Stpm Trial 2009 Phy Q&a Terengganu

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CONFIDENTIAL* JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN 960/1 TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN NEGERI TERENGGANU JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR PHYSICS JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR PAPER 1 JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR 1 3 hours 4 JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR

TRIAL 2009

Instructions to candidates :

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO. There are fifty questions in this paper. For each question four suggested answers are given. Choose one correct answer and indicate it on the multiple choice answer sheet provided. Read the instructions on the multiple choices answer sheet very carefully. Answer all questions. Marks will not be deducted for wrong answers. The total score for this paper is the number of correctly answered questions.

This question paper consists of 18 printed pages . 960/1 *This question paper is CONFIDENTIAL until the examination is over

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2 Physics 960 Constants value

speed of light in vacuum

c

=

3.00 x 108 m s-1

permeability of vacuum

µo

=

4π x 10-7 H m-1

permittivity of vacuum

εo

=

8.85 x 10-12 F m-1

=

[1/(36 π)] x 10-9 F m-1

magnitude of electron charge

e

=

1.60 x 10-19 C

Planck constant

h

=

6.63 x 10-34 J s

atomic mass unit constant

u

=

1.66 x 10-27 kg

electron rest mass

me

=

9.11 x 10-31 kg

proton rest mass

mp

=

1.67 x 10-27 kg

molar gas constant

R

=

8.31 J K-1 mol-1

Avogadro constant

L, NA

=

6.02 x 1023 mol-1

Boltzmann constant

k

=

1.38 x 10-23 J K-1

gravitational constant

G

=

6.67 x 10-11 N m2 kg-2

free fall acceleration

g

=

9.81 m s-2

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1.

3

The Young’s modulus E of a solid is given by

E= Where

stress F / A = strain e / lo

F= force A= area of cross section e = extension l0 = initial length

The dimensions of Young’s modulus are the same as A

2

pressure

B

force constant

C

force

D

impulse

A ball of weight W slides along a smooth horizontal surface until it falls off the edge at time T.

X

Y

Which graph represent how the resultant vertical force F, acting on the ball, varies with time t as the ball moves from position X to position Y? F

F

A

B W

T 0

t

0

t T

-W

F

F

C

D W

T 0

t

0

t T

-W

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4

3 Two block, X and Y of masses m and 2m respevticely are accelerated along a smooth horizontal surface by a force F applied to block X as shown in the diagram

X

Y

What is the magnitude of the force exerted by block Y on block X during this acceleration? A

F 3

B

2F 3

C

F 2

D

0

4 A small mass metal sphere of mass m is moving trough a viscous liquid. When it reaches a constant downward velocity ν , which of the following describes the change with time in the kinetic energy and gravitational potential energy of the sphere? Kinetic energy

Decrease at a rate of ( mgv −

C

1 2 mv 2 1 Constant and equal to mv 2 2 Increases at a rate of mgv

D

Increases at a rate of mgv

Decrease at a rate of ( mv 2 − mgv )

A B

5

Gravitational potential energy

Constant and equal to

1 2 mv ) 2

Decrease at a rate of mgv Decrease at a rate of mgv

1 2

A particle moves with constant speed in a horizontal circle. Which of the following quantities is zero? A B C D

Resultant force Angular acceleration Angular velocity Centripetal acceleration

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5

6 A mass of 0.05 kg is attached to one end of a piece of elastic of unstreched length 0.50 m. The force constant of the elastic is 40 Nm −1 . The mass is rotated steadily on a smooth table in a horizontal circle of radius 0.70 m as shown below

What is the approximate speed of the mass? A

20ms −1 2

B

24ms −1

C

11ms −1

D

15ms −1

7 A small ball of weight W is suspended by a light thread. When a strong wind blows horizontally, exerting a constant force F on the ball, the thread makes an angle θ to the vertical as shown

θ F

W Which equation correctly relates θ , F and W ? A B C D

F W F sin θ = W F tan θ = W W tan θ = F cos θ =

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6 8 The diagram below shows a piece of inextensible string which passes trough a pulley. The pulley has a radius of 0.05 m and moments of inertia 6.0 × 10 −4 kgm 2 . The end Q is held such that an object P of mass 0.2 kg is stationary at a height of 0.2 m from the floor.

P

Q 0.2 m

If end Q is released and the pulley is rough enough to prevent the string from slipping, find the speed of P when it touches the floor. A

1.35ms −1

1.23ms −1

B

C

1.01ms −1

D

1.48ms −1

9 The gravitational potential energy E P of a body varies with its distance r from the centre of a planet as shown in the diagram below

EP 0

r

What does the gradient at any point on the curve present? A B C D

the gravitational potential at that value of r the gravitational field strength at that value of r the acceleration of the body towards the planet the force pulling the body towards the planet

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7

10 On the ground, the gravitational force on a satellite is W. What is the gravitational force on the satellite when at a height R/50, where R is the radius of the earth A A

1.02 W

B

1.04 W

C

0.96 W

D

0.98 W

11 A particle of mass m performs simple harmonic motion with an amplitude a and a frequency f. The total energy of this simple harmonic motion is A

2π 2 ma 2 f

2

B

2ma 2 f

2

C

1 ma 2 f 2

2

D

2ma 2 f

2

12 Which of the following statements is true for a damped system of oscillation ? A B C D

The oscillating amplitude is maximum when forced frequency is almost the same as the natural frequency The system does not oscillate at its natural frequency When resonance takes place, the system does not lose energy Damping causes the oscillating frequency to become gradually smaller

13 Which of the following correctly summarizes what happens when light waves of frequency f and wavelength λ move from air to glass ? A B C D

λ decreases remains the same decreases remains the same

f increases remains the same remains the same decreases

14 To recieve FM waves of frequency 100 MHz, the total length of a bipolar half-wave aerial most suitable for use is A

0.75 m

B

1.00 m

C

1.50 m

D

2.25 m

15 The fact that light waves are transverse wave can be shown by A B C D E

diffraction polarization interference Doppler effect photoelectric effect

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8 16 A wire under increasing tension, undergoes extension as shown in the graph.

