Physics Paper 1 Questions and Answers - Sunrise Evaluation Pre Mock Exams 2021

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INSTRUCTIONS TO THE CANDIDATES:

  • Write your name and index number in the spaces provided above.
  • Answer all the questions both in section A and B in the spaces provided below each question
  • All workings must be clearly shown; marks may be awarded for correct steps even if the answers are wrong.
  • Mathematical tables and silent electronic calculators may be used.
    (Take acceleration due to gravity g= 10ms-2 Density of water 1g/m-3)

For Examiners’ Use Only

SECTION  QUESTION  MAXIMUM SCORE  CANDIDATE'S SCORE 
Section A 1-12  25  
Section B  13  09  
   14  14  
   15  14  
   16  09  
   17  09  
   18  10  
  Total 80  


QUESTIONS

SECTION A (25 MARKS)
Answer all questions in this section in the spaces provided

  1. State Pascal’s principle of transmission of pressure in fluids (1mk)
  2. Water is known to boil at 100oC . a student heated some water and noticed that it boiled at 101oC. State two possible reasons for this observation (2mks)
  3. Fig 1. Shows a flask filled with water. The flask is fitted with a cork through which a tube is inserted. When the flask is cooled, the water level rises slightly, then falls steadily.
    Q3 JGAYTGDA
    Explain the observation (3mks)
  4. A pipe of radius 4mm is connected to another pipe of radius 6mm. if water flows in the wider pipe at the speed of 5ms-1, what is the speed in the narrower pipe? (3mks)
  5. The system in figure2 is in equilibrium
    Q5 YRFDA
    When the temperature of the water is raised the system is observed to tilt to the right, state the reason for this observation (2mks)
  6. Explain why a glass container with glass walls is more likely to crack than one with a thin wall when a very hot liquid is poured into them. (2mks)
  7. State two ways in which the stability of a body can be increased (2mks)
  8. The object in figure 3 is placed on a horizontal table with a chain hanging from its centre of gravity. State the type of equilibrium for the object (1mk)
    Q8 JGVFHAGFDA
  9. A ball is thrown upwards and returns to its starting point after 6 seconds. Calculate the maximum height reached(g=10m/s2) (3mks)
  10. The figure below shows a force pump
    Q10 UJYGAUTGD
    Explain how the water gets past valve V2 (2mks)
  11. When calibrating a liquid in glass thermometer, it is normally not advisable to dip the bulb in boiling water when getting the upper fixed point. Explain why it is so (2mks)
  12. Convectional and diffusion both involve motion of fluid molecules. Distinguish between the two (2mks)

SECTION B (55 MKS)

  1.              
    1. Define the term heat capacity (1mk)
    2. You are provided with the apparatus shown in Fig 5 and stop watch
      Q13 JYGATGDA
      Describe an experiment to determine the specific latent heat of steam, using the set up. In your answers clearly explain the measurements to be made and how these measurements could be used to determine 1 (6mks)
    3. A block of metal of mass 150g at 100ºC is dropped into a lagged calorimeter of heat capacity 40JK-1 containing 100g of water at 25ºC. The temperature of the resulting mixture is 34ºC. (Specific heat capacity of water=4200JK-1)
      Determine:
      1. Heat gained by calorimeter; (2mks)
      2. Heat gained by water; (1mk)
      3. Heat lost by the metal block; (1mk)
      4. Specific heat capacity of the metal block (3mks)
  2.                          
    1. Distinguish between solid and liquid states of matter in terms of intermolecular forces (1mk)
    2. In an experiment to estimate the diameter of an oil molecule, an oil drop of diameter 0.05 spreads over a circular patch whose diameter is 20cm
      Determine
      1. The volume of the oil drop (2mks)
      2. The area of the patch covered by the oil (2mks)
      3. The diameter of the oil molecule (2mks)
    3. State
      1. Any assumption made in (b) (iii) above (1mk)
      2. Two possible sources of errors in this experiment (2mks)
  3.                    
    1. State what is meant by centripetal acceleration (1mk)
    2. A block of mass 200g is placed on a frictionless rotating table while fixed to the centre of the table by a thin thread. The distance from the centre of the table to the block is 15cm. if the maximum tension the thread can withstand is 5.6N. determine the maximum angular velocity the table can attain before the thread cuts. (4mks)
  4.                        
    1. Define the term velocity ratio as used in machines (1mk)
    2. Figure 6 shows a block and tackle pulley system lifting a load of 500N
      Q16 YRFAYTFDYA
      1. Determine the velocity ration of the machine (1mk)
      2. If an effort of 120N is required to lift the load using the machines determine the efficiency of the pulley system (3mks)
      3. In the space provided below, sketch a graph of efficiency against load for the system. (2mks)
  5. A car of mass 2000kg travelling at 5m/s collides with a minibus of mass 5000kg travelling in the opposite direction at 7m/s, the vehicles stick and move together after collision. If the collision lasts 0.1 seconds
    1. Determine the velocity of the system after collision to 3 decimal places (3mks)
    2. Calculate the impulsive force on the minibus (3mks)
    3.                          
      1. Calculate the change in kinetic energy of the system to 5 significant figures (3mks)
  6.        
    1.                   
      1. State the law of floatation (1mk)
      2. Explain why a hollow metal sphere floats on water while a solid metal sphere of the same material sinks in water. (2mks)
    2. The figure 7 below shows a uniform block of uniform cross-sectional area of 6.0cm2 floating on two liquids A and B. The lengths of the block in each liquid are shown. Given that the density of liquid A is 800kg/m3 and that of liquid B is 1000kgm-3 determine the:
      Q18 UTAG YTDA
      1. Weight of liquid A displaced (2mks)
      2. Weight of liquid B displaced (2mks)
      3. Density of block (3mks)


