physics stpm past year questions with answer 2006

8/11/2019 Physics STPM Past Year Questions With Answer 2006

http://slidepdf.com/reader/full/physics-stpm-past-year-questions-with-answer-2006 1/18

1 © Majlis Peperiksaan Malaysia 2006

2006 STPM Physics Papers 1 & 2

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 tothe unit of energy?A Electron volt (eV)B Volt coulomb (V C)C Newton metre (N m)D Watt per second (W s –1)

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

Which of the following is the displacement-time ( s-t ) graph for the car?

A C

B D

3. The diagram below shows two bodies ofmasses 2 m and 3 m connected by a lightinelastic string. The bodies are pulled by aforce F on a smooth surface.

If the tension in the string is T , the force F isA T C 3.0 T B 2.5 T D 5.0 T

4. The engine of a car has a maximum outputpower of 54 kW. The air resistance acting onthe car when it is moving with speed v is 2 v2.The maximum speed which can be achievedby the car on a level road isA 3.0 m s –1 C 30 m s –1 B 5.2 m s –1 D 164 m s –1

5. The diagram below shows two bodies ofmasses mP and mQ attached to the ends P andQ of a light rigid rod. The rod is then rotatedabout a vertical axis through the centre O ofthe rod.

The ratio of the tension in the section PO tothe section QO of the rod is

mPA 1 C — mQ

BmP —– mQ

1 — 2 D mP — –

2 mQ

6. A flywheel rotates at a rate of n rotations persecond. If the mass and moment of inertiaof the flywheel are M and I respectively, thework that should be done to stop the flywheelis

1A — Mn 2 C 2π 2 Mn 2

2 1B — In 2 D 2π 2 In 2

2

v

0 T —2T t

s

0 T —2T t

s

0 T —2

T t

s

0 T —2

T t

s

0 T —2

T t

2m 3m T

F

m QO

m P

P

Q

Time: 1h 45 min

Booklet 2006 STPM Phy P1 2(08).indd 1Booklet 2006 STPM Phy P1 & 2(08).indd 1 3/7/2008 4:52:19 PM3/7/2008 4:52:19 PM

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



2© Majlis Peperiksaan Malaysia 2006


7. A coin of mass m is placed on a disc whichis rotating about its vertical axis at constantangular velocity ω . The coefficient of staticfriction between the coin and the surface ofthe disc is µ . If the coin does not slip, themaximum distance between the coin and thecentre of the disc is

g mgA — C — ω 2 ω 2

µ g µ mgB — D —–

ω 2 ω 2

8. The accelerations due to gravity at the equatorand the pole are ge and gp respectively. Which ofthe 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.

9. Which of the following statements aboutan object which performs simple harmonicmotion 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 greaterthan the maximum potential energy.

10. The diagram below shows two coherentsources P and Q which produce waves ofthe same phase with wavelength λ .

If the two waves meet at the point R, thephase difference between the waves at R is

2π A PR – QR C ————— λ (PR – QR) λ (PR – QR) 2 π (PR – QR)B ————— D ————–— 2π λ

11. Which of the following statements about acritical damping system is true?A No energy is lost from the system.

B Only a few oscillations are possible.C The system takes an infinite time to return

to equilibrium.D 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 whichare 4 x apart. Three small pieces of paper areplaced at positions Q, R and S.

When the wire is vibrated at a certainfrequency, the small piece of paper at R

remains on the wire but the small piecesof paper at Q and S drop off. What is thewavelength of the wave which is producedat the wire?A x C 4 x B 2 x D 8 x

13. A student blows across the top of a hollowbamboo pipe. The bottom of the pipe is closedby a finger so that a fundamental tone offrequency f is produced. Then the studentblows again with the bottom of the pipe

open. The tone that the student hears has afrequencyA equal to f B lower than f C higher than f D fluctuating in the range of f

14. The graph below shows the variation ofpotential energy U with separation r betweentwo atoms.

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.

P

QR

r 0 r 0

–ε

U

x x x x TP

Q R S

LoadPieces of paper

Metal wire







C The equilibrium separation between theatoms is r 0.

D An energy ε is required to separate thetwo atoms.

