stpm physics 2008

Actual 2008 STPM Physics Examination Paper 1Majlis Peperiksaan Malaysia 2007

PAPER 1 Time: 1 h 45 min

1 A fluid flows through a pipe of diameter d and length l. The volume flow rate R is given by

cd 4∆pR = ————— ,

ηl

where c is a dimensionless constant, ∆p the pressure difference between the two ends of the pipe and η the viscosity of the fluid. The unit of η in terms of base units isA kg m B kg s–2 C kg m–1 s–1 D kg m–3 s–1

2 The figure shows two tugs P and Q pulling a ship R from a point O to X.1.1

P

OR X

Q

15o

P pulls with force 6.0 � 104 N at an angle of 15° to OX and Q pulls with force 3.6 � 104 N at angle θ to OX. The value of θ isA 8.9° B 15.0° C 24.1° D 25.6°

3 A gun fires 120 bullets per minute at a speed of 200 m s–1 relative to the gun. If the mass of each bullet is 10 g, the recoiling force acting on the gun isA 2 N B 4 N C 240 N D 4000 N

4 Two cars X and Y moving at the same speed are acted upon by equal braking forces until they come to a stop. If the mass of X is twice that of Y, the ratio of the distance travelled by X to the distance travelled by Y isA 1:2 B 1:1 C 2:1 D 4:1

5 The figure shows a car travelling at a constant speed along the road JKLMN.

NM

L

K J

Car

Ace Ahead Physics (Actual 08) 5t1 1Ace Ahead Physics (Actual 08) 5t1 1 2/11/09 10:47:30 AM2/11/09 10:47:30 AM

2 Actual 2008 STPM Physics Examination Paper Majlis Peperiksaan Malaysia 2007

At which part of the journey does the car have the greatest acceleration?A JK B KL C LM D MN

6 The figure shows a rod pivoted at point P on a smooth horizontal surface.

30o

30o

5.0 N

5.0 N

2.0 m

3.0 m

P

Two forces each of magnitude 5.0 N acting in opposite directions are applied at the two ends of the rod. The resultant torque on the rod isA 2.5 N m B 12.5 N m C 21.7 N m D 25.0 N m

7 A tyre is fixed to an alloy rim and another identical tyre is fixed to an iron rim. The two rims are of the same dimensions. When the wheels are rotated from rest, it is found that it is relatively easier to rotate the wheel with the alloy rim. This is becauseA the mass of the alloy rim is smallerB the mass of the alloy rim is more uniformly distributedC the material of the alloy rim is denserD the material of the alloy rim is stronger

8 The figure shows three point masses each with mass m at vertices of an equilateral triangle with the length of its side equal to x.

xx

x mm

m

The potential energy of the system is Gm2 Gm2 2Gm2 3Gm2

A –——— B –——— C –———— D –———— 3x 2x x x

9 The figure shows a uniform rod of weight W placed on a smooth hemisphere with one end of the rod in contact with the floor.

Which figure shows the correct directions of the forces acting on the rod?A B

W W



C D

W

10 The graph shows the variation of displacement x against time t produced by the oscillation of a simple pendulum.

x/cm

t /s

5

02 4 6 8 10

–5

Which statement is true?A The velocity at 6 s is zero. C The acceleration at 4 s is maximum.B The period of oscillation is 6 s. D The kinetic energy at 2 s is minimum.

11 The graph shows two progressive waves y1 and y

2 propagating in opposite directions at the time

t = 0 s.

The speed of both waves is 3 cm s–1. Which of the following graphs shows the resultant wave at the time t = 1 s?

A

B

y /cmy1 y2

1

0

–1

1 2 3 4 5 6 7 8 9 10 11 12 x /cm

y /cm

1

0

–1

–2

1 2 3 4 5 6 7 8 9 10 11 12 x /cm

y /cm

1

2

0

–1

–2

1 2 3 4 5 6 7 8 9 10 11 12 x /cm

W



C

D

12 The displacement of a particle moving in simple harmonic motion is given by x = 5 sin 2t, where x is in centimetres and the time t in seconds. If the period of the motion is T, the acceleration of the

Tparticle at t = — s is

6

A –17.3 cm s–2 B –10.0 cm s–2 C 10.0 cm s–2 D 17.3 cm s–2

13 The figure shows a standing wave in a string fixed at both ends.

Which statement is not true?A The string vibrates the air around it.B The air vibrates at the same frequency as that of the standing wave.C The sound produced has the same wavelength as that of the standing wave.D The standing wave is the superposition of a propagating wave and its reflection.

14 A cyclist approaches a songbird on a tree at a speed of 10 m s–1. If the bird sings at a frequency of 420 Hz, what is the frequency heard by the cyclist? [The speed of sound in air = 343 m s–1]A 396 Hz B 408 Hz C 432 Hz D 445 Hz

15 The figure shows the positions of particles in a medium at a particular instant when a sound wave of frequency 1500 Hz passes through it.

The speed of sound in the medium isA 75 m s–1 B 150 m s–1 C 300 m s–1 D 330 m s–1

16 A metal has molar mass M, density ρ and atomic separation d. The Avogadro’s number is given by

Md 3 Mρ d 3 MA ——— B ——— C ——— D ———

ρ d 3 Mρ ρd 3

17 Three wires P, Q and R of different materials are stretched. P undergoes plastic deformation before it snaps. Q snaps without undergoing plastic deformation. R shows only a small extension with a large tensile force. Which characteristics describe P, Q and R?

y /cm

1

0

–1

–2

1 2 3 4 5 6 7 8 9 10 11 12 x /cm

y /cm

1

2

0

–1

–2

1 2 3 4 5 6 7 8 9 10 11 12 x /cm

0 5 10 15 20 25 30 35 40 45 50 55 x /cm



A

B

C

D 18 The r.m.s. speed of the molecules of an ideal gas which has volume V at pressure p is directly

proportional to p

A pV B pV C —— D V

p—V

19 A box contains an ideal gas at 27 °C and 2.0 � 10–6 Pa. The number of gas molecules per unit volume isA 1.1 � 108 m–3 B 1.2 � 109 m–3 C 4.8 � 1014 m–3 D 5.4 � 1015 m–3

20 The graph shows the variation of pressure p with volume V for a mass of gas.

What happens to the heat transfer, internal energy and work done when the gas changes from states R to S?