Tension/ N

80 60 40 20 0

Extension / mm 5

10

How much work is done on the wire to cause an extension of 9 mm ? A

0.72 J

B

0.43 J

C

0.36 J

D

0.29 J

17 A steel wire has an original length of l , cross-sectional area of A, and Young modulus of E. The force constant can be expressed as A

AE l

B

Al E

C

E Al

D

l AE

18 Equation W = p (V2 – V1) represents the work done by a gas during A B C D

free expansion isothermal expansion adiabatic expansion expansion at constant pressure

19 Find the work done when 3 moles of an ideal gas is expanded from 4 dm 3 to 6 dm 3 at 400 K A

8 kJ

B

-4kJ

C

4kJ

D

6kJ

20 The specific heat capacity at constant volume for an ideal gas is 2.4 x 102 J kg-1 K-1. The change in the internal energy of 5.0 x 10-3 kg of the gas when the temperature of the gas is increase from 27o C to 327oC is A

32 J

B

49 J

C

180 J

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D

360 J

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21 A ln p (pressure) against ln V (volume) graph is sketched as shown.

ln p

ln V 0 What is the quantity represented by the gradient of the straight line ? A B C D

-γ γ Change in internal energy Change in heat energy

22 An ideal gas is expanded at constant pressure, then cooled at constant volume, and finally compressed adiabatically until it is returned to its original state. Which graph shows the changes which occur to the gas? A

P

B

P

V C

P

V D

P

V

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P

Q

M

T

N

23 A composite rod consists of a rod P and a rod Q where P is a better thermal conductor. The ends M and N are maintained at constant temperatures TM and TN respectively where TM > TN. If the composite rod is well insulated, which of the following graphs shows the variation of temperature, θ along the composite rod? A

θ

θ

B

TM

TN C

TM

M

T

TN

N

θ

D

TM

M

T

N

M

T

N

θ TM

TN

TN

M

T

N

24 Which of the following statements is true about electrically equipotentials surfaces? A B C D

The charge density is uniform No work is done to move a charge along an equipotential surface The electric field at any point on the equipotentials surface is zero There is no electric potential difference between 2 nearby electrically equipotential surfaces.

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11

C1 + + C2 25 Figure shows two capacitors are connected as shown. A capacitor C1 with a capacitance of 4µF is charged to 200µC and another capacitor C2 with a capacitance of 3µF is charged to 300µC. The total energy lost in the two capacitor are A

0

B

2.0 × 10 −3 J

C

1.8 × 10 −2 J

D

2.0 × 10 −2 J

26 Which graph best represent the variation in the drift velocity of the electrons in a uniform copper wire as the potential difference across the wire changed? (Assume temperature of the wire remains constant)

A

Drift velocity

B

Drift velocity

Potential difference

Potential difference

Drift velocity

Drift velocity

D

C

Potential difference

Potential difference

27 The resistance of a piece of pure silicon decreases rapidly with increasing temperature because A B C D

the charge carriers move rapidly the number of charge carriers increases the ratio of negative charge carriers to positive charge carriers increases the ratio of positive charge carriers to negative charge carriers increases

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12 28 Kirchhoff’s two laws for electric circuits can be derived by using certain conservation laws. On which conservation laws do Kirchhoff’s law depend? Kirchhoff’s first law charge charge current energy

A B C D

Kirchhoff’s second law current energy mass current

29 The diagram below shows a model of an atom in which two electrons move around a nucleus in a circular orbit. The electrons complete one full orbit in 1.0 × 10 −15 s .

What is the current caused by the motion of the electrons in the orbit? A

1.6 × 10 −34 A

B

3.2 × 10 −34 A

C

1.6 × 10 −4 A

D

3.2 × 10 −4 A

30 An electron moving with uniform velocity enters a magnetic field which is perpendicular to its direction. The electron will then move A B C D

in a straight line as before in a straight line parallel to the magnetic field in a circle at a plane normal to the magnetic field in a parabola at a plane normal to the magnetic field

31 Three straight conductors X, Y and Z are carrying currents of the same magnitude as shown. Which of the following represents the resultants force acting on the conductor X?

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13

32 The e.m.f induced in a coil of wire, which is rotating in a magnetic field, does not depend on A B C D

the area of the coil the resistance of the coil the angular speed of rotation the number of turns on the coil

33 When the speed of an electric motor is increased due to a decreasing load, the current flowing through it decreases. Which of the following is the best explanation of this? A B C D

The induced back e.m.f increases The resistance of the coil changes Frictional forces increased as the speed increases At high speeds it is more difficult to feed current into the motor

reactance P Q

frequency 34 Graph shows the variation of reactance and frequency of an alternating current flowing through two electrical components P and Q. What are the components P and Q ?

