Physics STPM Past Year Questions with answer 2006.pdf

2006 STPM Physics Papers 1 & 2

Time: 1h 45 min

There are fifty questions in this paper. Answer all questions. Marks will not be deducted for wrong answers.

1. Which of the following is not equivalent to the unit of energy? A Electron volt (eV) B Volt coulomb (V C) C Newton metre (N m) –1 D Watt per second (W s )

If the tension in the string is T , the force F is A T C 3.0 T B 2.5 T D 5.0 T 4. The engine of a car has a maximum output power of 54 kW. The air resistance acting on the car when it is moving with speed v is 2v2. The maximum speed which can be achieved by the car on a level road is –1 –1 A 3.0 m s C 30 m s –1 –1 B 5.2 m s D 164 m s

2. The velocity-time (v-t ) graph for a car is shown below. v

0

T — 2

5. The diagram below shows two bodies of masses mP and mQ attached to the ends P and Q of a light rigid rod. The rod is then rotated about a vertical axis through the centre O of the rod.

T t

Which of the following is the displacementtime (s-t ) graph for the car? A s

C s

Q O

m Q

m P

P 0

T

T — 2

0

t

B s

T — 2

T

t

The ratio of the tension in the section PO to the section QO of the rod is mP A 1 C — mQ

D s

B 0

T — 2

t

T

0

T — 2

T

t

T

3m

D

  m — — P– mQ

2

6. A flywheel rotates at a rate of n rotations per second. If the mass and moment of inertia of the flywheel are M and and I respectively, respectively, the work that should be done to stop the flywheel is 1 2 2 2 A — Mn C 2π Mn 2 1 2 2 2 B — In D 2π In 2

3. The diagram below shows two bodies of masses 2m and 3m connected by a light inelastic string. The bodies are pulled by a force F on on a smooth surface. 2m

 

— mP 1 2 —– —– mQ

F

1

http://e u.jo http://e u.joshuatl shuatly.co y.com m http://www.joshuatly.co

© Majlis Peperiksaan Malaysia 2006


7. A coin of mass m is placed on a disc which is rotating about its vertical axis at constant angular velocity ω . The coefficient of static friction between the coin and the surface of the disc is µ . If the coin does not slip, the maximum distance between the coin and the centre of the disc is g mg A —2 C —2 B

ω µ g —2 ω

D

B C D

Only a few oscillations are possible. The system takes an infinite time to return to equilibrium. The system does not oscillate but returns easily to its equilibrium position.

12. The diagram below shows a metal wire stretched between two supports P and T which are 4 x apart. Three small pieces of paper are placed at positions Q, R and S.

ω µ mg —– ω2

P

8. The accelerations due to gravity at the equator and the pole are ge and gp respectively. Which of the following statements about ge and gp is true? A ge < gp because the Earth rotates. B ge > gp because the Earth rotates. C ge < gp because the Earth rotates and it is not a perfect sphere. D ge > gp because the Earth rotates and it is not a perfect sphere.

x

x

R

Q

x

T

Metal wire

S

Pieces of paper

Load

When the wire is vibrated at a certain frequency, the small piece of paper at R remains on the wire but the small pieces of paper at Q and S drop off. What is the wavelength of the wave which is produced at the wire? A x C 4 x B 2 x D 8 x

9. Which of the following statements about an object which performs simple harmonic motion is true? A The acceleration is maximum when the velocity is maximum. B The acceleration is maximum when the displacement is maximum. C The maximum potential energy is greater than the maximum kinetic energy. D The maximum kinetic energy is greater than the maximum potential energy.

13. A student blows across the top of a hollow bamboo pipe. The bottom of the pipe is closed by a finger so that a fundamental tone of frequency f is produced. Then the student blows again with the bottom of the pipe open. The tone that the student hears has a frequency A equal to f B lower than f C higher than f D fluctuating in the range of f

10. The diagram below shows two coherent sources P and Q which produce waves of the same phase with wavelength λ . P

14. The graph below shows the variation of potential energy U with separation r between two atoms.

Q

U

R

If the two waves meet at the point R, the phase difference between the waves at R is 2π A PR – QR C ————— λ (PR – QR) 2π (PR – QR) λ (PR – QR) B ————— D ————–— 2π λ

0

r 0

r

–ε

Which of the following statements is not true? A The interatomic force is zero at r 0. B The total energy of the system is zero at r 0.

11. Which of the following statements about a critical damping system is true? A No energy is lost from the system. © Majlis Peperiksaan Malaysia 2006

x

2

http://e u.joshuatly.com http://www.joshuatly.co


The equilibrium separation between the atoms is r 0. An energy ε is required to separate the two atoms.

C D

18. The graph below shows the change in state of a mass of an ideal gas from X to Y and then isothermally from Y to Z, where p is the pressure and V the volume. The pressures at X and Z are equal.

15. Which of the following is the stress-strain graph for a glass rod which is stretched until broken? A Stress

Y

p

C Stress 0

0

B Stress

V

Which of the following graphs shows the variation of temperature T with volume V ?

0 Strain

Z

X

A T

Strain

D Stress

C

Y

Z

T

0 0 Strain

16. The pressure p of an ideal gas is related to the mean square speed of the gas molecules by the equation 1 p = —nm 3 What are represented by n and m? n m Mass of one A Number of moles mole gas Number of Mass of one B moles molecule Mass of one C Number of molecules molecule Mass of one D Number of molecules per molecule unit volume

0

V

B T

Strain

Z

X

X 0

Y

D

Y

T

V

Y

X Z

X

0 V

Z

0

V

19. The diagram below shows a metal rod PQ of length  and a diameter which increases uniformly from P to Q. The rod is wellinsulated and the ends P and Q of the rod are at temperatures θ 1 and θ 2 respectively, where θ 1 > θ 2. Insulation

Rod

Q P

Which of the following graphs shows the variation of temperature θ with distance x from P when a steady state is achieved?

