Physics_STPM_Past_Year_Questions_with_answer_2007 by stpmform6

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

                                                                                                     Time: 1h 45 min

            1. The tensile strength of a wire is the maximum          4. What is the energy change that occurs when
               stress on the wire just before it breaks. What            a ball rolls down an inclined plane without
               is the dimension of tensile strength?                     slipping?
               A ML–1T–1               C MLT–1                           A Potential energy → translational kinetic
               B ML–1T–2               D MLT–2                                                   energy
                                                                         B Potential energy → rotational kinetic energy
            2. The figure shows two blocks X and Y, of                   C Potential energy → translational kinetic
               masses m and 2m respectively, placed on a                     energy + rotational kinetic energy
               smooth horizontal surface.                                D Potential energy → translational kinetic
                                                                             energy + work against friction
                                         m   2m

                               F                                      5. A car of mass 1250 kg accelerates from 0 to
                                         X   Y                           100 km h–1 in 4.0 s. The average power of
                                                                         the car is
                                                                         A 121 kW
                If the blocks are accelerated by a force F, the
                                                                         B 341 kW
                force exerted on block X by block Y is
                      1                      2                           C 484 kW
                A —F                    C —F                             D 681 kW
                      3                      3
                B —F  1                 D F
                      2                                               6. For a particle moving in a circle with a
                                                                         constant speed, the physical quantity which
            3. The figure shows the displacement–time                    is always constant is
               graphs of two vehicles P and Q.                           A displacement
                                                                         B acceleration
                         Displacement
                                                                         C linear momentum
                                                  P                      D angular momentum
                                                      Q

                                                                      7. A firework in the shape of a disc can spin
                                                                         freely about its axis. The radius of the
                                                                         disc is 2.0 cm and its moment of inertia is
                               0             T
                                                          Time           1.0 × 10–6 kg m2. When the firework is ignited,
                                                                         gas is ejected from the rim in the tangential
                Which statement describes the motion of the              direction at a constant speed of 50 m s–1
                two vehicles at time T ?                                 relative to the rim. The gas is ejected at a rate
                A P and Q are momentarily at rest.                       of 1.0 × 10–5 kg s–1. The angular velocity of
                B The speeds of P and Q are the same.                    the disc after 10 s of burning is
                C The speed of P is greater than the speed               A 10 rad s–1
                   of Q.                                                 B 50 rad s–1
                D The accelerations of P and Q are equal                 C 100 rad s–1
                   to zero.                                              D 200 rad s–1

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Ace Ahead STPM Physic vol 1 4th.indd 1                                                                               3/7/2008 4:48:24 PM
            8. Which figure shows that coplanar forces,                                A                         C
               each of magnitude F acting on a disc, are in
               equilibrium?
               A          F
                                    C
                                                                                       B                         D

                                                          F            F
                        F                                          F

                B                        F        D
                                                          F            F
                    F                                                              11. The equation of motion of an object performing
                                                                                       simple harmonic motion is given by
                                                                                                                  π
                                                                                                        x = 5 sin —t,
                                 F                        F            F                                          6
                                                                                       where x is the displacement in metres and t is
            9. The figure shows two stars X and Y of masses                            the time in seconds. When t = 1 s, the speed
               m1 and m2 respectively and separated by a                               of the object is
               distance r.                                                             A 2.3 m s–1             C 4.3 m s–1
                                                                                                   –1
                                                                                       B 2.6 m s               D 4.6 m s–1
                                                  r
                                         m1   P                   m2               12. The figure shows a mass attached to a spring
                            X                                              Y
                                                                                       made to oscillate vertically by an oscillator.
                                         r
                                         −
                                         3
                                                                                                                 Oscillator

                If the gravitational field strength at point P
                             r
                at distance — from X is zero, the ratio of
                             3
                m1 to m2 is                                                                                      Mass
                A 1:4
                B 1:2                                                                  If the damping is slightly increased, which
                C 2:1                                                                  graph shows the variation of amplitude a with
                D 4:1                                                                  frequency f of the oscillation?
                                                                                       A                      C
          10. The figure shows an object in space moving                               a                         a
              near a planet of mass M.                                                               Increased                Increased
                                                                                                     damping                  damping
                                                      v
                                     Object
                                                                                                         f                        f
                                                                                       0                         0
                                              r
                                                                                       B                         D
                                                                                       a                         a
                                Planet                        M
                                                              M
                                                                                                                              Increased
                                                                                                    Increased
                                                                                                                              damping
                                                                                                    damping

                                                                                                         f                        f
                The shortest distance between the object and                           0                         0
                the planet is r. The speed v of the object at
                                                                                   13. What is the phase difference between two
                the nearest position is 1.5 GM , where G is the
                                            —–                                         points separated by a distance of 3.0 cm in
                                             r
                                                                                       a wave of wavelength 5.0 cm?
                gravitational constant. Which figure shows                             A 0.6 rad             C 1.9 rad
                the correct trajectory of the object?                                  B 1.7 rad             D 3.8 rad

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          14. The graph shows the displacement of the                                  F
              particles in a rope caused by a travelling wave
              at a particular time.
                            Displacement                                              0                                 r
                                                                                           r1   r2
                                         P