A B C D

Component P capacitor inductor inductor resistor

Component Q resistor resistor capacitor capacitor

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14

Vi

time

35 Figure shows an op-amp acting as an integrator with negative feedback. Choose V0 for the shown Vi . V0

V0

A

B

time

time

V0

V0

D

C

time

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15

36 Which of the following is true about the difference between an electromagnetic wave and a mechanical wave?

A B C D

Electromagnetic wave Cannot produce stationary waves Longitudinal waves only Cannot propagate in vacuum Travel with a speed almost the same or the same as speed of light

Mechanical wave Can produce stationary waves Longitudinal or transverse waves Propagates in vacuum Travel with a speed less than that of light

37 An object is placed 5 cm from a convex mirror with a radius of curvature of 20 cm. The image formed is A B C D

3.3 cm in front of the mirror and is diminished 3.3 cm behind the mirror and is enlarged 4 cm in front of the mirror and is diminished 4 cm behind the mirror and is enlarged

38 Two waves are said to be coherent if both the waves have A B C D

the same phase the same amplitude the same frequency the same wavelength

39 In a Young’s double slit experiment, when the distance between the slits and screen is increased A B C D

the fringe separation decreases the maximum intensity of bridge fringes decreases the fringe separation decreases the maximum intensity of bridge fringes increases the fringe separation increases the maximum intensity of bridge fringes decreases the fringe separation increases the maximum intensity of bridge fringes increases

40 Which of the following is true based on the photoelectric effect? A B C D

The emission of electrons will not occur for very low light intensity. The number of electrons produced per second does not depend on the intensity of light. The maximum velocity of the electrons increases when the wavelength of light decreases. The average kinetic energy of the electrons decreases when the frequency of light increases.

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16 41 When light of wavelength 350 nm is incidented on a metal surface, photoelectrons with maximum kinetic energy 2.0 eV are emitted from the surface. What is the maximum wavelength of light that can emit photoelectrons from the metal? A

200 nm

B

580 nm

C

620 nm

D

802 nm

42 What is the maximum possible number of emission spectral lines produced by an atom which has five distinct energy levels only? A

4

B

6

C

10

D

12

43 From Bohr’s theory for hydrogen atom, the n-th energy level in eV is given by En = −

13.6 n2

, n = 1, 2, 3, ...

What is the wavelength of a photon which is absorbed to excite an electron from ground state to the level n = 4? A 1.36 x 10-9 m B 9.15 x 10-8 m C 9.54 x 10-8 m D 1.46 x 10-6 m

44 When the potential difference of an X-ray tube increases, the intensity of the K characteristic line increases because A B C D

more electrons cause the transition which produces the K characteristic line the frequency corresponding to the K characteristic line has increased the number of electrons escapes from the filament has increased anode temperature in the X-ray tube has increased

45 Which of the following is not true of the charge of an electron ? A B C D

It is measured in Coulombs Its has the same magnitude as the charge of a proton It has the same magnitude as the charge of a beta particle It can only be determined by experiment when its mass is given

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17

46 In the mass spectrometer shown in the figure above, the ions which follow paths P1 and P2 always have A B C D

the same charge the same speed the same mass the same acceleration

47 The table below shows the count rate of a radioactive source at different times at a location.

Count rate (count per second )

Time(hours) 10 20 30

With source 60 30 20

Without source 20 20 20

Determine the half-life of the source based on the data given in the table. A C

5 hours 12 hours

B D

10 hours 18 hours

48 At time t = 0 minute, 16 mg of a radioactive element X of half life 4.0 minutes is inserted into a closed container. At time t = 8.0 minute, 8 mg of the radioactive element X is added into the container. What is the mass of the radioactive element X left in the container at time t = 12.0 minutes ? A C

2 mg 6 mg

B D

3 mg 8 mg

49 The energy of the sun is acquired through A B C D

the liberation of energy from unstable nuclei the liberation of helium nuclei to form hydrogen nuclei the fission of uranium nuclei by neutrons the fusion of hydrogen nuclei to form helium nuclei

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18 50

A strong nuclear force A B C D

abides by the inverse square law is short range is repulsive if the charge of the particles are of the same type is an electrostatic force

END OF QUESTION PAPER.

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TRIAL 2009

Instructions to candidates :

Answer all the questions in Section A in the spaces provided. Answer any four questions from section B. For this section, write your answers on the answer s heets pr ovided. B egin each answer on a f resh sheet of paper. Answers should be i llustrated by l arge, c learly labeled diagrams wherever suitable. Answers may be written in either Malay or English. Arrange y our ans wer i n num erical or der and tie the answer sheets to this booklet.

For examiner’s use Section Marks Marks

Obtained 1 2 A 3 4 5 6 7 8 9 10 B 11 12 13 14 TOTAL

5 5 5 5 5 5 5 5 15 15 15 15 15 15 100

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2 Physics 960 Constants value

speed of light in vacuum

c

=

3.00 x 108 m s-1

permeability of vacuum

µo

=

4π x 10-7 H m-1

permittivity of vacuum

εo

=

8.85 x 10-12 F m-1

=

[1/(36 π)] x 10-9 F m-1

magnitude of electron charge

e

=

1.60 x 10-19 C

Planck constant

h

=

6.63 x 10-34 J s

atomic mass unit constant

u

=

1.66 x 10-27 kg

electron rest mass

me

=

9.11 x 10-31 kg

proton rest mass

mp

=

1.67 x 10-27 kg

molar gas constant

R

=

8.31 J K-1 mol-1

Avogadro constant

L, NA

=

6.02 x 1023 mol-1

Boltzmann constant

k

=

1.38 x 10-23 J K-1

gravitational constant

G

=

6.67 x 10-11 N m2 kg-2

free fall acceleration

g

=

9.81 m s-2

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3

1. A water wheel has eight bucket equally spaced around its circumference as shown in diagram below.

Water

1

Empty bucket

Full, bucket

8 2

45 0 Direction of rotation

7 3

1.8 m 4

6

5

Water

The wheel makes six revolution per minute. Calculate (a) the total change in potential energy of the water in the buckets in one revolution of the wheel.