17. The air inside a hot-air balloon is heated up from a density of 1.25 kg m–3 and a temperature of 27 °C until the balloon begins to float. During this heating process, some air escapes from the balloon and the pressure inside the balloon remains constant at just above the atmospheric pressure. The mass and volume of air in the balloon just before it rises are 300 kg and 400 m3 respectively. The temperature of hot air is A 45 °C C 318 °C B 227 °C D 500 °C

A

θ θ 1

C θ θ 1

θ2

θ2

0



x

0

B θ θ 1

D θ θ 1

θ2

θ2

0

3




x

0



x



x



20. A diesel engine does not need spark plugs for the combustion of a mixture of diesel and air in a cylinder. Spontaneous combustion occurs when a high temperature is achieved through air compression. If air with an initial temperature of 27 °C is compressed adiabatically to a final temperature of 681 °C, what is the ratio of the initial volume to the final volume? [Assume that air is a diatomic ideal gas.] A 6 C 18 B 15 D 32

24. The diagram below shows a metal strip of cross-sectional area A and length  connected to a battery.

Metal strip

If the potential difference across the strip is V and the current flowing is I , the conductivity of the strip is I  V  A — C — VA IA VA IA B — D —– I  V 

21. Two bodies each of mass 3 × 105 kg are 200 km apart. The same amount of charge is placed on each body. What is the magnitude of the charge on each body so that the gravitational attractive force is balanced by the electric force? –10 C A 6.7 × 10 –9 B 1.0 × 10 C –8 C 4.7 × 10 C –5 D 2.6 × 10 C

25. Which of the following statements is true of the resistivity of a metal conductor? A The resistivity depends on the type of metal. B The resistivity decreases when the temperature increases. C The resistivity increases when the length of the conductor increases. D The resistivity is inversely proportional to the cross-sectional area of the conductor.

22. The diagram below shows two capacitors of 300 µ F and 500 µ F connected in series to a 6.0 V battery. 300 µ F

26. An electric circuit is shown below. 2.0 Ω

500 µ F

6.0 Ω

12.0 V

X

Y 2.0 Ω

6.0 Ω 6.0 V 2.0 Ω

The potential difference between the points X and Y is A 2.0 V C 6.0 V B 4.0 V D 8.0 V

The charge in the 300 µ F capacitor is A 0.68 mC C 1.8 mC B 1.1 mC D 4.8 mC

27. A galvanometer which has a resistance of 1 Ω gives a full-scale deflection when a current of 1 mA flows through it. This galvanometer can be modified to measure potential differences up to several volts. Which of the following modified arrangements is correct?

23. A parallel plate capacitor is charged until a certain voltage. With the power supply still connected, the plates of the capacitor are separated further until the distance between them is two times the original distance. If the energy stored in the capacitor before and after the plates are separated are U 1 and U 2 U respectively, the ratio —1 is U 2 1 A — C 2 2 B 1 D 4 © Majlis Peperiksaan Malaysia 2006

A

C

1Ω

10 k Ω G

G

B

G 1Ω

4


D

G 10 k Ω


28. A beam of protons moves in a straight path through a region which has uniform magnetic and electric fields. The electric field is produced by two parallel metal plates which are 2.0 cm apart with potential difference 1000 V. If the magnetic field is 1.0 × 10–2 T, the speed of the protons is 2 –1 5 –1 A 5.0 × 10 m s C 1.0 × 10 m s 4 –1 6 –1 B 5.0 × 10 m s D 5.0 × 10 m s

B

0

C

29. A proton enters perpendicularly into a uniform electric field and another proton enters perpendicularly into a uniform magnetic field. How is the force on each proton influenced by the speed v of the proton? A B C D

In electric field

In magnetic field

Proportional to v Proportional to v2 Not influenced by v Not influenced by v

Proportional to v Not influenced by v Proportional to v Proportional to v 2

X

f

X

0 f

D

X

0 f

33. The diagram below shows an operational amplifier with an open-loop gain of 105. The voltage supply is ±9.0 V. The inverting and non-inverting input voltages are V 1 and V 2 respectively.

30. An alternating current I = I 0 sin 2π ft flows through the primary coil of a transformer, where I 0 is the peak current and f the frequency. If M is the mutual inductance between the primary and secondary coils, the induced e.m.f. in the secondary coil at time t is A MI 0 sin 2π C 2π ft f MI 0 sin 2π ft B MI 0 cos 2π D 2π ft f MI 0 cos 2π ft

+9.0 V V 1

–

V 2

+

V o

–9.0 V

Which of the following produces a saturated output voltage of V o?

31. When the load decreases, the speed of rotation of an electric motor increases and the current flowing through it decreases. This happens because. A the back e.m.f. increases B the frictional force decreases C the resistance of the armature coil increases D the inductance of the armature coil increases

A B C D

V 1 +50 µ V –150 µ V –120 µ V +195 µ V

V 2 0 0 –120 µ V +240 µ V

34. The planes of vibration of the electric and magnetic fields in an electromagnetic wave are A parallel to each other and parallel to the direction of the propagation of the wave B parallel to each other and perpendicular to the direction of the propagation of the wave C perpendicular to each other and parallel to the direction of the propagation of the wave D perpendicular to each other and perpendicular to the direction of the propagation of the wave

32. Which of the following graphs shows the variation of the reactance X of a capacitor with the frequency f of an alternating potential difference which is applied across it? A X

0 f

5




35. Which of the following graphs shows the u variation of — with u, where u is the object v distance and v the image distance of a convex lens? A

C

u — v

Monochromatic light A B

u — v

C D

40. Light of frequency f radiates the surface of a metal. If the maximum kinetic energy of the emitted electron is K and h is Planck’s constant, the threshold frequency of the metal is A hf – K C K – hf

0

0

u

u

B

D u — v

u — v

B 0

0 u

θ

K – hf ——— h

42. A hydrogen atom absorbs a photon and as a result an electron is excited from the ground state to the first excited state. If the nth energy level of the hydrogen atom is given by 13.6 eV, the photon energy absorbed E n = – —— n2 is A 3.4 eV C 13.6 eV B 10.2 eV D 17.0 eV

I

If the intensity of the unpolarised light is I 0, the intensity I of the transmitted light is 1 2 2 A I 0 cos θ C — I 0 cos θ 2 1 2 2 B I 0 sin θ D — I 0 sin θ 2

43. Which of the following statements is not true of Bohr’s atomic model? A Radiation is emitted when an electron orbits in the allowed orbit. B The angular momentum of an electron in the allowed orbit is an integral multiple h of —–. 2π C The electron in the orbit nearest to the nucleus has the lowest energy. D The Coulomb force between the nucleus and electron maintains the electron in its orbit.

