with current I is given by,

Force per unit length on wire X = _____________ Nm-1

Force per unit length on wire Y = _____________ Nm-1 [2]

Fig 4.1

X Y

2

oIB

r

8


For Examiner’s

Use 5 A double slit with slit separation d = 0.400 mm is situated a distance 3.2 m from a

detector moved vertically along XY as shown in Fig. 5.1. The double slit is illuminated with coherent light of wavelength .

(a)

Fig . 5.1

A detector was moved along XY and the detected intensity is shown in Fig. 5.2. The relative positions of the peaks are with respect to O, the central maximum.

Fig . 5.2 (i) Deduce the wavelength of the light.

λ = m [2]

O

M

X

Y

D =3.2 m

S1

S2

light d = 0.400 mm mm

0.0 3.0 -3.0 6.0 -6.0

intensity

distance / mm

M

9


For Examiner’s

Use

(ii) With reference to Fig 5.1, calculate the phase difference between the two waves at point M.

Phase difference = ___________ π rad

[2]

(b) Both the slits are covered with two identical thin transparent boxes. The

one in front of S2 is filled with vacuum whereas the one in front of S1

is filled with gas of refractive index n as shown in Fig. 5.3. It is considered that the actual path length in the gas to be the product of its refractive index and actual distance travelled, d. The new light intensity pattern along XY is shown in Fig. 5.4.

Fig 5.3

Fig 5.4

O

S1

S2

light

Screen X

Y

0.0 3.0 -3.0 -6.0

intensity

distance / mm

Central Max

10


For Examiner’s

Use

With reference to Fig 5.2 and 5.4, the central peak has been shifted by mm towards M. By deducing the phase difference at the central position or otherwise, calculate the

refractive index, n of the gas if the gas is m thick.

Refractive index = ___________

[3]

11


For Examiner’s

Use

6 Fig.6 shows how emission and absorption spectra can be obtained.

Fig 6 (a) State the type of spectrum the eyes at position A and B are expected to

see: Position A : ______________ spectrum Position B : ______________ spectrum

[1]

(b) Explain how the emission spectrum is an evidence for discrete energy levels in an atom. ………………………………………………………………………………………. ………………………………………………………………………………………. ……………………………………………………………………………………….

[2]

(c) Suggest how absorption spectrum can be used to identify elements present in the sun. ………………………………………………………………………………………. ………………………………………………………………………………………. ……………………………………………………………………………………….

[2]

Hot light source producing continuous spectrum of light

Position A

Position B

Cooler cloud of gas

12


For Examiner’s

Use

Section B

Answer two questions for this section. 7 (a) Define acceleration [1]

(b) Fig 7.1 shows how the vertical component of the velocity of a parachutist changes

with time during the first 20 s of his jump. To avoid air turbulence caused by the aircraft, he waits a short time after jumping before pulling the cord to release his parachute.

Fig 7.1

(i) Using Fig 7.1, determine the total vertical distance fallen by the parachutist in

the first 11 s of the jump. Show your method clearly

Vertical distance fallen = m [3]

Vertical component of velocity / m s

-1

13


For Examiner’s

Use

(ii) Regions A (0 to 2 s), B (2 to 11 s) and C (11 to 12 s) of the graph show the

speed before the parachute was opened. With reference to the forces acting on the parachutist, explain why the graph has this shape in the region marked,

1. A

[2]

2. B

[2]

3. C

[2]

(iii) Determine the acceleration of the parachutist at 6.0 s.

acceleration = m s-2 [2]

14


For Examiner’s

Use (iv) Complete Fig 7.2 (including the blank column with a suitable ratio) and hence

deduce whether the air resistance experienced by the parachutist from 3.0 s to 9.0 s is directly proportional to the square of the velocity of the parachutist. Take the acceleration of free-fall as 10.0 m s-2.

t/s v/ m s-1 a/ m s-2 (g a) / m s-2

3.0 28.5 6.90 3.10

6.0 43.2

9.0 48.0 1.20 8.80

Fig 7.2

[4]

(c) (i) Show that the maximum deceleration of the parachutist when the parachute

opened at time 12 s is about 30 m s-2.

