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MERIDIAN JUNIOR COLLEGE Preliminary Examination Higher 2

_______________________________________________________________________

H2 Physics 9646/1

Paper 1 24 September 2014

1 hour 15 min

_______________________________________________________________________

Class Reg Number

Candidate Name _____________________________

READ THESE INSTRUCTIONS FIRST Do not open this booklet until you are told to do so.

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 Optical Mark Sheet (OMS).

Read very carefully the instructions on the OMS.

Write your name and class in the spaces provided on the OMS.

Shade your Index Number column using the following format:

1) first 2 digits is your index number in class (e.g. 5th student is shaded as “05”); 2) ignore the last row of alphabets.

Preliminary Examination Meridian Junior College 24 September 2014 JC2 H2 Physics 2014

2 [Turn over

Data speed of light in free space c = 3.00 x 10

8 m s

-1

permeability of free space o = 4 x 10-7 H m-1

permittivity of free space ε0 = 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 + 1

2at2

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 = xo sin t

velocity of particle in s.h.m. v = vo cos t

= 2 2

o-x x

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 = 04

Q

r

alternating current/voltage x = xo sin t

transmission coefficient T exp(-2kd)

where k =

2

2

8 ( )m U E

h

radioactive decay x = xo exp(-t )

decay constant

= 1

2

0.693

t


3 [Turn over

Answer all 40 questions in this paper and shade your answers on the answer sheet provided. 1 In a simple electrical circuit, the current in a resistor is measured as (2.50 ± 0.05) mA. The

resistor has a resistance of 4.7 Ω ± 2 %.

If these values were used to calculate the power dissipated in the resistor, what would be

the percentage uncertainty of the power?

A 4 % B 5 % C 6 % D 7 %

2 A metal sphere of radius r is dropped into a tank of water. As it sinks at speed v, it

experiences a drag force F given by F = krv, where k is a constant. What are the SI base units of k?

A kg m2 s-1 B kg m-2 s-2 C kg m-1 s-1 D kg m s-2

3 A stone is thrown at an angle and follows a parabolic path. The highest point reached is A

and the point just before it reaches the ground is B. Assuming that air resistance is negligible, the vertical acceleration of the stone is

A zero at A.

B greatest at A.

C greatest at B.

D the same at A and B.

4 A stone is dropped from the top of a tower of height 50 m. The stone falls from rest and air

resistance is negligible. What time is taken for the stone to fall the last 15 m to the ground?

A 0.52 s B 1.7 s C 2.7 s D 3.2 s

5 The diagram below shows 4 identical wooden blocks connected by inelastic strings, A, B

and C. A constant force accelerates the blocks to the right on a horizontal frictionless table.

Which string has the greatest tension? A String A B String B C String C D All have the same

tension

A B C Force


4 [Turn over

6 An elevator consists of a passenger car of mass 520 kg, supported by a cable that runs over

a light, frictionless pulley to a balancing weight of mass 640 kg. The balancing weight falls as the passenger car rises. The passenger has a mass of 80 kg.

Calculate the magnitude of the acceleration of the passenger car as it rises. A 0.032 m s-2 B 0.32 m s-2 C 0.65 m s-2 D 0.75 m s-2

7 The graph below shows the variation of the length of a spring with the load attached to it. For

a load W, the length of the spring is L.

The spring constant of the spring is given by

A the gradient of the graph

B the inverse of the gradient of the graph

C ratio of W to L

D ratio of L to W

Balancing weight

Not drawn to scale

Passenger Car

Length

L

Load W 0


5 [Turn over

8 A metal nugget floats in between some water and mercury. The densities of the metal

nugget, mercury and water are 7900 kg m-3, 13600 kg m-3 and 1000 kg m-3 respectively.

What is the ratio of the volume of the nugget submerged in water to that in mercury?

A 0.541 B 0.826 C 0.924 D 1.21

9 A rubber ball has a diameter of 10 cm. It is released from rest with the top of the ball 80 cm

above a horizontal surface. It falls vertically and then bounces back up so that the maximum height reached by the top of the ball is 40 cm, as shown in the diagram below.

If the kinetic energy of the ball is 0.80 J just before it strikes the surface, what is the kinetic energy just after it leaves the surface?

A 0.34 J B 0.37 J C 0.40 J D 0.46 J

10 The force resisting the motion of a car is taken as being proportional to the square of the

car’s speed. At a speed of 30 m s-1, the resistive force is 1800 N. What is the effective power required from the car’s engine to maintain a steady speed of 15 m s-1?

A 0.45 kW B 6.8 kW C 7.2 kW D 14 kW

Mercury

Metal nugget

80

40

Point of release

Maximum height after bounce

Water


6 [Turn over

11 A rock attached to a string swings in a vertical circle. At the highest point, A no forces act on the rock.

B only one force acts on the rock.

C two forces act on the rock and their resultant is zero.

D two forces act on the rock and their resultant is not zero.

12 A cyclist completes a circular path of circumference 343 m in 22 s on a rough horizontal

surface. Assuming she cycles at constant speed, what is the angle made by her with the vertical?

A 8.23 B 24.4 C 65.6 D 81.8

13 A satellite is moved from a low orbit to a higher orbit. Which of the following correctly

describes the energy of the satellite? Total energy

Gravitational potential energy

Kinetic energy

A stays the same decreases increases

B stays the same increases decreases

C increases decreases increases

D increases increases decreases


7 [Turn over

14 Two isolated masses M and m, of the same radius, are held a distance d apart, as shown below. Mass M is greater than mass m. The gravitational field strength g is measured on a line between the two masses. Which graph best shows the variation with distance x from the larger sphere of the magnitude of the field strength g?

A

B

C

D

g

x d 0

g

x d 0

g

x d 0

g

x d 0

x

d

M m


8 [Turn over

15 A housewife notices that her washing machine vibrates most vigorously whenever the r.p.m.

(revolutions per minute) of the spinning drum reaches a certain value. She sends it for repair and the repairman merely attached a heavy mass to the drum of the washing machine. Which of the following outcomes is most likely to happen to the washing machine?

A It does not vibrate vigorously anymore at all r.p.m. values.

B It vibrates most vigorously at the same r.p.m. value.

C It vibrates most vigorously at a higher r.p.m. value.

D It vibrates most vigorously at a lower r.p.m. value.

16 The graph below shows the variation of the potential energy Ep of a body with its

displacement s from a fixed point X.

Which feature of the graph best indicates that the body is undergoing simple harmonic motion?

A The graph passes through the origin.

B The potential energy does not have negative values.

C The graph is approximately linear for large values of s.

D The value of the potential energy increases as the body moves away from point X.

17 Alice is standing 15 m away from a speaker listening to a music broadcast. Due to a

technical fault, the power of the speaker is suddenly reduced by 25%. How far away from the speaker should Alice stand now in order for the music to be as loud as before? Assume that the speaker is a point source of sound.

A 13 m B 11 m C 7.5 m D 3.8 m

0 s

Ep


9 [Turn over

18 Humans are able to detect the general direction of a sound source because sound from the

source reaches the ear at slightly different times. The figure below shows a human head with the ears 22 cm apart. Sound waves of wavelength 1.7 m from a distant source reach the head at an angle of 25o.

What is the phase difference between the waves reaching the left and right ear?

A 0.34 rad B 0.58 rad C 0.74 rad D 0.81 rad

19 A sound source is placed several metres from a plane reflecting wall in a large chamber

containing a gas. A microphone connected to a cathode-ray oscilloscope (CRO) is used to detect nodes and antinodes along the line from the source to the wall. The microphone is moved from one node through 15 antinodes to another node, a distance of 1.2 m. On the CRO, the time-base is set to 0.50 ms cm-1, and the distance between 5 consecutive crests is 8.5 cm. What is the speed of sound in this gas?

A 150 m s-1 B 300 m s-1 C 320 m s-1 D 400 m s-1

20 A narrow beam of monochromatic light falls at normal incidence on a diffraction grating.

Second-order diffracted beams are formed at angles of 32° to the original direction. How many bright spots can be seen on the screen?

A 3 B 4 C 7 D 9

25o

ear

head Top view

(figure not to scale)

wavefronts

22 cm


10 [Turn over

21 The figure below shows two identical point charges Q which are fixed 6.0 × 10-3 m apart. The

two charges carry a charge of +1.45 × 10-15 C each. An electron is then placed 4.0 × 10-3 m below the two charges and along the perpendicular bisector between the two charges. The electron is then projected with a speed into the paper such that it performs a circular motion with its centre at point Z.

What is the speed of projection for the electron to perform such a circular motion?

A 2.0 × 104 m s-1 B 2.1 × 104 m s-1

C 2.4 × 104 m s-1 D 2.7 × 104 m s-1

22 A uniform electric field is set up between two parallel plates of length 60 mm and spaced

30 mm apart. A potential difference of 46 V is applied between the two plates. An electron is projected horizontally into the electric field with a speed of 8.0 × 106 m s-1.

What is the vertical displacement y of the electron when it exits the parallel plates? Ignore relativistic effects.

A 0.95 mm B 1.9 mm

C 3.8 mm D 7.6 mm

6.0 × 10-3 m

4.0 × 10-3 m

Q Q Z

electron

8.0 × 106 m s-1

y

60 mm

electron

parallel plates

30 mm


11 [Turn over

For questions 23 and 24, refer to the diagram below. 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.