38. The glass lens of a camera is coated with magnesium fluoride. The coating prevents the reflection of light of wavelength 565 nm. What is the minimum thickness of the coating? [Refractive index of magnesium fluoride = 1.38, refractive index of glass = 1.50] A 94 nm C 102 nm B 98 nm D 141 nm 39. Which of the following will produce the maximum separation of fringes in Young’s two-slit experiment? © Majlis Peperiksaan Malaysia 2006

D

41. A beam of laser with power P and wavelength λ is perpendicularly incident to a surface. What is the force on the surface? [c = speed of light; h = Planck’s constant.] Pλ A Pc C —– hc P hc B — D —– c Pλ

37. The diagram below shows two pieces of polarisers. The angle between the polarisation axes is θ . I 0

hf – K ——— h

u

36. The image of an object which is formed by a convex mirror is always A real and big C virtual and big B real and small D virtual and small

Unpolarised light

Red light Red light Green light Green light

Distance Distance between between slit two slits and screen 1 mm 2m 2 mm 1m 1 mm 2m 2 mm 1m

6



44. The diagram below shows a spectrum produced by an X-ray tube.

47. A radioac tive element X decays to a radioactive element Y which then decays to an element Z. If initially there is only X, which of the following influences the ratio of the number of nuclides of Y to that of X? A Type of Z B Half-life of Y C The initial total of X D Surrounding pressure

K β K α

Intensity

λ min

Wavelength

If the accelerating voltage is raised, A the value of λ min will decrease B the value of λ min will increase C the wavelength of the characteristic lines K α and K β will decrease D the wavelength of the characteristic lines increase K α and K β will

48. A nuclear reaction which involves carbon-13 and hydrogen is suggested as follows. 13 6

1

14

C + 1H

7

Which of the following is true of the nuclear reaction? 1

[Mass of 1 H = 1.007825 u, mass of

45. Which of the following is true of lasers? A Only red waves are produced. B The waves produced are longitudinal waves. C The output is produced by radioactive decay. D The output is produced by stimulated emission.

is true? A The number of protons does not change. B The number of neutrons does not change. 14 7

N are in the higher

energy states than the nucleons of D

50. Which of the following forces has the greatest magnitude outside a nucleus? A The gravitational force B The electromagnetic force C The weak nuclear force D The strong nuclear force

14

C. 6

14

The binding energy per nucleon of 7 N is greater than that of

C

49. A radioactive nuclide of mass number m1 decays by emitting a β -particle and γ -ray to produce a nuclide of mass number m2. Which of the following is true of the relation between m1 and m2? A m2 = m1 – 2 C m2 = m1 B m2 = m1 – 1 D m2 = m1 + 2

β -particle. Which of the following statements

The nucleons of

6

14

46. The isotope 6 C decays into 7 N by emitting a

C

13

= 13.003355 u, mass of 7 N = 14.003074 u.] A The reaction releases energy. B The reaction does not happen at all. C The reaction does not release energy. D The reaction must involve the liberation of an additional neutral atom.

14

14

N

14 6

C.

7




Time: 2h 30 min

Section A

Answer

all the

[40 marks] questions in this section.

1. (a) Determine the dimension of Young modulus. [2 marks] (b) The Young modulus can be determined by propagating a wave of wavelength λ with velocity v into a solid material of density ρ. Using dimensional analysis, derive a formula for Young modulus. [3 marks] 2. A system consists of an object attached to a spring with spring constant k . The system moves with simple harmonic motion of amplitude A. Calculate, in terms of k and A, the kinetic energy of the spring at a distance A — from its equilibrium position. [4 marks] 4 3. One end of a string of mass 1.45 g and length 50.0 cm is attached to a frequency generator and the other end to a weight holder which hangs over a fixed pulley. The part of the string between the generator and the pulley is horizontal. The string is set to vibrate. Several weights are added until a fundamental frequency of 120 Hz is achieved. [2 marks] (a) Describe the motion of the particles in the string. (b) Calculate the tension in the string when the fundamental frequency is achieved. [3 marks]

4. When a copper cube of side 2.0 cm is immersed into a perfectly insulated container filled with 1.0 kg of water at 5 °C, the temperature of water rises to 7 °C. Assuming no heat loss to the surroundings, calculate the original temperature of the cube. [5 marks] –3 –1 [Density of copper = 8900 kg m , specific heat capacity of water = 4180 J kg K–1 and specific heat capacity of copper = 385 J kg –1 K–1.] 5. A bird stands on a high voltage transmission wire with its feet 4.00 cm apart. The wire is made of aluminium with diameter 2.00 cm and carries a current of 100 A. (a) Calculate the resistance of the wire between the bird’s feet. [3 marks] –8 [Resistivity of aluminium = 2.65 × 10 Ω m.] (b) Calculate the potential difference between the bird’s feet. [2 marks] 6. (a) (i) What is a shunt ? [1 mark ] (ii) State the function of a shunt. [1 mark ] (b) Describe briefly, with the aid of a circuit diagram, how a shunt can be used in measuring a large current by using a galvanometer without damaging it. [4 marks] 7. (a) State the differences between the production of a continuous spectrum and a characteristic spectrum of X-rays. [2 marks] (b) Calculate the shortest wavelength of X-rays emitted by electrons striking the surface of a 20 kV television picture tube. [2 marks] 8. (a) Describe the thermonuclear fusion in the Sun. © Majlis Peperiksaan Malaysia 2006

8


[3 marks]


(b) The intensity of radiation of the Sun on the Earth is 1340 W m–2. The mean distance of the Earth from the Sun is 1.50 × 1011 m. Calculate the radiation energy of the Sun in a year. [3 marks] 7 [1 year = 3.15 × 10 s.]