15. Which of the following is the stress-straingraph for a glass rod which is stretched untilbroken?A C

B D

16. The pressure p of an ideal gas is related to themean square speed < c2> of the gas moleculesby the equation

What are represented by n and m?n m

A Number of

moles

Mass of one

mole gasB Number ofmoles

Mass of onemolecule

C Number ofmolecules

Mass of onemolecule

D Number ofmolecules perunit volume

Mass of onemolecule

17. The air inside a hot-air balloon is heatedup from a density of 1.25 kg m –3 and atemperature of 27 °C until the balloon beginsto float. During this heating process, someair escapes from the balloon and the pressureinside the balloon remains constant at justabove the atmospheric pressure. The massand volume of air in the balloon just beforeit rises are 300 kg and 400 m 3 respectively.The temperature of hot air isA 45 °C C 318 °CB 227 °C D 500 °C

18. The graph below shows the change in stateof a mass of an ideal gas from X to Y andthen isothermally from Y to Z, where p is thepressure and V the volume. The pressures atX and Z are equal.

Which of the following graphs shows thevariation of temperature T with volume V ?A C

B D

19. The diagram below shows a metal rod PQof length and a diameter which increasesuniformly from P to Q. The rod is well-insulated and the ends P and Q of the rod areat temperatures θ 1 and θ 2 respectively, whereθ 1 > θ 2.

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

B D

1 p = — nm<c2> 3

Stress

0Strain

Stress

0Strain

Stress

0Strain

Stress

0Strain

p

ZX

V 0

Y

Z

X

V 0

YT Z

X

V 0

YT

ZX

V 0

YT

ZX

V 0

YT

θ θ 1

θ 2

x 0

θ θ 1

θ 2

x 0

θ θ 1

θ 2

x 0

θ θ 1

θ 2

x 0

Insulation

RodP

Q







20. A diesel engine does not need spark plugsfor the combustion of a mixture of diesel andair in a cylinder. Spontaneous combustionoccurs when a high temperature is achievedthrough air compression. If air with aninitial temperature of 27 °C is compressedadiabatically to a final temperature of 681 °C,what is the ratio of the initial volume to thefinal volume?

[Assume that air is a diatomic ideal gas.]A 6 C 18B 15 D 32

21. Two bodies each of mass 3 × 10 5 kg are 200 km apart. The same amount of charge isplaced on each body. What is the magnitudeof the charge on each body so that thegravitational attractive force is balanced bythe electric force?A 6.7 × 10 –10 CB 1.0 × 10 –9 CC 4.7 × 10 –8 CD 2.6 × 10 –5 C

22. The diagram below shows two capacitors of300 µ F and 500 µ F connected in series to a6.0 V battery.

The charge in the 300 µ F capacitor isA 0.68 mC C 1.8 mCB 1.1 mC D 4.8 mC

23. A parallel plate capacitor is charged until acertain voltage. With the power supply stillconnected, the plates of the capacitor are

separated further until the distance betweenthem is two times the original distance. Ifthe energy stored in the capacitor before andafter the plates are separated are U 1 and U 2

U 1 respectively, the ratio — is U 2 1A — C 2 2

B 1 D 4

24. The diagram below shows a metal strip ofcross-sectional area A and length connectedto a battery.

If the potential difference across the strip is V and the current flowing is I , the conductivityof the strip is

I V A — C —

VA IA VA IAB — D —– I V

25. Which of the following statements is true ofthe 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.

26. An electric circuit is shown below.

The potential difference between the pointsX and Y isA 2.0 V C 6.0 VB 4.0 V D 8.0 V

27. A galvanometer which has a resistance of 1 Ω gives a full-scale deflection when a current of1 mA flows through it. This galvanometer can

be modified to measure potential differencesup to several volts. Which of the followingmodified arrangements is correct?A C

B D

300 µ F 500 µ F

6.0 V

Metal strip

G

1 Ω 10 k Ω

G

6.0 Ω

6.0 Ω

2.0 Ω

2.0 Ω

2.0 Ω12.0 V

Y

X

1 ΩG

10 k ΩG







28. A beam of protons moves in a straightpath through a region which has uniformmagnetic and electric fields. The electricfield is produced by two parallel metalplates which are 2.0 cm apart with potentialdifference 1000 V. If the magnetic field is1.0 × 10 –2 T, the speed of the protons isA 5.0 × 10 2 m s –1 C 1.0 × 10 5 m s –1

B 5.0 × 10 4 m s –1 D 5.0 × 10 6 m s –1

29. A proton enters perpendicularly into a uniformelectric field and another proton entersperpendicularly into a uniform magnetic field.How is the force on each proton influencedby the speed v of the proton?