[1]

(ii) Hence draw a graph on Fig 7.3 showing the variation with time of the

acceleration of the parachutist during the first 18 s of the jump. (No further calculations required.)

[3]

acceleration/ m s

-2

time/s

0

2

4

6

8

10

12

14

16

18

10

20

10

20

30

Fig 7.3

15


For Examiner’s

Use

8 A spring is attached to a sheet of aluminium and a mass, as illustrated in Fig. 8.1.

Fig 8.1

An electromagnet is placed near to the centre of the aluminium sheet. The mass is

displaced vertically and, with the electromagnet switched off, the mass is released. The variation with time t of the displacement x of the mass is shown in Fig. 8.2.

Fig 8.2

(a) The electromagnet is switched on and the experiment is repeated with the same initial displacement. Damped oscillations are observed.

(i) On Fig 8.2, sketch the new variation with time t of the displacement x of the

mass. [2]

15 V

16


For Examiner’s

Use 8 (a) (ii) State Faraday’s law of electromagnetic induction, and explain why the

oscillations of the mass are damped, using principle of conservation of energy.

[6]

(iii) Draw and label on Fig 8.1 the followings, as the aluminium sheet moves down

during the damped oscillation, the direction of 1. the current in the coils of the solenoid. [1] 2. the eddy currents in the aluminium sheet [1] 3. the direction of the induced magnetic field due to the eddy currents in

the aluminium sheet. [1] 4. the resultant force on the aluminium sheet. [1]

17


For Examiner’s

Use

8 (b) Suggest how critical damping can be demonstrated using the apparatus of Fig 8.1.

[4]

(c) The coil has a length 8.00 cm, radius of 1.00 cm, 50 turns and a resistance of

5.00 . The variable resistor has a range of 0 – 20 .

The magnetic flux density B at the centre of a solenoid is given by

oB nI

where n is the number of turns per unit length. Calculate the maximum magnetic flux density at one end of the solenoid.

Magnetic flux density = _____________ T [4]

18


For Examiner’s

Use

9 (a) Distinguish between a nucleon, a nucleus and a nuclide. [4]

(b) A nuclear power plant is capable of generating up to 4.7 gigawatts of power through

neutron-induced fission of Uranium-235. Fig 9.1 illustrates a typical nuclear fission reaction where a neutron is captured by a Uranium-235 nucleus.

Fig. 9.1 On average, 2.5 neutrons are emitted in fission reactions and large amounts of energy are released. When conditions are suitable, a chain reaction can occur and if it is controlled, a source of continuous power may be created.

(i) Explain what is meant by nuclear fission. [2]

(ii) Two possible fission products that can be formed from the fission of Uranium--

235 are Barium-141 ( ) and Krypton-92 (

). Write down the nuclear equation for this nuclear reaction.

[2]

19


For Examiner’s

Use

(c) Fig. 9.2 shows how the binding energy per nucleon of a nucleus varies with mass

number.

Fig. 9.2

(i) State what is meant by the binding energy per nucleon of a nucleus. [2]

(ii) Explain why fission of nuclei having low mass numbers (N < 60) is not

associated with a release of energy.

[2]

(iii) Using data from Fig. 9.2 and (b)(ii), show that the energy released in one

fission reaction is about 185 MeV. [3]

Source: https://nige.wordpress.com/2011/03/

20


For Examiner’s

Use (iv) Suggest one form of energy in which the energy released in (c)(iii) is

transformed to. [1]

(v) A typical nuclear reactor has about 100 tonnes (1 ton = 1000 kg) of Uranium

spread throughout the fuel assemblies to act as fuel to generate electrical energy. If a nuclear reactor is 30% efficient, estimate, in years, how long can a nuclear reactor supply electricity to the households in Singapore given that the total annual household electricity consumption is 6500 GWh. ……………….. years [4]

