Physics Paper 2 Questions and Answers - Cekana Mock Exams 2023

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Instruction to The Candidates

  1. This paper consists of two Sections A and B.
  2. Answer all the questions in sections A and B in the spaces provided.
  3. All working must be clearly shown in the spaces provided.
  4. Mathematical tables and electronic calculators may be used.



Answer all the question in this section in the space provided.

  1. Sketch the magnetic field pattern given by a current carrying conductor in the magnetic field shown in the figure below. Also show the direction of the forces produced on the conductor (2 mks)
  2. State any two ways in which energy is lost in a transformer, and briefly explain each loss can be minimised. (2 mks)
  3. Below is part of the electromagnetic spectrum in order of increasing wavelength
    A B C Visible light Infrared D C
    How are waves A Produced (1 mk)
    1. State one similarity and one difference between a camera and a human eye. (2 mks)
    2. The figure below is a concave mirror used to form a virtual, magnified image. Complete the ray diagram to show the position of the object (3 mks)
      PhycPP2no.4 b
  5. The figure below shows a ray of light incident on a glass prism
    Given that the critical angle of the glass is 390, sketch on the diagram the path of ray through the prism until it exits the prism (2mks)
  6. A sharp point of a pin is held over a positively charged electroscope. State and explain the observation made on the electroscope (2mks)
  7. The figure below is a circuit of 3 resistors connected to a 12v battery
    Determine the P.d across the 3 resistor (3mks)
  8. The figure below shows a magnet. Point A and B are in front of the magnet
    On the axis provided sketch a graph showing how the magnetic field strength changes the A to B (2mks)
  9. The figure below shows a beam of beta particles entering a magnetic field whose direction below
    Complete the diagram to show the path of the beta particles as they pass through magnetic field and out of it (1mk)
  10. A girl standing 400m from the foot of a high cliff claps her hand and the echo reaches her 2.32 seconds later. Calculate the velocity of sound in air using this observation. (2mks)
  11. Explain two of ways of dealing with defect of polarisation in a simple cell (2mks)
  12. The figure below shows an a.c source connected across diode D and Resistor R
    On the axis provided sketch the output voltage as observed in the CRO (1mk)
    PhycPP2no.12 b


    1. The figure below shows a system of capacitor connected to a 100v d.c supply.
      PhycPP2no.13 a
      From this circuit determine:
      1. Its effective capacitance (2mks)
      2. The charge through the 6μF capacitor (2mks)
      3. The p.d across 8μF capacitor (2mks)
    2. State any two factors that affects the capacitance of a parallel plate capacitor (2mks)
    1. The figure below shows the features of an x-ray tube
      PhycPP2no.14 a
      1. Name the parts labelled A and B (2mks)
      2. Explain the effect of increasing the potential P on x-rays produced (2mks)
      3. During the operation of the tube, the target become very hot. Explain how this heat is caused (2mks)
    2. In a certain x-ray tube, the electrons are accelerated by a p.d of 12kv. Assuming that all energy goes to produce x-rays, determine the frequency of x-ray produced (h=6.62×10-34Js and e-=1.6×10-19C) (3mks)
    1. The figure below shows a connection to a 3 pin plug as viewed from its back
      1. Name the terminal pin labelled A (1mk)
      2. State any two mistakes with wiring of his 3pin plug (2mks)
      3. Give any two reasons why the earth pin is normally longer than the other two pins (2mks)
    2. The figure below is graph of K.E of electrons against frequency of radiation in photo – electric effect in cathode metal plate