A

B

C

D

21 The graph shows the variation of pressure p with volume V for 0.04 mol of a diatomic gas.

Heat transfer

Supplied to gas

Supplied to gas

Released by gas

Released by gas

Internal energy

Increase

Decrease

Increase

Increase

Work done

By gas

On gas

On gas

By gas

P Q R

Ductile

Ductile

Rigid

Rigid

Brittle

Rigid

Brittle

Ductile

Rigid

Brittle

Ductile

Brittle

S

R

p

V

p /105 Pa

20.2X

Y1.1

0 1.0 8.0 V /10–4 m 3



The gas expands adiabatically from state X to state Y. The internal energy change in the process XY isA –7130 J B –4280 J C –285 J D –171 J

22 A wooden box of surface area 1.0 m2 and wall thickness 2.0 cm contains 20.0 kg of ice at 0 °C. The specific latent heat of fusion of ice is 3.3 � 105 J kg–1 and the thermal conductivity of wood is 0.50 W m–1 K–1. If the room temperature is 33 °C, the time required for all the ice in the box to melt isA 825 s B 863 s C 4000 s D 8000 s

23 The figure shows a metal rod wrapped with an insulating material.

Which graph shows the variation of temperature with distance along the rod if both ends of the rod are maintained at constant temperatures?A B

P Q R S Distance

Temperature

P Q R S Distance

Temperature

C D

P Q R S Distance

Temperature

Temperature

P Q R S Distance

24 The figure shows the electric field lines for a positive point charge and a nearby negative point charge that are equal in magnitude.

P Q

Insulation

R S

R

Q P+ –



The electric field strengths at P, Q and R are EP, E

Q and E

R respectively. Which arrangement

shows the correct relative electric field strengths?A E

P > E

Q > E

R

B EP > E

R > E

Q

C EQ > E

P > E

R

D EQ > E

R > E

P

25 The figure shows a spherical Gaussian surface of radius 20.0 cm enclosing a charge +2 C.

+2 C

Gaussian surface

The electric flux through the Gaussian surface isA 4.43 � 10–12 N m2 C–1

B 1.77 � 10–11 N m2 C–1

C 2.26 � 1011 N m2 C–1

D 4.50 � 1011 N m2 C–1

26 A charged capacitor is connected in series to a switch and a resistor. A voltmeter of high resistance is connected across the capacitor. The switch is closed, and a set of values for the voltage V and the time t is obtained. Graph ln V against t plotted is as follows:

The time constant of the circuit isA 0.1 s B 2 s C 10 s D 20 s

27 Which arrangement of four identical capacitors produces the biggest capacitance between points X and Y?

A X Y B

X Y

In V

2

1

010 20 t (s)



C D X Y

X Y

28 Which statement is not true of the model used to explain the electrical conduction in a metal at the microscopic level?A The kinetic energy for the random motion of the free electrons depends on the temperature of

the metal.B The excess energy acquired by the free electrons in the electric field is lost during collisions

with the lattice atoms.C The mean free time decreases as the temperature of the metal increases because the lattice

atoms vibrate more strongly.D At any instant, all the free electrons are moving with the same drift velocity because the

electrons are accelerated by the same electric field.

29 A battery with e.m.f. ε1 and internal resistance r

1 is connected to a battery with e.m.f. ε

2 and

internal resistance r2 in the circuit shown.

W

Z

X

Y

R1

R3I2

I1R2

Which of the following is true of the closed loop WXYZW?A ε

1 + ε

2 = I

1(R

1 + R

2 + r

1) + I

2(R

3 + r

2)

B ε1 – ε

2 = I

1(R

1 + R

2 + r

1) – I

2(R

3 + r

2)

C ε1 + ε

2 = I

1(R

1 + R

2 – r

1) + I

2(R

3 – r

2)

D ε1 – ε

2 = I

1(R

1 + R

2 – r

1) – I

2(R

3 – r

2)

30 Which of the following is true of the internal resistances of a voltmeter and an ammeter?

A

B

C

D

Voltmeter

High

High

Low

Low

Ammeter

High

Low

High

Low



31 The figure shows a 40 cm straight rod carrying a current of 20 A in a 2.0 T magnetic field.

The magnetic field acts in the horizontal plane at an angle of θ with the direction of the current. If the weight of the rod is 8.0 N and the magnetic force is just enough to balance the weight of the rod, what is the angle θ?A 30° B 60° C 120° D 150°

32 The figure shows the trajectories of two charged particles P and Q directed perpendicularly into a region of uniform magnetic field with the same velocity.

If the radii of the trajectories are the same, what are P and Q?

A

B

C

D

33 The figure shows a uniform magnetic field in a rectangular region and a copper ring moving through positions 1 to 4.

1

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

2 3 4

Which statement is not true of the changes experienced by the ring?A At position 1, there is no change in magnetic flux and no current is induced.B At position 2, there is an increase in magnetic flux and a current is induced.C At position 3, there is a constant magnetic flux and a constant current is induced.D At position 4, there is a decrease in magnetic flux and a current is induced.