In electric field In magnetic field A Proportional to v Proportional to vB Proportional to v2 Not influenced by vC Not influenced by v Proportional to vD Not influenced by v Proportional to v 2

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

31. When the load decreases, the speed of rotationof an electric motor increases and the currentflowing through it decreases. This happensbecause.A the back e.m.f. increasesB the frictional force decreasesC the resistance of the armature coil increasesD the inductance of the armature coil

increases

32. Which of the following graphs shows the

variation of the reactance X of a capacitorwith the frequency f of an alternating potentialdifference which is applied across it?A

B

C

D

33. The diagram below shows an operationalamplifier with an open-loop gain of 10 5. Thevoltage supply is ±9.0 V. The inverting andnon-inverting input voltages are V 1 and V 2 respectively.

Which of the following produces a saturatedoutput voltage of V o?

V 1 V 2A +50 µ V 0B –150 µ V 0C –120 µ V –120 µ VD +195 µ V +240 µ V

34. The planes of vibration of the electric andmagnetic fields in an electromagnetic wave areA parallel to each other and parallel to the

direction of the propagation of the waveB parallel to each other and perpendicular

to the direction of the propagation of thewave

C perpendicular to each other and parallel tothe direction of the propagation of the wave

D perpendicular to each other andperpendicular to the direction of thepropagation of the wave

X

f 0

X

f 0

X

f 0

X

f 0

–

+

V 1

V 2

–9.0 V

+9.0 V

V o







35. Which of the following graphs shows theu variation of — with u, where u is the objectv

distance and v the image distance of a convexlens?A C

B D

36. The image of an object which is formed bya convex mirror is alwaysA real and big C virtual and bigB real and small D virtual and small

37. The diagram below shows two pieces ofpolarisers. The angle between the polarisationaxes is θ .

If the intensity of the unpolarised light is I 0,the intensity I of the transmitted light is

1A I 0 cos 2 θ C — I 0 cos 2 θ 2 1B I 0 sin 2 θ D — I 0 sin 2 θ 2

38. The glass lens of a camera is coated withmagnesium fluoride. The coating preventsthe reflection of light of wavelength

565 nm. What is the minimum thickness ofthe coating? [Refractive index of magnesium fluoride

= 1.38, refractive index of glass = 1.50]A 94 nm C 102 nmB 98 nm D 141 nm

39. Which of the following will produce themaximum separation of fringes in Young’stwo-slit experiment?

Monochromaticlight

Distancebetweentwo slits

Distancebetween slitand screen

A Red light 1 mm 2 mB Red light 2 mm 1 mC Green light 1 mm 2 mD

Green light 2 mm 1 m40. Light of frequency f radiates the surface of

a metal. If the maximum kinetic energy ofthe emitted electron is K and h is Planck’sconstant, the threshold frequency of the metalis A hf – K C K – hf

hf – K K – hf B ——— D ——— h h

41. A beam of laser with power P and wavelengthλ 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 hcB — D —– c Pλ

42. A hydrogen atom absorbs a photon and as aresult an electron is excited from the groundstate to the first excited state. If the nthenergy level of the hydrogen atom is given by

13.6 E n = – —— eV, the photon energy absorbedn2

isA 3.4 eV C 13.6 eVB 10.2 eV D 17.0 eV

43. Which of the following statements is not trueof 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 multipleh 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 itsorbit.

u —v

u 0

u —v

u 0

u —v

u 0

u —v

u 0

Unpolarisedlight

I

θ I 0







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

If the accelerating voltage is raised,A the value of λ min will decreaseB the value of λ min will increaseC the wavelength of the characteristic lines

K α and K β will decreaseD the wavelength of the characteristic lines

K α and K β will increase

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. 46. The isotope 14

6 C decays into 14

7 N by emitting a

β-particle. Which of the following statementsis true?A The number of protons does not change.B The number of neutrons does not change.

