KCSE 2010 Physics Paper 2 Questions with Marking Scheme

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QUESTIONS

SECTION A (25 marks)
Answer ALL the questions in this section in the spaces provided.

  1. Figure 1 shows a ray of light indecent on a plane mirror at O. The mirror is then rotated anticlockwise about O from position M1 to position M2 through an angle of 100. The final reflected ray is OC.

    ray of light incidence kcse 2010

    Determine the angle BOC. (2 marks)

  2. Figure 2(a) shows the magnetic compass placed under the horizontal wire XY

    magnetic compass kcse 2010

    A large current is passed from X to Y. Draw the final position of the magnetic compass needle in figure 2(b). (1 mark)

  3. Figure 3 shows a diagram of current carrying wire wound on a U- shaped soft iron.

    electro magnet field pattern kcse 2010

    Draw the magnetic field pattern around P and Q. (2 marks)

  4. A positively charged sphere is suspended by an insulating thread. A negatively charged conductor is suspended near it. The conductor is first attracted, after touching the sphere is repelled. Explain this observation. (2 marks)

  5. Figure 4 shows a bright electric lamp placed behind a screen which has a hole covered with a wire gauze. A concave mirror of focal length 25cm is placed in front of the screen. The position of the mirror is adjusted until a sharp image on the gauze is formed on the screen.

    concave mirrors kcse 2010

    Determine the distance between the mirror and the screen. (2 marks)

  6. Explain why electric power is transmitted over long distances at high voltages. (2 marks)

  7. Figure 5, shows the displacement of a point varies with time as a wave passes it.

    displacement of waves kcse 2010

    On the same diagram, draw a wave which passes the point with a half the amplitude and twice the frequency as the one shown. (2 marks)

  8. A water wave of wavelength 18 mm is incident on a boundary of shallow water at right angles. If the wavelength in the shallow end is 14.4 mm, determine the refractive index of water for a wave moving from the deep to the shallow end. (3 marks)

  9. The initial mass of a radioactive substance is 20g. The substance has a half-life of 5 years. Determine the mass remaining after 20 years. (2 marks)

  10. A certain I flowing through a wire of resistance R was increased seven times. Determine the factor by which the rate of heat production was increased. (3 marks)

  11. Figure 6, shows a horizontal conductor in a magnetic field parallel to the plane of the paper.

    induced current kcse 2010

    State the direction in which the wire may be moved so that the induced current in the direction shown by the arrow. (1 mark)

  12. An x-ray tube produces soft x-rays. State the adjustment that may be made so that the tube produces hard x-rays. (1 mark)

  13. The wavelength of a radio wave is 1 km. Determine its frequency. (2 marks)
    (Take the speed of light as 3.0×108 ms -1. )

  14. Figure 7 shows a block diagram of p-n junction diode.

    p-n junction diode kcse 2010

    On the diagram, show how the battery may be connected so that the diode is reverse biased. (1 mark)

    SECTION B (55 marks)
    Answer ALL the questions in this section in the spaces provided.

    1. Figure 8, shows a circuit that may have been used to charge a capacitor.

      milliameter kcse 2010
      1. State the observation on the milliameter when the circuit is switched on. (1 mark)
      2. Explain the observation in (a) above. (2 marks)

    2. The circuit in figure 8 is left on for some time. State the value of the p.d. across:
      1. the resistor R: (1 mark)
      2. the capacitor C. (1 mark)

    3. Sketch the graph of potential difference (V) across R against time. (1 mark)

    4. Figure 9 shows three capacitors connected to a 10V battery.

      capacitance kcse 2010

      Calculate:
      1. The combined capacitance of the three capacitors; (3 marks)
      2. The charge of the 5.0 μF capacitor. (3 marks)

    1. Figure 10 shows an object placed in front of a converging lens of focal length 50mm.

      converging lens kcse 2010

      1. On the same figure, draw a ray diagram showing the location of the image. (3 marks)
      2. Use the ray diagram to determine:
        1. Image distance; (1 mark)
        2. Magnification. (2 marks)

      3. State the adjustment that should be done to obtain a larger virtual image using the same lens. (1 mark)
      4. State one application of the arrangement in figure 10. (1 mark)