P

α-particle

Electron

Electron

Positron

Q

Electron

α-particle

Positron

Electron

20 A

B = 2.0 T

P

Q

X X X

X X X

X X X

X X X

X X X



34 The figure shows the side view of a circular coil of N turns and area A in a magnetic field B.

Circular coil

B

If the normal to the plane of the coil makes an angle α with the magnetic field, the total magnetic flux through the coil isA BA sin α B BA cos α C NBA sin α D NBA cos α

35 In a circuit, an alternating current of r.m.s. value 1 A passes through a resistor. In another circuit, a steady current of magnitude I passes through an identical resistor. If the two resistors dissipate heat at the same rate, the value of I isA 0.71 A B 1.00 A C 1.41 A D 2.00 A

36 The figure shows an operational amplifier circuit.

Which graph represents the variation of the output voltage Vo with the input voltage V

i?

A B Vo /V

Vi /V

+10

–2.5 2.50

–10

Vo /V

Vi /V

+10

–2 20

–10

C D Vo /V

Vi /V

+10

–2.5 2.50

–10

Vo /V

Vi /V

+10

–2 20

–10

+

+10 V

–10 V

-

VoVi



37 Which figure shows an oscillator circuit?A B

+9 V

–9 V

V1

V2

Vo

-

+

+9 V

–9 V VoVi

-

+

C D

Vi Vo

+9 V

C

R

–9 V

-

+

Vo

+9 V

–9 V

-

+

C R

R

RCR

38 The figure shows the symbol of an operational amplifier.

Which terminal and label do not correspond?

A

B

C

D

39 A concave mirror has a radius of curvature r. If an object is placed at a distance 2r from the mirror, the linear magnification of the image is

1 1A — B — C 1 D 3 5 3

40 The distance between an object and a screen is 100 cm. When a lens is placed in between the object and the screen, an image which is real and of the same size as the object is formed on the screen. Which lens is used?A A concave lens with a focal length of 25 cmB A concave lens with a focal length of 50 cmC A convex lens with a focal length of 25 cmD A convex lens with a focal length of 50 cm

Terminal

P

Q

R

S

Label

Inverting input

Non-inverting input

Feedback output

Signal output

Q

P

R

S

-

+



41 The figure shows monochromatic light of wavelength λ which is illuminated nearly normal onto a thin soap film of thickness t and refractive index n.

t

Air

Air

Soap

The optical path difference between the reflected rays from the upper and lower surfaces of the soap film is

2t 2t λ λA — B 2nt C — + — D 2nt + —

n n 2 2

42 In the photoelectric effect, the energy of the photon is usedA to release the electron from the lattice onlyB as the kinetic energy of the electron onlyC to release the electron from the lattice and as the kinetic energy of the electronD by the electron to produce a new electron called photoelectron

43 If λ is the wavelength of a particle with momentum p, which graph has a gradient equal to the Planck constant?A B

C D

p

0

p

10

p

10

p

0



44 The figure shows some of the energy levels of an atom.

E4

E3

E2

E1

The maximum number of spectral lines produced by the transition of electrons from these four energy levels isA 3 B 4 C 5 D 6

45 The graph shows the X-ray spectra I and II produced by an X-ray tube when it is operated at two different voltages.

Relativeintensity

Wavelength

I

II

0

Which statement explains why the minimum wavelength of spectrum II is longer than that of spectrum I?A A lower voltage is used to produce spectrum II.B A higher voltage is used to produce spectrum II.C A lighter element is used as the target material to produce spectrum II.D A heavier element is used as the target material to produce spectrum II.

46 Which statement is not true of stimulated emission?A Photons emitted by stimulated emission are in phase.B Photons produced by stimulated emission are of different frequencies.C Radiation produced by stimulated emission in a laser is monochromatic.D Radiation produced by stimulated emission in a laser has high intensity.

47 The nucleus of iron 5626Fe is more stable than the nucleus of bismuth 209

83Bi. Which of the following is true of the more stable nucleus?A Higher mass per nucleonB Lower ratio of proton to neutronC Lower binding energy per nucleonD Higher binding energy per nucleon



PAPER 2 Time: 2 h 30 min

Section A [40 marks]

Answer all questions in this section.

1 The figure shows a ball kicked from a stage of height h with initial velocity u at angle θ with the horizontal.

(a) Write the equations of motion of the ball to show its horizontal and vertical positions at any time t after the ball is kicked. [2 marks]

(b) If h = 1.0 m, u = 8.0 m s–1 and θ = 30°, calculate the horizontal distance L of the ball from the stage when it strikes the ground. [4 marks]

2 A flywheel of radius 0.20 m with moment of inertia 0.15 kg m2 rotates at 180 revolutions per minute. A tangential force is applied on the rim of the flywheel and it stops after 12 revolutions.(a) Determine the initial angular velocity of the flywheel in rad s–1. [1 mark](b) Calculate the angular deceleration. [3 marks](c) Calculate the magnitude of the tangential force. [2 marks]

48 A radioactive element X decays to a radioactive element Y which then decays to element Z. If the sample initially contains X only, which of the following influences the ratio of the number of Y nuclei to the number of X nuclei after a certain time?A Surrounding pressure C Half-life of element YB Initial number of X nuclei D Half-life of element Z

49 When a slow neutron is captured by a stationary 11348Cd nucleus, a 114

48Cd nucleus is formed and a γ-ray photon is emitted. What is the wavelength of the γ-ray?[Mass of 1

0n = 1.0087 u, mass of 11348Cd = 112.9044 u, mass of 114

48Cd = 113.9034 u]A 2.05 � 10–23 m C 1.37 � 10–13 mB 6.63 � 10–16 m D 1.45 � 10–12 m

50 A strong force is responsible forA alpha decay C holding the nucleons togetherB beta decay D holding the electrons in the nucleus

h

L

x

y

u

Stage

0



3 A pipe of length 20 cm has one of its ends closed and the other end open. Determine its fundamental frequency and first overtone. [4 marks][Assume the speed of sound to be 345 m s–1]

4 The table shows two measured quantities for aluminium and copper.

Material Young’s modulus (N m–2) Maximum tensile strength (N m–2)

Aluminium

Copper

69 � 109

110 � 109

110 � 106

250 � 106

(a) Based on the quantities in the table, compare the physical properties of aluminium and copper. [2 marks]

(b) An aluminium wire of length 2.0 m and diameter 0.10 mm is subjected to a tensile force of 10.0 N. Calculate the extension produced, and state your assumption made in the calculation.