C The nucleons of14

7 N are in the higher

energy states than the nucleons of 14

6 C.

D The binding energy per nucleon of14

7 N is

greater than that of14

6 C.

47. A radioactive element X decays to aradioactive element Y which then decaysto an element Z. If initially there is only X,which of the following influences the ratio ofthe number of nuclides of Y to that of X?A Type of ZB Half-life of YC The initial total of XD Surrounding pressure

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

13

6 C +1

1H14

7 N

Which of the following is true of the nuclearreaction?

[Mass of

1

1 H = 1.007825 u, mass of13

6 C= 13.003355 u, mass of

14

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.

49. A radioactive nuclide of mass number m1 decays by emitting a β-particle and γ -ray toproduce a nuclide of mass number m2. Which

of the following is true of the relation betweenm1 and m2?A m2 = m1 – 2 C m2 = m1

B m2 = m1 – 1 D m2 = m1 + 2

50. Which of the following forces has the greatestmagnitude outside a nucleus?A The gravitational forceB The electromagnetic forceC The weak nuclear forceD The strong nuclear force

IntensityK β

Wavelength

K α

λ min







Answer all the 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 formulafor Young modulus. [3 marks ]

2. A system consists of an object attached to a spring with spring constant k . The system moveswith simple harmonic motion of amplitude A. Calculate, in terms of k and A, the kinetic energy

A of the spring at a distance — 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 generatorand the other end to a weight holder which hangs over a fixed pulley. The part of the stringbetween the generator and the pulley is horizontal. The string is set to vibrate. Several weightsare added until a fundamental frequency of 120 Hz is achieved.(a) Describe the motion of the particles in the string. [2 marks ]

(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 filledwith 1.0 kg of water at 5 °C, the temperature of water rises to 7 °C. Assuming no heat lossto the surroundings, calculate the original temperature of the cube. [5 marks ]

[Density of copper = 8900 kg m –3, specific heat capacity of water = 4180 J kg –1 K–1 andspecific 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 ismade 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 ] [Resistivity of aluminium = 2.65 × 10 –8 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 measuringa 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 characteristicspectrum of X-rays. [2 marks ]

(b) Calculate the shortest wavelength of X-rays emitted by electrons striking the surface ofa 20 kV television picture tube. [2 marks ]

8. (a) Describe the thermonuclear fusion in the Sun. [3 marks ]

Time: 2h 30 min

Section A [40 marks]







(b) The intensity of radiation of the Sun on the Earth is 1340 W m –2. The mean distance ofthe Earth from the Sun is 1.50 × 10 11 m. Calculate the radiation energy of the Sun in ayear. [3 marks ]

[1 year = 3.15 × 10 7 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. Calculatethe velocity of the body when it reaches the bottom. [3 marks ]

(d) If the body slides down the inclined plane in (c) with coefficient of kinetic friction equalto 0.2, calculate(i) the work done against friction, [3 marks ]

(ii) the net work done on the body, [3 marks ]

(iii) the velocity of the body when it reaches the bottom. [2 marks ]

10. (a) (i) What is meant by diffraction ? [2 marks ]

(ii) Describe how the Huygen’s principle is used to explain a single slit diffractionphenomenon. [2 marks ]

(iii) Sketch the variation of the intensity of light across the single slit diffraction patternwith diffraction angle. [2 marks ]

(b) (i) Using a ray diagram, show that the destructive interference of a single slit diffractionis given by

mλ sin θ = —–, a

where θ is the diffraction angle, λ the wavelength, a the width of the slit and m aninteger. [4 marks ]

(ii) If the diffraction angle is very small, estimate the optical resolution for light ofwavelength 500 nm. [Assume that the aperture of eyelens is 3.0 mm.] [3 marks ]

(iii) Using the equation in (b) (i) with m = 1, explain why an electron microscope resolvesbetter than an optical microscope. [2 marks ]

1.0 m

3.0 m







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

(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 icemelt in 10 minutes. [3 marks ]

(c) Half of the aluminium rod is then replaced with a brass rod of equal dimensions, withits free end connected to block C as shown in the diagram below. The system reachesequilibrium.

(i) Calculate the temperature at the aluminium-brass junction. [3 marks ]

(ii) Calculate the rate of heat flow through the aluminium-brass junction. [2 marks ]

(d) Calculate the length of a copper rod that has the same conducting ability as the 1.0 maluminium rod. [2 marks ]

[Thermal conductivity of aluminium = 205 W m –1 K–1, 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 × 10 5 J kg –1.]