      PhycPP2no.15 b
      From the graph, determine
      1. Threshold wavelength ()of the cathode metal surface (2mks)
      2. Plancks constant (h) of cathode (2mks)
      3. Work function of cathode in ev (2mks)
    1. The figure below shows a simple cathode ray tube
      PhycPP2no.16 a
      1. Explain how the electrons are produced in the tube (2mks)
      2. State one function of the anode (1mk)
      3. At what part of the cathode tube would the time base be connected? (1mk)
    2. The figure below represent a displacement time graph for a wave moving with a velocity of 80cm/s
      PhycPP2no.16 b
      Determine the wave
      1. Frequency (f) (2mks)
      2. Wavelength (2mks)
    3. A ray of light traveling through air in to medium 1 and 2 as shown in the figure below.
      PhycPP2no.16 c
      1. Calculate the refractive index of medium 1 (1mk)
      2. Angle of refraction r2 in medium 2 (2mks)
    1. State the lenz’s law of electromagnetic induction. (1mk)
    2. A bar magnet is moved into a coil of insulated copper wires connected to a centre zero – galvanometer (G) as shown in the figure below
      PhycPP2no.17 b
      1. With arrows show the direction of induced current in the coil (1mk)
      2. Explain clearly what is observed on the galvanometer when the south pole of a magnet is moved into the coil and then withdrawn (2mks)
    3. A transformer has 1600 turns in the primary coil and 80 turns in the secondary coil. If the transformer is 80% efficient and its primary coil is 240v with a current of 0.75A, determine
      1. The secondary e.m.f of its coil (2mks)
      2. The output power in its secondary coil (2mks)
    1. Define the term radioactivity (1mk)
    2. The figure below shows tracks formed in a diffusion cloud chamber given by a certain radioactivity sample
      PhycPP2no.18 b
      State the type of radiations emitted by this sample, explain your answer (2mks)
      1. The figure below shows radioactive decay of iodine
        PhycPP2no.18 c
      2. What fraction of this iodine will have decayed after 3 half – lifes (2mks)
    4. The average count rate of a radioactive material sample is 92 counts per second. After 420 seconds, the count rate had dropped to 29 counts per second. Given that the background count rate was 20 counts per second, determine the half life of this sample. (3mks)


  1. PhycPP2no.1 ms
  2. Eddy currents – By laminating the core
    Hysteresis – Use of a soft iron core
    Magnetic flux leakage – winding the secondary coil over primary coil
    Copper losses – use of thick copper wire on coil carrying higher current
  3. Gama Result from energy changes occurring in the nuclear of the radioactive atom
    1. similarity
      Eye                                                                                                                           Camera
      Crystalline convex lens
      Choloid layer is black
      Has retina where images are formed
      Iris controls the amount of light entering the eye
      Convex lens
      Camera box painted black
      Has light sensitive film where images are formed
      Diaphragm control the amount of light entering the camera

      1. focal length of the eye lens is variable while that of camera is fixed
      2. camera has variable image distance while eye has constant image distance
      3. camera takes one photograph at time while the eye is always open to form constantly changing images
    2. PhycPP2no.4 b ms

  5. PhycPP2no.5 ms
    correct 1mk
    Angle of l = r = 1mk
  6. Leaf fall The electron flow from the ground and discharges to the sharp point
    RT = 9                ∴ Pd across 3.2 = 1R
    v = 1R                                          = 1.333 x 3 
    12 = 9 x 1                                     = 3.999 v
    I = 1.333A                                    
  8. PhycPP2no.8b

  9. PhycPP2no.9 ms

  10. V = 2d 2 x 400  
            t         2.32
     = 344.83 m/s
  11.  Use of depolarizer eg potassium dichromate
    Brushing of hydrogen gas from the copper plate
  12. PhycPP2no.12 ms