23 What is the emf of the battery?

A 9.0 V B 10.0 V C 11.0 V D 12.0 V

24 What is the internal resistance r?

A 1.0 Ω B 2.0 Ω C 4.5 Ω D 9.0 Ω

25 The diagram below shows a simple potentiometer circuit for measuring a small e.m.f.

produced by a thermocouple. The potentiometer wire PQ is 1.0 m long and has a resistance

of 5.0 . The driver cell has an e.m.f. of 7.0 V. If a balance point is to be obtained at 20 cm along PQ, what is the emf E2 ?

A 0.53 V B 0.88 V C 2.6 V D 4.4 V

r E

R

Q

3.0 7.0 V

P

E2

J


12 [Turn over

26 A lighting circuit consists of four lamps connected as shown in the figure below. The

resistance of each lamp is 120 Ω. A fault is discovered in the circuit, so switch 1 is opened and the fuse is removed for safety. A resistance meter is connected between the point A and B and the following readings are obtained for different switch positions.

Switches Resistance Meter Reading/

1 2 3 4 5

open open open open open 14600000

open close open open close 0.4

open open open close close 60

open open close close open 60

open close close close open 0.2

Which of the lamp is most likely faulty?

A Z B Y C X D W

27 An electron enters the vacuum between two oppositely charged plates with velocity v. The

electron is followed by an alpha particle moving with the same initial velocity as the electron. A uniform magnetic field is directed into the plane of the paper. The electron’s path is undeflected. The path of the alpha particle will be

A undeflected

B deflected upward

C deflected downward

D deflected into the plane of the paper

1 2 3 4 5

Fuse

A

B y

Z Y X W

electron

charged plates

charged plates

v

alpha particle

v


13 [Turn over

28 The diagram shows a long solenoid of length L connected to a battery of negligible internal

resistance. The magnetic field strength at the centre of the solenoid is T.

The solenoid is now disconnected from the battery and cut in half and one of the halves is reconnected to the same battery as shown below.

Given that the magnetic field strength at the centre of a solenoid is equal to onI, where n is

the number of turns per unit length and I is the current through the coil.

The field strength of this solenoid is

A T/2 B T C 2T D 4T

29 A flat circular coil of 850 turns, each of area 0.055 m2, is placed with its plane aligned at 30°

to a uniform magnetic field as shown below. The flux density of the field is changed steadily from 80 mT to 30 mT over a period of 4.0 s.

What is the average e.m.f. induced in the coil during this time? A 0.29 V B 0.34 V

C 0.51 V D 0.58 V

30°

magnetic flux density

coil

L

L/2


14 [Turn over

30 A rectangular loop of wire is released from rest and falls vertically through a region of uniform

magnetic field under the influence of gravity. The loop and the region of magnetic field has the same height d as shown in the figure below.

Which of the following graphs shows the flow of the current in the loop of the wire as it falls and passes through the region of magnetic field?

A

B

C

D

Loop of wire

Region of uniform

magnetic field

d

d

current

time

current

time

current

time

current

time


15 [Turn over

31 An alternating voltage V varies with time t according to the equation given below

12 cos(50 )V t

What is the mean power dissipated in a resistive load of 10 ?

A 7.20 W B 12.0 W C 14.4 W D 28.8 W

32 A steady current dissipates a certain power in a variable resistor. The resistance has to be

halved to obtain the same power when a sinusoidal alternating current is used. What is the peak value of the alternating current?

A 1

2I B 2 I C I D 2 I

33 An metallic object of mass 10 g with a specific heat capacity of 890 J kg-1 K-1 is heated up by

a constant power electric heater. The rate of heat loss to the surrounding by the object depends on temperature as shown in figure below.

Given that the rate of increase in temperature of the object is 0.51 K s-1 at the instant when it is at 300 K, what is the power of the heater?

A 1.5 W B 4.5 W

C 6.0 W D 7.5 W

Rate of heat loss to surrounding / W

2.5

5.0

7.5

Temperature / K 0 300 150 450 600


16 [Turn over

34 In a system of n mol of ideal gas, the variation of the total kinetic energy associated with the

gaseous particles Ek with temperature θ is shown in the figure below.

What is the value of the vertical intercept of the straight line graph? Note: k is Boltzmann’s constant.

A 273k B 410k C 2270n D 3400n

35 In a simple model for a radioactive nucleus, an alpha particle (m = 6.64 x 10-27 kg) is trapped

by a square barrier that has width 2.0 fm and height 30.0 MeV. What is the tunnelling probability, if the proportionality constant is 1.0, when the alpha particle encounters the barrier if its kinetic energy is 1.0 MeV below the top of the barrier?

A 0.0147 B 0.0174 C 0.147 D 0.174

36 In a photoelectric experiment, the maximum kinetic energy of the ejected photoelectrons is

measured for various wavelength of incident electromagnetic radiation. A graph of this maximum kinetic energy, Kmax, as a function of the wavelength λ of the incident electromagnetic radiation falling on the surface of a metal is shown below. What is the work function for this metal?

A 4.97 eV B 6.22 eV C 7.96 eV D 24.9 eV

Ek / J

θ / °C 0

Kmax (eV)

λ (nm) 0 100 200 300


17 [Turn over

37 Which of the following best describes a hole in an intrinsic semiconductor? A A missing valence electron in a Group III atom.

B A missing valence electron in a Group IV atom.

C A missing valence electron in a Group V atom.

D An ionised acceptor atom.

38 Why is laser light monochromatic? A The excited electrons are in a metastable state.

B The system is in a state of population inversion.

C The emitted photon and incident photon are of the same phase.

D Photons of the same energy as that of the incident photons are emitted when the electrons transit down from a higher energy level.

39 A thin gold foil is bombarded with α-particles as shown.

The results of this experiment provide information about the

A wave property of a gold atom

B mass number of a gold atom

C size of a gold nucleus

D binding energy of a gold nucleus

incident α-particles

gold foil


18 [Turn over

40 Which pair of nuclei are isotopes of one another? nucleon number number of neutrons

A 182 186

108 112

B 184 186

110 108

C 184 186

110 110

D 186 186

110 112

End of Paper


MERIDIAN JUNIOR COLLEGE

Preliminary Examination Higher 2

_______________________________________________________________________

H2 Physics 9646/2


1 hour 45 min

_______________________________________________________________________

Class Reg Number

Candidate Name _____________________________

READ THESE INSTRUCTIONS FIRST This booklet contains 7 questions.

Do not open this booklet until you are told to do so. Answer all questions.

Write your answers on this question booklet in the blanks provided.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the end of each question or part question. Marks will be deducted if units are not stated where necessary or if answers are not quoted to the appropriate number of significant figures.

All working for numerical answers must be shown. You are reminded of the need for good English and clear presentation of your answers.

Examiner’s Use

Section A

Q1 /12

Q2 /14

Q3 /10

Q4 / 8

Q5 / 8

Q6 / 8

Q7 /12

Deductions

Total /72


2


8 m s

-1



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












s = ut + 1

2at2

v2 = u2 + 2as




r



= 2 2

o-x x


2kT




Q

r



where k =

2

2

8 ( )m U E

h


decay constant

= 1

2

0.693

t


3

Answer all the questions in the spaces provided.

1 A large metal ball is hung from a static crane by a cable of length 5.8 m as shown in Fig. 1.1.

In order to knock down a wall, a metal ball of mass 380 kg is pulled away from the wall and then released. The ball moves through an arc and makes contact with the wall when the cable is vertical. The graph below in Fig. 1.2 shows the variation of the speed of the ball v with time t.

Fig. 1.1

Fig. 1.2

v / m s-1

t / s

cable crane

metal ball

wall

5.8 m


4

(a) Just before the ball hits the wall, (i) Explain why the tension is not equal to the weight of the ball.

................................................................................................................................

................................................................................................................................

................................................................................................................................

.................................................................................................................................. [2] (ii) With reference to Fig. 1.2, estimate the tension in the cable.

tension = ………………………….. N [3]

(b) Taking rightwards as positive, when the ball collides with the wall,

(i) determine the change in momentum of the ball.

magnitude of change in momentum = ………………………….. N s

direction of change in momentum = ……………………………… [2]


5

(ii) sketch a clearly labelled graph to show the variation with time of the force on the ball during the collision.

[2]

(c) (i) State the principle of conservation of momentum.

................................................................................................................................

................................................................................................................................

.................................................................................................................................. [1] (ii) The ball has lost momentum in its collision with the wall. Explain whether the

principle of conservation of momentum is violated in this situation.

................................................................................................................................

................................................................................................................................

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

time / s

Force / N


6

2 (a) An astronomer is investigating the relationship between the radius R and mass M of eight different stars of the same species. The graph of the ratios R / Rs against M / Ms of these eight different stars are plotted in Fig. 2.1, where Rs and Ms are the Sun’s radius and mass respectively.

Fig. 2.1

0.0 0.4 0.8 1.2 1.6 2.0

M / Ms

2.00

1.75

1.50

1.25

1.00

0.75

0.50

0.25

0.0

R / Rs


7

(i) On Fig. 2.1, draw the best fit line for the relationship between R / Rs and

M / Ms for these stars. [1]

(ii) Determine the equation for this linear relationship. [2]

(iii) Estimate the maximum ratio of M / Ms for the mass of this species of star. maximum ratio of M / Ms = ……………..………… [1] (iv) Suggest why your answer to (a)(iii) is only an estimate.