12. (a) The diagram below shows an electron which enters a region of uniform electric fieldbetween two charged plates in a horizontal direction with velocity vo. Its position is0.5 cm from the negative plate. The electron traces a path and comes out of the fieldregion at the end of one of the plates. The two plates are each 5.0 cm long and separatedby a distance of 2.0 cm. The uniform field has magnitude 500 V m –1.

(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 fieldregion (ignore gravity). [5 marks ]

(iii) What path does the equation describe? [1 mark ] (iv) Determine the initial velocity of the electron. [2 marks ]

BC

Rod

B C

Brass

Aluminium

5.0 cm

0.5 cm

Electron v o 2.0 cm

–

+







(b) Consider a proton being accelerated from rest through a region of electric potential V 1 toV 2. The final speed of the proton is 2.0 × 10 6 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 ](c) If this page of this question paper is illuminated by 120 W m –2 of light with wavelength

550 nm, calculate (i) the power received by an area of 1.0 cm 2, [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 gasradon-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 ]





12


© Oxford Fajar Sdn Bhd (008974–T) 2007

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. 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 T 1 T 1 At t = —, s = —a —

2 = —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.

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

2m 2

3T 5T ∴ F = T + — = — = 2.5 T 2 2

4. C Power, P = F × v When maximum speed is reached, force

= air resistance.∴ P = 2v2 × v = 2v3

54 × 103 = 2v3

v = 30 m s –1

5. C For each mass, the centripetal force is provided bythe 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

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 broughtto a stop.

7. B Maximum frictional force =µ mg Maximum distance is r when frictional force is

maximum.

µ mg = m ω 2r µ g r = — ω 2

8. C At the equator, objects are rotating with the Earth andthey need a source of force to provide the centripetalforce needed. This source of force is the Earth’sgravitational pull. Part of the gravitational force isused to make the objects rotate, and therefore onlythe balance provides the acceleration due to freefall. At the pole, there is no rotation. Therefore thegravitational force fully provides the acceleration dueto free fall. So g e < g p. Also, the Earth, not being aperfect sphere, is flattened at the poles. The distancefrom the centre of the Earth to the surface is smallerat the pole. This also contributes to the increasedvalue of g p at the pole.

9. B For simple harmonic motion, acceleration isproportional to displacement.

a = –ω 2x Therefore when x is maximum, a is maximum.

10. D Path difference ∆x and phase difference ∆ϕ are relatedby

∆x ∆ϕ = — × 2π λ

(PR–QR) = 2π × ———— λ

11. D If there are oscillations, then the system is

underdamped. At critical damping, there will be nooscillations, and the system which is displaced willreach equilibrium in the shortest possible time.

12. C There is a node at R and antinodes at Q and S. Thisis possible for a wavelength which equals the fulllength of the string.

Therefore, λ = 4x . 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 fundamentalc frequency f ' 0 = —–

2L. ∴ f ' 0 = 2f 0

14. B At the equilibrium separation r 0, the total energy is(potential energy + kinetic energy).

Potential energy is –ε and kinetic energy will dependon the temperature. If the temperature is absolutezero, then the kinetic energy will be zero, otherwisethe system will have positive kinetic energy which isless than ε . Thus the total energy will have a negativenon-zero value.





13



15. A Glass is brittle. It obeys Hooke’s law and in its stress-strain graph there is no plastic region. It breaks whenthe elastic limit is exceeded.

16. D nm represents the density of the gas which ismass

——— volume. Mass of gas = number of molecules

× mass per molecule. Mass ∴ ——— = number of molecules per unit volume

Volume × mass per molecule.

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

18. A The change from X to Y takes place at constantvolume. For the pressure to increase at constantvolume, the temperature must increase. (Pressurelaw, p ∝ T).

In the T -V graph, constant volume is represented bya vertical line XY which is parallel to theT -axis.

The change YZ is isothermal, meaning that thetemperature remains constant. Therefore in the T -V graph, constant temperature is represented by ahorizontal line YZ which is parallel to theV -axis.

Graph A satisfies both these two conditions.

dQ dθ

19. B —– = – kA — dt dx dQ At steady state, — and k are constant throughout the

dt rod.

dθ 1 ∴ —–∝ – — (A = cross-sectional area of the rod) dx A Ignoring the negative sign and considering magnitudes

only, as A increases, the gradient of the graphdecreases.

