Physics Questions and Answers - Form 4 Mid Term 2 2022

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QUESTIONS

  1.      
    1. Define the term principal focus as used in thin lenses. (2mks)
    2. A four times magnified virtual image is formed by an object place 12cm from a converging lens. Calculate the position of the image and the focal length of the lens. (4mks)
  2.    
    1. Describe how x-rays are produced. (2mks)
    2. Differentiate between hard X-rays and soft X-rays. (2mks)
    3. An X-ray tube is operating with an anode potential of 25kV and a current of 20 m A.
      1. Calculate the number of electrons hitting the target per second. (3mks)
      2. Determine the average velocity with which the electron strike the target (e = 1.6 x 10-19c, mass of electron = 9.1 x 10-31kg) (3mks)
  3.    
    1. Define the term threshold frequently. (1mk)
    2. The figure below shows a set up used to demonstrate photoelectric effect using a photocell.
      1
      1. Explain why current flow when uv is shown on the part labeled A. (2mks)
      2. Explain why u.v and not infrared radiation is used. (1mk)
      3. Give one reason why the photocell is evacuated. (1mk)
    3. In an experiment to observe photo-electric emission from a clean caesium surface, the following readings were observed.
      Stopping potential (v) 0.6 1.0 1.4 1.8 2.2
      Frequency (x1014)Hz 6 7 8 9 10
      1. Plot a graph of stopping potential (vs) against frequency. (4mks)
        From the graph;
      2. Threshold frequency of the surface (1mk)
      3. Threshold wavelength of the surface (c = 3.0 x 108 m/s) (2mks)
      4. Planck’s constant (2mk)
      5. Work function of the surface in ev (1mk)
  4.      
    1. Give two properties of cathode rays. (2mks)
    2. The figure below shows the trace on the screen of an ac signal connected to the y – plates of a CRO with the time base on:
      2
      Given that the time base control is 5ms/div and y- gain is at 100v/div, determine
      1. The frequency of the a.c. (3mks)
      2. The peak voltage of the input signal. (2mks)
  5.      
    1. Define the term electromagnetic induction. (1mk)
    2. Describe two factors that affect the strength of an electromagnet (2mks)
    3. A transformer with primary coil of 400 turns and secondary coil 200 turns in connected to 240v a.c mains. Calculate the secondary voltage. (2mks)
  6.      
    1. Describe how the following factors affects the centripetal force of a body. (2mks)
      1. Mass of the body.
      2. Radius of the path.
    2. A car of man 1200kg moving round a bend of radius 50m. If the coefficient of friction between the road and then tyre is 0.8, calculate the maximum speed at which the car should be driven at for it not to skid on the bend. (3mks)
  7. The following reaction is part of a radioactive series. Identify the radiation X and determine the values C and Z. (3mks)


MARKING SCHEME

  1.      
    1. This a point on the principal axis where rays parallel and close to it coverage after refraction for a convex lens and appears to diverge from for a concave lens.
    2.      
      1. Image distance
        m = 4 = v/4
        U = 12cm
        4 = v/12
        = 48cm
      2. Since the image formed in virtual, then V is negative
        I/f = I/u + I/v
        1/f = 1/12 + 1/-48
         I  = - 1 + 4
         f        48
        1/f = 3/48 = 1/16
        f = 16cm
  2.        
    1. Fast moving electrons from the cathode which are thermionically emitted are stopped by a metal target. Part of their K.E is converted to X-rays.
    2.      
      1. Hard X-rays have very short wave length while soft X-rays have relatively longer wavelength.
      2. Hard are produced by a high acceleration voltage while soft are produced by low accelerating voltage.
      3. Hard has a high penetrating power while soft has low penetrating power.
    3.        
      1. Current I = 20 m A = 20 x 10-3
        Q = it also Q = en 
        20 x 10-3 x 1 = 1.6 x 10-19 x n
          20 x 10 -3=n
        1.6 x 10-19
        n = 1.25 x 1017 electrons
        Where n is the number of electron
      2. ev = ½ m 
        1.6 x 10-19 x 25, 000 = ½ x 9.1 x 10 -31 x
        4 x 10-15 = 4.55 x 10-31 x
           4 x 10-15 = V2
        4.55 x 10-31
        V2 = 8.791 x 1015
        V = 9.376 x 107m/s
  3.        
    1. This is the minimum frequency of the radiation that would cause omission of electrons from a metal surface.
    2.      
      1. The electrons on the surface of the cathode absorb the u.v light and are dislodged from the metal surface. The emitted electrons are then attracted to the anode causing the current to flow.
      2. Infrared has lower frequency than u.v therefore has low energy.
      3. To minimize the chances of the emitted electrons colliding with air which will reduce their kinetic energy.
    3. Plot a graph of stopping potential (vs) against frequency. (4mks)
      From the graph;     
      1. Threshold frequency of the surface (4.5 x 1014Hz) (1mk)
      2. Threshold wavelength of the surface (c = 3.0 x 108 m/s) (6.66 x 10-7m) (2mks)
      3. Planck’s constant (6.56 x 10-34 js) (2mk)
      4. Work function of the surface in ev (1.845ev) (1mk)
  4.      
    1.      
      • Travells in a straight line
      • Possesses kinetic energy
      • Deflected by both electric and magnetic fields
    2. The frequency of the a.c. (3mks)
      1. Time base control = 5ms/div
        Number of divisions covered = 8
        Total time = 5 x 8 = 40ms
        = 40 x 10-3 sec
        Number of cycles = 2
        Periodic time = 20 x 10-3 sec
        F = 1/T =     1      = 50Hz
                      20 x 10-3
      2. Y – gain = 100 v/div
        Dif lection = 3 div from zero level
        Peak voltage = Y – gain x number of division
        = 100 x 3
        = 300 v
  5.        
    1. This is the production of electricity/ voltage in a conductor situated in a changing field or a conductor moving through a stationary magnetic field.
    2.      
      • Strength of the magnet
      • Number of turns in the coil
      • Speed at which the wire cut the field.
    3. VP = NP
      Vs    Ns
      240 = 400 Vs = 120V
       Vs     200
  6.      
    1.         
      1. Mass of the body.
        The bigger the mass the bigger centripetal force.
      2. Radius of the path.
        The smaller the radius of the path, the bigger the centripetal force.
    2. U = fr/R =   fr/mg Fr = mv2/r v2 = 400
      Fr = ymg   9600 = 1200 x v2     v = 20m/s
                                      50
      = 0.8 x 1200 x 10
      = 9600N 
      = v2 = 9600 x 50
                    1200
  7. 21083A → 210 84B → czQ
    X – Beta practice
    C – 206
    Z - 82

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