Section B

[60 marks]

Answer any four questions in this section.

9. (a) Define the work done on an object and state the work-energy theorem. [2 marks] (b) Describe the process in which mechanical energy is converted to thermal energy. [2 marks]

(c) The diagram below shows a body of mass 5.0 kg placed 1.0 m high on an inclined plane with base length 3.0 m. The body slides down the inclined plane without friction. Calculate the velocity of the body when it reaches the bottom. [3 marks]

1.0 m

3.0 m

(d) If the body slides down the inclined plane in (c) with coefficient of kinetic friction equal to 0.2, calculate [3 marks] (i) the work done against friction, [3 marks] (ii) the net work done on the body, [2 marks] (iii) the velocity of the body when it reaches the bottom. 10. (a)

[2 marks] (i) What is meant by diffraction? (ii) Describe how the Huygen’s principle is used to explain a single slit diffraction [2 marks] phenomenon. (iii) Sketch the variation of the intensity of light across the single slit diffraction pattern [2 marks] with diffraction angle. (b) (i) Using a ray diagram, show that the destructive interference of a single slit diffraction is given by

mλ sin θ = —–, a

where θ is the diffraction angle, λ the wavelength, a the width of the slit and m an [4 marks] integer. (ii) If the diffraction angle is very small, estimate the optical resolution for light of [3 marks] wavelength 500 nm. [Assume that the aperture of eyelens is 3.0 mm.] (iii) Using the equation in (b) (i) with m = 1, explain why an electron microscope resolves [2 marks] better than an optical microscope.

9




11. (a) The diagram below shows an aluminium rod 1.0 m long and 1.0 cm in diameter with their ends connected to two blocks B and C. Blocks B and C are maintained at 100 °C and 0 °C respectively. The side of the rod is perfectly insulated. The system reaches equilibrium. Rod C

B

(i) Calculate the rate of heat flow in the rod. [3 marks] (ii) Calculate the temperature at a point 25 cm from block C. [2 marks] (b) If block C is instantaneously replaced by a huge block of ice, calculate the amount of ice melt in 10 minutes. [3 marks] (c) Half of the aluminium rod is then replaced with a brass rod of equal dimensions, with its free end connected to block C as shown in the diagram below. The system reaches equilibrium. Brass

B

Aluminium

C

[3 marks] (i) Calculate the temperature at the aluminium-brass junction. [2 marks] (ii) Calculate the rate of heat flow through the aluminium-brass junction. (d) Calculate the length of a copper rod that has the same conducting ability as the 1.0 m [2 marks] aluminium rod. –1 –1 [Thermal conductivity of aluminium = 205 W m K , thermal conductivity of brass = 109 W m–1 K–1, thermal conductivity of copper = 385 W m–1 K–1, latent heat of fusion of ice = 3.34 × 105 J kg–1.]

12. (a) The diagram below shows an electron which enters a region of uniform electric field between two charged plates in a horizontal direction with velocity vo. Its position is 0.5 cm from the negative plate. The electron traces a path and comes out of the field region at the end of one of the plates. The two plates are each 5.0 cm long and separated by a distance of 2.0 cm. The uniform field has magnitude 500 V m –1. 5.0 cm

– 0.5 cm Electron

v o

2.0 cm

+

(i) Sketch the electric field lines and trajectory of the electron in the field region. [2 marks]

(ii) Derive an equation to represent the motion of the electron while it is in the field region (ignore gravity). [5 marks] (iii) What path does the equation describe? [1 mark ] (iv) Determine the initial velocity of the electron. [2 marks] © Majlis Peperiksaan Malaysia 2006

10



(b) Consider a proton being accelerated from rest through a region of electric potential V 1 to V 2. The final speed of the proton is 2.0 × 106 m s–1. (i) Describe the energy change in the proton. [2 marks] (ii) Determine the potential difference between V 1 and V 2. [3 marks] 13. (a)

(i) Define the stopping potential in photoelectric effect. [1 mark ] (ii) State four most important results from a photoelectric effect experiment. [4 marks] (b) The work function for cesium is 2.14 eV. (i) Calculate the maximum wavelength of light that ejects electrons from a cesium target. [3 marks] (ii) If light of wavelength 452 nm is illuminated on cesium, calculate the maximum kinetic energy of photoelectrons. [2 marks] –2 (c) If this page of this question paper is illuminated by 120 W m of light with wavelength 550 nm, calculate (i) the power received by an area of 1.0 cm2, [2 marks] (ii) the number of photons striking each square centimetre per second. [3 marks]

14. (a) Define the half-life and decay constant of a radioactive substance. [2 marks] (b) Naturally occurring radium-226 with atomic number 88 produces the radioactive gas radon-222 by alpha decay. (i) Write an equation for the decay process. [1 mark ] (ii) Calculate the total kinetic energy of the decay products in MeV. [3 marks] [Mass of radium = 226.025402 u, mass of radon = 222.017570 u and mass of helium = 4.002603 u.] (c) Give three reasons why some radioisotopes are useful for medical diagnoses. [3 marks] (d) A doctor uses 1.49 µ g of iodine-131 to treat thyroid disorder. Iodine-131 has half-life of 8.0 days. Calculate (i) the number of nuclei initially present, [3 marks] (ii) the initial activity. [3 marks]

11




µ mg = m ω2 r µ g r = —2 ω

PAPER 1

1. D W is unit for power, J s –1. ∴ W s–1 = J s–2 This is not the unit for energy.

8. C At the equator, objects are rotating with the Earth and they need a source of force to provide the centripetal force needed. This source of force is the Earth’s gravitational pull. Part of the gravitational force is used to make the objects rotate, and therefore only the balance provides the acceleration due to free fall. At the pole, there is no rotation. Therefore the gravitational force fully provides the acceleration due to free fall. So g e < g p. Also, the Earth, not being a perfect sphere, is flattened at the poles. The distance from the centre of the Earth to the surface is smaller at the pole. This also contributes to the increased value of g p at the pole.

