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Physics Paper 2 Questions and Answers - MECS Cluster Joint Mock Examinations 2022

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QUESTIONS
SECTION A: (25MARKS)

  1. What is observed when the hole of a pinhole camera is enlarged? (1mk)
  2. State one use of a charged electroscope (1mk)
  3. The chart below shows an arrangement of a section of the electromagnetic spectrum
    P Q R UV Light S Gamma rays
    Name the radiation represented by letter Q   (1mk)
  4. Draw a circuit diagram to show P-N junction diode in the forward biased mode. (2mks)                                                                                                       
  5. Explain why the walls of studio are padded with woolen materials (1mk)                           
  6.      
    1. Define half- life as used in radioactivity (1mk)
    2. The initial mass of a radioactive substance is 20g.The substance has half-life of 5yrs.Detemine the mass remaining after 20yrs.(2mks)
  7. Give a reason why it is necessary to leave the caps of the cells open when charging lead-acid accumulator(1mk)
  8.      
    1. State one property of soft iron that makes it suitable for use as a transformer core.(1mk)
    2. The primary coil of a transformer has 1200 turns and the secondary coil has 60 turns. The transformer is connected to a 240V a.c source. Determine the output voltage.(3mks)
  9. State two ways of minimizing electrical power losses during transmission (2mks)
  10. A convex mirror is preferred to a plane mirror for use as a driving mirror. Explain why. (1mk)
  11. An electric bulb is rated 60W, 240V. Determine the current that flows through it when it is connected to a 240v supply (2mks)
  12. The figure below shows a defect of vision
    1
    1. Name the defect. (1mk)
    2. List two possible causes of the defect. (2mks)
  13. A broadcasting station produces radio waves of wavelength 600m. Determine their frequency in MHz (speed of air is 3X108m/s)(3mks)                        

SECTION B (55MARKS)

  1.      
    1. Define the refractive index of a medium (1mk)                                                                       
    2. The figure below shows a ray of light incident on a glass-air interface
      2
      Given that the refractive index of the glass is 1.5, determine angle θ(3mks)
    3. State one condition for total internal reflection to occur  (1mk)
    4. The diagram below shows a narrow beam of white light shone onto a glass prism
      3
      1. What is the phenomena represented in the diagram? (1mk)
      2. Name the colours at A and at B (2mks)
      3. Explain the reason for your suggestion of the colours named above.  (1mk) 
      4. What is the purpose of the slit. (1mk)
  2.      
    1. State one factor that affect the resistance of a metallic conductor (1mk)
    2. The figure below shows resistors in a circuit. The internal resistance of the battery is negligible
      4
      1. Calculate the effective resistance of the circuit  (2mks)
      2. Find the total current in the circuit  (2mks)
      3. Find the P.d between P and Q  (2mks)
    3. What is the effect of decreasing the distance between the plates of a parallel plate capacitor on the capacitance (1mk)     
    4. The figure below shows electrical circuit with three capacitors A, B and C of capacitance 5µF, 6µF and 4µF respectively connected to a 12V battery
      5
      Determine
      1. The combined capacitance of the three capacitors  (2mks)
      2. The potential difference across the capacitor B (3mks)
  3.      
    1. Name two factors which determine the frequency of sound from a stretched wire at room temperature(2mks)
    2. The figure below shows two loud speakers S1 and S2 connected to a signal generator
      6
      1. An observer walks along B1B2. State what is observed.(1mk)
      2. Give reasons for observation above (2mks)
      3. Another observer walks along AA1, state and explain what he observed (2mks)                                                                                             
    3. A stretched string is vibrating between two fixed ends. The figure shows how the string is vibrating
      7
      1. State the name of:  (2mks)
        1. Distance a: 
        2. Distance b:
      2. On the diagram, label the node and the antinode  (2mks)
  4. The figure below shows a block diagram of a cathode ray oscilloscope(CRO)
    8
    1.          
      1. State the names of the parts labelled B1 and B2   (2mks)
      2. State and explain the function of the part marked A  (2mks)
      3. Why is the tube highly evacuated? (1mk)    
    2. Give a reason why the target in an X-ray tube is made of tungsten or molebdnum  (1mk) 
    3. X-rays are used for detecting cracks inside metal beams.State with a reason which type of X- rays is used. (2mks) 
    4. In a certain X-ray tube the electrons are accelerated by a p.d of 12kV.Assuming all the energy goes to produce X-rays, determine the frequency of the X-rays produced .(Planks constant h=6.62x10-34 Js and charge of an electron =1.6x10-19C)(2mks)    
  5.        
    1. state two factors that affect photoelectric emission  (2mks)
    2. Light of wavelength 4.0x10-7 m is incident on two different metal surfaces ,nickel and potassium (Take speed of light as 3.0x108 m/s and planks constant h=6.63x10-34Js)
      1. Determine the energy of the incident radiation (3mks)
      2. If the work function of nickel is 8.0x10-19 J and that of potassium metal is 3.68x10-19J,state with a reason which of the two metals given light will eject electrons(2mks) 
      3. Determine the velocity of the emitted electrons from the metal surface in b(ii).(Take mass of an electron as 9.1x10-31 kg)(3mks) 