    2. Figure 11 shows a pin 60 mm long placed along a principal axis of the lens used in part (a). The near end of the pin is 80 mm from the lens.

      converging lens kcse 2010

      Determine the length of the image. (5 marks)

    1. Figure 12 shows an electrical circuit including S1, S2, S3, and three identical lamps L1, L2, L3. A constant potential difference is applied across X and Y.

      potential difference kcse 2010
      1. Other than L1, state the lamp that will light when S1 and S2 are closed. (1 mark)
      2. How does the brightness of L1 in (i) above compare to its brightness when all the switches are closed? (1 mark)
      3. Explain the observation in (ii) above. (1 mark)

    2. Figure 13 shows a cell in series with a 3Ω resistor and a switch. A high resistance voltmeter is connected across the cell.

      electromotive force kcse 2010

      The voltmeter reads 1.5V with the switch open and 1.2V with the switch closed.
      1. State the electromotive force of the cell. (1 mark)
      2. Determine the current through the 3Ω resistor when the switch is closed. (2 marks)
      3. Determine the internal resistance of the cell. (2 marks)

      1. Another resistor R is connected in series with the 3Ω resistor so that a current of 0.15A flows when the switch is closed. Determine the resistance of R. (3 marks)

  15. Figure 14a, is a diagram of a cathode ray tube, M and N are parallel vertical plates.

    cathode ray tube kcse 2010

    1. When switch S is open, a spot is seen at the centre of the screen as shown in figure 14(b).
      1. State what happens to the spot when S is closed. (1 mark)
      2. State what would happen to the spot if the potential difference across MN is increased. (1 mark)
      3. State what would be seen on the screen if the battery is replaced with an alternating emf of:
        1. A low frequency about 1 Hz: (1 mark)
        2. A high frequency about 50Hz. (1 mark)

    2. Explain the process by which electrons are produced at F. (2 marks)
    3. State with a reason how the brightness of the spot can be increased. (2 marks)
    4. The accelerating voltage of the tube is 1000V and the electron current in the beam is 1.5mA. Determine the energy conveyed to the screen per second. (2 marks)

    1. State the property of radiation that determines the number of electrons emitted when a radiation falls on a metal surface. (I mark)

    2. Figure 15 is a graph of the supporting potential Vs against frequency in an experiment on photoelectric effect.

      stopping potential kcse 2010

      1. What is meant by stopping potential? (1 mark)
      2. Given that the stopping potential Vs is related to the frequency by the equation.

        kcse10pp2q19bii where e is the charge of an electron, (e = 1.6 x 10-19C)

        Determine from the graph:
        1. Plank’s constant, h; (4 marks)
        2. The work function ωo for the metal in electron volts (eV). (3 marks)