[2]

(b) the average input power to the wheel

[2] (c) Suggest why a larger number of small bucket is preferred to a smaller number of large bucket containing the same total mass of water

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4

The figure below illustrates a mass which can be made to vibrate vertically between two springs.

mass

variable frequency vibrator

The vibrator itself has constant amplitude. As the frequency is varied, the amplitude of vibration of the mass is seen to change as shown in Fig 3A

Fig 3A

(a) Name the phenomenon which is illustrated in Fig 3A.

[1] (b) For the mass vibrating at maximum amplitude, calculate the angular frequency.

[2] (c) A light piece of card is fixed to the mass with its plane horizontal. On Fig 3A, draw a curve to show the new variation with frequency of the amplitude of vibration of the mass. [1] (d) State one situation in which the phenomenon illustrated in Fig 3A is used to advantage.

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5

A uniform copper rod with thermal conductivity 380 W m-1 K-1 which is perfectly insulated has a cross-sectional area of 2.50 cm2 and length 20.0 cm. Heat is conducted by the copper rod. When the steady state is achieved, the temperatures at the ends of the rod are 130 oC and 20 oC. Calculate (a) the rate of heat flow in the rod.

[3] (b) the temperature 15.0 cm from the hot end.

[2]

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6

(a) Electrical conduction in a metal can be explained in terms of motion of free electrons. State the estimated values at room temperature for mean random velocity of the free electrons.

[1] (b) A 5.0 A current flows through a wire of length 1.50 m and crass sectional area of 1.2 mm2 where the potential difference across the wire is 0.24 V. (i)

Calculate the power dissipated from the wire

[2]

(ii) Explain what will happen to the drift velocity of the free electrons if the power produced in the wire is increased.

[2]

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7

5. 330 kΩ

+15 V

33 kΩ V1

V0 V2 10 kΩ

+

-15 V

(a) State the name of the amplifier circuit which is connected to various input as shown in the diagram above.

[1] (b) Calculate the output voltage V0 if V1 is 0.50 V and V2 is 0.20 V.

[3] (c) What happen to the output voltage if the supply ±15 is replaced with ±9V ?.

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8

(a) Write the lens maker’s formula. Explain the symbols that you use.

[2]

(b) The diagram below shows a lens that has a curved surfaces with radius of curvature

20.0 cm and 60.0 cm. The refractive index of the lens material is 1.65.

Determine the focal length of the lens when the lens is immersed in water of refractive index 1.33.

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9

7 Monochromatic light with wavelength 365 nm illuminates a metallic surface with work function 2.30 eV. (a) Calculate the maximum speed of the photoelectrons emitted.

[3]

(b) Estimate the de Broglie wavelength of the photoelectrons.

[2]

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10

8

The graph above represents the decay of a sample of a specific radioactive element. Find the half-life of the element. Hint :

N = N 0 e − λt Count rate

dN = − λ N 0 e − λt dt

[5]

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11 Section B [60 marks] Answer any four questions in this section

9 (a) Define the terms moment of a force and draw a sketch to illustrate the meaning. (b) State the condition for a body to be in equilibrium

[2] [1]

(c) During the construction of many modern bridges, section are added from both banks until the two halves meet at the centre. Figure 1 below shows a new section S, of weight 3.0 × 10 5 N , after it has been attached to an existing part B of a bridge Support cable

S

B

3.0 × 10 5 N The support cable which keeps section S in equilibrium is at an angle of 25 0 to the horizontal. The existing part B of the bridge provides a horizontal force on S. (i) Draw a labeled vector diagram and show the three forces on S

[2]

(ii) Use your diagram to determine the tension in the cable and the horizontal force which B exerts on S [2] (d) In the sport of clay pigeon shooting, a clay disc is launched into the air by a spring, and the contestant fires a shot at the moving disc. A launching device has been modified to project the disc vertically upwards. The spring in the launching device has a spring constant of 2000 Nm −1 and the clay disc has mass 80g. In use the spring obeys Hooke’s law and the extension decreases from 13 cm to 7.0 cm. (i) Sketch a graph of force against extension for the spring and shade the area which represent the loss of energy stored by the spring when launching the clay disc. [2] (ii) Calculate (a) the loss of elastic potential energy of the spring [2] (b) the initial speed of the clay disc, assuming all the energy lost by the spring becomes kinetic energy of the disc [2] (c) the height to which the disc will rise, assuming that air resistance is negligible. [2] 960/2 *This question paper is CONFIDENTIAL until the examination is over

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12

10 (a) State three differences between a progressive wave and a stationary wave.

[3]

(b) In large auditoriums, the walls are covered with thick curtains. State the effect of thick curtains on the sound waves and hence explain the advantage of this practice. [2] (c) Two sound waves have frequencies 890 Hz and 894 Hz respectively. The intensity of the sound waves at a given point is 6.3 x 10-9 W m-2. (i)

What are the conditions required for beats to be heard from two separate sound sources?

[2]

(ii) What is the frequency of the resultant wave and what is the beat frequency?

[2]

(iii) Determine the maximum intensity level of the beats heard at the point concerned. [The minimum intensity of audible sound is 1.0 x 10-12 W m-2]

[2]

(iv) Another sound sources with intensity level 98.0 dB is located at the point. What is the ratio of the intensity of the first sound to that of the second sound? [2] (d) A high-speed train is travelling at a speed of 47.4 m s-1 when the engineer sounds the 418 Hz warning horn. The speed of the sound is 343 m s-1. What is the frequency of the sound as perceived by a person standing at a crossing when the train is approaching? [2]

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13

11 (a) State Hooke’s Law

[2]

(b) Explain the main difference between elastic deformation and plastic deformation

[3]

(c) The graph shows a simplified version of the variation of the load applied against the extension of the material. The original length of the wire is 1.0 m and its diameter is 1.5 mm.

