Physics Paper 1 Questions and Answers - Londiani Joint Mock Exams 2022

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QUESTIONS

SECTION A: (25 MARKS)

  1. A ball bearing is held between the anvil and spindle of a micrometer screw gauge as shown in the Figure 1 below.
    1
    What is the diameter of the ball bearing? (1 mark)
  2. State two properties of a liquid that is suitable for use in a thermometer. (2marks)
  3. A column of air 15cm is trapped by mercury thread of 10cm as shown below. Calculate the length of the trapped air when the tube is laid horizontally given that atmospheric pressure is 76cmHg. (3 Marks) 
    2
  4. Why are gases more compressible while liquids and solids are almost incompressible? (1mark)
  5. The graph shows variation of extension and stretching force F for a spring which obeys Hooke’s law.
    3
    1. Determine the spring constant in SI units. (1mark)
    2. The energy stored when the extension is 20cm. (2marks)
  6. The figure 2 below shows a rod made of wood on one end and metal on the other end suspended freely with a piece of thread so that it is in equilibrium.
    4
    The side made of metal is now heated with a Bunsen flame. State with a reason, the side to which the rod is likely to tilt. (2 marks)
  7. State one factor that would increase the surface tension of pure water in a beaker of water. (1 mark)
  8. The figure below (figure 3) shows a uniform metal rod balanced at its Centre by different forces.
    5
    Determine the value of T. (3 marks)
  9. Two rods of copper A and B of the same length but different thickness with candle wax attached to either end are heated as shown below.
    6
    State and explain the observation made. (2 marks)
  10. Figure 4 shows a manometer attached to a gas supply. If the atmospheric pressure is 1.0336 x 105Pa. Calculate the pressure of the gas supply.
    (Density of mercury = 13600kg/m3) (2 marks)
    7
  11. A block of wood measuring 0.8m by 0.5m by 2m floats in water. 1.2m of the block is submerged. (Density of water = 1000kg/m3 , g=10N/kg) Determine the weight of the water displaced. (3 marks)
  12. The figure 5 below shows two light sheets of paper arranged as shown.
    8
    Its observed that the papers move away from each other when strong air is blown at the same time behind paper Q and in front of paper R as shown. Explain (2 marks)

SECTION B (55 MARKS)

  1. A block and tackle is made up of three pulley wheels on top and two pulley wheels at the bottom in figure 6
    9
    1. Complete the diagram by drawing the chain which passes over the wheels and indicate where the effort is applied (2 marks)
    2. What is the velocity ratio (V.R) of the machine (1 mark)
    3. A load of 1120N is lifted by an effort of 250N
      Determine
      1. The mechanical advantage (M.A) of the system (2 mark)
      2. The efficiency, E, of the system (2 marks)
    4. Using the axes given below, sketch a graph of efficiency, E, against load (1 mark)
      10
  2.      
    1. What is meant by the term specific latent heat of fusion of a substance? (1 mark)
    2. Water of mass 200g at a temperature of 60ºC is put in a well lagged copper calorimeter of mass 80g. A piece of ice at 0ºC and mass 20g is placed in the calorimeter and the mixture stirred gently until all the ice melts. The final temperature of the mixture is then measured (Latent heat of fusion of ice = 334000Jkg-1, specific heat capacity of water = 4200Jkg-1K-1)
      Determine:
      1. The heat absorbed by the melting ice at 0ºC (2 marks)
      2. The heat absorbed by the melted ice (water) to rise to temperature T (2 marks)
      3. The heat lost by the warm water and the calorimeter (Specific heat capacity of the calorimeter = 900Jkg 1K-1) (3 marks)
    3. The final temperature T of the mixture (2 marks)
  3. A lead shot of mass 40g is tied to a string of length 70cm. It is swung vertically at 5 revolutions per second. (Take g=10m/s2)
    1. Determine;
      1. Periodic time, (1 mark)
      2. Angular velocity (2 marks)
      3. Linear velocity (2 marks)
      4. Maximum tension in the string. (2 marks)
    2. The figure 7 below shows a container with small holes at the bottom in which wet clothes have been put. When the container is whirled in air at high speed as shown, it is observed that the clothes dry faster. Explain how the rotation of the container causes the clothes to dry faster. (2 marks)
      11
  4.      
    1. Give a reason why the inside of a helmet is lined with sponge. (1 mark)
    2. The figure below shows a balloon filled with air.
      12
      When the mouth is suddenly opened, the balloon moves in the direction shown above by the arrow. Explain that observation. (2 marks)
    3. A rock of mass 150kg moving at 10m/s collides with a stationary rock of mass100kg. They fuse after collision. Determine the
      1. Total momentum before collision. (2 marks)
      2. Their common velocity after collision. (2 marks)
  5.        
    1. On the axis below, sketch a graph to show how the pressure of a fixed mass of a gas varies with volume at constant temperature. (1 mark)
      13
    2. The set-up below shows an arrangement that can be used to Verify Charles’ law.
      14
      1. State any one use of sulphuric acid index in the above set up. (1 mark)
      2. What is the use of the stirrer? (1 mark)
      3. State two measurements that should be taken in this experiment. (2 marks)
      4. Describe how the set up can be used to verify Charles’ law. (4 marks)
    3. The volume of a gas enclosed with a movable piston is 300 cm3 when the temperature is 290K. Determine the temperature at which the volume of the gas increases to 355 cm3 (Assume pressure does not change) (3 marks)
  6.      
    1. The section of the tape shown below was produced when a tape running down an incline plane was attached to a ticker-tape timer of frequency 50Hz.
      15
      1. Indicate above the tape the direction in which the trolley was moving. (1 mark)
      2. What type of current was used to operate the ticker timer? (1 mark)
      3. Find the acceleration of the trolley in SI units. (3 marks)
    2. A stone is projected vertically upwards with initial velocity of 40m/s from the ground.
      Calculate:
      1. Time taken to reach maximum height (2 marks)
      2. Maximum height reached (2 marks)