2. D s = ∫ v dt T From t = 0 to — , v = at. 2 1 ∴ s = ∫ at dt = —at 2 2 This is a quadratic function starting from zero. T From t = — to T , v = –at + aT. 2 1 ∴ s = ∫ (–at + aT ) dt = – —at 2 + aTt + C 2 2 T 1 T 1 At t = —, s = —a — = —aT 2 2 2 2 8 This is a quadratic function in t with the coefficient of t 2 being negative. Therefore it is an inverted U . Option D fits these conditions best.

 

9. B For simple harmonic motion, acceleration is proportional to displacement. a = –ω 2x Therefore when x is maximum, a is maximum.

3. B Let the common acceleration be a . T 2m T T For the mass 3m , F – T = 3ma = 3m × —– = 3— 2m 2 3T 5T ∴ F = T + — = — = 2.5 T 2 2 For the mass 2m , T = 2ma

10. D

∴ a = —–

4. C Power, P = F × v When maximum speed is reached, force = air resistance. ∴ P = 2v 2 × v = 2v 3 54 × 103 = 2v 3 v = 30 m s –1

11. D If there are oscillations, then the system is underdamped. At critical damping, there will be no oscillations, and the system which is displaced will reach equilibrium in the shortest possible time. 12. C There is a node at R and antinodes at Q and S. This is possible for a wavelength which equals the full length of the string. Therefore, λ = 4x .

5. C For each mass, the centripetal force is provided by the tension in the rod. m P ω 2r = T P m Q ω 2r = T Q ω and r are the same for the two masses. T P m P ∴ —– = —– T Q m Q

c 13. C In a closed pipe, the fundamental frequency f 0 = —, 4L where c is the speed of sound and L is the length of the pipe. However, for the open pipe, the fundamental c frequency f 0' = —– 2L. ' ∴ f 0 = 2f 0

1 6. D Kinetic energy of the flywheel = — I ω 2; ω = 2π n 2 1 ∴ K.E. = — I (2π n) 2 = 2π 2 I n 2 2 This amount of work is done by the flywheel against friction or any other retarding force when it is brought to a stop.

14. B At the equilibrium separation r 0, the total energy is (potential energy + kinetic energy). Potential energy is –ε and kinetic energy will depend on the temperature. If the temperature is absolute zero, then the kinetic energy will be zero, otherwise the system will have positive kinetic energy which is less than ε . Thus the total energy will have a negative non-zero value.

7. B Maximum frictional force = µ mg Maximum distance is r when frictional force is maximum. © Oxford Fajar Sdn Bhd (008974–T) 2007

Path difference ∆x and phase difference ∆ϕ are related by ∆x ∆ϕ = — × 2 π λ (PR–QR) = 2 π × ———— λ

12



15. A Glass is brittle. It obeys Hooke’s law and in its stressstrain graph there is no plastic region. It breaks when the elastic limit is exceeded.

21. D Electric force = gravitational force Q2 GM 2 –——2 = —– 4πε r r2

√ 4πε G × 3 × 105 = √ 4 × π × 8.85 × 10 –12 × 6.67 × 10 –11 × 3 × 105

Q =

16. D nm represents the density of the gas which is mass ——— volume. Mass of gas = number of molecules × mass per molecule. Mass ∴ ——— = number of molecules per unit volume Volume × mass per molecule.

= 2.6 × 10 –5 C 22. B The combined capacitance is given by 1 1 1 1 1 — = — + — = —— + —— C C 1 C 2 300µ 500µ C = 187.5 µ F Q = CV = 187.5 µ × 6 = 1.1 mC This is the charge that flows out of the cell. This is the amount of charge in each of the two capacitors because they are in series.

m 17. B pV = — RT M m RT p = — —– ∝ ρT = constant V M 300 —— T 2 = (1.25)(300) 400 T 2 = 500 K = (500 – 273) = 227 °C

      

ε A 23. C C = —– d 1 U = —CV 2 2 ε A = —–V 2 2d V is constant because the capacitor is connected to the power supply throughout. It is assumed that the material in the capacitor continues to completely fill the space between the plates as the plates are separated. Hence ε is constant. 1 ∴ U ∝ — d Since d 2 = 2d 1, U1 d2 —= — = 2 U2 d1

18. A The change from X to Y takes place at constant volume. For the pressure to increase at constant volume, the temperature must increase. (Pressure law, p ∝ T ). In the T -V graph, constant volume is represented by a vertical line XY which is parallel to the T -axis. The change YZ is isothermal, meaning that the temperature remains constant. Therefore in the T -V graph, constant temperature is represented by a horizontal line YZ which is parallel to the V -axis. Graph A satisfies both these two conditions. 19. B

dQ dθ —– = – kA — dt dx

V ρl 24. A R = — = — A I 1 I l Conductivity = — = — ρ VA

dQ At steady state, — and k are constant throughout the dt rod.

25. A Resistivity is dependent on the type of material and the temperature. For a metal, the higher the temperature, the higher the resistivity. It does not depend on the length or the cross-sectional area of the material.

d 1 dx A Ignoring the negative sign and considering magnitudes only, as A increases, the gradient of the graph decreases. From the diagram, as x increases, A increases, and therefore the gradient of the graph should decrease. B is the only graph that shows a continuously decreasing magnitude of its gradient.