                                             Q       S
                                 0
                                                             Position           Which statement is not true of the graph?
                                                                                A Hooke’s law is true around r1.
                                                 R                              B The equilibrium separation is r2.
                Which statement is not true of the motion of                    C The force is an attractive force when r > r1.
                the particles in the rope?                                      D The force is an attractive force when r > r2.
                A The speed of the particle at P is                         18. When a metal wire is stretched over a limit,
                    maximum.                                                    it undergoes plastic deformation. Which
                B The acceleration of the particle at Q is                      statement is true of plastic deformation?
                    zero.                                                       A Stress is proportional to strain.
                C The energy of the particle at R is entirely                   B The metal is not a crystalline solid.
                    potential energy.                                           C The atomic planes slide over each other.
                D The energy of the particle at S is entirely                   D The atoms are displaced a little from their
                    kinetic energy.                                                  equilibrium positions.
          15. The figure shows a sound source S moving                      19. A cylinder of volume 0.09 m3 contains an
              away from observer P towards observer Q.                          ideal gas at 100 °C and 8 × 104 Pa. If the
                                                                                mass of one mole of the gas is 0.034 kg, the
                            P            S           Q                          r.m.s. speed of the gas molecules is
                                                                                A 2.71 × 102 m s–1 C 7.33 × 104 m s–1
                What is the effect of the motion on the                         B 5.23 × 102 m s–1 D 2.73 × 105 m s–1
                wavelength of the sound received by P and
                by Q?                                                       20. At what temperature is the r.m.s. speed of
                                 P                       Q                      oxygen molecules the same as that of helium
                                                                                molecules at 300 K?
                A         Increases              Decreases                      [Relative molecular mass of oxygen = 32,
                B         Increases              No change                       relative molecular mass of helium = 4]
                C         Decreases              Increases                      A 38 K               C 1440 K
                                                                                B 849 K              D 2400 K
                D         No change              Decreases
                                                                            21. The work done to adiabatically compress one
          16. A person can hear a sound of intensity                            mole of a diatomic gas is 50 J. What is the
              1.0 × 10–11 W m–2. When a hearing aid is used,                    temperature rise of the gas?
              the sound level increases by 35 dB. What is                       A 0K                  C 2.41 K
              the new intensity of the sound heard?                             B 1.72 K              D 4.10 K
              A 3.50 × 10–11 W m–2
              B 3.50 × 10–10 W m–2                                          22. Which statement is true of an adiabatic
              C 3.16 × 10–9 W m–2                                               process?
              D 3.16 × 10–8 W m–2                                               A Boyle’s law is obeyed.
                                                                                B The temperature is always constant.
          17. The graph shows the variation of the                              C The internal energy always increases.
              interatomic force F with the separation r                         D No heat is transferred into or out of the
              between two atoms.                                                    system.

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          23. In an experiment to measure the thermal                               q2                     q2
                                                                               A –——                   C —–—
              conductivity of a good conductor, a long test                       4πε0r                  2πε0r
              sample is used so that                                                 q2                      q2
                                                                               B – –——                 D – –——
              A the heat loss can be neglected                                      4πε0r                   2πε0r
              B the steady state is easily achieved
                                                                           27. The circuit diagram shows a 10 kΩ resistor
              C the heat flow is parallel throughout the
                                                                               and a 40 kΩ resistor connected across a
                  sample
                                                                               charged capacitor C.
              D the temperature difference between the
                  two ends of the sample is sufficiently                                              V
                  large
                                                                                                     C         S

          24. The figure shows a uniform metal rod which
                                                                                                    10 kΩ
              is not insulated.
                                                                                                    40 kΩ
                            θ1        X                       Y   θ2
                                                                               After switch S is closed, the potential
                                                                               difference V across the capacitor varies with
                Heat flows through the rod in a steady state,
                                                                               time t. The graph of ln V against t is given
                and the temperatures at the ends of the rod
                                                                               as follows.
                are θ1 and θ 2, where θ1 > θ 2. Which statement
                is true of the two positions X and Y on the                             In V
                rod as shown?
                A No heat is lost to the surroundings at the                            2
                     area between X and Y.
                B The rate of heat loss to the surroundings
                     at X is greater than that at Y.
                C The rates of heat flow at X and Y are the                             0                           t (s)
                                                                                                             20
                     same.
                                                                               The capacitance of the capacitor is
                D The temperature gradients at X and Y are
                                                                               A 1.25 × 10–5 F      C 1.25 × 10–3 F
                     the same.
                                                                               B 2.00 × 10 F–4
                                                                                                    D 2.50 × 10–3 F
          25. The figure shows three point charges +q, +q                  28. A parallel-plate capacitor is charged by a
              and –q at the vertices X, Y and Z respectively                   battery. The space between the plates is
              of an equilateral triangle.                                      then completely filled with a material of
                                                                               dielectric constant εr while the battery remains
                                              X +q                             connected. What is the ratio of the energy
                                                                               stored after the material is inserted to the
                                          T          S
                                                                               energy stored before the material is inserted?
                                               O                               A 1: ε 2 r              C ε r: 1
                                                                               B 1: ε r                D ε2: 1r

                                 Y
                                     +q        R
                                                              Z            29. A lamp lights up immediately when the
                                                         −q
                                                                               switch in a circuit is closed, although the
                                                                               drift velocity of electrons is small. This is
                The electric field strength at the centroid O
                                                                               because
                of the triangle is in the direction of
                                                                               A the current flowing in the circuit is large
                A OS                    C OY
                                                                               B the potential difference supplied in the
                B OT                    D OZ
                                                                                    circuit is large
          26. The work done to bring two point charges                         C the random velocity of the electrons is large
              +q and –q from infinity to separation r is                       D all the free electrons drift simultaneously
                                                                                    in the circuit

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Ace Ahead STPM Physic vol 1 4th.indd 4                                                                                      3/7/2008 4:48:34 PM
          30. The graph shows the variation of potential                         33. The figure shows a positive charge moving
              difference V across a resistor with current I                          through a uniform magnetic field at velocity
              flowing through it.                                                    v in the positive z-direction.
                            V(V)                                                                          y


                           8
                           6                                                                              +             x
                                                                                                      v
                           4
                           2                                                                      z

                            0                                        I (A)
                                   0.1 0.2 0.3 0.4                                  If the magnetic field is in the positive
                                                                                    x-direction, the magnetic force acting on the
                When the current is 0.23 A, the power supplied
                                                                                    positive charge is in
                to the resistor is
                                                                                    A the positive x-direction
                A 0.53 W               C 2.12 W
                                                                                    B the negative x-direction
                B 1.06 W               D 4.23 W
                                                                                    C the positive y-direction
          31. The figure shows a circuit consisting of two                          D the negative y-direction
              batteries and three resistors.
                                                                                 34. Hall effect cannot be used to measure
                                   12 V         4.0 Ω
                            P                                                        A the charge density
                                                                I1                   B the magnetic flux density
                                                                                     C the mass of charge carriers
                                                2.0 Ω
                                                                                     D the type of charge carriers

                                                                                 35. The figure shows a square coil of sides
                                           I2
                                                                Q                    10 cm and resistance 2.00 Ω placed in a
                                   8.0 V        6.0 Ω                                uniform magnetic field of 10.0 T directed
                If I1 and I2 are 2.56 A and 1.64 A respectively, the                 perpendicularly to the coil.
                potential difference between points P and Q is
                A 1.8 V                    C 9.8 V
                                                                                                          10 cm
                B 4.0 V                    D 20 V

          32. The figure shows a circuit, where XY is a                                                             10 cm
              uniform wire.