[3 marks]

5 In an a.c. circuit, the supply voltage is given by V = 240 sin 5000t and the current in the circuit is π

given by I = 0.480 sin �5000t – —�. 2(a) Determine the reactance of the circuit. [2 marks](b) Sketch the phasor diagram of V and I. [1 mark](c) State the electrical component in the circuit. [1 mark]

6 The figure shows an operational amplifier.

-

+

V1

Vo

V2

(a) State the typical value of the open-loop gain of an operational amplifier. [1 mark](b) State the condition for the output voltage V

o to vary linearly with the input voltage difference

(V2 – V

1). [1 mark]

(c) What are the advantages for an operational amplifier having high input impedance and low output impedance? [2 marks]

7 In a photoelectric experiment, light of wavelength 450 nm is incident on a metallic surface with work function 2.3 eV.(a) Determine the velocity of the most energetic electrons ejected from the surface. [4 marks](b) Calculate the stopping potential. [2 marks]

8 The initial number of atoms in a 2.0 g radioactive element is 6.0 � 1021. The half-life of the element is 10 hours.(a) Calculate the number of atoms which decay in 24 hours. [4 marks](b) Determine the mass of the radioactive element left after 24 hours. [1 mark]

Section B [60 marks]

Answer any four questions in this section.

9 (a) Define simple harmonic motion. [2 marks](b) The displacement x from the equilibrium position of a mass undergoing simple harmonic

motion is given by x = 0.50 cos (2t + φ), where x is in metres and t in seconds. The initial velocity is – 0.20 m s–1. Calculate (i) the maximum velocity [2 marks] (ii) the maximum acceleration [2 marks] (iii) the phase angle φ . [3 marks]



(c) The bob of a simple pendulum is displaced and then released. (i) Show that the shortest time when the kinetic energy of the system equals its potential

energy is one eighth of the period. [4 marks] (ii) Deduce the subsequent time when the kinetic energy of the system equals its potential

energy again. [2 marks]

10 (a) (i) State two assumptions of an ideal gas. [2 marks] (ii) State two physical conditions under which a gas behaves as an ideal gas. [2 marks] (iii) A 0.35 m3 gas tank contains 7.0 kg of butane gas. Assuming that the gas behaves as an

ideal gas, calculate its pressure at 27 °C. [3 marks][Molecular mass of butane is 58 g mol–1]

(iv) Butane gas normally behaves as a real gas. The actual pressure of the butane gas is higher than the calculated value in (a)(iii). Give a reason. [1 mark]

(b) (i) What is meant by the degrees of freedom of a gas molecule? [1 mark] (ii) Write an expression relating the total kinetic energy E of a gas molecule to the number

of degrees of freedom f. Explain any other symbols used. [2 marks] (iii) The escape velocity of Mars is 5.0 � 103 m s–1. If the temperature of Mars is 300 K,

determine whether oxygen gas can exist on the planet. [4 marks][Molecular mass of oxygen is 32 g mol–1]

11 (a) Two thin conducting plates have an area of 0.50 m2 each. They are placed parallel to each other and 25 mm apart. One plate is maintained at +75 V while the other at –75 V by a d.c. supply. (i) Determine the amount of charge stored on each plate. [4 marks] (ii) Calculate the energy stored in the electric field between the plates. [2 marks]

(b) The figure shows a simple circuit of the photographic flash used in a camera.

Battery

Switch

Flash bulb

The capacitance of the capacitor is 40.0 μF, and the resistance of the resistor is 45.0 kΩ. (i) Explain the function of the capacitor in this application. [4 marks] (ii) Calculate the time required to charge the capacitor to 63% so that a good flash can be

obtained. [4 marks] (iii) Suggest a way to reduce the charging time of the capacitor. [1 mark]

12 A thin biconvex lens made of glass with refractive index 1.5 has surfaces with equal radius of curvature 15.0 cm. An object is placed 20.0 cm on the principal axis to the left of the lens.(a) Determine

(i) the focal length of the lens [2 marks] (ii) the image position [2 marks] (iii) the magnification [2 marks] (iv) the nature of the image. [2 marks]

(b) A plane mirror is then placed perpendicularly to the principal axis of the lens with its reflecting surface in contact with the right hand side of the lens. (i) Show, on the same diagram, the image formed by the lens and the subsequent image

formed by the mirror. [2 marks] (ii) Determine the final position of the image formed by the combination of the lens and the

mirror. [3 marks] (iii) Calculate the overall magnification. [2 marks]



13 (a) Explain briefly the stimulated emission process and the population inversion in the production of a laser. [5 marks]

(b) State two differences between the laser and fluorescent light. [2 marks](c) A laser produces a bright beam of light of wavelength 450 nm. Calculate the energy of a

photon in eV. [3 marks](d) Calculate the wavelength of the light emitted when a hydrogen atom makes a transition from

the energy level n = 4 to n = 2 according to the Bohr model. [5 marks]

14 (a) Define the binding energy of a nucleus and the mass defect of a reaction. [2 marks](b) Determine the binding energy per nucleon of tritium 3

1H. [4 marks][Mass of proton = 1.007276 u, mass of neutron = 1.008665 u, mass of tritium = 3.016049 u, 1 u = 931.5 MeV]

(c) The graph shows the variation of the binding energy per nucleon against the mass number.