From the diagram, as x increases, A increases,and therefore the gradient of the graph shoulddecrease.

B is the only graph that shows a continuouslydecreasing magnitude of its gradient.

20. C For an isothermal change, TVγ–1 = constant. For diatomic gas, γ = 1.4, γ – 1 = 0.4

(V 1)0.4 T 2 —–— = — (V 2)0.4 T 1

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

21. D Electric force = gravitational force Q2 GM2 –—— = —– 4πε r 2 r 2

Q = 4πε G √ × 3 × 105

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

= 2.6 × 10–5 C

22. B The combined capacitance is given by1 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 capacitorsbecause they are in series.

ε A

23. C C = —– d 1 U = —CV2

2

ε A = —–V2

2d V is constant because the capacitor is connected to

the power supply throughout.It is assumed that the material in the capacitorcontinues to completely fill the space betweenthe plates as the plates are separated. Hence ε isconstant.

1 ∴ U ∝ —

d

Since d 2 = 2d 1, U1 d 2 — = — = 2 U2 d 1

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

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

26. A Let the potential of the negative terminal of the cellbe 0 V and that of the positive terminal be 12 V. Inthe loop consisting of the cell and the three 2 Ω resistors in series, the potential drops by 4 V foreach 2 Ω resistor. Therefore the potential at Y is12 – 4 = 8 V.

In the loop with the cell and two 6 Ω resistors, thepotential drops by 6 V for each 6 Ω resistor. Thereforethe potential at X is 12 – 6 = 6 V.

Therefore the potential difference between X and Y= 8 – 6 = 2 V.

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





14



28. D For the charge to move at uniform velocity, themagnetic force = electric force and B and E must beperpendicular to each other.

Bev = Ee potential difference 1000 E = ———————— = ———— = 5 × 104 V m–1

separation of plates 2.0 × 10–2

5 × 104 v = ———— = 5 × 10 6 m s –1

1.0 × 10 –2

29. C Electric force =Eq and magnetic force = Bqv.

30. D Induced e.m.f. in the secondary coil by mutualinduction is given by

d I E = M— = M 2π f I 0 cos 2π f t dt

( I = primary current = I 0 sin 2π f t)

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. dependson the speed of rotation. Thus in the equation above,with E and r being constant, when E b increases, Idecreases.

132. D Reactance of a capacitor, X c = ——.2π fC

1 Thus, X ∝ —. D shows this inverse proportion. C

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.34. C Electromagnetic waves are transverse waves in which

the electric and magnetic fields and the velocity ofpropagation are mutually perpendicular.However, the velocity lies in the plane of vibration ofthe electric field, and also in the plane of vibration ofthe magnetic field. This situation is best describedby option C.

1 1 135. C The lens formula is — + — = —. u v f Multiplying throughout byu ,

u u 1 + — = — v f u u — = — – 1

v f u Thus the graph of — against u is a straight line with av u negative intercept on the — axis. v Option C is the best answer.

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

37. C After passing through the first polaroid, the intensity1 becomes I ' = — I 0. 2

On passing through the second polaroid, theamplitude of the emerging wave is A cos θ , whereA is the amplitude of the incident wave. Thereforethe intensity of the emerging wave which isproportional to the square of the amplitude becomes

1 I ' cos 2 θ = — I 0 cos 2 θ 2

38. C For a non-reflective coating when the refractive indexn of the coating material is less than the refractiveindex of the glass,

λ

nt = — (t = minimum thickness) 4 λ 565 nm ∴ t = — = ———— = 102 nm 4n 4 × 1.38

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, — is more for optiona 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, — is more for optiona 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

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 ∴ — = —. λ c Force = change of momentum per second.

If we assume that each photon that strikes the surfacestops and does not bounce back, the change of

h momentum per photon = —. λ P Total change of momentum per second, F= nh —–

λ = —. c

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

–13.6 E 2 = –—— = –3.4 eV 22





15



Energy absorbed = energy of photon = E 2 – E 1 = –3.4 – (–13.6) = 10.2 eV.