                                100 Ω       P   R       Q
                                                                                    If the magnetic field is reduced to zero at a
                                           G        G                               constant rate in 40 ms, what are the magnitude
                      X                                     Y                       and direction of the induced current?
                                40 cm
                                44 cm                                                       Magnitude             Direction
                                                                                    A        1.25 A               Clockwise
                When the galvanometer is connected at point
                P and then at point Q, the balanced lengths                         B        1.25 A           Anticlockwise
                are 40 cm and 44 cm respectively. The value
                of R is                                                             C        2.50 A               Clockwise
                A 10 Ω               C 110 Ω                                        D        2.50 A           Anticlockwise
                B 91 Ω               D 400 Ω

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          36. Which statement is not true of the back e.m.f.                                          Rf
                                                                                      A The value of — is 4.
              of an electric motor?                                                                   Ri
              A The back e.m.f. opposes the applied                                   B The voltage gain is – 4.
                  voltage.                                                            C When the input voltage is –1 V, the output
              B The back e.m.f. is caused by                                            voltage is + 4 V.
                  electromagnetic induction.                                          D When the input voltage is 2 V, the
              C The back e.m.f. helps increase the                                      potential at point P is 2 V.
                  magnitude of the current flowing through
                                                                                  40. Which figure shows the correct directions of
                  the coil of the motor.
                                                                                      the magnetic field B and electric field E in
              D The magnitude of the back e.m.f. increases
                                                                                      relation to the direction of propagation of an
                  if the speed of rotation of the coil of the
                                                                                      electromagnetic wave?
                  motor is increased.
                                                                                      A E                     C     E

          37. An a.c. power supply with a constant r.m.s.
              voltage and variable frequency is connected in
              series to a pure inductor. Which graph shows                                B         Direction of                      Direction of
              the variation of the r.m.s. current I with the                                        propagation B                     propagation
                                                                                      B                               D
              frequency f ?                                                                                               E
              A                       C                                                                    Direction of
                     I                                       I                       B                                            B
                                                                                                           propagation
                                                                                              E
                                                                                                                                  Direction of
                                                                                                                                  propagation
                                                 f                        f
                    0                                     0                       41. Which statement is true of an electromagnetic
                B                                     D                               spectrum?
                         I                                   I                        A Radio waves are longitudinal waves.
                                                                                      B Only the visible spectrum can be plane-
                     0                            f                                       polarised.
                                                                          f           C X-rays travel at a higher speed than
                                                          0
                                                                                          ultraviolet waves.
                                                                                      D Gamma rays have a higher frequency than
          38. The alternating voltage supplied to a 200 µF
                                                                                          infrared waves.
              capacitor is given by V = 5 sin 300t, where V
              is in volts and t is in seconds. The maximum                        42. The figure shows two lenses of focal
              current in the circuit is                                               lengths f1 and f2 placed in contact with each
              A 0A                     C 0.30 A                                       other.
              B 0.21 A                 D 83 A
                                                                                                  Object                              Image
          39. The figure shows a circuit of an inverting
              amplifier.
                                                 Rf
                                                                                                              u               v
                                                      +9 V
                             Ri          P
                     Vin                     −
                                                                   Vout               If the object distance is u and the image distance
                                             +                                                     1     1
                                         Q                                            is v, then — + — equals
                                                      −9 V                                         u     v
                                                                                      A —   1 +—  1             C ——–    1
                An input voltage of +2 V produces an output                                 f1   f2                  f1 + f2
                voltage of –8 V. Which statement about the                            B —–— 1     1             D ——–    1
                circuit is not true?                                                        f1   f2                  f 1 – f2


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          43. A diffraction grating with 2400 lines per             47. Stimulated emission occurs in the laser when
              centimetre is used to produce a diffraction               a photon P causes the emission of a photon
              pattern that is recorded by a detector. The               Q. Which statement is true of the photons?
              first-order fringe is 8.94 cm from the central            A The phase of P is the same as that of
              fringe. If the distance between the grating and               Q.
              the detector is 0.625 m, what is the wavelength           B The frequency of P is less than that of
              of the electromagnetic wave used?                             Q.
              A 295 nm                 C 1180 nm                        C The wavelength of P is less than that of
              B 590 nm                 D 1190 nm                            Q.
                                                                        D The energy of P is more than that of Q.
          44. An experiment is performed to show the
              photoelectric effect. The frequency of the            48. If the masses of the proton, neutron,
              light used is kept constant while the intensity           electron and 12C atom are mp, mn, me and
                                                                                       6
              is increased. Which quantity will increase?               mc respectively, the mass defect ∆m of the
              A The momentum of the photoelectrons                      nucleus 12C is given by
                                                                                 6
              B The emission rate of the photoelectrons                 A ∆m = 6mp + 6mn – mc
              C The maximum kinetic energy of the                       B ∆m = 6mp + 6mn – 6me – mc
                   photoelectrons                                       C ∆m = 6mp + 6mn + 6me – mc
              D The minimum de Broglie wavelength of                    D ∆m = 6mp + 6mn + 6me – 6mc
                   the photoelectrons
                                                                    49. A sample consists of 2 g of a radioactive
          45. The figure shows the energy levels of the                 element. The molar mass of the element is
              hydrogen atom.                                            67 g. If the half-life of the element is 78 hours,
                                                                        the activity of the sample after 48 hours is
                     n   =   ∞            0 eV                          A 2.7 × 1016 s–1
                     n   =   5           −0.54 eV
                     n   =   4           −0.85 eV
                                                                        B 2.9 × 1016 s–1
                     n   =   3           −1.51 eV                       C 4.4 × 1016 s–1
                     n   =   2           −3.39 eV                       D 1.0 × 1020 s–1
                     n=1                 −13.58 eV
                                                                    50. A nuclide decays by emitting x alpha particles
                Which transition produces radiation of                  and y beta particles to form an isotope of the
                wavelength 436 nm?                                      original nuclide. What are the values of x and
                A n = 4 to n = 1   C n = 5 to n = 1                     y?
                B n = 4 to n = 2   D n = 5 to n = 2
                                                                                   x            y
          46. In an X-ray tube, the minimum wavelength                  A          1            2
              produced is 4.0 × 10–11 m. If the potential               B          1            4
              difference between the cathode and anode is
                                                                        C          2            1
              decreased to half of the original value, the
              minimum wavelength becomes                                D          2            2
              A 2.0 × 10–11 m       C 8.0 × 10–11 m
              B 4.0 × 10–11 m       D 1.6 × 10–10 m




                                                QUESTION PAPER ENDS

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                                                                                                Time: 2 h 30 min

               Section A [40 marks]

          Answer all questions in this section.