(i) Determine the most stable nucleus. [1 mark] (ii) Estimate the energy released for the fusion reaction

21H + 2

1H → 42He [4 marks]

(d) Explain the working principle of a mass spectrometer to measure the masses of ions. [4 marks]

9Binding energyper nucleon/MeV

8

7

6He

C Ni

5

4

3

2

1

0

1

2H

50 100 150 200 250

Mass number

24

6 28

62

U92

238

12



SUGGESTED ANSWERS

PAPER 1

cd 4∆p 1. C: R = ————— ηl

cd 4∆pη = ————— Rl 1

Unit of η = m4(N m–2)——————— (m3 s–1)m 1

= m4(kg m s–2 m–2)—————— (m4 s–1)

= kg m–1 s–1

2. D:

Since there is no motion in the OY direction,Q sin θ = P sin 15°

Psin θ = — sin 15°

Q

6.0 � 104

= �———————— � sin 15° 3.6 � 104

θ = 25.6° 3. B: Recoiling force

= rate of change of momentum= (number of bullet s–1) � (change of

momentum per bullet) 120

= �——— �(0.010 � 200 – 0.010 � 0) N 60

= 4 N 4. C: Work done against braking force

= loss of kinetic energy 1

Car X: Fs1 = —(2m)(u2) (1)

2

1Car Y: Fs

2 = —(m)(u2) (2)

2

(1) s1—— : —— = 2

(2) s2

v2

5. B: Centripetal acceleration = —— r

Since speed v is constant, the centripetal acceleration is the greatest when radius of curvature of the track is the smallest, that is along KL.

6. B:

3.0 m

2.0 m

5.0 cos 30o

5.0 cos 30° P

5.0 sin 30°30°

5.0 sin 30°5.0 N

5.0 N

30°

No torque is produced by the horizontal component, (5.0 cos 30°). Resultant torque by the vertical component (5.0 sin 30°) of both the forces about the point P in the clockwise direction= (5.0 sin 30° N)(3.0 m) + (5.0 sin 30° N)(2.0 m)= 12.5 N m

7. A: Moment of inertia I is the resistance to change in rotational motion and depends on the mass and (radius)2 of the wheel.A wheel with a smaller moment of inertia is easier to set into rotation.The moment of inertia of the alloy wheel is smaller.Since both wheels have the same radius, the mass of the alloy wheel must be smaller.

8. D: Gravitational potential energy of the system

Gm2 Gm2 Gm2

= –——— + �(–——— ) + (–——— )� x x x 3Gm2

= –———— x

9. C:

B

R

O

F

A

W

Since the forces W, R and F are in equilibrium, their lines of action must pass through a common point, O.

10. C: Acceleration = –ω 2x Acceleration is maximum at

t = 4 s when x = –5 cm 11. A: After 1 s, y

1 is displaced 3 cm to the right,

and y2 is displaced 3 cm to the left as shown

in the graph below.

O

P sin 15o

Q sin

Y

15o

X

P

Q



Apply the principle of superposition of waves to obtain the resultant wave y = y

1 + y

2

12. A: x = 5 sin 2t dx

v = —— = 10 cos 2t dt

dv Acceleration = —— = – 20 sin 2t dt 2π�ω = 2 = —— � T

T When t = —,

6 T acceleration = – 20 sin 2�—� 6 2π T

= – 20 sin �—— � �—� T 6 = – 17.3 cm s–2

13. C: Wavelength of sound produced speed of sound

= ————————————— frequency, f

Wavelength of standing wave speed of travelling wave in the string= ——————————————————————————————

frequency, fSpeed of sound is not equal to the speed of travelling wave in the string.Hence the wavelengths are different.

14. C: Frequency heard by the cyclist, v + uof ′ = �————— � f v 343 + 10

= �———————— �(420) Hz 343

= 432 Hz 15. C: Wavelength, λ = distance between two

successive compressions= (25 – 5) cm

Speed = fλ = (1500)(0.20) m s–1

= 300 m s–1

16. D: There are A (Avogadro’s number) of atoms in 1 mole.

MVolume of 1 mole, V = ——

ρ M

A(d 3) = —— ρ M

A = ——— ρd 3

17. A: Ductile materials undergo plastic deformation before breaking.Brittle materials do not undergo plastic deformation before breaking.Rigid materials require a large force to produce a small extension.

18. B: R.m.s. speed is directly proportional to T , and pV = nRT.

Therefore r.m.s. speed is directly proportional

to pV . N 19. C: pV = �—— �RT N

A

N pNA—— = ———

V RT

(2.0 � 10–6)(6.02 � 1023)= ———————————————————— m–3

(8.31)(273 + 27)

= 4.8 � 1014 m–3 20. A: Since the volume of the gas increases, work

is done by the gas.On the graph, the point S is higher than R. Temperature of the gas increases, and its internal energy increases.Heat must be supplied to the gas to increase its internal energy and for the gas to do work.

21. C: Internal energy of a diatomic gas, 5

U = —nRT, pV = nRT 2

Change in the internal energy 5

= —nR(T2 – T

1)

2

5 5= —(pV)

2– —(pV)

1 2 2

5= —[(1.1 � 105)(8.0 � 10–4)

2– (20.2 � 105)(1.0 � 10–4)]

= – 285 J 22. D: (Rate of flow of heat) � (time taken)

= heat required to melt all the ice

dθkA�—— �(t) = ML

dx

MLTime taken, t = ————————

(dθ/dx)kA

(20.0)(3.3 � 105)= ————————————————————— s

[(33 – 0)/0.020](0.50)(1.0)

= 8000 s 23. D: Along PQ and RS, the rod is insulated,

temperature decreases linearly.Along QR, the rod is not insulated, temperature decreases non-linearly.

24. B: Electric field is stronger where the lines of forces are closer.

y1y /cm

x /cm

1

0

–11 2 3 4 9 10 11 12

y2

y = y1 + y2

x /cm

1

0

–2

1 2 3 4 5 8 9 10 11 126 76 7

5 6 7 85 6 7 8



25. C: Applying Gauss’ law, algebraic sum of electric charge enclosed

electric flux = ————————————————————— ε

0

2.0 = ——————————— N m2 C–1

8.85 � 10–12

= 2.26 � 1011 N m2 C–1

26. C: V = V0e–t/CR

tln V = ln V

0 – ——— which is the equation of

CRa straight line.