(i) The material is brittle or ductile ?

[1]

(ii) What is the Young’s Modulus of the material ?

[3]

(iii) If the wire snaps when the load reaches the value at P, how much energy is required to snap the wire ?

[3]

(iv) Assume that the cross-section of the wire remains constant throughout the extension. What is the stress in the wire when the load reaches the value at P ? [3]

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14

12. (a) Explain why a charge Q will move in a circular path in a uniform magnetic field, B. Write an expression for the force, F exerted on the charged particle. Explain any quantity you have given in your expression. [4] (b) An electron is moving in a circular path in a uniform magnetic field. If the radius of the path is 2.5 × 10 −2 m and the magnetic field strength is 2.0 × 10 −2 T and the specific charge of electron (

e ) is − 1.76 × 1011 Jkg −1 . m

(i) Derive an expression for the angular velocity of the electron [2] (ii) Calculate the orbital period if the electron [2] (iii) If the kinetic energy of the electron is reduced by half of its initial value, what is the value of the orbital period? [2]

+ X

X

X

X

X electron path

X

X

X

X

X

(c) Figure shows an electron moving with constant velocity enters an electric field which is perpendicular to a magnetic field. If the magnetic field strength, B is 0.008 T and the electric field strength, E is 6.0 × 10 4 Vm −1 , (i)

show the direction of the electrostatic force and the magnetic force exerted on the electron. [2] (ii) calculate the velocity of the electron [3]

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15

13 (a) State Bohr’s second postulate

[2]

(b) The energy of the electron in a hydrogen atom is given as En = −

13.6 n2

eV ,

n = 1, 2, 3, …

A hydrogen atom is excited by a photon and makes a transition from energy level n = 1 to energy level n = 4. (i) Explain how the hydrogen atom is excited. (ii) What is the energy absorbed by the excited atom? (iii) Calculate the wavelength of the photon.

[2] [2] [3]

(c) (i)

Explain the difference between the process of production of continuous X-rays with the process of production of line X-rays from an X-ray tube. [4] (ii) The accelerating potential difference across an X-ray tube using copper as target is 5.0 kV. Calculate the minimum wavelength of the X-rays produced. [2]

14 (a)

Give definition for each of the following terms (i) isotope (ii) nuclear fission (iii) nuclear fusio

[3]

(b) (i) State the quantities that are conserved in the nuclear reaction.

[3]

(ii) Copy and complete the equations below : 10 5

B + 24He  →136C + ………………………..

…………………………. 10 5

+ 37Li  → 2 24He

B + 01n  → 37 Li + …………………………..

(c) (i) Calculate the energy released when

28 13

Al decays to

[3]

28 13

Si

[2]

(ii) If the γ - ray emitted has a wavelength of 6.99 x 10-13 m , calculate the energy of γ -ray photon. Hence, deduce the total kinetic energy of the products of decay. [ Mass of

28 13

Mass of

28 13

Al atom = 27.98191 mu ,

Si atom = 27.97693 mu, where mu is the atomic unit.]

[4]

END OF THIS PAPER

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JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR 960/1/2 JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANU JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR

TRIAL 2009

MARK SCHEME PAPER 1 AND 2 PHYSICS 960 JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN TERENGGANUJABATANPELAJARANTERENGGANU JABATANPELAJAR JABATAN PELAJARAN ERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJAR

1

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ANSWER PAPER 1 Question

Answer

1

A

2

B

Explanation Strain= e / l0 is dimensionless [stress] = F/A= [pressure] During the time from 0 to T, the body slides along the horizontal surface with zero resultant vertical force since its weight is balanced by the reaction force from the surface. After time T, the resultant force acting on the ball is its own weight W under free fall situation.

a= 3

B

F F = m + 2m 3m

To block X alone, F − FY = ma = (2m)(

F ) 3m

Hence, FY = force exerted by block Y on block X=

2F 3

1 2 mv = constant since v is constant. 2 dx Potential energy of sphere = mg = −mgv dt Kinetic energy of sphere =

4

B v

5

B

x

v is constant because v and r are constant. Hence angular acceleration = 0 r The force constant of the elastic string is 40 Nm −1 . Hence, the force exerted by w=

the string on the mass to keep it in circular motion is

F = kx = 40(0.70 − 0.50) = 8 N If v is the linear velocity of the mass, it is related to this centripetal force by

F =m 6

C

v2 r

8 = (0.05)

v2 0.7

v = 10.58 = 11ms −1

2

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θ 7

C

T F

W If T is the tension in the light thread, then

T sin θ = F

tan θ =

T cos θ = W

mgh =

1 2 1 2 mv + Iω 2 2

F W

v = rw

v 2mgh = mv + I   r 1 2  = m + v r    6.0 × 10 −4 2(0.2)(10)(0.2) =  0.2 + 0.05 2  v = 1.35ms −1 2

2

8

A

From E P = U = −G

 2 v 

Mm r

E P is the gravitational potential energy of a body of mass m at distance r from

9

D

the centre of a planet of mass M Force of attraction =

dU Mm = +G 2 dr r

Thus, the gradient at any point on the gravitational potential energy curve represent the force pulling the body towards the planet.