            1. A cylinder of mass 10 kg and radius 0.2 m rolls down an inclined plane of height 1.0 m
               without slipping from rest. Calculate the linear velocity of the cylinder when it reaches the
               ground.                                                                              [4 marks]
               [Moment of inertia of a cylinder of mass m and radius r is —  1 mr2.]
                                                                             2

            2. (a) On the same axes, sketch graphs of displacement against time to show underdamped,
                   critically damped and overdamped oscillations. Label your graphs.         [3 marks]
               (b) What is meant by resonance in forced oscillations?                        [2 marks]

            3. (a) In each of the following cases, state any change in the interference pattern of a two-slit
                   arrangement.
                    (i) One slit is covered by an opaque material.                                   [1 mark]
                   (ii) The distance between the slits is decreased.                                 [1 mark]
               (b) Explain how interference can be used to determine the flatness of lens surface. [2 marks]

            4. A cylindrical brass rod with Young’s modulus 9.7 × 1010 Pa and original diameter 10.0 mm
               experiences only elastic deformation when a tensile load of 200 N is applied.
               (a) Calculate the stress that produces the deformation.                                [3 marks]
               (b) If the original length of the rod is 0.25 m, calculate the change in the length of the rod.
                                                                                                      [2 marks]

            5. A 1.5 µF capacitor and a 2.0 µF capacitor are connected in series across a 24 V source.
               (a) Calculate the equivalent capacitance across the source.                         [2 marks]
               (b) Calculate the charge stored on each capacitor.                                  [2 marks]
               (c) If a dielectric is added to each capacitor, explain what happens to the charge stored in
                   each capacitor.                                                                 [2 marks]

            6. An ideal solenoid consists of 1000 turns of wire per cm wound around an air-filled cylindrical
               structure. The solenoid is of 2.0 cm long and cross-sectional area 1.8 cm2. A current of
               2.0 A passes through the wire.
               (a) Calculate the magnetic flux in the solenoid.                                     [3 marks]
               (b) Calculate the self-inductance of the solenoid.                                   [2 marks]

            7. A light beam of wavelength 0.110 nm collides with an atom. After the collision, an electron
               is emitted with kinetic energy 180 eV.
               (a) Calculate the energy absorbed by the atom.                                     [3 marks]
               (b) Calculate the velocity of the electron emitted.                                [2 marks]

            8. In a nuclear reactor, a very slow moving neutron is absorbed by a stationary boron atom. The
               equation for the nuclear reaction is
                                                  1   10      7     4
                                                  n + 5B → 3Li + 2He
                                                  0



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                After the reaction, the speed of the helium atom is 9.10 × 106 m s–1 and the kinetic energy
                of the neutron is approximately zero.
                [mn = 1.0087 u, mB = 10.0130 u, mLi = 7.0160 u, mHe = 4.0026 u]
                (a) Calculate the kinetic energy of the lithium atom after the reaction.            [4 marks]
                (b) Calculate the reaction energy.                                                  [2 marks]


               Section B [60 marks]

          Answer any four questions in this section.

            9. (a) An object is projected with initial speed v at angle θ to the horizontal. Derive an equation
                   for the path of the projectile.                                                     [3 marks]
               (b) The figure shows a jeep travelling at a constant speed of 10 m s–1 towards a security post.

                                                         Bullet
                                         Security
                                                  15 m                                    Jeep
                                         post

                                                                  0.7 km
                      A security guard holding his rifle 15 m above the ground shoots horizontally when the
                      jeep is 0.7 km from the security post. The bullet of the rifle strikes the jeep. Calculate
                        (i) the time taken by the bullet to strike the jeep                              [2 marks]
                       (ii) the distance of the jeep from the security post when it is struck            [2 marks]
                      (iii) the initial speed of the bullet                                              [2 marks]
                      (iv) the speed and direction of the bullet when it strikes the jeep.               [6 marks]

          10. (a) What is meant by Doppler effect?                                                 [2 marks]
                                                                                              –1
              (b) A toy car with the siren on is approaching a wall at a speed of 2.0 m s . The siren
                  produces a sound wave in the form
                                               y = 2.5 × 10–5 sin 2π(500t – 1.4x)
                  where x is in metres and t is in seconds. Calculate the speed and frequency of the sound
                  wave.                                                                            [4 marks]
              (c) The sound wave in (b) is then reflected from the wall. A detector is fixed on the ground
                  behind the source of the sound wave to detect any frequency changes. Determine
                   (i) the frequencies detected by the detector                                    [4 marks]
                  (ii) the beat frequency                                                          [2 marks]
                  (iii) the equation of the reflected wave                                          [1 mark]
                  (iv) the intensity ratio of the approaching wave to the reflected wave.          [2 marks]

          11. The following is a p–V diagram for a 0.2 mol ideal monatomic gas.
                                               p(Pa)


                                                     A
                                          3 × 105


                                                    C                  B

                                                0 2.0                12.0      V (× 10−3 m3)

                The gas expands isothermally from A to B and is compressed from B to C at constant pressure.
                It finally undergoes a constant volume process from C to A. Calculate

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Ace Ahead STPM Physic vol 1 4th.indd 9                                                                                       3/7/2008 4:48:39 PM
                (a) the temperatures at A, B and C                                                      [5 marks]
                (b) the net work done during the cycle                                                  [5 marks]
                (c) the net heat absorbed by the gas.                                                   [5 marks]
                                                    3
                [For an ideal monatomic gas, CV = —R and Cp = CV + R]
                                                    2

          12. (a) A 0.3 µF capacitor is connected to a source of alternating current with output voltage
                                                  V = 240 sin 120πt
                   (i) Calculate the reactance of the capacitor.                                        [3 marks]
                  (ii) Determine the r.m.s. current flowing through the capacitor.                      [3 marks]
              (b) A rectifier is a device which conducts electric current in one direction only such as a
                  diode.
                   (i) Describe briefly a full-wave rectifier.                                          [4 marks]
                  (ii) With the aid of diagrams, describe briefly the process of smoothing rectified alternating
                       current voltages.                                                                [5 marks]

          13. (a) State Bohr’s postulates for an atom.                                       [2 marks]
              (b) The figure shows an electron of mass m and charge –e moving at speed v in a circular
                  orbit of radius r round a nucleus.
                                                           v