................................................................................................................................

........................................................................................................................ [1]

(v) Suggest and explain whether the Sun belongs to this species of stars.

................................................................................................................................

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

(b) Fig. 2.2 shows the variation of the total gravitational potential with distance from the

centre of one of the eight stars. R is the radius of the star which is 3300 km.

Distance from centre of the star Gravitational potential / MJ kg-1

1.0 R -13.5

1.5 R -9.4

2.0 R -7.2

2.5 R -6.0

3.0 R -5.2

3.5 R -4.6

4.0 R -4.2

4.5 R -3.8

5.0 R -3.5

5.5 R -3.3

6.0 R -3.0

Fig. 2.2


8

A rocket of mass 1.2 x 104 kg lifts off from the surface of this star. Use Fig. 2.2 to (i) estimate the change in gravitational potential of the rocket at a distance 3R above

the surface of the star,

change in gravitational potential = ............................ J kg-1 [1] (ii) estimate the minimum launch speed that the rocket must have in order to reach a

height of 3R above the surface of the star,

minimum launch speed = ……………………… m s-1 [2] (iii) estimate the magnitude of the gravitational field strength at a distance 3R above

the surface of the star.

gravitational field strength = ................................. N kg-1 [2]


9

(iv) Student A claims that the gravitational field strength at any point above the

surface of this star is not inversely proportional to the square of the distance from the centre of this star. Discuss the validity of this claim.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

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


10

3 (a) Define electric potential at a point in an electric field.

...........................................................................................................................................

...........................................................................................................................................

................................................................................................................................... [1] (b) Two charged spheres of radius 2.0 mm are held 4.00 cm apart (centre-to-centre).

Fig. 3.1 shows the variation with displacement x, of the electric potential V, between the two charged spheres.

V / V

x / cm

Fig. 3.1

Position of charged sphere


11

(i) Using Fig. 3.1, determine the magnitude of the electric field strength at

x = 1.00 cm and state its direction.

electric field strength = ………………………… N C-1 [2]

direction: ……………………………………. [1] (ii) Using Fig. 3.1, determine the charge on each sphere.

charge = …………………………….. C [2] (iii) State an assumption made for the calculation in part (ii).

................................................................................................................................

................................................................................................................................ [1]


12

(c) The scenario in part (b) can be used as a one-dimensional potential well. A potential well

is a region which can be used to “trap” particles. (i) Suggest one type of particle which can be “trapped” within such a well and explain

briefly how the well “traps” the particle.

Type of particle: ......................................................................................................

How it works: ..........................................................................................................

................................................................................................................................

................................................................................................................................

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

(ii) “Trapped” particles can escape the well if it possesses sufficient energy. However

it is found that some particles which do not have sufficient energy can still escape from the well. State the phenomenon associated with this.

................................................................................................................................ [1]

4 (a) Define magnetic flux in a coil.

...........................................................................................................................................

...........................................................................................................................................

.......................................................................................................................................... [1] (b) Electrical transformers are widely used in industries and there are various types of

transformers suited for different applications. Fig. 4.1 illustrates a modified version of a 3-legged core transformer.

Fig. 4.1

The terminals of coil B is connected to a 240 V r.m.s. AC source. The ratio of the number of turns of wire in the coils of coil A : B : C is 2 : 13 : 1.

Coil C

Soft Iron Core Coil B Coil A

X


13

(i) With reference to the laws of electromagnetic induction, explain why a potential

difference is produced across the terminals of coil C.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

............................................................................................................................... [2] (ii) The magnetic flux produced by coil B is B and the magnetic flux through coil C

is C . State the relation between B and C .

........................................................................................................................ [1]

(iii) Determine the peak output potential difference across the terminals of coil C.

potential difference = …………………………… V [2] (iv) Due to impact on the transformer, a gap is formed in the soft iron core along the

dashed line labelled X as shown in Fig. 4.1. State and explain how the gap affects the maximum e.m.f. induced in coil C.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

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


14

5 (a) Fig. 5.1 below shows an alternating voltage supply given by 0

2cosV V t

T

connected

to four diodes numbered 1, 2, 3 and 4 and a resistor R.

Fig 5.1

(i) On the axes given below, sketch and label the voltage-time graph of the resistor

for at least two cycles when all four diodes are in use. Clearly label the period of

each cycle. [2]

(ii) If diode 4 is removed, leaving a break in the circuit, sketch and label the new

voltage-time graph of the resistor for at least two cycles. [2]

V / V

t / s

V / V

t / s

R

1 2

3 4


15

(b) A power station produces 20.0 MW of power for delivery to a town some distance away.

This power is generated at 32.0 kV and then stepped up to 240 kV using an ideal

transformer before transmission. The total resistance of the transmission cables is 5.0 Ω.

(i) State the turns ratio of the secondary coil to the primary coil in this transformer.

turns ratio = .................................... [1]

(ii) Explain why it is more economical to step up to 240 kV before transmitting

electrical power to the town. Justify your answer in terms of power loss in the

transmission cables.

................................................................................................................................

................................................................................................................................

........................................................................................................................

[3]


16

6 (a) Explain the origin of the depletion region in a p-n junction.

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

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

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

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

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

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

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

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

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

[4]

(b) Sketch the I-V characteristic of a p-n junction in the diagram below [1]

(c) Explain how a p-n junction operates in a forward biased connection.

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

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

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

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

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

………………………………………………………………………………………………... [3]

V / V

I / A


17

Name: ________________________ Class: _____________

Cover Page Question 7


18

It is recommended that you spend about 30 minutes on this question.

Name: ________________________ Class: _____________

7 It is common to observe condensation of water droplets on the glass windscreen or windows of an air-conditioned vehicle. One such observation is indicated in the photograph shown below.

A car manufacturer needs to ensure that the effects of condensation does not affect driver’s

sight of vision. A 5% reduction in visibility is generally acceptable. Design an experiment to investigate how the thickness of the glass affects the time taken to reach the 5 % reduction in visibility. The following equipment is available:

- Cool air blower

- Laser pointer

- Intensity meter

- Six similar square sized glass panels (20 cm by 20 cm) of thicknesses ranging from 8 mm to

18 mm, and any other equipment available in a school laboratory.

You should draw a labelled diagram showing the arrangement of apparatus you would use. In your account, you should pay particular attention to

(a) the equipment you would use,

(b) the procedure to be followed,

(c) the control of variables,

(d) how you determine the 5% reduction in visibility,

(e) any precautions that would be taken to ensure safety and improve the accuracy of the experiment.

[Total marks for this question: 12]

Cool air blowing towards the windscreen

Condensation


19

Name: ________________________ Class: _____________

Diagram:

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End of Paper


MERIDIAN JUNIOR COLLEGE Preliminary Examination Higher 2

_______________________________________________________________________

H2 Physics 9646/3


2 hours

_______________________________________________________________________

Class Reg Number

Candidate Name _____________________________

READ THESE INSTRUCTIONS FIRST This booklet contains Sections A and B of the Preliminary Examination Paper 3. Do not open this booklet until you are told to do so. Section A Answer all questions.

Section B Answer any two questions. You are advised to spend about one hour on each section. Write your answers on this question booklet in the blanks provided. INFORMATION FOR CANDIDATES

The number of marks is given in brackets [ ] at the end of each question or part question. Marks will be deducted if units are not stated where necessary or if answers are not quoted to the appropriate number of significant figures.

All working for numerical answers must be shown. You are reminded of the need for good English and clear presentation of your answers.

Examiner’s Use

Section A

Q1 / 6

Q2 / 12

Q3 / 8

Q4 / 10

Q5 / 4

Section B

Circle the questions you

have attempted

Q6 /20

Q7 /20

Q8 /20

Deductions

Total /80


2


8 m s

-1



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












s = ut + 1

2at2

v2 = u2 + 2as




r



= 2 2

o-x x


2kT




Q

r



where k =

2

2

8 ( )m U E

h


decay constant

= 1

2

0.693

t


3

Section A Answer all the questions in this section.

1 A small wooden block is projected up a frictionless ramp with an initial speed of 4.0 m s-1 as

shown in Fig. 1 below. The ramp is inclined at 20° and the block was projected from a point 6.0 m from the bottom of the ramp.

Fig. 1

(a) Calculate the distance moved by the block before it comes to instantaneous rest.

distance moved = ............................ m [2] (b) Determine the time taken for the block to reach the bottom of the ramp from the time it

was projected.

time = .................................... s [2]

6.0 m

20°

4.0 m s-1


4

(c) Sketch the displacement-time graph of the entire motion from the time the block was

projected to the time it reaches the bottom of the ramp.

[2]

time / s

displacement / m


5

2 Five identical resistors along with a NTC thermistor, a filament lamp and a voltmeter are

connected to an e.m.f. source of 15.0 V as shown below in Fig. 2.1. The e.m.f source has negligible internal resistance.

Fig. 2.1

(a) (i) Determine the effective resistance of the network of resistors between points B

and C.

effective resistance = ............................ Ω [3] (ii) Hence, determine the voltmeter reading.

voltmeter reading = .................................... V [2]

V

15.0 V

B C

V

6.00 Ω

6.00 Ω

6.00 Ω

6.00 Ω

6.00 Ω 170 Ω

210 Ω

A D


6

(iii) Determine the power dissipated in the filament lamp

power = ....................................W [1] (b) (i) On the axes below, sketch and label the I-V characteristics of the thermistor and

the filament lamp. Explain how the resistance of each of the devices can be obtained from their respective graphs.