MARKING SCHEME

  1. The image becomes brighter and blurred /Not clear.
  2. Detect type of charge
    Identify conductors and insulator
  3. Infrared radiation

  4. 9
    diode symbol (1mk)
    forward biasing(1mk)
  5. To absorb incident sound waves hence preventing the formation of an echo
  6.        
    1. Time taken for half the number of nuclides initially present in a radioactive sample to decay.
    2. No. of  halflives =20/5 =4 
      20g -------10g-------5g----2.5g-----1.25g
      OR
      N=NO(½)T/t =20(½)20/16 =1.25g
  7. To allow hydrogen gas produced at the electrodes to escape
  8.        
    1. Iron is easily magnetized and demagnetized/it’s a soft magnetic material.
    2. Vs/Vp =Ns/Np
      Vs =240x 60/1200
      Vs=12V
  9. Stepping-up the voltage
    Using thicker aluminum cables for transmission
  10. Convex mirror provides a wider field of view
  11. I=P/V = 60/240 =0.25A
  12.      
    1. Long sightedness (hypermetropia)
    2. Long focal length
      Short eyeball         
  13. V=fλ
    f=V/λ
    f =3 X 108=50 000Hz =5.0X105Hz =0.5MHz
          600
  14.      
    1. It is the ratio of sine of the angle of incidence to the sine of angle of refraction for a ray of light that is travelling from air to a medium.
    2. ang = sin90    1.5 =   1     sin θ =  1   = 0.6667        θ=41.810
               sinθ              sin θ             1.5
    3. Light must travel from optically denser medium to a less dense medium/rare medium OR The angle of incidence in the optically denser medium must be greater than the critical angle
    4.      
      1. Dispersion of white light
      2. A: Red
        B: Violet
      3. Red light is the least deviated colour while violet is the most deviated colour. This is because red light travels at a greater velocity in glass than violet
      4. It ensures the beam of the white light is narrow and direct
  15.        
    1. length of the metal
      cross-sectional area of the metal
      temperature of the metal
      type of the metals
    2.    
      1. RE=4 X 6 + 5.6 RE = 2.4+5.6 = 8.0 Ω
               4 + 6
      2. V=IR
        I=V/R I=6/8 =0.75A
      3. P.d at 5.6 Ω                        OR 
        V=0.75X5.6 = 4.2V           V=IR
        P.d between P and Q     0.75 X 2.5
        6 - 4.2                              =1.8V
        = 1.8V
    3. Capacitance increases
    4.      
      1. Capacitors in parallel
        = 6+5
        =11µF
        Capacitors in series
        = 4X11 =44/15 =2.933µF
            11+4
      2. Q=CV
        =2.933 X 10-6 X 12
        =3.5196 X 10-5C
        V=Q/C
        =3.5196 X 10-5
              11 X 10-6
        =3.1996V     
  16.    
    1. Tension of the wire
      Mass per unit length / thickness of the wire
      Length of the wire
    2.      
      1. The observer experiences alternate loud sound and soft sound
      2.      
        • Loud sound was due to constructive interference where crests of one wave meets the crest of another wave/
        • Trough of one wave meets the trough of another wave/
        • Two waves arrive at a point in phase
          Soft sound was due to destructive interference where crest of one wave meets the trough of another wave.
      3. The observer will hear loud sound all through
        This is the locus of point equidistant from the two sources where the path difference is zero and constructive interference occurs all through
    3.      
      1.    
        1. Amplitude
        2. Wavelength
      2.      
  17.    
    1.      
      1.  B1 – Y-plates
        B2---X- plates                                                                                                
      2. A is Grid –It controls the brightness of the spot on the screen. When made more negative ,less electrons flow and brightness reduces.When grid made less negative ,more electrons flow and the brightness increases.
      3. To prevent the cathode rays from losing energy due to collision  with air molecules in the tube
    2. Tungsten or molebdnum have very high melting point hence they can withstand high temperatures
    3. Hard X-rays – They have higher penetrating power.
    4. ev =hf 
      1.6x 10-19 x12000 = 6.62x10-34xf
      f =1.6x10-19 x 12000
                 6.62x10-34
      f =2.9x 1018Hz
  18.      
    1. Type of the metal
      Frequency/Energy of the radiation 
      Intensity of the radiation
    2.      
      1. E = hc/λ
        E = 6.63 X 10-34 X 3.0 X 108
                        4.0 X 10-7
        E = 4.9725 X 10-19J
      2. Potassium 
        This is because the energy of the incident radiation is greater than the work function of potassium.
      3. E =Wo +K.E
        K.E =E –Wo
        =4.9725x10-19 -3.680x10-19
        =1.2925x10-19J 
        1.2925X10-19 = ½ X 9.1X 10-31 XV2
        V2= 1.2925 X 10-19
                  4.55 X 10-31
        =2.84066 X 1011
         V =5.330m/s