θ ∴ —– ∝ – — (A = cross-sectional area of the rod)

 

26. A Let the potential of the negative terminal of the cell be 0 V and that of the positive terminal be 12 V. In the loop consisting of the cell and the three 2 Ω resistors in series, the potential drops by 4 V for each 2 Ω resistor. Therefore the potential at Y is 12 – 4 = 8 V. In the loop with the cell and two 6 Ω resistors, the potential drops by 6 V for each 6 Ω resistor. Therefore the potential at X is 12 – 6 = 6 V. Therefore the potential difference between X and Y = 8 – 6 = 2 V.

20. C For an isothermal change, TV γ –1 = constant. For diatomic gas, γ = 1.4, γ – 1 = 0.4 (V 1)0.4 T 2 —–— 0.4 = — (V 2) T 1

27. C To convert a galvanometer to a voltmeter, a high resistance is connected in series with the galvanometer.

V 1 (T 2)2.5 (681 + 273)2.5 — = ——2.5= –————— = 18 V 2 (T 1) (27 + 273)2.5

13

© Oxford Fajar Sdn Bhd (008974–T) 2007



28. D For the charge to move at uniform velocity, the magnetic force = electric force and B and E must be perpendicular to each other. Bev = Ee potential difference 1000 = E = ———————— = ———— 5 × 104 V m–1 –2 separation of plates 2.0 × 10 5 × 10 4 = v = ———— 5 × 10 6 m s–1 –2 1.0 × 10

37. C

29. C Electric force = Eq and magnetic force = Bqv. 30. D Induced e.m.f. in the secondary coil by mutual induction is given by d I E = M— = M 2 π f I 0 cos 2π ft dt ( I = primary current = I 0 sin 2π f t )

38. C For a non-reflective coating when the refractive index n of the coating material is less than the refractive index of the glass, λ nt = — (t = minimum thickness) 4 565 nm λ ∴ t = — = ———— = 102 nm 4n 4 × 1.38

31. A For a motor, E = E b + I r, where E is the applied voltage, E b is the back e.m.f., I is the current and r is the armature resistance. When the load decreases, the armature will accelerate and its speed will increase. This will cause the back e.m.f. to increase because the back e.m.f. depends on the speed of rotation. Thus in the equation above, with E and r being constant, when E b increases, I decreases. 1 32. D Reactance of a capacitor, X c = ——. 2π fC 1 Thus, X ∝ —. D shows this inverse proportion. C

39. A Fringe separation x is given by λ D x = —. a Wavelength of red light is more than wavelength of green light. D Comparing options A and B, — a is more for option A. ∴ x is bigger for A. Comparing options A and C, λ is more for option A. ∴ x is bigger for A. D Comparing options C and D, — a is more for option

33. B The output voltage V o = 105(V 2 – V 1) Saturation is reached if this value exceeds 9 V. In option B, V o = 105[0 – (–150 µ V)] = 15 V. This exceeds the supply voltage of 9 V. Therefore saturation is reached. In all the other options, V o is less than 9 V.

C. ∴ x is bigger for C. Based on these comparisons, option A has the largest value of x . 40. B Einstein’s equation for the photoelectric effect is hf = K + hf 0 (f 0 = threshold frequency) hf – K ∴ f 0 = ——— h

34. C Electromagnetic waves are transverse waves in which the electric and magnetic fields and the velocity of propagation are mutually perpendicular. However, the velocity lies in the plane of vibration of the electric field, and also in the plane of vibration of the magnetic field. This situation is best described by option C.

41. B Power = energy per second. hc Energy of each photon = —. λ If n photons strike the surface per second, total nhc energy per second, P = ——. λ nh P ∴ — = —. c λ Force = change of momentum per second. If we assume that each photon that strikes the surface stops and does not bounce back, the change of h momentum per photon = —. λ nh P Total change of momentum per second, F = —– = —. λ c

1 1 1 35. C The lens formula is — + — = —. u v f Multiplying throughout by u , u u 1+—=— v f u u — = — – 1 v f u Thus the graph of — against u is a straight line with a v u negative intercept on the — axis. v Option C is the best answer.

42. B Ground state, n = 1. First excited state, n = 2. E 1 = –13.6 eV, –13.6 E 2 = –—— = –3.4 eV 22

36. D Convex mirror produces only a virtual and diminished image of a real object.


After passing through the first polaroid, the intensity 1 becomes I ' = — I 0. 2 On passing through the second polaroid, the amplitude of the emerging wave is A cos θ , where A is the amplitude of the incident wave. Therefore the intensity of the emerging wave which is proportional to the square of the amplitude becomes 1 I ' cos 2 θ = — I 0 cos 2 θ 2

14



[Y ] = [k λx v y ρz ] ML–1T–2 = Lx (LT–1) y (ML–3)z = L x + y –3z T– y Mz Equating indices of M: z = 1 T: y = 2 L: x + y – 3z = –1 x = –1 + 3(1) – 2 = 0 Hence, Y = kv 2ρ

Energy absorbed = energy of photon = E 2 – E 1 = –3.4 – (–13.6) = 10.2 eV. 43. A Radiation is emitted when electrons undergo transitions from higher energy levels to lower energy levels, and not when they are in an allowed orbit. hc 44. A λ min = –—. eV When V increases, λ min decreases.

m 2π 2. Period, T = — = 2π — ω k k ω 2 = — m Velocity, v = ω A2 – x 2 A 1 1 A2 When x = —, kinetic energy = —mv 2 = —m ω 2 A2 – — 4 2 2 16 1 k 15A2 15 2 = —m — —— = —kA 2 m 16 32

45. D Laser is Light Amplification by Stimulated Emission of Radiation. 46. D The radioactive decay takes place to increase the stability of the nucleus. Since C decays into N, the N nucleus is more stable than the C nucleus. If a nucleus is more stable, the binding energy per nucleon is higher. Therefore N has a higher binding energy per nucleon.