1Gradient of graph = – ———

CR 1

Time constant, CR = –—————————————— gradient of graph (20 – 0)

= –—————— s = 10 s (0 – 2) C 27. D: C

A = —

4 1 1 1

——— = ——— + ——— , CB = 0.4C

CB C/2 2C

C C C

C = — + — = C

2 2 C

D = 4C

28. B: Energy transferred to the lattice atoms is not lost but is dissipated as heat.

29. B: In the loop WXYZW,ε

1 is positive, but ε

2 is negative.

I1 is positive, but I

2 is negative.

Applying Kirchhoff’s Second Law,ΣE = Σ(IR)

ε1 – ε

2 = I

1(R

1 + R

2 + r

1) – I

2(R

3 + r

2)

30. B: A voltmeter has high resistance, and an ammeter has low resistance.

31. A: Magnetic force, FM

= W, weight of rod BIL sin θ = W

8.0 sin θ = ———————————— (2.0)(20)(0.40)

θ = 30° mv2

32. D: FM

= Bqv = ——— r q v

— = —— m rB

Since v, r and B for both the particles are the qsame, then the magnitude of — must be the msame for the particles.Positive charge is deflected upwards and negative charge downwards.P must be positively charged and Q negatively charged.

33. C: At position 3, since the magnetic flux is constant, there is no current induced.

34. D: Magnetic flux through the coil = N(B cos α)A 35. B: The r.m.s. current of 1 A is the effective

value of the a.c. and equals the steady current of 1.00 A.

36. A: Circuit is that of an inverting amplifier. R

fOutput voltage, Vo = – �—— �Vi R

i

100= – �——— �Vi 25

= – 4Vi

37. D: Oscillator circuit uses• C and R in series for feedback.• C and R in parallel for generating a.c.

38. C: Terminal R is connected to voltage supply. r 39. B: Focal length, f = —, object distance u = 2r 2 1 1 1

— = — + — f u v 1 1 1

— = ——— – —— v r/2 2r 2r

v = —— 3 v 2r 1

Linear magnification, m = — = ———— = — u 3(2r) 3 40. C: Real image is formed by a convex lens.

When image and object are of the same size, u = v = 2fu + v = 4f = 100 cm f = 25 cm

λ 41. D: Path difference of — occurs when light is 2

reflected by a denser medium, on the upper surface. The optical path of the ray reflected at the lower surface is greater by 2nt.

λTotal path difference = 2nt + —

2 42. C: Apply Einstein’s equation for photoelectric

effect. h 43. B: Momentum of photon, p = — λ 1

Gradient of the graph of p against — = h, λ

Planck’s constant 44. D: All the possible transitions are as shown in

the figure below.

45. A: Minimum wavelength λmin

is given by hc ——— = eV λmin

hc λmin = —— eV

λmin for II > λmin for I

VI < V

II

E4

E3

E2

E1



46. B: Photons emitted by stimulated emission have the same frequency.

47. D: Stable nuclei have higher binding energy per nucleon.

48. C: The ratio of the number of nuclei of Y to that of X depends on the half-lives of X and Y. The best option is C.

49. C: 11348Cd + 1

0n → 11448Cd + γ

hc112.9044 u + 1.0087 u = 113.9034 u + ——

λ hc

—— = 0.0097 u λ

= (0.0097)(1.66 � 10–27)c2

(6.63 � 10–34)λ = ———————————————————————————— m (0.0097)(1.66 � 10–27)(3.00 � 108)

= 1.37 � 10–13 m 50. C: Strong force holds the nucleons together.

PAPER 2

Section A

1.

h = 1.0 m

P

L

Q

u

y

xu cos

u sin

0

(a) At any time = t,• horizontal displacement, x = (u cos θ)t 1• vertical displacement, y = (u sin θ)t – —gt2

2(g = acceleration due to gravity)

(b) When the ball lands at the point Q, vertical displacement, y = – 1.0 mConsider vertical motion 1Using y = (u sin θ)t – —gt 2, 2

1 –1.0 = (8.0 sin 30°)t – —(9.81)t 2

2

9.81t 2 – 8.0t – 2.0 = 0

– (–8.0)± (8.0)2 – (4)(9.81)(–2.0)t = ——————————————————————————— s 2(9.81)

= 1.02 sHorizontal displacement,L = (8.0 cos 30°)(1.02) m

= 7.07 m 2. (a) Initial angular velocity,

ω0 = 180 rpm

180= �——— �(2π) rad s–1 = 18.8 rad s–1

60

(b) Using ω 2 = ω 20 + 2αθ,

0 – (18.8)2

angular acceleration, α = ———————— 2(2π)(12)

= – 2.34 rad s–2

Angular deceleration = 2.34 rad s–2

(c) Torque = Fr = Iα IαTangential force, F = —— r

(0.15)(2.34)= ————————— N

0.20

= 1.76 NAlternative methodLoss of rotational kinetic energy = work done by torque 1—Iω 2 = (Fr)(2πN) 2 (0.15)(18.8)2

F = —————————— N 4π(12)(0.20)

= 1.76 N 3.

λ0For fundamental mode, —— = 0.20 m

4 v —— = 0.20 m 4f

0

345Fundamental frequency, f

0 = ————— Hz

4(0.20)= 431 Hz

Frequency of first overtone, f1 = 3f

0

= 1293 Hz stress 4. (a) Young’s modulus = ———— strain

• Young’s modulus of copper is greater than Young’s modulus of aluminium.

• A piece of copper wire requires a greater stress to produce the same strain as a piece of aluminium wire.

• Therefore copper is more rigid than aluminium.

4

0.20 m

Fundamental mode

First overtone

4



Maximum tensile strength of copper is greater than that of aluminium.Hence the elastic limit of copper is higher than that of aluminium.