                                                           −e m
                                                       r

                                                  +e




                    If the force of attraction between the electron and the nucleus provides the centripetal
                    acceleration of the electron, derive an expression for the radius of the nth orbit of the
                    electron.                                                                        [5 marks]
                (c) An electron in a Bohr orbit has kinetic energy 8.64 × 10 J.
                                                                             –20

                     (i) Calculate the speed of the electron.                                        [3 marks]
                    (ii) Determine the allowed orbit.                                                [3 marks]
                    (iii) Calculate the radius of the orbit.                                         [2 marks]

          14. (a) The bombardment of a beryllium nucleus by an α-particle produces a fundamental particle
                  X, as follows.
                                                 α + 9Be ⎯⎯→ 12C + X
                                                     4        6

                     (i) Complete the equation above by giving the proton and nucleon numbers to the
                          α-particle and X.                                                 [2 marks]
                    (ii) What are the α-particle and X?                                     [2 marks]
                    (iii) State two important properties which make X difficult to detect.  [2 marks]
                (b) Determine the equivalent energy in MeV of a mass of 1 u.                [5 marks]

                (c) An element of unknown atomic mass is mixed with 12C atoms in a mass spectrometer.
                                                                     6
                                                                                           12
                      The radii of curvature of the tracks of the ions of the element and 6C are 26.2 cm and
                      22.4 cm respectively. What is possibly the element? State any assumption you make.
                                                                                                        [4 marks]

                                                 QUESTION PAPER ENDS

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Ace Ahead STPM Physic vol 1 4th.indd 10                                                                             3/7/2008 4:48:40 PM
          PAPER 1                                                           14. A: P at the amplitude, its speed = 0.
                                                  F    MLT–2
           1. B: [Tensile strength] = [stress] = [—] = ——– = ML–1T–2        15. A: See figure.
                                                  A     L2
           2. C: If a = acceleration, F = (m + 2m)a = 3ma,
                                                                2F                      P             S            Q
                    force of Y on X = force of X on Y = (2m)a = —
                                                                3
           3. C: Speed = gradient of graph.
                 When t = T, gradient of P > gradient of Q.                 16. D: Intensity level = 35 dB = 10 lg ———–—I
                                                                                                                   1.0 × 10–11
           4. C: Ball has translational and rotational K.E.                                        I = 3.16 × 10–8 W m–2
                                                    1
                                                    — mv 2                  17. B: F = 0 at the equilibrium separation.
                                    gain in K.E.    2
           5. A: Average power = ————— = ——— = 121 kW
                                        time           t                    18. C: Plastic deformation, stress is not proportional to
           6. D: Angular momentum = Iω = (mr 2)ω = constant.                       strain.

                                                          dm                       1       3        3RT     3(8.31)(373)
           7. C: Force produced by the ejected gas, F = v —–                19. B: —Mc 2 = —RT, c = —— = ————— m s–1
                                                           dt                      2       2         M         0.034
                                         dm r                                                           = 5.23 × 102 m s–1
                 Torque = Fr = Iα, α = v —– —
                                          dt I
                                                                                        3RT1      3RT2
                                                  0.020                     20. D: c1 = —— = c2 = ——
                           = (50)(1.0 × 10–5) ——–—— rad s–2                              M1        M2
                                                1.0 × 10–6
                                                                                          M1     32
                           = 10 rad s–2                                              T1 = —–T2 = —(300) K = 2400 K
                                                                                          M2      4
                    Angular velocity after 10 s,
                    = αt = (10)(10) rad s–1 = 100 rad s–1                                              5             2W
                                                                            21. C: Adiabatic, W = ∆U = —R(∆T ), ∆T = —– = 2.41 K
                                                                                                       2             5R
           8. D: Resultant force = 0,
                 resultant torque about centre of disc = 0.                 22. D: ∆Q = 0
                           Gm1    Gm2                                       23. D: Temperature gradient is small. A longer length is
           9. A: EX = EY, —— = —–—
                          (r/3)2 (2r/3)2                                           required to produce a sufficiently large temperature
                             m1 1                                                  difference.
                             —=—
                             m2 4                                           24. B: Temperature gradient at X is greater.

          10. B: For object to move in circular orbit, velocity = GM
                                                                  —–        25. D:                        X
                                                                                                              +q
                                                                   r

                    Escape speed = 2GM
                                   ——                                                                         EY
                                    r                                                                     O
                                                                                                                   EZ
                    Speed of 1.5 GM > 2GM > GM
                                 —–   ——    —–                                                 Y
                                                                                                              EX
                                                                                                                          Z
                                  r    r     r                                                +q                        −q
                    Therefore trajectory is as in B.                                 The electric field strengths EX, EY and EZ at O due to
                                                                                     the charges at X, Y and Z are of the same magnitude
                           π
          11. A: x = 5 sin —t                                                        and in the directions shown in the figure. Therefore
                           6
                                                                                     the resultant electric field at O is in the direction
                            dx      π      π                                         OZ.
                 Speed, v = — = 5(—) cos —t
                             dt     6      6
                                      π cos — = 2.3 m s–1
                                             π                              26. B: Work done = electric potential energy of the system
                 When t = 1 s, v = 5 —
                                      6      6                                                 (+q)(–q)
                                                                                             = ——–—
          12. C: Increased damping: Lower peak and lower resonant                                4πε0r
                                     frequency.                             27. C:     V = V0e–t/CR
                                    2π                                                             t
          13. D: Phase difference = —–(3.0) rad = 3.8 rad                            lnV = lnV0 – –—
                                    5.0                                                           CR

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Ace Ahead STPM Physic vol 1 4th.indd 11                                                                                             3/7/2008 4:48:40 PM
                    Gradient of graph = –2 = – —
                                        —       1                            43. B:
                                        20     CR
                     1      1       1
                     — = —— + —–— = —–—      5                                                                           8.94 cm
                     R 10 kΩ 40 kΩ 40 kΩ                                                  Grating
                                                                                                           θ1
                                              5
                    Capacitance, C = (10) ———— F = 1.25 × 10–3 F
                                          40 × 103                                                           62.5 cm

