PHYSICS PAPER 2 - 2020 KCSE PREDICTION SET 1 (QUESTIONS AND ANSWERS)

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INSTRUCTIONS TO CANDIDATES

  1. This paper consists of two sections, A and B
  2. Answer all the question in both sections
  3. All working must be clearly shown 
  4. Mathematical table and silent electronic calculators may be used.

SECTION A (25 MARKS)

Answer all questions in this section

  1. In a certain pinhole camera, the screen is 4cm from the pinhole. When the camera is placed 6m away from a tree, a sharp image of the tree 16cm high is formed on the screen. Determine the height of the tree (2 mks)
  1. A student set up the apparatus shown below to magnetise a steel bar
    phyc2fret
    State two mistakes in the set- up (2 mks)
  1. When a bulb is connected between two plates of a simple cell the bulb lights up. However, the brightness of the bulb fades after a while
    1. Explain why the bulb lights up (1 mk)
    2. Why does the brightness fall (1 mk)
  1. An object is placed infront of a diverging lens and forms an image at the position shown in the diagram below.
    phyc4fret
    Use a ray diagram to locate the principal focus of the diverging lens above (2 mks)
  1. An electric heater rated 240v, 3kw is to be connected toa 240v mains supply through a 10A fuse. Determine whether the fuse is suitable or not (2 mks)
  1. The figure below shows a current carrying coil in a magnetic field. The direction of current and the resultant force are shown. Study the figure and answer the questions below
    phyc6fret
    1. Label the poles marked A and B
    2. Explain the purpose of the split ring in the diagram (2 mks)
  1. The figure below shows a displacement time graph for a wave motion
    phyc7fret
    What is the frequency of the wave (2 mks)
  1. The figure below shows a series of wave fronts one wavelength apart approaching a gap between barriers in a ripple tank
    phyc8fret
    On the same diagram, show what happens when the waves pass through the gap (1 mk)
  1. A thunderclap is heard 3.2 seconds after the lighting is seen. If the speed of sound is 340m\s, determine the distance of the lighting flash (2 mks)
  1. State one cause of power loss in a long distance transmission wires and how this loss can be minimized (2 mks)
  1. For a given source of x-rays, how would the following be controlled
    1. Penetrating power (1 mk)
    2. Intensity (1 mk)
  1. In a certain radioactivity uranium- 238 undergoes decay to become Thorium – 234 as shown below
    phyc12fret
    Identify the radiation X emitted (1 mk)
  1. What is doping as applied in electronics (1 mk)

 SECTION B: (55MKS)

Answer all questions in this section

  1.  
    1. The figure below shows a modern X-ray tube
      phyc14fret
      1. Name the part labeled C (1 mk)
      2. Name a metal that can be used as B in the diagram and state the reason why it is appropriate (2 mks)
      3. Why is the tube evaluated (1 mk)
      4. Given that 99% of the energy of the electrons to converted into heat while the rest is converted into x-rays, determine the frequency of the x-rays produced given that the accelerating potential is 100KV (plank constant = 6.63 x 10 -34 Js, e = 1.6x 10-19c ) (3 mks)
      5. State one application of x-rays in medical field (1 mk)
    2.  
      1. state two characteristics of electromagnetic radiation (2 mks)
      2. Arrange the following electromagnetic radiations in order of their increasing wavelength infrared, ultra violet, gamma rays, X- rays (1 mk)
  1.  
    1. State the Ohm’s law (1 mk)
    2. the figure below shows an electric circuit
      phyc15bfret
      1. calculate the effective resistance of the circuit (2 mks)
      2. The current flowing through the 8Ω resistor (2 mks)
      3. The current through the 6Ω resistor (2 mks)
      4. In large current circuits, large resistors in parallel are preferred to low resistors in series Explain (2 mks)
    3. A 2 µF capacitor is connected in series with a 3µF capacitor as shown below.
      phyc15cfret
      Determine
      1. The effective capacitance of the circuit (2 mks)
      2. The charge stored in the 2µF capacitor (2 mks)
  2.  
    1. state two factors that affect photoelectric effect (2 mks)
    2. The diagram below shows a graph of kinetic energy KE against frequencyfof incident radiation for a given metal surface 
      phyc16fret
      Given the equation of the graph is KE = hf –W0
      Use the graph to determine
      1. Plank’s constant h  (3 mks)
      2. threshold frequency f0 (1 mk)
      3. Work function  W0 (1 mk)
    3. A surface whose work function w0 = 6.4 x 10-19 J is illuminated with light of frequency 3.0 x 10-15 Hz. Find the minimum KE of the emitted electrons ( h= 6.4 x 10-34 Js) (4 mks)
  3.  
    1. State the lenz’s law of electromagnetic induction ( 1 mk)
    2. A bar magnet is moved into a coil of an insulated copper wire connected to a zero centre galvanometer as shown below
      phyc17bfret
      1. show on the figure above the direction of the induced current in the coil (1 mk)
      2. State and explain what is observed on the galvanometer when the south pole of the magnet is moved into and then withdrawn from the coil (2 mks)
    3. A transformer has 800 turns in the primary and 40 turns in the secondary winding. The alternating voltage connected to the primary is 240V and current of 0.5.A. Determine
      1. the power in the secondary if transformer is 95% efficient (2 mks)
      2. explain how power losses in a  transformer  are reduced by having
        1. soft iron  core (1 mk)
        2. lamination (1 mk)
  4. The figure below shows the electric wiring of an electric iron box
    phyc18fret
    1. identify wires P and R (2 mks)
    2. The heating element of the iron box in (i) above is made of a wire of resistance 28.8 Ω and is connected to a 240V mains supply.
      Determine
      1. The power rating of the iron box (3 mks)
      2. The current flowing in the circuit (2 mks)
      3. The cost of using the iron box for half  an hour per day for 30days if the cost per unit is sh 12.50 (3 mks)