Section B 55 marks

    1. Effective capacitance (2mks)
      1. Ct = C1C2 = 4 + 8 = 12μF
        Cr = CtC3 = 12 x 6 = 4μF
               Ct+C3   12 x 6 
      2. charge through 6μF capacitor
        Qr = Q1 = Q2 for seyies
        Qr = CTVr = 4 x 100 = 400μF
        ∴ Q6μF = Qr = 4μC
      3. The p.d across 8μF the capasitor (2mks)
        mtdI                                                                  mdtlII     
        p.d across 6μF                                                p.d across parallel capacitor
        v = Q = 400μF = 66.67                                    V = Q = 400μF
              C      6μF                                                          C     12μF
                                                                                           = 33.33v
        p.d across parallel section / 8μF
                        100v - 66.67
         = 33.33v
      4. Factors affecting capacitance of parallel plate capacitors
        • area of overlap of plates
        • distance of separation of plates
        • nature of dielectric material used
      1. Name part A and B
        A – cathode ray✔1
        B - Anode✔1
      2. Effects of increasing potation P 2mks
        more heat is produced by increasing current increasing number of electrons ✔1 emitted the emionically leading to increase in intensity of x-ray produced at the target
      3. causes of heat produced in the x-ray tube 1mk
        most of K.E of electrons hitting the anode is converted to heat✔1
    2. frequency (f) of x-ray product
      ev = hf
      1.6 x 1-19 x 1200 = 6.62 x 10-34 x f
      f  = 1.6 x 10-19 x 1200
      = 2.9 x 1018H2
      1. part labeled A of 3 pin – plug 1mk
        A – Earth wire pin
      2. any two mistakes of this 3 pin plug 2mks
        • green yellow wire which is the earth wire is connected to live pin
        • Blue wire which is the neutral wire is connected to earth pin
        • Brown wire which is live wire is connected to neutral pin
      3. why the earth pin is longer than others 2mks
        • To open the live and neutral socket holes by pressing down their shutter✔1
        • To ensure the device is earthed even before it has started working✔1
      1. threshold wavelength
        f0 = 2.5 x 1014
        λ0 = c =   3 x 108    = 1.2 x 10-6 m
                f0    2.5 x 1014

      2. plancks constant(h)
        h = slope (m)
         ( 2 - 1 ) x 10-19  = 6.6667 x 10-34 Js
            ( 5.5 - 4 ) x 1014   

      3. work function(W0)
        = hf0 = h c/λ0
        6.6667 x 10-34 x 2.5 x 1014 = 1.6667 x 10-19 Js
        = 1.6667 x 10-19 = 1.04167 eV
               1.6 x 10-19
      1. how electrons are produced in C.R.T 2mks
        Filament heat cathode, electron are emitted by thermionic emission by cathode.
      2. functions of the anode 1mk
        accelerating the electron beam
        focusing the beam
      3. where the time base is connected
        across the x-plate
    2. determine the wave
      1. frequency (t)
        T = 20 x 10-35
        f =  1      1         =  50Hz
              T     20 x 10-3
      2. wavelength
        λ = v = 0.8 m/s = 0.016m
              f      50Hz
    3. refractive index of m1 2mks
      1. Ωm= sinC0 = sin40 = 1.5803
                   sinr       sin24   
      2. Ω1sinθ1 = Ω2sinθ
        1.58035sin24 = 2sinr
        r = 18.747
    1. The direction of induced emf is such that the induced current which it causes the flow produces a magnetic effect that opposes the change producing it
      1.  PhycPP2no.17 b ms

      2. Deflection to one side/ left and back to zero then on opposite direction and back to zero.
        • Change in magnetic flux induces current in opposite direction, a stationary magnet does not cause a change in flux hence no induced emf.
      1. Vs  Ns
        Vp  Np
        Vs = VpNs
        = 240 x 80 = 12v
      2. n = power output    x 1000
               power input
        power output = n x power input
    1. Spontaneous random disintegration of heavy unstable nuclides as they try to stabilize giving radiation
    2. Beta radiation
      Because they are lighter and faster they cause less ionization of air molecules hence there irregular track or they are easily deflected by electrons of the air molecules.
      1. 8.1days
      2. 400-200-100-50
    4. 92-20 = 72
      29- 20 = 9
      72 t 36 t 18 t 9
      3t = 420
      t = 140s
      Method II
      N = No(½) T/t
      9 = 72(½)420/t
      9/72 = (½) 420/t
      t = 140seconds
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