 

47. B When equilibrium is reached, the rate of decay of X will equal the rate of decay of Y. If half-life of Y is large, a very large quantity of Y is needed so that its rate of decay equals that of X. Thus the ratio of the number of nuclides of Y to X will be large. On the other hand, if the half-life of Y is small, a smaller quantity of Y is enough for its rate of decay to be equal to that of X. Then the ratio of Y to X will be small.

2

Tension, T = 4f 0 l 2m = 4(120)2(0.500)2(2.90 × 10–3) = 41.8 N

4. Let θ ° C = initial temperature of the copper cube. Heat loss by the cube = heat gained by the water (L3ρ)c 1(θ – 7) = (1.0)(4180)(7 – 5) (m = L 3ρ) 8360 (θ – 7) = ———————— (0.020)3 (8900)(385) θ = 312 °C

49. C Mass number means nucleon number. During a β -decay, the nucleon number remains the same.

ρl (2.65 × 10 –8)(0.0400) 5. (a) Resistance, R = — = ————————— π (0.0100)2 A

50. B At distances exceeding the diameter of a nucleus, the nuclear forces are very weak because nuclear forces are short-ranged forces. Electromagnetic and gravitational forces are long-ranged forces. Between these two, electromagnetic forces are stronger.

= 3.37 × 10–6 Ω (b) V = I R = (100)(3.37 × 10–6) = 3.37 × 10–4 V 6. (a)

PAPER 2 Section A

stress F/A 1. (a) Young modulus, Y = ——— = —– strain e/l F/A Dimension of Y = —— e / l MLT–2 / L2 = ————— L/L = ML–1T–2





3. (a) Transverse stationary wave is set up in the string. At the nodes, displacement of the particles = 0. At the antinode, the particle vibrates with the maximum amplitude. Amplitude of vibration increases from a node to an antinode. 1 1.45×10–3 (b) Fundamental frequency, f 0 = — T — ( m = —–—— 2l m 0.500 = 2.90 × 10–3 kg m–1)

48. A The total mass before reaction = 13.003355 u + 1.007825 u = 14.01118 u. The mass after the reaction = 14.003074 u. There is a mass defect of 14.01118 u – 14.003074 u = 8.106 × 10 –3 u. Thus energy will be released when the reaction takes place.

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(i) Shunt: A conductor connected across the terminals of a galvanometer. (ii) Function: So that most of the current bypasses the galvanometer.

(b) R 1

I



I G

G R

(b) Young modulus, Y = k λx v y ρ z , where k , x , y and z are non-dimensional constants.

I S

15




Connect the shunt of resistance R in parallel to the galvanometer of resistance R 1, where R <


(d)

µ R



mg sin θ

θ 3.0 m

7. (a) Continuous spectrum is produced when fast electrons from the cathode are decelerated on collision with the target. The decrease in energy of the decelerated electrons is radiated as photons in the continuous spectrum. Characteristic X-ray is produced when a vacancy in the inner shell (e.g. K-shell) of the target atom is filled by an electron from a higher shell. The difference in energy of the electron is radiated as a characteristic X-ray photon. hc (b) eV = —— λ min

mg

1.0 m

mg cos θ

 



= 29.4 J (ii) Net work done = net force × displacement = (mg sin θ – µ mg cos θ ) 10 1.0 = (5.0)(9.81) —– 10 3.0 – (0.2)(5.0)(9.81) —– ( 10 ) 10 = 19.6 J (iii) Gain in kinetic energy = net work done 1 — mv 2 = 19.6 J 2 2 × 19.6 v = ———– = 2.80 m s –1 5.0





8

8. (a) Due to the very high temperature (>107 K) in the Sun, • hydrogen atoms travel at high speed (or have sufficiently high kinetic energy) • to overcome repulsive electrostatic force • fusion between hydrogen atoms occurs. During each fusion, the energy equivalent of the mass defect is radiated. (b) If P = power radiated from the Sun, then P intensity, I = —— 2 4π R P = (4π R 2) I Energy radiated in a year = Pt = (4π R 2) I t = (4π )(1.50 × 1011)2(1340)(3.15 × 107) = 1.19 × 1034 J

10. (a)

Section B

9. (a) Work done = F .s F = force on the object, and s = displacement of the object. Work-energy theorem: Work done on an object = increase in the mechanical energy of the object. (b) Example: When an object travels on a rough surface, the decrease in the kinetic energy equals the increase in the random kinetic energy of the atoms or heat in the object. (c) Gain in kinetic energy = loss in potential energy 1 —mv 2 = mgh 2 Velocity, v = 2gh =

θ

Length of incline = 12 + 32 = 10 m (i) Friction = µ R = µ mg cos θ Work done against friction = ( µ mg cos θ )( 10 ) 3.0 = (0.2)(5.0)(9.81) —– 10 10

(6.63 × 10 )(3.00 × 10 ) λ min = ————————–—— = 6.22 × 10–11 m –19 3 (1.60 × 10 )(20 × 10 ) –34

R

   



(i) Diffraction: Spreading of a wave after passing a narrow slit or obstacle. (ii) Points on the wavefront that arrive at the slit behave as point sources emitting spherical wavelets in phase. Along directions where constructive interference occurs, light of high intensity is detected. No light is detected along directions where destructive interference occurs. (iii) Intensity

(b) (i)

0 Diffraction angle, θ

θ A

a —– 2m

B

N

P, mth minimum C

Slit

2(9.81)(1.0)

If P is the m th minimum, divide the slit into 2 m equal parts. a Distance between two point sources A and B = –—. 2m

= 4.43 m s –1


16



Wavelets from A and B produce destructive interference at P when 1 BP – AP = BN = —λ 2 Angle PBC = angle BAN = θ BN In triangle ANB, sin θ = —– AB AB sin θ = BN a 1 —–sin θ = — λ 2m 2 m λ sin θ = —– a λ (ii) If θ is small, sin θ = θ radians = — a λ 500 × 10–9 Optical resolution, θ = — = ——–—— a 3.0 × 10–3 = 1.7 × 10–4 rad (iii) de Broglie’s wavelength of electron beam, λ e << λ , wavelength of light. λ For optical microscope, resolution θ = — a λ For electron microscope, resolution θ e = —e a Since λ e << λ , θ e<< θ. Hence compared to the optical microscope, the electron microscope is able to distinguish clearly two points that are much closer.