MARKING SCHEME

  1.  
    PHYC1DRE
  1. Used an a.c instead of d.c  √1
    Placed the steel bar in East- West direction√1
  1.  
    1. Due to flow of charge/ current through it√1
    2. Polarization occurs/ bubbles covers the positive plate
  2.  
    PHYC4DRE       
  1. P = IV
    3000 = I x 240
    I = 12.5A √1
    hence not suitable  √1
  1.  
    1. A= North 
      B= South √1
      Deny if north–north or south – south
    2. Helps reversing the direction of the current in the coil  √1
      This ensures that the force on the coil remain the same direction.√1
  2.  
    PHYC7DRE 
  3.  
    PHYC8DRE
  1. Speed =distance/time
    340m/s =distance/3.2
    Distance = 340x 3.2 = 1088m  √1
  1. Heating
    High resistance   √1
    Transmitting at high voltage /low current   √1
    Use of thicker cables
  1.  
    1. Accelerating voltage √1
    2. Heating current √1
  2. Alpha radiation√1
  3. Doping is introduction of a foreign atom into crystal of pure – semi conductor .√1
  4.  
    1.  
      1. copper anode 
      2. molybdenum / tungsten
        – has high melting point 
      3. To maintain high speed of the electrons/ to maintain high energy of the electrons. 
      4.  
        PHYSICSP214  
      5. -Treatment of cancer
        -x- ray photograph                          any one 
        -sterilizing medical equipment
    2.  
      1. Travels with velocity of light
        -Carry no charge                                                                    any two
        -Possess energy                                                                      @√1
        -Do not get deflected by magnetic or electric field
      2. Gamma ray, x-rays ultraviolet, infrared .
  5.  
    1. Current flowing through a metallic conductor is directly proportional to the p.d across its ends provided temperature and other physical factors are kept constant. √1
    2.  
      1. RE =12×6 + 8 
               12+6
        = 12 Ω
      2. I = V/R
        24/12
        = 2A

        P.d across 8 Ω resistor = 2 x 8 = 16V √1
        Current across the 8 Ω resistor

        I = V/R
        16/8
        = 2A
      3. P.d across 6Ω resistor = 24-16= 8V    √1
        OR

        V = IR
        2 x 12×6  = 2 x 4 = 8V
              12+6
        I = V/R1
        8/6
        = 1.333A √1
        OR

        12/18 X 2A = 1.333A
      4. The effective resistance reduces when large resistors are arranged in parallel. √1
        Since current is divided this also reduced the overheating effect due to high resistance    √1
    3.  
      1.  
        PHYC3456DRE
      2. Q = CV
        = 1.2 x 10-6 x 3
        = 3.6 x 10-6C   
  1.  
    1. Energy of incident radiation
      Type of metal      any two @ √1
      Intensity of incident radiation
    2.  
      1. h = gradient of graph √1
        = 6.667 x 10-34 JS √1
      2. fo= x- intercept √1
        = 3 x Hz  
      3. Y – intercept = wo
        = 2 x J
    3. wo = hfo
      f0 = W0 = 6.4 × 10-19  1
              h     6.4 × 10-34

                         
      = 1.0 x 10-15 Hz   √1
      k.e = hf – hfo
      =h (f – fo)

      = 6.4 x 10-34 ( 3.0 x 1015 – 1.0 x 1015) √1
      = 6.4 x 2 x 10-19J

      = 12.8 x 10-18J √1
  1.  
    1. The direction of the induced e.m.f is such that the current which it causes to flow causes magnetism which oppose the change producing it. √1
    2.  
      1. induced current flows in the clockwise direction. √1
      2. pointer of the galvanometer moves anticlockwise when south poles moves inward; end becomes south to oppose the motion of the magnet; √the pointer moves clockwise when south pole moves outward, end becomes north to oppose the outward motion of the magnet. √1
    3. power Ps =95/100 x IPx VP √1
      = 95/100x 240 X 0.5

      =114W √1                    
    4.  
      1. for quick magnetization and demagnetization √1
      2. to reduce eddy currents 
  2.  
    1. P – line/ live cable 
      R - Neutral
    2.  
      1. P= V2
             R

        P = 240x240
                 28.8
        P = 2000 W
      2. P = IV
        2000 = 240 x I 
        I = 8.333A √1
      3. Cost = number of units x cost per unit√1
        = 2000/1000 x 1/2  x 30 x 12.50√1
        = sh 375.√1

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Read 4700 times Last modified on Monday, 18 May 2020 06:15

2 comments

  • Comment Link PANASHE Tuesday, 02 February 2021 07:22 posted by PANASHE

    REALLY HELPED ME

  • Comment Link Eminnoh Sunday, 16 May 2021 06:23 posted by Eminnoh

    It's good

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