          28. C: C1 = εrC0, V remains the same because the battery
                                          1            1
                 remains connected. E0 = —C 0V 2, E1 = —C1V 2 = εrE 0
                                          2            2
                                                                                                        1         8.94
                                                                                      λ = d sin θ1 = ———–— —————— m = 590 nm
          29. D                                                                                     2400 × 102 8.942 + 62.52
                                       0.23
          30. B: When I = 0.23 A, V = ——(4 V)                                44. B: When the intensity is increased, the rate of photons
                                        0.2
                                                                                    incident on the cathode increases. Rate of emission
                                     0.23
                 Power = IV = (0.23) ——(4) W = 1.06 W                               of photoelectrons increases.
                                      0.2
          31. A: Current in 2.0 Ω resistor,                                             hc           (6.63 × 10–34)(3.00 × 108)
                                                                             45. D: E = — =         ——————————— eV
                 I = I1 – I2 = (2.56 – 1.64) A = 0.92 A                                 λ            (436 × 10–9)(1.60 × 10–19)
                 VPQ = IR = (0.92)(2.0) V = 1.84 V                                         =        2.85 eV = E5 – E2

          32. A: P.d. across 40 cm = I(100)                                                 hc
                                                                             46. C: λmin = —
                 P.d. across 44 cm = I(100 + R),                                            eV
                              44                                                              hc
                 100 + R = —– (100) Ω, R = 10 Ω                                     λ'min = ——— = 2λmin = 8.0 × 10–11 m
                              40                                                            e(V/2)
          33. C:                          y                                  47. A: Laser photons are coherent and in phase.
                                              F=vxB                          48. C: Mass of C-12 atom = 6mp + 6mn + 6me
                                                                                                                   2          23            22
                                                                             49. D: Initial number of atoms, N0 = — (6.02 × 10 ) = 1.80 × 10
                                                                                                                  67
                                          +           B
                                                  x                                                             dN          In 2
                                                                                      Initial activity, A 0 = – —– = λN 0 = —–– N0
                                   z                                                                            dt           T—1
                              v                                                                                                2

                                                                                                   48 1
                   VH                I                                                t = 48 h = –—T — = 0.6154T —    1
          34. C: e — = Bev = Be ——                                                                 78 2               2
                    d               nAe
                                                                                                                 A0
                 which is not in terms of the mass m of the charge                    Activity after 48 h, A = –——
                 carriers.                                                                                      20.6154
                                                                                           1 In 2
                                                                                      = ——– ––— (1.80 × 1022) s–1 = 1.0 × 1020 s–1
                       dφ    0 – BA                                                      20.6154 78
          35. A: E = – — = – —–——
                       dt    40 ms
                                                                             50. A: A P →
                                                                                    Z
                                                                                             A –4
                                                                                                Z
                                                                                                                  0
                                                                                                    Q + 4 He + 2 –1 e
                                                                                                        2
                                         E    (10.0)(0.10)2
                    Induced current, I = — = ——————— A = 1.25 A
                                         R (40 × 10–3)(2.00)
                                                                             PAPER 2
                    Clockwise direction: Apply Lenz’s law.
                                                                             Section A
          36. C: Because of the back e.m.f., the current decreases.
                                                                              1. Gain in translational K.E. + rotational K.E. = loss of P.E.
                     V     V   1
          37. C: I = —– = —— ∝ —                                                            1        1
                     XL   2πfL f                                                           —mv 2 + —Iω2 = mgh
                                                                                            2        2
                        Vmax                                                      1 2 + — —mr 2 — 2 = mgh
                                                                                            1 1         v
          38. C: Imax = —— = ωCVmax = (300)(200 × 10–6)(5) A = 0.30 A            —mv
                         XC                                                       2         2 2         r
                                                                                               2 1     1
          39. D: VP = 0 V                                                                     v — + — = gh
                                                                                                  2    4
          40. C: E × B in the direction of propagation.                                4
                                                                                 v = —gh
                                                                                       3
          41. D
                                                                                      4
                                                                                    = —(9.81)(1.0) m s–1
                 1   1   1   1   1                                                    3
          42. A: — = — + — = — + —
                 F   f1  f2  u   v
                                                                                    = 3.6 m s–1

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Ace Ahead STPM Physic vol 1 4th.indd 12                                                                                                3/7/2008 4:48:42 PM
           2. (a)               Displacement                                          (b) φ = LI
                                            Critically damped                                 0.090
                                                 Underdamped                              L = —–— H = 0.045 H
                                                                                               2.0
                                                        Overdamped
                                                                                   7. (a) Energy absorbed by the atom
                                          0                                               = energy of photon – K.E. of electron
                                                             Time
                                                                                            hc
                                                                                          = — – 180 eV
                                                                                             λ
                (b) •       Resonance occurs when the amplitude of a                         (6.63 × 10–34)(3.00 × 108)
                                                                                          = ——————————— – 180 eV
                            forced oscillation is maximum.                                  (0.110 × 10–9)(1.60 × 10–19)
                      •     Resonance occurs when the frequency of the                    = 1.11 × 104 eV
                            driver system equals the natural frequency of                                    1
                                                                                      (b) Kinetic energy, —mv 2 = 180 eV
                            the system that is forced into oscillation.                                      2
                                                                                                                    2(180)(1.60 × 10–19)
                                                                                          Velocity of electron, v = ———–————— m s–1
           3. (a)         (i) The single-slit diffraction pattern as shown
                                                                                                                        9.11 × 10–31
                              below is obtained.
                                                                                                                   = 7.95 × 106 m s–1
                                                                                   8. (a) Using the principle of conservation of linear
                     (ii) The fringe separation increases.                                momentum,
                (b) Place the lens on a mirror and allow monochromatic                    MLiVLi + MHeVHe = 0
                    light to be incident on the lens.
                                                                                                                            MHe
                    If the lens surface is perfect, the interference                      Velocity of lithium atom, VLi = – –— VHe
                                                                                                                            MLi
                    pattern observed (Newton’s rings) would be perfect                          4.0026
                                                                                          = – —–—– (9.10 × 106) m s–1
                    concentric circles.                                                         7.0160
                                  Monochromatic light                                     = – 5.192 × 106 m s–1
                                                                                          Kinetic energy of lithium atom
                                                                                             1
                                                                                          = —MLi (VLi)2
                                                                                             2
                               Lens                                                          1
                                                                                          = —(7.0160 × 1.66 × 10–27)(5.192 × 106)2 J
                                                                                             2
                                          Mirror        Newton s rings                    = 1.57 × 10–13 J
                           F         200
           4. (a) Stress = — = —————– N m–2                                           (b) Total mass before reaction = 1.0087 u + 10.0130 u
                           A π(5.0 × 10–3)2                                                                          = 11.0217 u
                              = 2.5 × 106 N m–2
                                                                                          Total mass after reaction = 7.0160 u + 4.0026 u
              (b) Young’s modulus, E = –—F/A                                                                        = 11.0186 u
                                         e/l
                                                                                          Reaction energy,
                  Change in length, e = —Fl                                               E = (∆m)c 2
                                        EA
                                                                                            = (11.0217 – 11.0186)(1.66 × 10–27)(3.00 × 108)2 J
                                               (200)(0.25)                                  = 4.63 × 10–13 J
                                      = ——————————2 m
                                        (9.7 × 1010)π(5.0 × 10–3)
                                      = 6.6 × 10–6 m                              Section B
                   1     1     1
           5. (a) — = –— + –—, C = 0.86 µF                                         9. (a)
                   C    2.0 1.5
                                                                                                 v sin θ v                   P (x, y)
              (b) Charge stored in each capacitor, Q = CV
                                                     = (0.86)(24) µC
                                                                                                                         y
                                                     = 21 µC                                           θ
              (c) With the dielectric                                                                v cos θ
                  • Capacitance of each capacitor increases.                                                   x
                  • Charge stored in each capacitor increases.                                When the object is at P,
           6. (a) Magnetic flux density, B = µ0nI                                             • horizontal displacement, x = (v cos θ)t,
                  Magnetic flux,                                                                                          t = ———x
                  φ = NBA                                                                                                     v cos θ
                                                                                                                                        1
                    = N(µ0nI)A                                                                • vertical displacement, y = (v sin θ)t – —gt 2
                                                                                                                                        2
                                                                                                                x        1
                                                                                                = (v sin θ)(—–——) – —g(—–——)2    x
                                                      1000                                                   v cos θ     2 v cos θ
                           = (4π × 10–7)(1000 × 2.0) ——–— (2.0)
                                                     1 × 10–2
                             × (1.8 × 10–4) weber                                                                gx 2
                                                                                                = x tan θ – ——–——
                           = 0.090 weber                                                                     2v cos2 θ
                                                                                                                2