12. (a) (i)



– E

+

Path

(ii) At any time = t, x • horizontal displacement, x = v ot, t = — v o 1 • vertical displacement, s = ut + —a t 2 2 1 x y = 0 + — a (—)2 2 v o eE (Acceleration = a = — = constant) m eEx 2 y = ——– 2mv 02 (iii) Parabolic path (iv) When x = 5.0 cm = 0.050 m, y = 1.50 cm = 0.0150 m eEx 2 y = —–—2 2mv o eE Initial velocity, v 0 = x —— 2my (1.60 × 10 –19)(500) = (0.050) —————————– 2(9.11 × 10 –31)(0.0150)

dQ (i) Rate of heat flow, –— dt dθ 100 – 0 = –kA – — = (205)π (0.005)2 ———– dx 1.0 = 1.61 W (ii) Temperature gradient = – 100 °C m –1 Temperature 25 cm from C or 75 cm from B = 100 + (–100 × 0.75) = 25 °C (b) Heat transferred in 10 minutes, Q = (1.61)(10 × 60) = 966 J Q 966 Mass of ice melted = — = ————–5 L 3.34 × 10 = 2.89 × 10–3 kg (c) (i) Let θ = temperature at the junction. 1 Temperature gradient ∝ — k 100 – θ 1 For aluminium, —–—— ∝ —— (1) 0.50 205 1 θ – 0 For brass, —–— ∝ —— (2) 0.50 109 (1) 100 – θ 109 — , ——–— = —— (2) 205 θ θ = 65.3 oC dQ dθ (ii) Rate of heat flow, — = –kA — dt dx 100 – 65.3 = (205)π (0.005)2 ————— = 1.12 W 0.50 (d) Rate of heat flow in copper rod = rate of heat flow in aluminium rod

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

 

11. (a)





100 – 0 100 – 0 (385)π (0.005)2 ——— = (205)π (0.005)2 ——— x 1.0 385 Length of copper rod, x = —– 1.0 = 1.88 m 205



(b)

13. (a)



17

= 2.71 × 106 m s–1 (i) Kinetic energy increases, electric potential energy decreases. Gain in kinetic energy = loss in electric potential energy (ii) Loss in electric potential energy = gain in kinetic energy 1 e (V 1 – V 2) = —mv 2 2 mv 2 (V 1 – V 2) = —– 2e (1.67 × 10 –27)(2.0 × 106)2 = —————————— = 2.09 × 104 V 2(1.60 × 10–19) (i) Stopping potential: Minimum reverse potential that is required to stop all photoelectrons from reaching the anode. (ii) • Existence of threshold frequency, which is the minimum frequency of the electromagnetic radiation that would eject photoelectrons from a metal, or the existence of work function which is the minimum energy required for a photoelectron to escape from a metal surface. • Instantaneous emission of photoelectrons. • Photoelectric current increases when the intensity of electromagnetic radiation increases. © Oxford Fajar Sdn Bhd (008974–T) 2007



•

(b)

(i)

(ii) (c)

(i) (ii)

Maximum kinetic energy of photoelectron does not depend on the intensity of the electromagnetic radiation but increases when the frequency of the radiation increases. hc Work function = —— λ max (6.63 × 10 –34)(3.00 × 10 8) λ max = —————————–— = 5.81 × 10–7 m 2.14(1.60 × 10–19) hc Maximum kinetic energy, K max = — – W λ –34 8 (6.63 × 10 )(3.00 × 10 ) = —————————–— – 2.14 = 0.610 eV (452 × 10 –9)(1.60 × 10 –19) Power = intensity × area = (120)(1.0 × 10–4) = 0.0120 W hc Power = N — λ (0.0120)(550 × 10 –9) N = —————————— (6.63 × 10 –34)(3.00 × 10 8) = 3.32 × 1016 photons s–1

(b) (i)

Ra →

4

222 86

Rn + 2He

(ii) Mass defect, ∆m = 226.025402 u – (222.017570 + 4.002603) u = 0.005229 u Total kinetic energy = (∆m )c 2 = (0.005229)(1.66 × 10–27 )(3.00 × 108 )2 J = 4.88 MeV (c) • High penetration power of γ -ray: Emissions from a radioisotope in the body can be detected by a detector that is outside the body. • Produce ionisation: Used in radiotherapy to destroy cells. • A radioactive isotope retains its activity whatever compound its atoms are attached to: Used as radioactive tracer.

 

(d) (i)

14. (a) Half-life: Time taken for half the number of radioactive atoms in a sample to decay, or the time taken for the number of radioactive atoms in a sample to decay to half of its initial number. rate of decay of a dN –— radioactive sample dt Decay constant = —— = —————————— number of radioactive N atoms in the sample


226 88

Number of I-131 nuclei 1.49 × 10–6 = ————–— 131 (6.02 × 1023) = 6.85 × 1015 dN In 2 (ii) Initial activity, – — = λ N and λ = —– dt T 1— 2 dN In 2 – — = —— N dt T —1 2 In 2 = ——————— (6.85 × 1015) 8.0 × 24 × 3600





  

= 6.87 × 109 Bq

18


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Physics STPM Past Year Questions with answer 2006.pdf

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