MARKING SCHEME

  1. Pressure at a point in a fluid is transmitted equally to all points of the fluid and to the walls of the container.
  2. Atmospheric pressures is higher than normal/standard or boiling was below Pressure of impurities
  3. When flask is cooled it contracts/its volume reduces but due to poor conductivity of the glass/materials of the flask water falls as it contraction is greater than the of glass (3mks) marks are independent unless there is contradiction
  4.  π x 42 x v1=π x 62 x 5
  5. Heated water has lower density hence lower up thrust
  6. Glass is a poor conductor of heat
    For the thick glass inner wall gain heat and expands while the outer wall does not. The tension between the two walls breaks the glass
  7. Increasing the base area the centre of gravity
  8. Unstable equilibrium
  9. S=ut-1/2 gt2v=u-gtv=0
    S=60-1/2 x 10x 36 u=10x6
    360-180
    =180m
    =60
  10. During the down stroke V1 closes down due to its own weight. Pressure is increased in the chamber forcing V2 to open hence water flows past V2
  11. Water contains impurities which raise points
    Only steam is pure enough to give the exact value of the boiling point
  12. Diffusion occurs in all directions, molecules move in all directions
    Convetion occurs in one direction-upwards or downwards
  13. Heat energy required to raise the temperature of a body by 1 degree Celsius/centigrade of Kelvin
    Measurements or
    Initial mass of water and calorimeter M1
    Final mass of water and calorimeter M2
    Time taken to evaporate (M1-M2),t
    Heat given out by heater=heat of evaporation = ML
    Pt(m1-m2)1
    L=pt
    M1-M2
    1. C∆T=40x(34-25)=40x9=360J
    2. MWCW∆ T
      100x10-2x4.2x103(34-25)=3780J
    3. MmCm∆T or sum of (i) and (ii)
      =150x103xCm6 360+3780
      =9.9CmJ =4140J
    4. 150x10-3xCmx66=4140 heat lost = heat gained by water+ heat gained
      Cm = 4140                             9.9cm=360+3780
       150x10-3x60                           Cm =4140 
                                                          0.15x60
      418J/kgk                                  418J/Kgk
  14.                      
    1. In solids the molecules are held in position by intermolecular forces that are very large. In liquids the molecules are able to roll over one another since the forces are smaller
    2.                    
      1. Volume = 4/3πr3
        =4/3πx 0.253
        = 6.54x 10-5cm3 (2mks)
      2. Area = πr2
        π x 102
        =314cm2 (2mks)
      3. Ax diameter of molecule=volume
        314xd=6.54x10-5
        D=2.1x10-7cm (3mks)
    3.                        
      1. The soil is assumed to have spread to thickness of one molecule (1mk)
      2. Sources of errors
        • Getting the right oil
        • Measuring drop diameter
        • Measuring diameter of patch
        • Getting drop of a right size (any 2x1=2mks)
  15. Rate of change of velocity towards the centre
    • Acceleration directed towards the centre of the motion
    • Acceleration towards the centers orbit/nature of surface (1mk)
  16.                      
    1. The ratio of the distance moved by the effort to the distance moved by the load;( 1mk)
    2.                
      1. V.R=5
      2. Efficiency =(M.A)/(V.R )x100%
        500/120 x 1/5 x 100%
        =83.33%
        memo 15 jygadga
  17.                  
    1. 2000x5+5000x (-7)=v(2000+5000) √1
      V=(-25000)/7000√1
      =-3.571m/s √1 (3D.P a must)
    2. Ft=m(v-u) √ or F=(m(v-u))/t
      (5000(-3.571))/0.1
      =171,450N√
    3.            
      1. Initial K.E = ½ x 2000x 52 + ½ x 5000x (-7)2
        =147,500J√1
        Final = ½ x 7000(-3.571)2
        =44,632J√1
        Change= 44632-147,500
        =-102,868J√1
        Kinetic energy is converted to heat sound and deformation
  18.                
    1.                        
      1. A floating body displaces its own weight of fluid in which floats√1
      2. The weight of the solid sphere is more than the weight of the volume of water it displaces hence it sinks √ 1while the weight of the hollow sphere is equal to the weight of the volume of water it displaces hence it floats √1
    2.                  
      1. Weight = vol x density x J
        =6x10-4x2x10-2x10√1
      2. Weight =V xℓxJ
        6x10-4x2x10-2x10√1
        =1.2x10-1N√1
      3. Weight of block=weight of fluid displaced
        =1.2x10-1+9.6x10-2
        =2.16x10-1N
        mass of block= 2.16x10-1
                                 10√1
        =2.16x10-2
        Density = mass/Vol
              2.160x10-2
          6x10-4 x6x10-2
        =600kg/m3
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