43. A Radiation is emitted when electrons undergotransitions from higher energy levels to lower energylevels, and not when they are in an allowed orbit.

hc 44. A λ min = –—.

eV When V increases, λ min decreases.

45. D Laser is Light Amplification by Stimulated Emissionof Radiation.

46. D The radioactive decay takes place to increase thestability of the nucleus. Since C decays into N, theN nucleus is more stable than the C nucleus. Ifa nucleus is more stable, the binding energy pernucleon is higher. Therefore N has a higher bindingenergy 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 Yis needed so that its rate of decay equals that of X.Thus the ratio of the number of nuclides of Y to Xwill be large.On the other hand, if the half-life of Y is small, asmaller quantity of Y is enough for its rate of decayto be equal to that of X. Then the ratio of Y to X willbe small.

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 of14.01118 u – 14.003074 u = 8.106 × 10–3 u.

Thus energy will be released when the reaction takesplace.

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

50. B At distances exceeding the diameter of a nucleus,the nuclear forces are very weak because nuclearforces are short-ranged forces. Electromagnetic andgravitational forces are long-ranged forces. Betweenthese two, electromagnetic forces are stronger.

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

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

[Y] = [k λ x v y ρ z]

ML–1T–2 = Lx (LT–1) y (ML–3)z

= Lx + 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 = kv2

ρ

2π 2. Period, T = — = 2π m — k ω

k

ω 2 = — m Velocity, v = ω A2 – x 2

A 1 1 A2

Whenx = —, kinetic energy = —mv2= — m ω2 A 2 – — 4 2 2 16

1 k 15A2 15 = —m — —— = —kA 2

2 m 16 32

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 themaximum amplitude.

Amplitude of vibration increases from a node to anantinode.

1 1.45× 10–3

(b) Fundamental frequency, f 0 = — T — m (m = —–——

2l 0.500 = 2.90 × 10–3 kg m–1) Tension, T = 4f 2

0 l2m = 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 = L3ρ) 8360 (θ – 7) = ———————— (0.020)3 (8900)(385) θ = 312 °C

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

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

6. (a) (i) Shunt: A conductor connected across theterminals of a galvanometer.

(ii) Function: So that most of the current bypasses

the galvanometer. (b)

I GR 1

G

R

I S

I





16



Connect the shunt of resistance R in parallel to thegalvanometer of resistance R 1, where R <<R 1.

R Current in galvanometer, I G = —–— I << I .

R + R

7. (a) Continuous spectrum is produced when fastelectrons from the cathode are decelerated oncollision with the target. The decrease in energy ofthe decelerated electrons is radiated as photons inthe continuous spectrum.

Characteristic X-ray is produced when a vacancyin the inner shell (e.g. K-shell) of the target atomis filled by an electron from a higher shell. Thedifference in energy of the electron is radiated asa characteristic X-ray photon.

hc (b) eV = —— λ min

(6.63 × 10–34)(3.00 × 108) λ min = ————————–—— = 6.22 × 10–11 m (1.60 × 10–19)(20 × 103)

8. (a) Due to the very high temperature (>107 K) in theSun,

• hydrogen atoms travel at high speed (or havesufficiently high kinetic energy)

• to overcome repulsive electrostatic force• fusion between hydrogen atoms occurs.

During each fusion, the energy equivalent of themass defect is radiated.

(b) If P = power radiated from the Sun, then

P intensity, I = —— 4π R2

P = (4π R2) I Energy radiated in a year

= Pt = (4π R2) I t = (4π )(1.50 × 1011)2(1340)(3.15 × 107) = 1.19 × 1034 J

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 roughsurface, the decrease in the kinetic energy equals

the increase in the random kinetic energy of theatoms or heat in the object.(c) Gain in kinetic energy = loss in potential energy

1 —mv2 = mgh

2 Velocity, v = 2gh

= 2(9.81)(1.0)

= 4.43 m s –1

(d)

Length of incline = 12 + 32 = 10 m (i) Friction =µ R = µ mg cos θ Work done against friction

= (µ mg cos θ )( 10 )

= (0.2)(5.0)(9.81)3.0—–10 10

= 29.4 J (ii) Net work done = net force× displacement = (mg sin θ – µ mg cos θ ) 10

= (5.0)(9.81)1.0—–10

– (0.2)(5.0)(9.81)3.0—–10

(10

)

= 19.6 J (iii) Gain in kinetic energy = net work done

1 — mv2 = 19.6 J 2

v = 2 × 19.6———–

5.0 = 2.80 m s –1

10. (a) (i) Diffraction: Spreading of a wave after passinga narrow slit or obstacle.