(b) Assuming that the extension is below the limit of proportionality,

stressYoung’s modulus, Y = ————

strain F /A

= ———— e / l

Extension, Fle = ——

YA (10.0)(2.0)

= —————————————————————— m (69 � 109)π(0.05 � 10–3)2

= 3.69 cm 5. (a) V = 240 sin 5000t and π

I = 0.480 sin (5000t – —) 2

Therefore, V0 = 240 V and I

0 = 0.480 A

V0Reactance, X = ——

I0

240= ———— Ω

0.480= 500 Ω

(b) Phasor diagram

(c) Since the voltage V leads the current I by π

— radians, the component is an inductor. 2 6. (a) Open-loop gain, A

0 = 105

(b) Condition for output voltage, V

o = A

0(V

2 – V

1):

A0(V

2 – V

1) � voltage of power supply

or voltage of power supply(V

2 – V

1) � ———————————————————

A0

(c) Advantage of high input impedance is that the op-amp requires a small current for it to operate.The value of the input voltage after connection to the op-amp is the same as before it is connected to the op-amp.Advantage of low output impedance is that the voltage across the component which is connected to the output is the same as the output voltage V

O.

7. (a) Using Einstein’s equation,maximum kinetic energy,

1 hc —mv 2

max= —— – W

2 λ (6.63 � 10–34)(3.00 � 108)

= —————————————————————— – 2.3(1.60 � 10–19) 450 � 10–9

= 7.4 � 10–20 J

vmax

= 2�——————————— �(7.4 � 10–20) m s–1

9.11 � 10–31

= 4.03 � 105 m s–1

(b) If Vs is the stopping potential, then

1 eV

s = —mv 2

max 2= 7.4 � 10–20 J

7.4 � 10–20

Vs = —————————— V

1.60 � 10–19

= 0.463 V 8. (a) Time, t = 24 h, T

1/2 = 10 h

24 x = —— = 2.4

10After 24 hours, number of atoms,

N0N = ——

2x

6.0 � 1021

= ———————— 22.4

= 1.14 � 1021

Number of atoms which decay in 24 hours= (6.0 – 1.14) � 1021

= 4.86 � 1021

(b) After 24 hours, number of atoms= 1.14 � 1021

Mass of (6.0 � 1021) atoms = 2.0 g

Mass of (1.14 � 1021) atoms

1.14 � 1021

= �—————————— �(2.0) g 6.0 � 1021

= 0.38 g

Section B

9. (a) A body is in simple harmonic motion if its acceleration is directly proportional to its displacement x from a fixed point andis constantly directed towards the fixed point.Acceleration = –ω 2x (ω 2 = constant)

(b) (i) x = 0.50 cos (2t + φ)

dxVelocity, v = —— = – 2(0.50) sin (2t + φ)

dt= – 1.00 sin (2t + φ)

Maximum velocity = 1.00 m s–1

dv (ii) Acceleration = ——

dt= – 2.00 cos (2t + φ)

Maximum acceleration = 2.00 m s–2

2rad

I

V



(iii) At time t = 0, velocity = – 0.20 m s–1

– 1.00 sin [2(0) + φ] = – 0.20 sin φ = 0.20 φ = 11.5°

(c) (i) If the bob is displaced and then released, the displacement of the bob after a time t is given by

x = x0 cos ω t

Kinetic energy = potential energy 1 1 —mω 2(x 2

0 – x2) = —mω 2x2

2 2 x

0 x = —— = 0.707x0 2

x0When x = x

0 cos ω t = —— ,

2 π ω t = — 4 π π t = ——— = —————— 4ω 4(2π/T) T

= — 8

(ii)

From the graph above, the kinetic energy and the potential energy are again equal when the distance of the bob from the equilibrium position is again 0.707x

0, which is at time

3T T Tt = —— which is — after t = —

8 4 8 10. (a) (i) Assumptions (any two)

• Internal energy of the ideal gas consists only of the total kinetic energy of the gas molecules.

• There is no force between gas molecules except during collisions.

• Collisions between gas molecules are perfectly elastic.

• Volume of the gas molecules is negligible compared to volume of the gas.

(ii) A gas behaves as an ideal gas when• the pressure is very low, or

approaches zero.• the temperature is low.

(iii) V = 0.35 m3, mass of gas, m = 7.0 kg, T = 27 °C = 300 K, molecular mass, M = 58 g mol–1

mUsing pV = �—— �RT,

M mRT

pressure, p = ———— MV (7.0)(8.31)(300)

= —————————————— Pa (58 � 10–3)(0.35)

= 8.60 � 105 Pa (iv) Because of the high pressure,

8.60 � 105 Pa, butane gas deviates from the behaviour of the ideal gas.

(b) (i) Degrees of freedom – independent modes of motion or independent modes of acquiring kinetic energy by the gas molecule.

(ii) Kinetic energy of a gas molecule, f

E = —kT 2

where k is Boltzmann’s constant and T is the temperature of the gas in kelvin.

(iii) Translational kinetic energy of oxygen molecule at 300 K,

1 3—mv2 = —kT

2 2Speed of oxygen molecule,

Mv = �m = ——— , kN

A = R�

3kT——— m N

A

= 3RT——— M

= 3(8.31)(300)—————————— m s–1

0.032= 483 m s–1

< 5.0 � 103 m s–1, the escape speed.Hence, oxygen molecules will not escape and can exist on Mars.

ε0A

11. (a) (i) Capacitance of capacitor, C = ——— d

Charge on each plate of capacitor,Q = CV

ε0A

= �——— �V d (8.85 � 10–12)(0.50)

= ——————————————— [75 – (–75)] C (25 � 10–3)

= 2.66 � 10–8 C (ii) Energy stored in the capacitor,

1U = —QV

2 1= —(2.66 � 10–8)(150) J 2

= 2.00 � 10–6 J(b) (i) The capacitor is charged by the battery.