...................................................................................................................................

......................................................................................................................... [3]

(ii) Explain the I-V characteristic of the thermistor and how it changes with increasing

temperature.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

......................................................................................................................... [3]

I / A

V / V


7

3 (a) A long vertical rigid conductor passes through a sheet of cardboard that is held

horizontal. A small compass is placed at the point P and the needle points in the direction shown in Fig. 3.1 (a) and Fig. 3.1 (b). Fig. 3.1 (a) Side view

Fig. 3.1 (b) Top view A current is then passed through the conductor and the compass needle now points in a

direction that makes an angle of 30 to its original direction as shown in Fig. 3.2. Fig. 3.2 Top view

P

P

30

cardboard sheet

long rigid conductor


8

(i) With reference to Fig. 3.2, state the direction of the current in the conductor.

Explain your answer.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

......................................................................................................................... [3]

(ii) The compass is 2.0 cm from the wire and the current I in the wire is 4.0 A.

Calculate the magnetic flux density before the current is turned on.

The magnetic flux density B at a distance d from a wire is given by 2

oBd

I.

B = ............................ T [2] (b) The long rigid conductor is now placed horizontally between the poles of a magnet and

the current flowing through it is increased until the wire just lift off the ground as shown in Fig. 3.3. The mass of the conductor is 0.42 kg and its length is 5.0 m. The magnitude of the magnetic field strength is 0.30 T and the width of the magnet is 0.80 m.

Fig. 3.3

magnet magnet

current

0.80 m


9

(i) On Fig. 3.3, label the letter ‘N’ the north pole of the magnet. Explain your choice.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

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

(ii) Determine the current in the wire.

current = …………………. A [1] 4 (a) Fig. 4.1 shows a fixed mass of ideal gas that undergoes a cyclic change from

A B C D A.

Fig. 4.1

(i) State the thermodynamic processes for AB and DA.

process AB : ..........................................................

process DA : ..........................................................

[1]

P / ×104 Pa

V / m3

13.0

9.0

1.9

A B

C D

1.0


10

(ii) Given that the gas in state A has a temperature of 300 K, determine the number of

moles of gas in the system.

number of moles = .................................. mol [1]

(iii) Determine the net work done on the gas in one cycle ABCDA.

net work done on gas = ...................................... J [1]

(iv) State the First Law of Thermodynamics.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

....................................................................................................................... [1] (v) Determine the total heat supplied during process AB.

total heat supplied = ...................................... J [2]


11

(vi) A student claims that an adiabatic expansion of a closed system of gas shows no

change in temperature. With reference to the First Law of Thermodynamics, explain why the claim is incorrect.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................

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

(b) Explain why atoms cannot have zero kinetic energy even when cooled to absolute zero

Kelvin.

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

................................................................................................................................ [2] 5 Explain what is meant by population inversion and stimulated emission and how they are

essential in the production of laser.

.......................................................................................................................................................

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............................................................................................................................................. [4]


12

Section B Answer two questions in this section.

6 (a) (i) Explain what is meant by 1. diffraction;

………………………………………………………………………………………..

………………………………………………………………………………………..

………………………………………………………………………………… [1] 2. interference; and

………………………………………………………………………………………..

………………………………………………………………………………………..

………………………………………………………………………………… [1] 3. coherence.

………………………………………………………………………………………..

………………………………………………………………………………………..

………………………………………………………………………………… [1] (ii) State two conditions for observable interference of two waves.

……………………………………………………………………………………………….

……………………………………………………………………………………………….

……………………………………………………………………………………………….

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


13

(b) To help guide large ships berth properly into docks, an engineer proposed using

interference of electromagnetic (EM) waves. The proposal suggests installing two EM wave emitters P and Q positioned 95 m apart at the edges of the dock gates. The two emitters can be taken to be point sources and they emit EM waves of frequency f1 in phase. The ship can be guided by searching for the strong signal radiated along the lines of constructive interference, also known as anti-nodal lines. For safety, it is important for the ship to ensure that it is sailing along the centre-line of the gates, as such the ship needs to “lock on” to the central anti-nodal line.

(i) Explain why the centre-line will always be an anti-nodal line regardless of the frequency of the EM waves used.

……………………………………………………………………………………………

……………………………………………………………………………………………

…………………………………………………………………………………….. [1]

(ii) State and explain which type of EM waves is suitable for such a system.

……………………………………………………………………………………………

……………………………………………………………………………………………

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

Top view (figure not to scale)

P

Anti-nodal lines

Q

dock gate

Fig. 6.1


14

(iii) Assuming that the ship is sailing along the centre-line, state and explain how the

intensity of the resultant signal varies as it approaches the dock gates.

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………… [2] (c) One particular large cargo ship strays off the centre-line as shown in Fig. 6.2.

Explain quantitatively whether the ship is on an anti-nodal line, given that f1 is 23.5 MHz.

……………………………………………………………………………………………………

…………………………………………………………………………………………… [3]

Top view (figure not to scale)

P

Q

dock gate

Fig. 6.2

895 m

95 m

250 m


15

(d) As an additional precaution to ensure that the ship “locks on” to the central anti-nodal

line, the emitters can simultaneously emit another EM wave of a different frequency f2.

(i) Explain how this precaution helps to prevent the ship from “locking on” to the wrong anti-nodal line.

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

…………………………………………………………………………………… [1]

(ii) Discuss why this additional precaution may still not be fool proof.

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

…………………………………………………………………………………… [1] (e) The large cargo ship is carrying loads of new cars. As the cargo ship is cruising, a car

which is not secured properly starts to move. A small piece of chewing gum is stuck to the edge of the wheel as shown in Fig. 6.3. A camera records the motion of the car’s wheel from the rear view as it is rotating.

(i) State the type of motion exhibited by the chewing gum from this viewpoint as shown in Fig. 6.3 when the wheel rotates.

……………………………………………………………………………………… [1]

Fig. 6.3 (rear view)

chewing gum

wheel


16

(ii) The car moves at a speed of 5.0 km h-1. Determine the period of the chewing

gum, given that the wheel has a diameter of 0.45 m.

period = ………………………. s [2] (iii) Hence determine the maximum vertical acceleration of the chewing gum.

maximum vertical acceleration = ………………………… m s-2 [2]


17

7 (a) Two beams of monochromatic light have similar intensities. The light in one beam has

wavelength of 400 nm and the light in the other beam has wavelength of 700 nm. The two beams are incident separately on three different metal surfaces. The work function of each of these surfaces is shown in Fig. 7.1.

Metal work function / eV

magnesium potassium caesium

3.68 2.26 1.88

Fig. 7.1

(i) Explain what is meant by the work function of the surface.

...................................................................................................................................

...................................................................................................................................

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

(ii) Using appropriate calculation, explain which combination/s, if any, of

monochromatic light and metal surface would give rise to photo-electric emission.

...................................................................................................................................

...................................................................................................................................

.........................................................................................................................

[4]


18

(iii) A student argued that the intensities of the two beams of light must be the same to compare the maximum kinetic energy of the emitted electrons. Explain the validity of his claim.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

.........................................................................................................................

[2]

(b) A parallel beam of electrons, all travelling at the same speed, is incident normally on a

carbon film. The scattering of the electrons by the film is observed on a fluorescent screen, as shown in Fig 7.2.

Fig. 7.2

(i) By considering the particulate nature of electrons, predict what would be seen on

the screen.

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

......................................................................................................................... [1]

(ii) However in this experiment, the particulate nature of the electrons was not

observed. Describe briefly the pattern that is actually observed on the screen. You may draw a sketch if you wish.

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

.........................................................................................................................

[1]

vacuum

flourescent screen

glass envelope

carbon film

beam of

electrons


19

(iii) The speed of the electrons is gradually increased.

State and explain what change, if any, is observed in the pattern on the screen.

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

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

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

......................................................................................................................... [3]

(c) In another experiment light is passed through a small opening. A photon changes

direction as shown in Fig. 7.3 below.

Fig. 7.3 (i) Using classical Physics, explain why the photon could not possibly have changed

direction.

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

......................................................................................................................... [1]

(ii) Suggest how this phenomenon could be explained using Heisenberg’s Uncertainty

Principle.

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

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

......................................................................................................................... [1]

(d) An X-ray spectrum is observed in an X-ray tube.

(i) Describe how the X-ray continuous spectrum is produced in an X-ray tube.

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

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

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

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

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

......................................................................................................................... [3]

photon


20

(ii) Suggest two reasons why X-ray is suitable for medical imaging.

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

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

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

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


21

8 (a) The isotopes Radium-224 and Radium-226 both undergo spontaneous α-particle decay.

(i) Explain what is meant by spontaneous.

...................................................................................................................................

......................................................................................................................... [1]

(ii) Describe how you would demonstrate that the radioactive decay is a random

process.

...................................................................................................................................

......................................................................................................................... [1]

(iii) A stationary Radium-224 nucleus decays to a Radon (Rn) by emitting an α-particle

with no gamma emission. The daughter nucleus is an isotope of Radon (Rn).

224

88Ra Rn + A

Z

1. Determine the number of neutrons and protons in a nucleus of Rn.

...................................................................................................................................

.........................................................................................................................