1 ACJC Prelim 2014 H2 P3

Qn Suggested MS

1(a)

x v xt ------ (1) find time to pass through parallel plate

a Eq

m--------(2) find vertical acceleration by e-field

v y at ------(3) solve for vy and substitution

Solving (1), (2) and (3) gives

y

x

Eq xv

m v

x

x

19

31 5

10 1.6 10 0.05

9.1 10 1.5 10

-1 m s55.8(54) 10

1(b)

v 5.854 105 2

1.5105 2

6.0(43)105m s -1

1(c)

(i) Quadratic graph as shown

initial KE=1.0(25) x 10-20 J and KE at x = 5.0 cm is 16.6(1) x 10-20 J

(ii) reflection of KE graph. (ecf from (i)) (KE + EPE must be seen to be approx. const)

2(a) A radian is the angle subtended by an arc where the arc length is equal to the radius.

2(b(i) Resultant force downwards provides centripetal force

2(b) (ii)

T+W = 2.63 N T= 2.63 - W T =2.63 - 1.4715 = 1.1585 = 1.16 N

0 Horizontal displacement

5.0 cm from O

1.0 x 10-20

16.6 x 10-20


2(b)(iii)

2(c) At the bottom of its path Tension = 2.63N + W Tension =2.63N + 1.4715 = 4.096= 4.10 N Point M =4.10 N

Shape as shown

M labeled (value not reqd)

3(a) (i)

The potential difference Vout across R is

Vout = V RR

R

T

(ii) Vout increases as RT decreases

As RT decreases as the temperature increases.

Hence Vout will increase to 8.0 V when the room temperature reached the temperature at which the fire alarm will be triggered to switch on. thus the circuit can be used to switch on the fire alarm when the pre-determined temperature is reached.

(iii) Vout = V

RR

R

T ; 12 = (8)

R3

3

T

RT = 1.5 k

From the graph when RT = 1.5 k

The temperature is 83(1)0C

(b) Is to set the value of the trigger voltage

P

Weight Tension

force / N

time / s 0

1.16

4.10

T


4(a)

Line density stronger next X; ignore shape

Direction correct

Shape as shown with neutral point nearer Y, no credit if drawn circular fields near conduct conductors

(b) Correct direction of F on diagram (attractive forces) (and equal length)

(c) The two forces are action and reaction pairs according to N3L, hence Magnitude of FY on X = Magnitude of FX on Y

4(d)

2

( )2

2

(2 )

2

X

X o

o yx

o

o

F BIl

F II

l r

II

d

I I

d

I

d

2

(by N3L)Y oF I

l d

Magnitude equal

Value correct

5

5(a)(i) By applying the equation, ax =λD

(ii)

path difference x

x

3

3

5 10

0.4 10 3.2

Phase difference =

path differencex2 3.3

FY on X

X Y

FX on Y


5(b)(i)

additional path difference x

x

3

3

1 10

0.4 10 3.2

(n-1) (1.0 x 10-6) = 1.25 x 10-7

Hence, the refractive index of the gas is 1.125 (1.13)

6(a) A: absorption spectrum; B: emission spectrum

(b) Emission Line spectrum implies only light of certain frequencies are emitted.

These frequencies corresponds to only specific energy differences.

(c) Each element only absorb certain frequencies of light. The set of dark lines correspond to a particular element.

7(a) Change of velocity with time / Rate of change of velocity (and acts in the direction of change of velocity)

(b)(i) Knows that dist is area under graph

Suitable method to find area

Range accepted: 380 – 400 m

(ii)1 Uniform acceleration

Only weight/gravitational force acting

(ii)2 Acceleration decreasing /speed increasing at decreasing rate

Air resistance increases as speed increase hence resultant force downwards decreases

(ii)3 Uniform speed

Air resistance = weight / resultant force zero

(iii) Clear Workings shown with tangent drawn with Tangent drawn at 6.0 s

Ans: 2.84 m s-2

(iv)

t/s v/ m s-1

a/ m s-2

(g a) / m s-1

(g a)/v2 /10-3 m-1

3.0 28.5 6.90 3.10 3.81

6.0 43.2 2.84 7.16 3.84

9.0 48.0 1.20 8.80 3.82

(g-a) at 6.0 s determined correctly

Resultant force = mg – kv2 = ma

Hence if Fv is proportional to v2, (g a)/v2 is a constant.