W = 10

C

GMm 1 ⇒ Wα 2 2 R R

Gravitational force on the satellite when at the highest R/50

R+

R2 R from the centre of the earth = W 2 50 R  R +  50  

3

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=

R2 1   R 1 +   50 

2

W

2

= 0.96 W

11

A

12

D

13

C

E=

1 1 ma 2ω 2 = ma 2 (2πf ) 2 = 2π 2 ma 2 f 2 2

2

The frequencyis determined by the source and not the medium through which it flows. In a denser medium, the velocity will decrease and as a result the wavelength will decrease.

Dipole aerial

14

C

15

B

l V=fλ (100 x 106) λ = 3 x 108 λ=3m l = λ / 2 = 2/3 = 1.5 m

Work = area under the graph

1 2

1 (60 + 80)(9 − 5)] × 10 −3 2 = [(150) + (70×4) x 10 −3 ] = 430 x 10 −3 = [ (60 × 5) +

16

B

= 0.43 J

F e Fl E = eA EA k = l

k= 17

A

18

D

19

C

V2

W =∫

V1

pδV

4

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nRT δV V1 V V = nRT ln 2 V1

=

V2

∫

( pV = nRT )

= 3(8.31)(400) ln

20

D

21

A

22

C

23

A

24

B

6 = 4043.30 J = 4kJ 4

Since ∆V = 0, W = p ∆V = 0, ∆Q = ∆U + W ∆U = ∆Q = mcv∆V = 5.0 x 10-3 x 2.4 x 102 x (600 – 300) J = 360 J γ pV = k p = kVγ ln p = ln k – γln V = - γ lnV + ln k (gradient = - γ )

w = q∆V

since ∆V = 0 ⇒ w = 0

Initial energy 1 Q12 1 Q22 + 2 C1 2 C 2

(

1  200 × 10 −6 =  2  4 × 10 −6 

)  +  (300 ×10 )  = 0.02 J −6 2

2

   

3 × 10 −6

 

Equivalent capacitance 25

B

C = C1 + C 2

(

) (

)

= 4 × 10 −6 + 3 × 10 −6 = 7 × 10 −6 F

Final energy 1 Q 2 1  500 × 10 −6  =   = 0.018 J 2 C 2  7 × 10 −6 

Energy lost, 0.02 − 0.018 = 2.0 × 10 −3 J

5

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The resistance of a copper wire remains unchanged at constant temperature and is given by R = ρ

l A

Potential difference, V 26

B

 l  V = IR =  ρ (nvAe) = ρ ln ve  A

Since I is directly proportional to drift velocity of electrons in the wire, thus directly proportional to potential difference. 27

A

28

B

When temperature increases, electrons are freed from covalent bonds. Hence the number of electron-holes increases. Kirchhoff’s first law is related to current and depends on the conservation of charge Kirchhoff’s second law is related to electric potentials and depends on the conservation of energy.

Q = It 29

D

30

C

31

A

(

Q (2 ) 1.6 × 10 −19 = t 1.0 × 10 −15 = 3.2 × 10 −4 A I=

)

Induced e.m.f ξ is given by

ξ = BANω

32

B

33

A

34

B

35

A

36 37

where N number of turns A area B flux density ω angular velocity

P is inductor, X L = wL Q is resistor

D B

r = 20cm, f = -10 cm, u = +5 cm

6

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1 1 1 = + f u v 1 1 1 = − v − 10 5 v = −3.3 cm

38 39

A C

From the formula of fringe separation y =

λD a

, y∝D

From Einstein’s equation,

40

1 2 mv max = hf − W 2

K max =

C

c

f =

λ If f increases or λ decreases, Kmax or vmax increases hf = K max + hf o

41 D

fo =

λo =

c

λ

−

K max 3.0 × 10 8 2.0 × 1.6 × 10 −19 = − = 0.374 × 1015 − 9 − 34 h 350 × 10 6.63 × 10

c = 8.02 × 10 −7 m fo

42

C

43

D

1   1 E 4 − E1 = ∆E = −(13.6) 2 − 2 (1.6 × 10 −19 ) = 1.36 × 10 −19 J 1  4 hc ∆E =

λ

λ=

A

The K characteristic line is produced by electronic transition in the target atom. The intensity is determined by the number of electrons hitting the target and the number of collisions determines the number of electronic transition.

D

The charge of an electron can still be determined if its mass is not given

44

45

hc 6.63 × 10 −34 × 3.0 × 10 8 = = 1.46 × 10 −6 m −19 ∆E 1.36 × 10

7

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B1 qv = qE speed , v = 46

B

E always constant because it does not depend on B1

charge q and mass M.

N = N 0 e − λt

47

B

10 = 40e −λ (30−10 ) ln 4 = 20λ λ = 0.0693 ∴ half life =

48

C

49

D

50

B

ln 2

λ

=

ln 2 = 10hours 0.0693

0 min 4.0 min 8.0 min 12.0 min 16 mg → 8 mg → 4 mg+ 8 mg → 6 mg Nuclear fission : 2 1

H + 12H → 23 He+ 01n

A strong nuclear force is experienced by nucleons such as protons and neutrons in the nucleus and are of very short range such as 1.7 x 10-15 m

8

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Paper 2 Question Structure 1

(a)

Answer

Mark

Total change in potential energy of water in the buckets in one revolution of the wheel = 8 × mg∆h = 8 × 40 × 9.81 × (1.6 × 2) = 1.0 × 10 4 J

1

1 1

Average input power to wheel = number of revolutions made per unit time X change in potential energy of the water in the buckets per revolution of the wheel (b)

6 × 1.0 × 10 4 60 = 1.0 × 10 3 W

=

=1kW

(c)

1

A large number of a small buckets is preferred because the rotation of the wheel would be smoother than the case would be when a smaller number of large buckets is used.