[1]

2. The mass of the Radium nucleus is greater than the combined mass of the

nuclei of Radon and the α-particle. Use a conservation law to explain how this decay is possible.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

.........................................................................................................................

[2]


22

3. The kinetic energy of the Rn is 0.1037 MeV. Determine the energy of the

α-particle given the following information:

rest mass of 224

88Ra = 224.0202 u

rest mass of Rn = 220.0114 u rest mass of α-particle = 4.0026 u

energy of α-particle = ............................ MeV [3] 4. As the α-particle travels in air, ionisation of air occurs, whereby a neutral atom

becomes an ion-electron pair. The energy required to produce an ion-electron pair at atmospheric pressure is 31 eV. Estimate, to two significant figures, the number of ion-electron pairs produced per unit length as the α-particle is stopped in air at atmospheric pressure.

number per unit length= ............................ m-1 [3] (iv) The total energy of the products formed for the decay of Radium-226 is lower than

that of the decay of Radium-224. Suggest, with a reason, which of the two isotopes has a longer half-life.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

.........................................................................................................................

[2]


23

(b) Radium-224 has a half-life of 3.6 days. A freshly prepared sample of Radium-224 has an

activity of 1.6 x 1014 Bq.

(i) Explain what is meant by the term half-life.

...................................................................................................................................

......................................................................................................................... [1]

(ii) Determine the mass of the sample.

mass = .................................... kg [3] (iii) After a period of time, 5.0 mg of the sample is found to have an activity of

2.1 x 1010 Bq. Determine the age of the sample.

age= .................................... days [3] End of Paper

MJC 2014 JC2 : H2 Physics : Preliminary Examination : Solutions

Meridian JC 2014 1

Answers to 2014 Prelim Paper 1 (H2 Physics)

1 C 11 D 21 C 31 A

2 C 12 B 22 D 32 D

3 D 13 D 23 A 33 C

4 A 14 C 24 B 34 D

5 C 15 D 25 B 35 D

6 B 16 D 26 D 36 A

7 B 17 A 27 A 37 B

8 B 18 A 28 C 38 D

9 A 19 A 29 A 39 C

10 B 20 C 30 C 40 A

MCQ 1: ( C )

2

2

0 052 0 02

2 50

0 06 6

P I R

P I R

P I R

..

.

. %

MCQ 2: ( C )

1

2

1

1 1

units of

Fk

rv

Nk

m ms

kgms

m ms

kgm s

MCQ 3: ( D ) Vertical acceleration is equal to gravitational acceleration which is constant. MCQ 4: ( A )

2

2

1

2

2

2 1

1

2

135 0 9 81

2

150 0 9 81

2

0 52 s

s ut at

. t

. t

t t t .

MCQ 5: ( C )

Acceleration is the same for each block. For the left most first block, TA = ma 2nd block, TB - TA = ma, TB = 2ma 3rd block, TC - TB = ma, TC = 3ma Hence, TC is the greatest.


Meridian JC 2014 2

MCQ 6: ( B ) Taking the balancing weight, the passenger car and the passenger as a system, Using N2L, Fnet = ma 640g – 520g – 80g = (640+520+80) a a=0.316 m s-2

MCQ 7: ( B ) Recall: To get spring constant, we need the gradient of load (Y-axis) against extension of spring (X-axis). Note that the graph is length of spring (Y-axis) against load (X-axis). Hence, from the graph given, the inverse of the gradient of the line would be the spring constant MCQ 8: ( B ) Uw + Um = W ρwVw g + ρmVm g = ρnVn g = ρn (Vw+Vm) g 1000 Vw + 13600 Vm = 7900 (Vw+Vm) 5700 Vm = 6900 Vw

Vw/ Vm= 57/69 = 0.826 MCQ 9: ( A ) By COE, Gain in KE = Loss in GPE KE just before ball strikes surface = initial GPE KE1 = mgh1 --- (1) KE just after ball strikes surface = final GPE KE2 = mgh2 --- (2) Taking the bottom of the ball as reference, (2)/(1): KE2 = h2 / h1 (KE1) = 30/70 x 0.80 = 0.34 J MCQ 10: ( B )

2

2

151800

30

151800 15

30

6750

F

P Fv

W

MCQ 11: ( D ) Object is moving in circular motion, hence there must be a resultant force which is known as the centripetal force. The two forces acting are tension and weight.


Meridian JC 2014 3

MCQ 12: ( B )

-1343 34354.6 m, 15.6 m s

2 22r v

2

2 2

cos , sin

15.6tan 0.454

54.6(9.81)

24.4

mvF mg F

r

v

rg

MCQ 13: ( D )

P

GMmE

r

1

2K

GMmE

r (Check derivation from notes)

T P K

1

2

1

2

E E E

GMm GMm

r r

GMm

r

When moved to higher orbit, r increases, hence ET increases, EP increases, EK decreases.

MCQ 14: ( C )

2 2( )net

GM Gmg

x d x

, hence gm-surface < gM-surface

There will be a point between the 2 masses where gm = gM and are in opposite direction, hence gnet = 0 MCQ 15: ( D ) By attaching a heavy mass, the natural frequency of the drum will be lowered. Hence resonance will now occur at a lower driving frequency. Note that attaching a heavy mass does not change the damping. MCQ 16: ( D ) The value of the potential energy increases as the body moves away from point X suggests that the force is directed towards X. Hence the body will oscillate about point X.

Note that force pdE

Fdx

. E.g. on the RHS of curve, gradient is positive, hence F will be negative

(since pdE

Fdx

), hence the force is directed towards the left (i.e. to the centre).

r

Rsource 0

GPE

KE

TE

Energy


Meridian JC 2014 4

MCQ 17: ( A ) For Alice to hear the music as loud as before, the intensity of the sound reaching her ears should remain the same.

2 2

0.75

4 15 4

13 m

P P

x

x

MCQ 18: ( A )

Phase difference o0.22 sin25

2 0.34 rad1.7

MCQ 19: ( A ) From a node, through 15 antinodes, to another node is equivalent to 7.5 wavelengths.

1.2000.160 m

7.5

On the CRO, 5 consecutive crests represents 4 periods.

3 38.50.50 10 1.0625 10 s

4T

-11150 m sv f

T

MCQ 20: ( C )

o

sin

23.7

sin sin 32

d n

d n

Hence max order is 3, and 7 bright spots can be seen (including the zeroth order).

MCQ 21: ( C )

3.0 × 10-3 m

4.0 × 10-3 m

Q Q Z

electron

5.0 × 10-3 m

FE

1 o

15 19

20

2 23

00

19

31 2

19

3

4 -1

4tan 53.13

3

1.45 10 1.6 108.3444 10 N

4 4 5.0 10

2 sin 1.3351 10 N

For circular motion,

9.11 101.3351 10

4.0 10

2.4 10 m s

E

net E

net c

QeF

x

F F

F ma

v

v


Meridian JC 2014 5

MCQ 22: ( D )

-1

19

14 -2

31

9

6

22 14 9

y

461533.33 V m

0.030

1.6 10 1533.332.693 10 m s

9.11 10

0.0607.5 10 s

8.0 10

1 1s 0 2.693 10 7.5 10 0.0076 m

2 2

y

x

x

y y

VE

d

F qEa

m m

st

v

u t a t

MCQ 23: ( A )

= 9.0 V

MCQ 24: ( B )

MCQ 25: ( B ) At balance length 20.0 cm, V

PJ = E

2

PQ

PQ

PJPJ V

L

LV

Using potential divider rule,

ER

RV

total

PQPQ

V

PQ= [5.0 / (5.0 + 3.0)] 7.0 = 4.375 V

Therefore, E

2 = V

PJ

= (20.0 / 100) (4.375) = 0.875 V

MCQ 26: ( D ) Lamp W is most likely faulty (fused), hence it forms a closed circuit across the bulb. The resistance

across a short circuit is very small and thus yielding the <1.0 reading in the resistance meter each time switch 2 is turned on.

10005.0

4500

Q

workdoneE

2 2

2 2

3

P=I R+I r

4500(0.50) (16) (0.50) r

1.0 x10

r = 2


Meridian JC 2014 6

MCQ 27: ( A ) FE = FB qE = Bqv Independent of charge, independent of mass. MCQ 28: ( C )

B I (n remains the same) When the solenoid length is halved, its resistance also decreases by half. Using V=IR, with V being constant and R decreasing by half, I will double, hence B will double.