MARKING SCHEME

  1.                
    • Reflected ray rotates 2 x 20 = 20°
    • Find deviation = (80° +20") = 100°
  2. Any slight deviation of the N-pole to the right
  3. Correct poles  √1 Correct direction + pattern√1
    q3 ajgdytag
  4. Initially attracted because of opposite charge √1
    (+ve or -ve)
    Then neutralised and charged positive and hence repel √1
    Charging by contact and law of electrostatics. √1 
  5. Distance = 2f = 2 x 25 √1/2= 50cm√1/2
    Alternative Just 50cm√1
    2 x 25 = 50cm√1/2
    Or 
  6. implies low current √1 So reduces √1 heat losses/ power loss Or
    I2R loss reduced
    P = I2R should be accompanied by power loss
    NB Heat losses/ Power Loss
  7. Mare practicel relationship between f and t
    Displacement
    q7 aygudyagd
  8. V1 =  FT1   or     η = T0
                                   T1
    V2 = FT2            η = 18
                                 14.4
    =1.25
    Accept all expressions
  9. 20g → 10g → 5g → 2.5g → 1.25g
    Mass remaining 
  10. lo - Initial current          I2 = 710
    P= I2R = I20R              PE (710)2R = 4912OR
    Power is 49 times the initial value
    Apply the power formula
  11. Motion out of paperi moves upwards.  Or Increases in p.d increases heating effect 
  12. Increasing the accelerating voltage OR Increase the P.d between a anode and cathode.
    Accept extra high tension increased
  13. f = = c
         λ     λ
    = 3.0 x 108         3.0 x M05Hz
         1000
  14. Look for biasing only. (any other device that does not affect the working should be ignored eg. diodes/ resists.
    q14 jagtaugduad
  15.                  
    1.                      
      1. Current fails off to zero falling to zero/ deflects to max. then zero
        Reducing gradually or after sometime
      2. current flows when the capacitor is charging 
        When fully charged capacitor stops (No current) and P.dis equal to the charging voltage
    2. Vc = 5V
    3. Touch both axis, Award for no labelled axis
      q15 aytfdtad
    4.                          
      1. 1 = 1  + 1  = 5 + 4  = 9
        cs   4     5         20     20
        Cs = 20
                 9
        C120 +3  = 5.22μF
                  9
        Accept 5.22μF only
      2. Change on series section = Q = Cv
        =20  ×  10 μc
           9 
        =22.2μc or
        Q series = QT - Q3μF
        = (5.22-3) x 10   μC1
        = 22.2 μC1
        Charge is the same on series section hence charge on 5.0μF is 22.246641
        Accept 22.2μC only
  16.              
    1.                
      1. Each rayamk indipendenih imkon dotted extrapolation
        1mk for datted image
        q16 aytfy6tfad
      2.                
        1. 50mm                 ±5mm 
        2. M = v = h1 = 50 = 2mm  ±0.2
                 h    ho    25
      3. Move the object towards F but not beyond
        Move obiect away from lens 
      4. no answer
    2. 1 1 + 1
      f      u    v
      11 + 1
      50   80   v
      V = 400/3
  17.                        
    1.                      
      1. L2
      2. Brighter
      3. Total resistance is less/reduced
    2.                
      1. 1.5V
      2. Ir = 1.5 - 1.2 = 0.3
        0.4r = 0.3
        V = 0.75R
        P.d and E.M.FI more practice and practical approach
    3. Rr = 3 +0.75 + R                0.15(R+3.75) = 1.5
      RT = R+3.75                      R +3.75 = 1.5 =  10
                                                              0.15
      E=IR                                 R = 10 -3.75
      1.5 = I(R+ 3.75) = 6.25Ω
      Or
      R = E = 1.5 = 10
            f     0.15
      R = RT - (V  + 3)  + 3.75
      R = 6.25Ω
      1.5 - 0.75 x 0.15 = I(3 + R)
      1.5 -0.1125 = 0.15(3 + R)
      1.3875  = 3 + R
        0.15
      R = 6.25Ω
  18.                        
    1.                        
      1. Deflected towards +ve plate (N)
      2. Deflection will be greater 
      3.              
        1. Spot moves back and forth
          To and fro (Not along across) 
        2. There will be a horizontal line 1 
    2. Electrons are given off as a result of heat produced by current
      So that electrons gain energy to break off from the surface
      Thermionic emission
    3. Increasing the filament current so that more electrons are released p.d across filament is increased so that more electrons are released
    4. P= IV
      = 100 x 1.5 x 10-3
      = 1.5J/s
      Accept J, W, J/S 
  19.              
    1. Intensity of radiation 
    2.                      
      1. (Min p.d)
        Negative potential sufficient to just stop the movement of electrons. 
      2.                    
        1. Gradient =
                          e
          h =            3.0 - 0                  =     3       
                  (12-14) x 76 x1014             7.6 x 1014
          Gradient = 0.3947 x 10-14 
          h = 0.3947 x 10-14 x 1.6 x 10-19
          = 0.6316 x 10-33 = 6.316 x 10-34
        2. =w=1.751.
            e
          W0 = y-intercept x e
          = 1.75 x 1.6 x 10-19
                  1.6 x 10-19
          =1.75ev
          Alternative
          = W0 = hf0
          =6.32 x 0.4 x 1014 x 10-34
                     1.6 x10-19
          =1.74ev
          OR
          Range 1.7 →1.8ev
          wo = y intercept =  1.75
            e
          Wo =  - 1.75  or we  = 1.75
          a                       e
          Wo =  1.75ev     OR   Wo = 1.75v  x  e
                                            = 1.75ev
          OR
          -Wo = 1.75 -1.75ev
            e
          OR
          Wo = 1.75  →  rej: 1.75v = Wo
          penalise -ve and units in
          Wo =  Y intercept
          = - 1.75

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