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Ace Ahead STPM Physic vol 1 4th.indd 13                                                                                                   3/7/2008 4:48:43 PM
                (b)    (i) Consider vertical component of motion of the                                         360
                                                                                                        = ———— 500 Hz
                           bullet                                                                           360 – 2.0
                           u = 0, a = g, s = 15 m                                                       = 503 Hz
                                          1               1                                             This frequency of 503 Hz is also detected by
                           Using s = ut +—at 2, 15 = 0 + —(9.81)t 2
                                          2               2                                             the detector.
                                                                                                   (ii) Beat frequency = f2 – f1
                                            2(15)
                            Time taken, t = —— s = 1.7 s                                                                  = (503 – 497) Hz = 6 Hz
                                            9.81
                                                                                                         1     f2 503
                       (ii) Distance of the jeep from the security post                           (iii) — = — = —– = 1.4
                                                                                                        λ2     v 360
                            = (700 – 1.7 × 10) m = 683 m                                                Equation of the reflected wave is
                      (iii) Horizontal distance travelled by the bullet,                                y = (2.5 × 10–5) sin 2π(503t + 1.4x)
                               v(1.7) = 683 m                                                     (iv) Intensity is directly proportional to (amplitude)2.
                            Speed, v = 402 m s–1                                                        Since the approaching wave and the reflected
                      (iv) Vertical component of the bullet’s velocity,                                 wave have the same amplitude,
                            vy = 0 + (9.81)(1.7)                                                        intensity of approaching wave
                               = 16.7 m s–1                                                             ———————————–— = 1
                                                                                                           intensity of reflected wave
                            Horizontal component of velocity,
                            vx = 402 m s–1                                               11. (a) At A,       pV = nRTA
                                               2   2                                                               (3 × 105)(2.0 × 10–3)
                            Speed = v + v      x   y                                                         TA = ———————— K = 361 K
                                                                                                                        (0.2)(8.31)
                                    =      4022 + 16.72 m s–1                                    At B, TB = TA = 361 K
                                                                                                 B to C, p = constant, VC ∝ TC , and VB ∝ TB
                                   = 402.3 m s–1
                            Direction of velocity is at an angle                                                                    VC
                                                                                                                             TC = –— TB
                                  vy                                                                                                VB
                                              16.7
                            tan–1 — = tan–1 —–—
                                  vx          402.3                                                                                 2.0 × 10–3
                                                                                                                                = ————— 361 K
                                      = 2.4° to the horizontal                                                                      12.0 × 10–3
                                                                                                                                = 60.2 K
          10. (a) Doppler effect is the apparent change in the                                                                                    VB
                                                                                             (b) From A to B, work done by the gas, W1 = nRTA ln —–
                  frequency of sound heard by an observer when                                                                                    VA
                  there is relative motion between the source of sound                                                     12.0 × 10–3
                                                                                                 = (0.2)(8.31)(361) ln ————— = 1075 J
                  and the observer.                                                                                         2.0 × 10–3
              (b) Comparing y = (2.5 × 10–5) sin 2π(500t – 1.4x)                                                         pAVA
                                                                                                 Pressure at B, pB = ——
                                           2π                                                                             VB
                  with y = A sin 2πft – —x ,
                                            λ                                                       (3 × 105)(2.0 × 10–3)
                                                                                                 = ———————–— Pa = 5.0 × 104 Pa
                  frequency, f = 500 Hz, λ = —– m 1                                                      12.0 × 10–3
                                                 1.4                                             From B to C, work done on the gas,
                  Speed of sound wave, v = f λ                                                   W2 = (5.0 × 104)(12.0 –2.0) × 10–3 J = 500 J
                                                       1
                                              = (500) —– m s–1                                   Net work done by the gas = (1075 – 500) J = 575 J
                                                      1.4
                                                                                             (c) From A to B, isothermal expansion, ∆U = 0.
                                              = 357 m s–1
                                                                                                 From                ∆Q = ∆U + W,
              (c) (i)
                                                                                                 heat absorbed, ∆Q1 =W1 = 1075 J
                                                                                                                                                5
                                                                    f2                           From B to C, ∆Q2 = nCp(∆T )2            CP = —R
                             Wall                  uS = 2.0 m s−1                                                                               2
                                                                         Detector                            5
                                          f2           Source       f1                           = (0.20)(—)(8.31)(60.2 – 361) J = –1250 J
                                                                                                             2
                                                                                                 From C to A, V = constant, ∆Q3 = nCV(∆T )3
                            The source is moving away from the detector.                                   3
                            Frequency detected by the detector,                                  = (0.2) — × 8.31 (361 – 60.2) J = 750 J
                                                                                                           2
                                    v                                                            Net heat absorbed = (1075 – 1250 + 750) J = 575 J
                            f1 = —–— f
                                  v + us                                                                     1
                                   360                                                   12. (a) (i) XC = —–                 (ω = 120π)
                            = ———— 500 Hz                                                                   ωC
                                360 + 2.0                                                                            1
                            = 497 Hz                                                                   = ———————— Ω = 8.8 × 103 Ω
                                                                                                          (120π)(0.3 × 10–6)
                            The source is approaching the wall.                                                Vrms
                            Frequency reflected by the wall                                       (ii) Irms = –—              Vrms = 240 V
                                                                                                                                     —–
                                                                                                                XC                     2
                            = frequency incident on the wall, f2
                                                                                                                 240
                                    v                                                                  = ————–—— A = 0.019 A
                            f2 = —–— f                                                                     2 × (8.8 × 103)
                                  v – us