MCQ 29: (A)

3

sin sin

0.055 850 sin30 (30 80) 10

1.17 Wb

BAN AN B

1.17e.m.f. 0.29 V

4.0t

MCQ 30: (C) By Lenz’s law, an induced current will flow in the loop such that it opposes the increase in the magnetic flux linkages producing it. When loop starts to enter the region of magnetic field, loop experiences an upwards retarding force larger than its weight and hence decelerate. By Faraday’s law, the induced emf is proportionate to the rate of cutting of magnetic flux. As the loop decelerates, the induced emf decreases and given that resistance is constant, induced current decreases in magnitude. When the loop is leaving the region of magnetic field, the magnetic flux linkages through the loop decreases, hence the induced current is such that it opposes the decrease in the magnetic flux linkages. induced current is now in the opposite direction but force is still upwards. As the upwards force is still larger than its weight, it continues to decelerate and magnitude of induced current will decrease. When the upwards magnetic force and weight become equal and opposite, the loop experience zero net force and reaches terminal velocity. Thus the rate of cutting of magnetic flux linkage and the induced current is constant. MCQ 31: (A)

Vo 12Vrms = 8.48

2 2

Mean Power = (Vrms)2/R =

28.487.20W

10

MCQ 32: (D) Average power in DC = Average power in AC I2 R = Irms

2 (R/2) Irms

2 = 2 I 2

Irms = 2 I

Hence Io = Irms (2=2I (22 I


Meridian JC 2014 7

MCQ 33: (C) Net power supplied to object = Pheater – Pheat loss

mcQnet

tmcPnet

At 300 K, Pheat loss = 1.5 W, m = 0.010 kg, c = 890 J kg-1 K-1 and 1-sK 51.0

t

51.0890010.05.1heater P

W04.6heater P

MCQ 34: (D)

nnRnnRnRnRTEk 8.34042

3)31.8)(15.273(

2

3

2

3)15.273(

2

3

2

3

MCQ 35: ( D )

Tunnelling probability , 2kdT e

k =

2 2 27 6 1914 -1

2 34 2

8 ( - ) 8 (6.64 10 )(1.0 10 )(1.6 10 )4.37 10 m

(6.63 10 )

m U E

h

Therefore, for constant of proportionality of 1.0,

14 152 2(4.37 10 )(2.0 10 ) 0.174kdT e e

MCQ 36: ( A ) Kmax = 0 eV occurs when λ = 250 nm. Therefore, the threshold wavelength is 250 nm.

By photoelectric equation, max

hcK

Hence,

34 8

max 9

19

6.63 10 (3.0 10 )0

250 10

7.956 10 J = 4.97 eV

hcK

MCQ 37: (B) An intrinsic semiconductor is made up of Group IV atoms. MCQ 38: (D) When the incident photon and the emitted photons have the same energy, they will have the same wavelength and frequency. This results in a monochromatic laser light.


Meridian JC 2014 8

MCQ 39: ( C ) 1. Most particles passed through undeflected or with a small deflection the nucleus occupies

only a small proportion of the available space.

2. Few particles are reflected backwards, through an angle close to 180o The nucleus is small and very massive.

MCQ 40: ( A ) Isotopes are atoms that have the same number of protons but different number of neutrons. For option A: Number of protons number of neutrons 182-108 = 74 108 186-112 = 74 112


Meridian JC 2014 9

Suggested Solutions to 2014 Prelim Paper 2 (H2 Physics) 1 (a) (i) Since the ball moves along the arc of a circle,

there will be a resultant force providing for the centripetal force. Hence the tension is larger than the weight.

M1 A1

(ii) From the graph v = 2.6 m s-1 at 1.2 s Using N2L, Fnet = ma Just before the ball hits the wall, it is almost vertical, hence T-mg = mv2/r T-(380 x 9.81) = 380 x 2.62 / 5.8 T = 4170 N

M1 M1 A1

(b) (i) Change in momentum = pf - pi = m (vf - vi) = 380 (0 - 2.6) = -988 N s The change in momentum of the ball is directed to the left, hence must have –ve .

M1 A1

(ii) By N2L,

F = 988/0.15 (time interval read off from graph) = 6590 N

Correct shape and labels for force and time -1m for positive force.

A1 M1

(c) (i) From notes:

The principle of conservation of momentum states that the total momentum of a system of objects remains constant provided no resultant external force acts on the system.

B2

(ii) Momentum is conserved: The system is the ball, the wall and the Earth. Loss in momentum of the ball equals to the gain in momentum of the wall and Earth. OR

M1 A1

Force/N Time/s

-6590

0.15


Meridian JC 2014 10

Momentum is not conserved: There is an external force acting on the ball (system), so the ball’s momentum is not conserved.

M1 A1

2 (a) (i)

Suitable BFL drawn

[B1]

0.0 0.4 0.8 1.2 1.6 2.0

M / Ms

2.00

1.75

1.50

1.25

1.00

0.75

0.50

0.25

0.0

R / Rs


Meridian JC 2014 11

(ii) Gradient = -0.94 0.02

Y-intercept = 1.80 0.02 R / Rs = -0.94 (M / Ms) + 1.80

[M1] [A1]

(iii) 1.92 0.02

[B1]

(iv) This value is obtained by extrapolation outside the data range and hence uncertainties are present. or The trend may not be a straight line beyond the set of data points given. or The maximum ratio corresponds to a star with zero radius hence this is not possible in reality.

[B1]

(v) No, the Sun does not belong to this species of star.

The data point which represents the Sun on the graph is found at coordinates (1,1), which does not lie in the trend presented by the data points.

[A1] [M1]

(b) (i) Change in gravitational potential = -4.2 – (-13.5) = 9.3 x 106 J kg-1 [B1] (ii) Using conservation of energy,

Energy at surface = Energy at 3R above surface

4 4

2

4

2 6 6

3 -1

10

2

14.2 10 ( 13.5 10 )

2

4.3 10 m s

R R R R

R R

KE PE KE PE

mv m m

v

v

Allow ECF from (b)(i)

[M1] [A1]

(iii)

6

3

-1

3.8 ( 4.6) 10

4.5 3.5 3300 10

0.24 N kg

dg

dr r

[M1] [A1]

(iv) The claim is valid.

The gravitational field strength at any point is a resultant of all the gravitational field strengths of this star and its neighboring stars.

[A1] [M1]

3 (a) The electric potential at a point in an electric field is defined as the work done per unit

positive charge by an external agent in bringing a positive test charge from infinity to that point without acceleration. [B1]

(b) (i)

gradient of graph

dVE

dr

E


Meridian JC 2014 12

2

2.50 0.00=

0.00 2.50 10

[M1]

-1100 N C [A1]

Acceptable range: 90 110E

Direction: in positive x-direction / towards right / from left to right / [B1]

(ii) At x = 0.20 cm,

0 04 4 0.040

Q QV

x x

0

1 16.00

4 0.002 0.040 0.002

Q

121.27 10 CQ

Accept also if student used other suitable points for substitution.

(iii) Charge is evenly distributed over the sphere OR Charged sphere is taken to be a point charge OR No other surrounding charges OR Centre of charge is taken to be at centre of sphere [B1]

(c) (i) Type of particle: Any positively charged particle / proton / positive ion / cation [B1]

The two ends of the well form a potential barrier such that particles without sufficient energy will not be able to overcome it to escape OR The force on the particle always points to the centre and hence keeps the particle trapped between the two charged spheres. [B1]

(ii) Quantum tunneling [B1] 4 (a) Magnetic flux through a coil is defined as the product of the magnetic flux

density normal to the coil surface, BN and the area A of the coil surface.

[B1]

(b) (i) Coil B, connected to AC source, produces a magnetic field that is varying

with time sinusoidally. This results in a varying magnetic flux through coil B and hence a varying magnetic linkage through coil C. By Faraday’s law, an emf is induced in coil C due to changing magnetic flux linkages through coil C.

[M1] [A1]

(ii) 2B C

[A1]


Meridian JC 2014 13

Notes:

1) The magnetic flux produced by coil B will distribute and flow equally

through coil A and coil C. (By geometry of the soft iron core)

2) Students wrongly state the turn ratio of coil B to coil C.

(iii)

B BB B

C CC C

d dN

dt dt

d dN

dt dt

2

2

2402

13 1

9.23 V

Peak output 9.23 2 13.1 V

CB

CB

B C

C

C

C

dd

dt dt

N N

V

[M1] [A1]

(iv) Since soft iron has a gap, the magnetic flux through coil C will be reduced

and hence the rate of change of magnetic flux linkage through coil C will be lowered. By Faraday’s law, induced emf in coil C is proportional to the rate of change of mangetic flux linkages, hence the induced emf will be lower.

[M1] [A1]


Meridian JC 2014 14

5 (a) (i)

[M1] for correct shape

- V = 0 when t = 0

- sharp edges at T/4

- at least 2 cycles should be drawn

[A1] for correct label of Vo and T

(ii)

[M1] for correct shape (ecf is given if student show understanding that only half of the cycle exists and the graph can only be in either the positive or negative axis). [A1] for correct label of Vo and T

(b) (i)

Turns ratio 240

32

s

p

N

N =15:2

[A1]

(ii) Since power in the primary coil = power in secondary coil

At 240 kV,

A 3.8310240

1020

W1020

W1020

3

6

6

6

out

outout

outin

I

VI

PP

2 2 83.3 5.0 =34.7 kWcablesP I R

At 32.0 kV, 6

3

20 10625 A

32.0 10outI

[M1] [M1]

V

t/s 0

V0

- V0

T/2 T 2T 3T/2

V

t/s 0

V0

- V0

T/2 T 2T 3T/2


Meridian JC 2014 15

2 2 625 5.0 1.95 MW lossP I R

Since less power is lost transmitting at 240 kV, it is more economical to

transmit at 240 kV.

[A1]

6 (a) A p-n junction is formed when a p-type semiconductor is joined to an n-type

semiconductor. When the junction is formed, the difference in concentration of electrons between the n-side and the p-side leads to mobile n-side donor electrons diffusing across the junction and filling the holes in the p-side.. [B1] As the n-side donor electrons diffuse across the junction, they leave behind immobile positive ions (recall that both an n-type and a p-type semiconductor on their own are electrically neutral). Similarly, the filling of a hole in the p-type makes a negative ion. [B1] Since the two sides of the depletion region each carry a net charge, an internal electric field is set up within the depletion region. [B1] As diffusion of the n-side donor electrons to the p-side continues, the depletion region widens and the difference in charge between the n-side and the p-side builds up. Eventually, the internal electric field becomes so strong (104 – 106 V cm-1) that it prevents further diffusion of the electrons across the junction and ensures that there is no current when no external potential difference is applied. [B1]

(b) Current increases exponentially with increasing forward bias passing through the origin.