Calculated value of (g a)/v2 all correct

Since values are almost the same, hence true.

(c)(i) Clear Workings shown with tangent drawn, Derivations showed that a = - 30 m s-2


(ii)

Graph for 1st 10 s correct trend (a decreasing from 10 to 0)

Graph from 12 to 18 s correct trend

Acceleration at t = 0, 11-12, 12, 16-18 s plotted correctly

8 (a) (i)

Shows decreasing amplitude throughout

Period same or slightly longer (up to 1.5 T)

(ii) Faraday’s law state that the magnitude of the induced emf in a conductor (or circuit) is directly proportional to the rate of change of magnetic flux linkage experienced by the conductor (or linking the circuit).

When the electromagnet is switched on, as current flows through the coil, it generates a magnetic field.

The oscillating aluminium sheet experience a change in magnetic flux linkage / cuts magnetic field lines

According to Faraday’s law, an emf will be induced in the aluminium sheet.

Since the aluminium sheet is a conductor / metal, induced (eddy) currents, circulates within it.

The mechanical energy of the oscillating system has been converted to electrical energy, and dissipated as heat, hence damping occurs.

(Since amplitude of the oscillation is proportional to the mechanical energy, amplitude decreases continuously.)


(iii)

Correct drawing and direction for,

I Iinduced or eddy current (as shown)

Binduced (top section - towards electromagnet, bottom section – same direction)

FB upwards

(b) The strength of the magnetic field produced by the electromagnet depends on the current following through it. (The current can be adjusted by the variable resistor.)

As the resistance of the variable resistor decrease, current increases, hence the strength of the magnetic field increase and the degree of damping increase / greater amt of electrical energy dissipated / more heat dissipated.

It will reach a point when the aluminium sheet (or mass) return to its equilibrium position in the shortest time, without overshooting (or crossing) the equilibrium position. This is called critical damping.

To detect the point of critical damping, a marker is attached to the aluminium sheet which traces out the displacement vs time curve on a scrolling paper, mounted vertically.

(c) For max B, needs max current, hence resistance of variable resistor = 0 .

15 (5.00)

3.0

V IR

I

I A At the centre of the solenoid,

7 504 10 3.0

0.0800

oB nI

3 32.36 10 ( 2.4 10 )T or T

At one end of the solenoid,

3

3 3

1(2.36 10 )

2

1.18 10 ( 1.2 10 )

B

T or T

9


(a) Nucleon: constituent of a nucleus, can be either a proton or neutron Nucleus: positively charged centre of an atom, made up of protons and neutrons Nuclide: an atom with a particular proton-neutron combination (nuclear structure)

Note: Isotopes are nuclides having the same number of protons but different number of neutrons

(b)(i) Disintegration of a massive nucleus into smaller fragments of comparable masses with the release of a large amount of energy

as the total mass of the daughter products is lower than the parent nuclei

(b)(ii) 235 1 141 92 1

92 0 56 36 03U n Ba Kr n energy

Correct balanced equation Inclusion of energy

(c)(i) Minimum energy required to break the nucleus into its separate nucleons such that the nucleons are separated to infinity per nucleon/divided by total number of nucleons

(c)(ii) Daughter nuclei has lower binding energy per nucleon than parent nuclei (as can be seen from the BE/nucleon graph).

and the total mass of the daughter nuclei formed is more than the parent nucleus which means that energy has to be supplied for the fission process to occur.

(c)(iii) Energy released = Total BE of Ba and Kr - Total BE of U

= [141(8.3) + 92(8.7)] – [235(7.6)]

= 184.7 MeV

= 185 MeV

(c)(iv) KE of daughter nuclei / KE of neutrons / γ rays

(c)(v) Number of nuclear reactions = (100 x 103/0.235)(6.02 x 1023) = 2.56 x 1029

185 MeV = (185x106)(1.6 x 10-19) = 2.96 x 10-11 J

Time = (0.30)(100 x 103/0.235)(6.02 x 1023)(185x106) (1.6 x 10-19)/ (6500 x 109 x 60 x 60)

= 97.2 = 97 yrs (2 s.f.)

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