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Ace Ahead STPM Physic vol 1 4th.indd 14                                                                                                            3/7/2008 4:48:44 PM
                (b)    (i) A full-wave rectifier that consists of four diodes                                           mv 2        e2
                                                                                                    •centripetal force, —— = ———
                           W, X, Y and Z is as shown below.                                                                r      4πε0r 2
                                                                                                                                    e2
                                                     Va.c.                                                                 v 2 = ———–            (2)
                                                                                                                                  4πε0mr
                                                                                                                              e2          n 2h 2
                             W           X                                                        From (1) and (2), m 2 ——–— r 2 = ——
                                                     0                   Time                                             4πε0mr          4π2
          A.c.
          supply
                                  Z Y                                                                                            ε0n h
                                                                                                                                    2 2
                                                                                                  Radius of the nth orbit, r = ———
                                                                                                                                 πme 2
                                                    V
                            R                                                                                           1
                                                                                              (c) (i) Kinetic energy, —mv 2 = 8.64 × 10–20 J
                   Full-wave rectifier                                                                                  2
                                                     0                 Time
                                                     Full-wave rectified voltage                                                     2(8.64 × 10–20)
                                                                                                              Speed of electron, v = ————–—– m s–1
                                 During the first half cycle of the a.c. voltage,                                                     9.11 × 10–31
                                 current flows through X, R and Z. During the                                                  = 4.36 × 105 m s–1
                                 second half cycle, current flows through Y, R                      (ii) Total energy of the electron in the nth orbit,
                                 and W. In both the half-cycles, current flows                            En = – K, the kinetic energy
                                 in the same direction through R. Therefore,                                           13.6
                                 the voltage across the resistor R is full-wave                           Also, En = – –—– eV
                                                                                                                        n2
                                 rectified as shown.
                                                                                                                       13.6
                      (ii)                                                                                Therefore, – –— eV = –8.64 × 10–20 J
                                                                                                                        n2
                                     W       X               Smoothened rectified                               (13.6)(1.60 × 10–19)
                A.c.                                    V                                                 n = ———————— = 5
                                         Z   Y
                                                             voltage                                                8.64 × 10–20
                supply
                                                                                                          Allowed orbit n = 5
                                                                                                                             h
                                         R           0                                              (iii) From (mvr ) = n –—
                                                                         Time                                               2π
                                                                                                                                 nh
                                    Capacitor C                                                           Radius of orbit, r = ——–
                                                                                                                                2πmv
                                 The rectified voltage is smoothened by                                             5(6.63 × 10–34)
                                                                                                          = ———————————— m
                                 connecting a capacitor C in parallel with the                               2π(9.11 × 10–31)(4.36 × 105)
                                 load R as shown in the circuit. During the first                         = 1.33 × 10–9 m
                                 half cycle, the capacitor C is charged until the
                                 peak voltage. Current stops flowing from the             14. (a)       (i)   α + 9Be → 12C + 1X
                                                                                                              4
                                                                                                              2   4      6    0
                                 a.c. supply, and the capacitor discharges until
                                 the potential difference across the capacitor is                   (ii) α-particle: Nucleus of helium.
                                                                                                                 1
                                 less than the voltage of the a.c. supply. Current                               0X: Neutron

                                 again flows from the a.c. supply until the                       (iii) X or neutron
                                 capacitor is charged to the peak voltage again.                        • is not charged.
                                 The process is then repeated. The voltage V                            • does not cause ionisation.
                                 across the load R varies with time as shown                  (b) Energy equivalent of 1 u, E = mc 2
                                 in the graph.                                                    = (1.66 × 10–27)(3.00 × 108)2 J
                                                                                                     (1.66 × 10–27)(3.00 × 108)2
          13. (a) Bohr’s postulates                                                               = ———————–——— MeV
                                                                                                          (1.60 × 10–19)106
                  • An electron can only orbit the nucleus in discrete
                                                                                                  = 934 MeV
                     allowed orbits such that
                                                                                                  mv 2
                                                                    h                         (c) —– = qvB
                     angular momentum of the electron = n —–                                        r
                                                                   2π
                     h is Planck constant, and n = 1, 2, 3, …                                             qB
                                                                                                  m = (–—)r
                  • When an electron drops from a higher energy                                            v
                     level E2 to a lower energy level E1, the difference                          Assumption: Ions of the element and ions of C-12
                     in energy of the electron is radiated as a photon                            have the same charge, then m ∝ r.
                     of frequency f.                                                              Hence mass number, A ∝ r
                                      E2 – E1 = hf                                                For X,       AX ∝ 26.2 cm
              (b) From the first postulate of Bohr,                                               For C-12, 12 ∝ 22.4 cm
                                                             h                                                          26.2
                  • angular momentum, (mvr ) = n —–                                               Mass number, AX = ——(12) = 14
                                                            2π                                                          22.4
                                                        n 2h 2
                                            m 2v 2r 2 = ——         (1)
                                                        4π2                                         The element is nitrogen, N.


                                                                                     15                           © Oxford Fajar Sdn Bhd (008974–T) 2008

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Ace Ahead STPM Physic vol 1 4th.indd 15                                                                                                           3/7/2008 4:48:45 PM

								
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