(ii) Points on the wavefront that arrive at the slitbehave as point sources emitting sphericalwavelets in phase. Along directions where

constructive interference occurs, light of highintensity is detected. No light is detected alongdirections where destructive interferenceoccurs.

(iii)

(b) (i)

If P is the m th minimum, divide the slit into 2m equal parts.

a Distance between two point sources A and B = –—.

2m

Intensity

Diffraction angle, θ 0

1.0 m

3.0 m

mg sin θ θ

mg cos θ θ

R

µ R

mg

θ

a —– 2m

P, m th minimum

B

Slit

CN

A





17



Wavelets from A and B produce destructiveinterference at P when

1 BP – AP = BN = —λ 2 Angle PBC = angle BAN =θ BN In triangle ANB, sinθ = —– AB AB sinθ = 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 e For electron microscope, resolution θ e = — a Since λ e << λ , θ e<< θ. Hence compared to the optical microscope,

the electron microscope is able to distinguishclearly two points that are much closer.

dQ 11. (a) (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 = — = ————–L 3.34 × 10 5

= 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 θ – 0 1 For brass, —–— ∝ —— (2) 0.50 109 100 – θ 109

(1)— ,(2) ——–— = ——

θ 205 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

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

12. (a) (i)

(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

eEx2 y = ——– 2mv 02

(iii) Parabolic path (iv) When x = 5.0 cm = 0.050 m, y = 1.50 cm

= 0.0150 m

eEx2

y = —–— 2mv o2

Initial velocity,v 0 =eE

x —— 2my

= (0.050) (1.60 × 10–19)(500)—————————–2(9.11 × 10 –31)(0.0150)

= 2.71 × 106 m s –1

(b) (i) Kinetic energy increases, electric potentialenergy 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) = —mv2

2 mv2 (V 1 – V 2) = —– 2e (1.67 × 10–27)(2.0 × 106)2

= —————————— = 2.09 × 104 V 2(1.60 × 10–19)

13. (a) (i) Stopping potential: Minimum reverse potentialthat is required to stop all photoelectrons fromreaching the anode.

(ii) • Existence of threshold frequency, which is theminimum frequency of the electromagneticradiation that would eject photoelectronsfrom a metal, or the existence of workfunction which is the minimum energyrequired for a photoelectron to escape froma metal surface.

• Instantaneous emission of photoelectrons. • Photoelectric current increases when the

intensity of electromagnetic radiationincreases.

E Path

–

+





18



• Maximum kinetic energy of photoelectrondoes not depend on the intensity of theelectromagnetic radiation but increaseswhen the frequency of the radiationincreases.

hc (b) (i) Work function = —— λ max

(6.63 × 10 –34)(3.00 × 108) λ max = —————————–— = 5.81 ×10–7 m 2.14(1.60 × 10

–19

) hc (ii) Maximum kinetic energy,K max = — – W λ (6.63 × 10–34)(3.00 × 10 8) = —————————–— – 2.14 = 0.610 eV (452 × 10 –9)(1.60 × 10 –19) (c) (i) Power = intensity × area = (120)(1.0 × 10–4) = 0.0120 W

hc (ii) Power = N —

λ (0.0120)(550 × 10 –9) N = —————————— (6.63 × 10 –34)(3.00 × 108) = 3.32 × 1016 photons s –1

14. (a) Half-life: Time taken for half the number of radioactiveatoms in a sample to decay, or the time taken forthe number of radioactive atoms in a sample todecay to half of its initial number.

dN – — dt

rate of decay of aradioactive sample

Decay constant = —— = —————————— N number of radioactive

atoms in the sample

(b) (i)22688Ra →

22286Rn +

4

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 ofγ -ray: Emissions froma radioisotope in the body can be detected by adetector that is outside the body.

• Produce ionisation: Used in radiotherapy todestroy cells.

• A radioactive isotope retains its activity whatevercompound its atoms are attached to: Used asradioactive tracer.

(d) (i) 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

http://e u.joshuatly.com

physics stpm past year questions with answer 2006

Documents