(c) p-side connected to positive terminal and n side connected to negative terminal of voltage source [B1] The internal potential difference (due to the difference in charge between the n-side and the p-side) decreases as its polarity is opposite to the polarity of the external voltage source. The depletion zone becomes narrower and no longer effectively inhibits the flow of electron [B1] This allows current to flow


Meridian JC 2014 16

Proposed solutions to Q7 Planning

Item Description Comments:

Basic

Procedure

(BP)

[1]

Correct general procedure to determine how time

taken for 5% reduction of intensity varies with

thickness (e.g. idea of experiment is correct and

experiment is repeated with glass panels of

difference thicknesses)

No mark given if student investigate any other

dependent or independent variables.

Mark will be given if student

describe correct general

procedures to meet the

objective of the experiment

even if diagram shown is not

workable.

Student also needs to write

about using different

thicknesses glass and

repeating experiment.

Diagram (D)

[2]

Diagram of workable arrangement that includes

glass panels, cool air-blower, stopwatch, metre

rule, thermometer, laser pointer, intensity meter,

vernier callipers or micrometer screwgauge. [D1]

Approriate retort stands/supports used to hold

equipment firmly (e.g. laser pointer and intensity

meter, air-blower) [D2]

Method of (M)

Varying and

Measuring/

Determining

the IV.

Measuring/

determining

the DV

Analyzing

the data

[3]

Measure and record the thickness of each glass

panel using vernier callipers or micrometer

screwgauge. [M1]

Workable method to determine the 5% reduction

in intensity of light (use of readings registered

from the intensity meter). [M2]

Workable method to determine time taken (using

a stopwatch). [M3]

Repeat experiment for 6 sets of readings and

plot a graph of minimum time taken vs thickness

of glass.

Independent variable:

- Need to explicitly measure

the thickness of each glass

panel.

Dependent variable

- Must explicitly say how is

the 5% reduction of

intensity of light is

determined.

Control of

other Variables

(CV)

[max 2]

Fix the distance at which the cool air-blower,

laser pointer (x1) and intensity meter (x2) from the

glass panel (using a metre rule) [CV1]

Check that the humidity of the room is kept

constant during the experiment (e.g. using

electronic humidifier). [CV2]

Check that the room temperature is kept

constant (using an air-conditioned room). [CV3]

Check that the initial intensity of the laser pointer

is similar (using intensity meter) by conducting

the experiment in an enclosed room with curtains

drawn and lighting reduced. [CV4]

- Students must explicitly

discuss HOW the variables

are controlled or kept

constant.

- For example if the

distances are to be kept

constant, must mention

about making markings

and using ruler to measure

distance.


Meridian JC 2014 17

Other details

(Reliability)

(R)

[max 3]

Steps taken to reduce random error/uncertainty &

represent data

Experiment repeated by testing the glass panels

again and results (i.e. time taken) averaged

under same conditions (to ensure repeatability).

[R1]

Other Good Design features/reliability measures

Before starting experiment, check if range of

independent variable can give an observable

significant change for dependent variable by

testing with the thinnest and thickest glass

panels before carrying out the rest of the

experiment. [R2]

Before repeating the experiment, needs to make

sure that the surface of the glass panel is

completely dried (e.g. wipe with clean cloth) [R3]

Check that the temperature at which the cool air

is consistent with a thermometer placed in the

path of the cool air. Or that the power setting of

the air blower is set to a constant value. [R4]

Check the thickness of the glass panel at various

positions of the glass (to determine the average

thickness correctly). [R5]

The glass panels to be clamped with insulation

sheets so that there is no temperature variance

at the clamped area. [R6]

Ensure that the glass panels have the same

initial temperature before start of each repeated/

new experiment (by placing a thermocouple on

the glass panel). [R7]

Other accepted answers include:

Ensure laser pointer and intensity meter are in

line or aligned to each other (and not simply “at

the same height/level”).

Ensure surrounding light intensity is constant (by

conducting in a dark room for example)

Account for initial light intensity due to ambient

light and subtract from intensity meter reading.

Ensure there are no external wind blowing on the

glass panel other than that from the air blower by

conducting in a closed room with fans switched

Need to mention explicitly what

conditions results are repeated

under, method to repeat

experiment in this case.


Meridian JC 2014 18

off.

Safety

Precautions

(S)

[max 1]

The glass panels may have sharp edges and

also the glass panels may be broken during the

experiment. Precaution is needed in handling the

glass panels and the broken pieces (e.g. handle

it with a thick glove). [S1]

Ensure that the laser pointer is not pointing into

the eyes of the experimenter to avoid damage to

eyes. [S2]

Other accepted answers include:

Ensure that the glass panel is clamped properly

so that it does not topple and break which can

cut and cause injuries.

Sample Diagram

Laser pointer above the cool air-blower (clamped)

x1 x2

metre rule

Clamped Intensity meter

Clamped glass

thermometer

Top view Clamped glass


Meridian JC 2014 19

Suggested Solutions to 2014 Prelim Paper 3 (H2 Physics) Section A

1 (a) Taking up the slope as positive,

2 2

2

3 355

2

0 4 2 3 355 [M1]

2 38 m / 2.4 m [A1]

a g sin .

v u as

. s

s .

2 2

2

1 1

2 2

10 9 81 20 4 [M1]

2

2 38 m / 2.4 m [A1]

mv mgh mu

. d sin

s .

(b)

2

2

2

1

2

16 4 3 355 [M1]

2

1 6775 4 6 0

3 43 s / 3.4 s [M1]

s ut at

t . t

. t t

t .

(c)

Correct shape [B1] Correct labels [B1]

2 (a) (i)

Effective resistance due to AC and ADC, R` =

11 1

4.06.0 12.0

Effective resistance due to AB and R` = 4.0 + 6.0 = 10.0 Ω

Total effective resistance between BC =

11 1

3.756.0 10.0

[M1] [M1] [A1]

3.4

s/m

2.4

t/s

-6.0


Meridian JC 2014 20

(ii) Effective resistance across thermistor and filament lamp =

11 1

93.9170 210

Using potential divider principle,

Voltmeter reading = 93.9

15.0 14.4V93.9 3.75

(If method is correct but calculation is wrong, M1 mark still given).

[M1] [A1]

(iii) Power dissipated in the filament lamp =

214.40.987W

210

[A1]

(b) (i)

1mark for the correct shape and label of the thermistor I-V graph

1mark for the correct shape and label of the filament lamp I-V graph

The resistance of the device can be obtained from the ratio of potential difference and its corresponding current value read from a particular point on the graph.

[A1]

(ii) The thermistor is made of semiconductor material such as silicon or

germanium.

Increased p.d. across the thermistor results in an increased current which

in turn, causes the temperature to rise.

As the temperature rises, the lattice ions vibrations increase and reduce

[M1]

I / A

V / V

Thermistor

Filament lamp


Meridian JC 2014 21

the drift velocity of the charged particles.

However, the effect of the increase in the number of free electrons and

holes due to temperature increase is more significant than the reduction

in drift velocity of the charged particles.

Hence, the resistance of the NTC thermistor decreases as temperature

increases.

[M1] [A1]

3 (a) (i) The direction of the compass needle is the resultant of the Earth’s

magnetic field and the field produced by the wire when current passes through it. The direction of the Earth’s magnetic field is given in Fig. 3.1b, pointing horizontally to the right. Hence in order to get the resultant field as shown, using Fig. 3.2, the magnetic field produced by the wire must be pointing upwards. Using right hand grip rule, the current in the wire points out of the cardboard (or paper).

[M1] [M1] [A1]

(ii) 7

5

with current 2

5 5

w/o current Earth

4 10 (4)4.0 10 T

2 2 (2.0 10 )

4.0 106.9 10 T

tan30 tan30

o

wire

Br

BB B

I

[M1] [A1]

(b) (i)

For the wire to lift off the ground, the magnetic force acting on the wire must point upwards since weight of the wire points downwards. Using Fleming left hand rule, since force points upwards, current points into paper, the magnetic field must point from right to left. Hence the right magnet is the North pole.

[M1] [A1]

(ii)

0.42(9.81)

0.30(0.80)

17 A

B mg

I

I

[B1]

Current into

plane of page N


Meridian JC 2014 22

4 (a) (i) A B: isobaric expansion D A: isochoric / isovolumetric heating

[A1]

(ii)

mol 5215.52

30031.80.1100.13 4

n

n

nRTPV

[A1]

(iii)

4

4

4

Work done on gas

9.0 (1.0 1.9) 13.0 (1.9 1.0) 10

9.0 13.0 1.9 1.0 10

3.6 10 J

P V

[A1]

(iv) First law of Thermodynamics states that the internal energy is a

function of state; the increase in the internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system.

[B1]

(v)

J 109.210925.29.0100.132

5

)0.19.1(100.130.19.1100.132

3

)(2

3

2

3

554

44

Q

Q

VPQPV

VPQTnR

WQU

[M1] [A1]

(vi) Adiabatic expansion means that there is no heat supplied to the

system, 0Q .