1

Structure 2 2(a) 2(b)

resonance From graph, fresonance = 12.5Hz ω = 2πf = 2π (12.5) =78.5 rads-1

1 1

1

2(c) mass without card mass with card

2(d) Structure 3 (a)

1

Microwave oven/ radio signal receiver

1

dQ dθ = kA dt dx

1

1

9

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= (380)(2.5 x10-4)

= 52.25 W

(130 − 20) 20 × 10 − 2

1

The temperature at 15.0 cm = ( 15 × 10 − 2 )( = 82.5 oC

(b)

130 − 20 ) 20 × 10 − 2

The temperature 15.0 cm from the hot end = 130 oC – 82.5 oC = 47.5 oC

1

1

Structure 4 (a)

(b-i)

(b-ii) Structure 5 (a)

(b)

velocity is a vector quantity. The velocities cancel out each other in any directions, since the number of free electrons is very large. So mean random velocity is zero. Power, P P = IV = (0.24)(5.0) = 1.2W

1 1

Power, P = I 2 R ⇒ PαI 2 Since, I = nev , therefore Pαv 2 Hence, when power increases drift velocity increases.

1 1

Adder operational amplifier

1

 R f   R f  V0 = −  V1  +  V2   Ri   R2   330   330  0.5  +  0.2  = −    10   33

1+1 1

= −11.60V

(c) Structure 6 (a)

Output voltage will become −9V only as saturation occurs. 1 1 1 = (n − 1) −  f  r1 r2 

In water applying the formula

1

1

where f = focal length, n = refractive index of the material of the lens, r1 = radius of curvature of the front surface receiving the incoming rays, r2 = radius of curvature of the hind surface where the outgoing rays emerges. [Rubric: Formula – 1 mark; Defining symbols – 1 mark] (b)

1

 1 1  1  n2 =  − 1 −  f  n1  r1 r2 

1  1.65  1 1  − = − 1  f  1.33  + 60 + 20  f = −125 cm f = −1.25 m

1

1 (formula)

1 (sign of r)

1 (answer)

10

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Structure 7 (a)

Using Einstein’s equation, maximum kinetic energy is 1 2 hc mvmax = −W λ 2 =

1

− 34

(6.63 × 10 )(3.00 × 10 ) − 2.30(1.6 × 10 −19 ) 365 × 10 − 9 8

= 1.77 × 10 −19 J vmax =

2(1.77 × 10 −19 ) 9.11 × 10 − 31

1 1

= 6.23 × 105 m s −1

(b)

Momentum of the photoelectrons, mv = λ= =

h mv

h

λ

1

6.63 × 10 − 34 (9.11 × 10 − 31 )(6.23 × 105 )

= 1.17 × 10 − 9 m

1

Structure 8

8

N = N 0 e − λt Count rate

dN = − λ N 0 e − λt dt

20000 = −λN 0 e −λ (10 ) .......(i ) 12000 = −λN 0 e i : ii

− λ ( 28 )

1 1

........(ii )

20000 e -10 λ = 12000 e -28λ 5 ln = - 10λ + 28λ = 18λ 3 λ = 0.0284

Half life T1 = 2

ln 2

λ

=

ln 2 = 24.4 min 0.0284

11

1

1

1

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Essay Essay 9 (a)

The moment of a force, F is the force multiplied by the perpendicular distance, d from the point about which the moment is being measured example the axis of rotation

1

d

Axis of rotation

1 F

(b)

For a body to be in a equilibrium, there must be no resultant force and no resultant torque

(c)(i)

FS

FB

25 0

1

All correct 2 or Two correct 1

Weight of section S, W= 3.0 × 10 N 5

(c)(ii)

Resolving forces vertically

FS sin 25 0 = W for equilibrium

Fs =

3.0 × 10 5 W = = 7.10 × 10 5 N 0 0 sin 25 sin 25

Resolving forces horizontally,

FB = FS cos 25 0 = 7.10 × 10 5 × cos 25 0

1

= 6.43 × 10 5 N

1

(d)(i) F/N

Draw graph correct 1

e/cm 7 (d)(ii)(a)

13

Loss of elastic potential energy of the spring =

12

1 1 2 2 kx 2 − kx1 2 2

Area shaded 1 1

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=

1 × 2000 × (0.13 2 − 0.07 2 ) 2

1

= 12 J (d)(ii)(b)

Kinetic energy of the disc = Energy lost by the spring

1 2 mv = 12 2 12 × 2 v= 0.080 (d)(iii)(c)

1

Initial speed = 17.3ms −1 Gain in gravitational potential energy of the disc = lose in kinetic energy

1

1 2 mv 2 12 h= 0.080 × 9.81

1

mgh =

=15.3 m 1 Essay 10

(a)

(b)

c(i)

c(ii)

c(iii)

c(iv)

Progressive wave Stationary wave 1. Wave profile moves - Wave profiles does not move 2. Adjacent particles of a - Particles between two adjacent medium vibrate in a different nodes of a medium vibrate in the same phase phases 3. The amplitude in constant for - Particles between two adjacent all particles of the medium nodes vibrate with different amplitudes Thick curtains will absorb the sound. This will reduce echo and interference of sound waves in the hall and hence the audience will be able to hear a performance clearly. Conditions: - the frequency of the two sound must be almost the same - the amplitudes from the two sound sources must be the same or almost the same - waves from the two sound sources must be propagated in the same direction Frequency of beats = 894 – 890 = 2 Hz Frequency of resultant wave = (894 + 890) ÷ 2 = 892 Hz Intensity level = log10