MARKING SCHEME

  1. 2.50 + 0.33 = 2.83mm
  2.    
    • does not wet the glass
    • has a wide range of temperature
    • expand and contract uniformly
  3. P1V1 = P2V2
    (76 + 10) x 15 = L x 10
    86 x 15 = 10L
    10L = 1290
    L = 129cm
  4. gases have larger intermolecular spaces than liquids and solids
  5.      
    1. K = F/e
      = 24 - 0
         12 - 0
      = 2N/cm
      = 200N/M
    2. E = ½Fe
      = ½Fe2
      = ½ x 200 x 0.22
      = 45
  6. It tilts in anticlockwise directions when the metal was heated it expands hence the position of CO
  7. lowering the temperature of water or cooling the water
  8. F1d1 + F3d3
    (4 x 0.35) + ( T x 0.5) = (8 x 0.4)
    1.4 + 0.5T = 3.2
    0.5T = 1.8
    T = 3.6N
  9. The wax on rod A takes off first then rod B later. A has a larger cross-sectional area hence conducts heat faster than B
  10. Patm + P4g = Pgas
    103360 + (0.5 x 13600 x 10)pg
    = 103360 + 68000
    = 171360Pa
  11. Volume of water displaced = 0.8 x 0.5 x 1.2
    = 0.4m3
    mass = ev = 1000 x 0.48 = 480kg
    weight = mg = 480 x 10 = 4800N
  12. when air is blown, it moves at high velocity on the outer sides of the paper producing a low pressure zone. The higher atmospheric pressure/between them pushes them out
  13.        
    1.    
      16
    2. 5
    3. M.A = 4E
      = 1120
          250
      = 4.48
    4. Efficiency = M.A x 100%
                         V.R
      = 4.48 x 100%
           5
      = 89.6%
    5.   
      17
  14.        
    1. This is the quantity of heat requiredd to change a unit mass of a substance from solid to liquid completely at a constant temperature
    2.         
      1. 0.02 x 334,000
        = 6680J
      2. Q = MCℑ
        0.02 x 4200 x T
        = 84TJ
      3. heat lost by water = 0.2 x 4200 x (60 - T)
        = 50400 - 840T
        heat lost by caloriemeter = 0.8 x 900 x (60 - T)
        = 4320 - 72T
        total heat lost = 50400 - 840T + 4320 - 72T
        (54720 - 912T)J
      4. heat lost = heat gained
        54720 - 912T = 6680 + 84T
        54720 - 6680 = 912T + 84T
        996T = 480
        T = 48.23C
  15.      
    1.      
      1. T= 1/t 
        = 1/5 = 0.25
      2. ω = 2πf
        = 2 x π x 5
        = 31.42rad/s
      3. v = wr
        = 31.42 x 0.7 
        = 21.99m/s
      4. Tmax = mv2/r + mg
        = 0.04 x 25.992 + 0.04 
        0.7
        = 27.64 + 0.4
        = 28.04n
    2. water being denser and massive, the clothes occupy the furthest end of the container hence spiling off through the holes
  16.      
    1. it increases the time of impact hence reducing the impulsive force produced during an accident
    2. leaving air exerts an action which produes an equal but opposite reaction free
    3.      
      1. momentum = M1U1 + M2U2
        = 150 x 10 + 100 x 0
        = 1500kgm/s
      2. initial momentum = final momentum
        1500 = (100 + 150)V
        260V = 1500 
        v = 6m/s
  17.         
    1.  
      18
    2.         
      1. indicate the level of trapped air or dry the trapped air
      2. to distribute the heat in the water evenly
      3. temperature of water
        height of trapped air
      4.    
        • initial in length of air column and temperature is recorded
        • water bath is heated and new height of air column is recorded with corresponding thermometer reading
        • this is replaced several times at suitable temperature intervals to obtain several pairs of values
        • a graph of volume against absolute temperature is plotted.  the graph is a straight line with a positive gradient 
        • this shows that the volume is directly porportional to absolute temperature
    3. V1/T1 = V2/T2 
      300/290 = 355/T2
      T2 = 355 x 290
      300
      = 343.17k
  18.        
    1.        
      1.    
        19
      2. alternating current
      3. u = 8/0.02 = 400cm/s
        = 4m/s
        v = 56 = 2800cm/s
        0.02 x 1
        = 28m/s
        a = v - u
                t
        =  28 - 4  
          0.02 x 6
        = 24/012 = 200m/s2
    2.      
      1. v = u - gt
        = 40 - 10t
        10t = 40
        t = 4s
      2. v2 = u2 - 2gs
        O2 = 402 - 2 x 10 x 5
        O = 1600 - 20s
        20s = 1600 
        s = 80M

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