An expansion leads to negative work done on the system, 0W

By First Law of Thermodynamics, WU , there is a decrease in

internal energy , 0U

This implies that there is a decrease in temperature, 0T

Hence it is not an isothermal expansion. Only used symbols to explain: minus 1 mark

[M1] [A1]

(b)

By Heisenberg’s uncertainty principle, 4

hE t

This implies that the uncertainty in energy E cannot be zero and hence the energy will not be zero even when it has been cooled to absolute zero Kelvin.

[M1] [A1]


Meridian JC 2014 23

5 Population inversion is required as there is a need to amplify the number of photons emitted by stimulated emission and this can only occur if more stimulated emission take place compared to stimulated absorption [B1] when the number of atoms in the excited state is higher than the number of atoms in the ground state[B1]. Stimulated emission occurs when an external photon triggers the emission of a photon from an excited atom[B1]. The emitted photon is in every way identical to the stimulating photon. [B1] This is critical to ensure the characteristic of laser which is highly monochromatic, highly coherent and highly directional.

Section B

6 (a) (i) 1. Diffraction refers to the bending or spreading out of waves when they travel through a small opening or when they pass round a small obstacle. [B1]

2. Interference refers to the superposing of two or more coherent waves to

produce regions of maxima and minima in space, according to the principle of superposition [B1]

3. Coherence refers to having a constant phase difference and same

frequency (between waves/sources/particles). [B1] (ii) Any two [B1 each]:

1. The waves must be coherent. 2. The waves must have approximately the same amplitude. 3. The waves must be unpolarised or polarised in the same plane (for transverse

waves). (b) (i) Since the two radio waves are in phase, along centre-line,

path difference is always zero [B1] Hence constructive interference occurs.

(ii) Radio waves have long wavelengths, hence the [A1]

anti-nodal lines will be far apart enough for the ship to differentiate [M1] (iii) Since the intensity of each individual wave is inversely proportional to the square

of the distance, the intensity of each individual wave will increase as the ship goes nearer, hence the resultant intensity will increase. [M1] OR Since the amplitude of each individual wave is inversely proportional to distance, the amplitude of each individual wave will increase as the ship goes nearer, hence the resultant amplitude of the superposed wave will increase. As intensity is proportional to the square of the amplitude, intensity increases.[M1] Hence, the intensity of the resultant increases as the ship approached the gate [A1]

(c) 8

1 61

3.0 1012.77 m

23.5 10

c

f

[C1]

Path difference 2 2

2 295 95895 250 895 250

2 2

[M1]

25.527 m

12 [A1]


Meridian JC 2014 24

Since the path difference is approximately 21, the ship is on an anti-nodal line.

Note: using D

xa

will earn no credit.

(d) (i) If ship is on the central anti-nodal line,

it should detect maximum signals from both frequencies / the maximum signal will be stronger [B1] OR If ship is on wrong anti-nodal line, only one of the frequencies will show a strong signal. [B1]

(ii) Higher orders of maxima from both frequencies may still coincide/overlap [B1]

Hence the ship could still detect maximum signals from both frequencies even though it is not on the central anti-nodal line.

(e) (i) Simple harmonic motion [B1] (ii) -1 -15.0 km h 1.39 m s

2

2

v r

rv

T

rT

v

2 0.225

1.39

[M1]

1.02 s [A1] (iii) 2

0 0a x

2

20.225

1.02

[M1]

-28.57 m s [A1]

Alternatively, 2v

ar


Meridian JC 2014 25

7 (a) (i) The work function of the surface is defined as minimum amount of the work

necessary to remove a free electron from the surface of the material. [B2] (ii)

By photoelectric equation, max

hcK

For any photo-electric emission to occur, hc

[B1] – this written explanation

is needed to get marks.

For wavelength of 400 nm, 34 8

19

9

6.63 10 (3.0 10 )4.97 10 J = 3.11 eV

400 10

hc

[B1]

For wavelength of 700 nm, 34 8

19

9

6.63 10 (3.0 10 )2.84 10 J = 1.78 eV

700 10

hc

Since 3.11 eV > 2.26 eV and 3.11 eV > 1.88 eV, and 1.78 eV < 1.88eV < 2.26 eV < 3.68 eV [B1]- for method only Therefore, the possible combinations of light and metal surface are:

1) Wavelength of 400 nm and potassium

2) Wavelength of 400 nm and caesium [B1- getting both correct]

(iii) Student’s claim is not valid. [A1] Intensity of light only affects the rate of photons

incident on the target metal. Each photon energy hc

remains the same for the

same monochromatic light. [B1]

By the photoelectric equation, max

hcK

the maximum kinetic energy of the emitted electrons remains unchanged if the target metal remains the same.

(b) (i) If the electrons behave as particles, a central bright spot would be observed on the

screen. [A1]

(ii) Pattern observed:

The electrons behave as wave and the pattern observed would be concentric rings of particles with increasing distances away from centre. [A1]


Meridian JC 2014 26

(iii) When the speed of electrons increases, the momentum of the electrons increases. [M1].

Hence, de Broglie’s wavelength, h

p will decrease. [ M1]

With a smaller wavelength, the extent of diffraction is reduced, hence leading to reduced diameters for the rings. [A1]

(c) (i) The photon does not experience any force acting on it. Hence, by Newton’s 1st

law, the photon should continue its state of motion without any change in direction. [B1] Or By conservation of momentum, the momentum of photon should remain the same since there is no resultant external force acting on the photon. Hence, photon should be undeflected. Or By Newton’s 2nd law, since there is no resultant force acting on the photon, there is no change in photon’s momentum. Hence, photon should be undeflected.

(ii) By Heisenberg’s Uncertainty Principle, as the photon passes through the small

opening, the uncertainty of its possible position is reduced at the opening. Hence, the uncertainty of momentum of the photon will increase. This leads to the change in direction of the photon. [B1]

(d) (i) High speed (or high energy) electrons are bombarded (or collided) against the

nucleus of the target metal. As the electron approaches a nucleus, it undergoes acceleration due to the attractive force between the nucleus and the electron. [B1] As the electron accelerates (due to a change in direction), it will emit electromagnetic energy in the form of a photon (X-ray) which is equivalent to a loss in kinetic energy of the electron. [B1] This scattered electron (with less energy) may have subsequent collisions with other target atoms and thus generating X-ray photons of different energies that gives rise to a continuous spectrum. [B1]

(ii) 1) X-ray penetrates matter.

2) X-ray penetrates to different degrees depending on the density of matter.

Denser matter (e.g. bone) cannot be penetrated as much (or absorbs

more)

3) X-ray affects photographic plate (so image can be formed)

4) Sharp shadows cast because of short X-ray wavelength.

B2- Any 2


Meridian JC 2014 27

8 a i The process cannot be speeded up or slowed down by (or independent of) chemical conditions or physical means such as changes in pressure, temperature, magnetic and electric fields. [1]

ii The random nature of decay may be demonstrated by placing a detector such as a Geiger-Mϋller tube connected to a counter near a radioactive source. The fluctuations in count rate show that decay is random. [1]

iii1. The Rn nucleus consists of 134 neutrons and 86 protons [1]

2. Mass-energy is conserved. [B1] The difference in mass manifests as energy in the form of kinetic energy of the products. [B1]

3.

Energy released = [Mass reactants – mass products]c2 = [224.0202– (220.0114+4.0026)] x 1.66 x 10-27 x (3.0 x108)2

[M1] = 9.2628 x 10-13 J = 5.7893 MeV Energy of α-particle = 5.7893 – 0.1037 [M1] = 5.686 MeV [A1] Allow alternative solution of using Conservation of Momentum

4.

Maximum number of ion pairs produced be each alpha particle

= 65686 10

31

. [M1]

= 1.834 x 105 Range of alpha particles in air = 3 cm (accept 3 to 5 cm) [C1] Number of ion pairs per unit length = 1.834 x 105 / (3.0 x 10-2) = 6.1 x 106 [A1]

- answers may vary, depending on the range of alpha particles stated by the student

iv Comparing the two decays, since the products produced from the decay of Ra-224 has a higher energy, it is more likely to decay to form more stable products. [M1] Hence, Ra-226 has a longer half-life. [A1]

b i Half-life is defined as the time taken for half the original number of radioactive nuclei to decay. [A1]

ii A = λN

14 ln216 10

3.6 24 3600. N

[M1]

N = 7.18 x 1019 Mass = 7.18 x 1019 x 224.0202 x 1.66 x 10-27 [M1] = 2.67 x 10-5 kg [A1]


Meridian JC 2014 28

iii For 5.0 mg of sample,

A0 = (5.0 x 10-6 / 2.67 x 10-5) x 1.6 x 1014 [M1] = 2.996 x 1013 Bq [Alternative: N0 = (5.0 x 10-3 / 224) x 6.02 x 1023 A0 = λN0 = 2.99 x 10 13 ]

0

1

2

nA

A

10

13

2.1 x 10 1

2.996 10 2

n

[M1]

Number of half-lives, n = 10.5 Age = 10.5 x 3.6 = 38 days [A1]

h2 physics 9646/1 - weeblyscore-in-chemistry.weebly.com/uploads/4/8/7/1/48719755/mjc_2014_pr… ·...

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