 I1  I2

β 1 − β 2 = 10 log

 I   I0

10

−9   = 10 log10  6.3 × 10  1.0 × 10 −12   = 38.0 dB

  

  

1 1 1

1 1

1 1 or 1 1 1 1

1

1

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I 38.0 − 98.0 = 10 log 1  I2 I  log 1  = −6  I2 

 I1   I2 d

 v f ' = f   v − vs

  

  = 10 − 6 

1

 343   = 418   343 − 47.4   = 485 Hz

1 1

Essay 11

(a)

(b)

(c)

Hooke’s Law : States that the extension is directly proportional to the stress (force) applied in an object, if the elastic limit is not exceeded. Elastic Deformation Wire can return to original shape & size when the stress has been removed

(i)

(ii)

Plastic Deformation Wire does not return to its original shape and size when the stress has been removed. (Permanent deformation occurs)

2

3 1

Brittle from graph : x = 1.0 x 10 -3 m for F = 250 N (250 )(1.0 ) FL E= = = 1.41 x 10 11 Pa 2 Ax π 0.75 x10 − 3 1.0 x10 − 3

(

)(

3

)

3 (iii) Energy = area under the curve = 26 J

(iv) at P :

stress =

F 350 = A π 0.75 x 10 -3

(

14

)

2

= 1.98 x 10 8 Pa

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Essay 12

(a)

When charge, q is moving, there will be a current, I will produce. By using Fleming left hand rule for charge, q moving in a uniform magnetic field, B a force, F will be exerted on the charge, q.

1

B, I and F must perpendicular to each others. Thus q will move in a circular path.

1

An expression for the force, F exerted on the charged particle is given by, F = Bqv(sin θ ). 1 where F is the force exerted on the charge B is the magnetic field q for charge v for velocity θ angle between v and B

1

Centripetal force = Magnetic force

1

mv = Bev r Ber v= m Ber v = rw = m Be w= m

1

2

(b-i)

w=

2π Be = T m

The period, (b-ii) T= =

2πm  2π  m   2π =   =  Be  B  e   B 2π

2.0 × 10

−2

   1    e   m

1

1     11  1.76 × 10 

= 1.78 × 10 −9 s

From equation T = (b-iii)

1 2πm , the period is independent of the velocity and thus independent Be

of the kinetic energy of the electron.

1

So the period is still same// equal to 1.78 × 10 −9 s

1

(c-i)

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FE 2 ν

FB

Electrostatic force = magnetic force eE = Bev

(c-ii)

Essay 13 13(a)

13(b)(i)

13(b)(ii)

1

E 6 × 10 = B 0.008 = 7.5 × 10 6 ms −1 4

v=

1 1

- An electromagnetic radiation is given out when an electron makes a transition from one state of higher energy level to another of lower energy level. - The energy of a photon of the electromagnetic radiation is given by ∆E = hf - The electron in the ground state gains energy that is exactly equal to the energy difference between the initial energy level of the electron and the final energy level. - Electron move up to the higher energy level than the ground state

1 1

1 1

Energy absorbed = E4 − E1

1

13.6 13.6 − (− 2 ) 42 1 = 12.8 eV =−

13(b)(iii)

1

Photon energy E=

hc

1

λ

12.8 × (1.6 ×10 −19 ) =

6.63 × 10 − 34 × 3.00 × 108

λ

−8

13(c)(i)

λ = 9.71 × 10 m Continuous X-rays: - When an electron strikes a metal, it can lose any portion of its energy. This energy loss of the incident electron is converted into energy of a X-ray photon. Hence, the energies of X-ray photons are different. Since the wavelength of a photon is inversely proportional to its energy, the wavelengths of the photons emitted are different. - If all the energy of the incident electron is lost as energy of a photon, X-ray photon with the minimum wavelength is produced.

1 1

1

1 Line X-rays: - The incident electrons may penetrate deep into the inner-most shell of the target atoms, causing the electron from the inner K or L shells to be excited to higher energy levels. - When an electron from a higher energy level falls to fill up these vacancies, the

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difference in energy is emitted as energy of line X-rays 1 13c(ii)

When all the energy of the incident electron is converted into energy of an X-ray photon, eV =

hc

1

λmin

(1.6 × 10 −19 ) × (5.0 × 103 ) =

(6.63 × 10 − 34 ) × (3.00 × 108 )

λmin

λmin = 2.49 × 10 −10 m

1

(i)

Nuclides which have the same number of proton but different number of neutrons

1

(ii)

The disintegration of a heavy nucleus to lighter nuclei with the release of a lot of energy

1

(iii)

The combination of lighter nuclei at very a high temperature to produce a heavy nucleus with release a lot of energy

1

(i)

Mass number, A Atomic number , Z Mass- energy

1 1 1

(ii)

10 5

Essay 14 14 a

14b

1

1 1

B + 24He →136 C + 11H

H + 37Li → 2 24He

10 5

1

B + 01n → 37 Li + 24He

1

14c (i)

Energy released E

E = mc 2

1

= [27.98191 - 27.97693] × 1.66 × 10 -27 × 3.00 × 10 8 = 7.44 × 10 -13 J

1

The energy of γ -ray photon , E = (ii)

E=

hc

λ

hc

λ

1

1

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(6.63 × 10 −34 )(3.00 × 10 8 ) 6.99 × 10 −13 = 2.85 × 10 -13 J Total kinetic energy of decay products E=

= 7.44 × 10 -13 − 2.85 × 10 −13 = 4.59 × 10

−13

1

1

J

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