Energy is stored in the electric field between the plates of the capacitor.When the switch is closed, the capacitor discharges.Energy stored in the capacitor is used to produce a flash in the flash bulb.

0

0.707x0

x0

x

T/8 T/4 T t

3T/8

–0.707x0



(ii) Voltage across the capacitor during charging,

V = V0(1 – e–t /CR)

When the capacitor is 63% charged,V = 0.63V

0

0.63V0 = V

0(1 – e–t /CR)

1et /CR = ————

0.37Time taken, 1t = (CR) ln �———— � 0.37

1= (40.0 � 10–6)(45.0 � 103) ln �——— � s 0.37= 1.79 s

(iii) The charging time can be reduced by using a resistor of resistance smaller than 45.0 kΩ.

12. (a) (i) r1 = r

2 = 15.0 cm

Focal length f of the lens is given by 1 1 1— = (n – 1)�— + —� f r

1 r

2

2= (1.5 – 1)�———— � 15.0

Focal length, f = 15.0 cm (ii) f = 15.0 cm, u = 20.0 cm

1 1 1Using — = — + —,

f u v 1 1 1

— = — – — v f u 1 1

= ——— – ——— 15.0 20.0

Image distance, v = 60.0 cm v

(iii) Magnification, m = — u 60.0

= ——— = 3.0 20.0

(iv) Image is real, inverted and magnified.(b) (i)

(ii) With the mirror behind the lens, light reflected by the mirror passes through the lens.The image I1 acts as a virtual object for the lens. u = – 60.0 cm, f = 15.0 cm

1 1 1— = — – —

v f u 1 1= ——— – ————— 15.0 – 60.0

Image distance, v = 12.0 cmFinal image I2 is 12.0 cm from the lens as shown in the above figure.

(iii) Overall magnification 60.0 12.0

= �———— ��———— � 20.0 60.0= 0.60

13. (a)

Stimulated emission occurs when • a photon passes by an excited atom.• an electron in the excited state is stimulated

to transit to a lower energy level.• a photon is produced by the transition of

the electron.• Condition: frequency of the photon that

stimulates the transition should be the same as the frequency of the emitted photon.

Population inversion occurs when • excited atoms of the pumping agent

collide with atoms of the lasing agent.• atoms of the lasing agent are raised to a

metastable state.• this results in more atoms in the excited

state than atoms in the ground state. (b) Differences between laser and fluorescent

light (any two)

Fluorescent lightLaser

Monochromatic

Coherent or in phase

Beam does not diverge

High intensity

Not monochromatic

Not coherent

Beam diverges

Low intensity

(c) Energy of photon

hc= ——

λ (6.63 � 10–34)(3.00 � 108)

= —————————————————————— eV (450 � 10–9)(1.60 � 10–19)

= 2.76 eV(d) In the hydrogen atom, for an electron to

orbit the nucleus, mv2 e2

centripetal force, ——— = ————— r 4πε

0r2

e2

v2 = —————— 4πε

0mr

O I2I1

First imageformed bylens

Lens

Final image

Mirror

Photon

Electron in excited state

Electron transition Photon

Stimulated photon



According to Bohr’s model, angular nhmomentum, (mv)r = ——

2π nh

m 2v 2r 2 = �——— �2

2π e2 nh

m 2�——————— �r 2 = �——— �2

4πε0mr 2π

h2ε0Radius of electron orbit, r = �————— �n2

πme 2

Energy of electron,E

n = kinetic energy + electric potential energy

1 –e2 e2

= —mv2 + �————— � �v2 = —————— � 2 4πε

0r 4πε

0mr

– e2 h2ε0= ————— �r = �———— �n2� 8πε

0r πme2

– me4

= —————— 8ε 2

0h2n2

For the transition from n = 4 to n = 2,energy of photon emitted, hc—— = E

4 – E

2 λ – me4 1 1

= ————— �—— – —— � 8ε 20h2 42 22

(9.11 � 10–31)(1.60 � 10–19)4

= ————————————————————————— 8(8.85 � 10–12)2(6.63 � 10–34)2

1 1��—— – —— � J

22 42

= 4.064 � 10–19 J

Wavelength, (6.63 � 10–34)(3.00 � 108)λ = ————————————————————— m

4.064 � 10–19

= 4.89 � 10–7 m 14. (a) Binding energy of a nucleus is the

energy required to completely separate the nucleons of the nucleus.Mass defect of a nucleus= (total mass before reaction) – (total mass after reaction)

(b) Mass defect = (mp + 2m

n) – m

N

= [1.007276 + 2(1.008665)] u – 3.016049 u

= 0.008557 uBinding energy = (0.008557)(931.5) MeV

= 7.97 MeV

7.97Binding energy per nucleon = ——— MeV

3= 2.66 MeV

(c) (i) The most stable nucleus is the nucleus with the highest binding energy per nucleon, that is 62

28Ni.

(ii) 21H + 2

1H → 42He

Energy difference in the =released binding energy

= 7 – 2(2 � 1) MeV

= 3 MeV

(d)

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

v

FE

S1

S2

S3

-FM +

Velocity selector– combined electricand magnetic fields

Photographic plate

Collimator slits

Source of positive ions

d1

d2

B Uniformmagnetic field

d3

Bainbridge mass spectrometer

• Ions from the source are collimated by the narrow slits S1 and S

2.

• In the velocity selector, mutually perpendicular electric field E and magnetic field B act. The selected velocity v is given by



FM

= FE

qBv = qE E v = —— B

• Ions with the selected velocity emerge from the slit S

3 and are deflected by

the magnetic field B. Ions of different masses follow different semicircular paths to strike a photographic plate.

mv2

——— = qBv r qB 2r Mass of ion, m = ———— EThe radii of the paths are obtained by measuring the diameters d

1, d

2 and d

3.


stpm physics 2008

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