Physics Paper 1 Questions and Answers - Form 4 Mid Term 2 Exams 2021

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PHYSICS
PAPER 1
FORM 4 MID TERM 2

INSTRUCTIONS

  • This Paper has Section A nd B. Answer all questions in Section A nd B
  • Working must be clearly shown.

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

  1. Figure 1 below shows a part of Vernier calipers used to determine the length of a metallic cube. If the cube has a mass of 1533g.
    1phyf4mt221p1q1
    Determine the density of the cubein g/cm3 (3marks)
  2. Figure 2 below shows a mercury manometer. Some dry gas is trapped in one of the limbs.
    1phyf4mt221p1q2
    Given that the atmospheric pressure is 76cm of mercury. Determine the pressure of the gas in mmHg (3 marks)
  3. When it is raining, it is advisable not to touch a canvas tent from inside. Explain. (1 mark)
  4. State the reason why it is easier to separate water into drop than to separate a piece of solid into smaller pieces. (1 mark)
  5. Figure 3 below shows a simple fire alarm.
    1phyf4mt221p1q5
    Explain how it works (2marks)
  6. It feels hotter to sit on a metallic chair that has been left in the sun for a long time than wooden bench at the same temperature. Explain(2 marks)
  7. State the principle of moments. (1 Mark)
  8. A uniform meter rule of mass 10g is balanced by masses 24 g and 16g suspended at 0 cm mark and 100cm mark respectively. Determine the position of the pivot. (2 marks)
  9. The figure below shows a compression spring, before and after a mass of 5kg was placed on it. Use it to answer questions 9, 10, and 11.
    1phyf4mt221p1q9
    Find the spring constant of the spring. (3marks)
  10. Sketch a graph of force against length if different masses were used in the above set up. (1 mark)
  11. Explain the shape of the graph above (1 mark)
  12. State Bernoulli’s effect (1 mark)
  13.      
    1. Water flows through a pipe of different cross-section areas as shown in the diagram below. Indicate in the diagram the levels of water in tubes A, B and C. (1 mark)
      1phyf4mt221p1q13a
    2. Give a reason for your answer in 13(a) above. (1 mark)
  14.      
    1. An electric heater is placed at equal distances from two similar metal cans A and B filled with water at room temperature. The outer surface of can A is shiny while that of can B is dull black. State with reasons which can will be at a higher temperature after the heater is switched on for some time. (2marks)
    2. Sketch a graph of temperature against time for can A and B after the heater is switched off. (1 mark)

SECTION B: (55 marks)

  1.      
    1. State Charles’ law. (1 mark)
    2. The Set up below was used to verify Charles’ law. Use it to answer the following questions.
      1phyf4mt221p1q15b
      1. State two measurements taken from the above set up. (2marks)
      2. Describe briefly how the set up above can be used to verify Charles’s law. (3marks)
      3. State the function of sulphuric acid index. (1 mark)
      4. Pressure of the trapped air remains unchanged throughout the experiment. Explain how this is possible. (2marks)
      5. A mass of 1200 cm3 of oxygen at 270 C and a pressure 1.2 atmosphere is compressed until its volume is 600 cm3 and its pressure is 3.0 atmospheres. Find the temperature of the gas after compression (2 marks)
  2.          
    1. Define the term heat capacity. (1 mark).
    2. In experiment to determine the specific latent heat of vaporization Lv of water, steam was passed into cold water in a copper calorimeter. The following data was obtained:
      Mass of calorimeter 105.2g 
      Mass of calorimeter + water =228.8 g 
      Mass of calorimeter + water + steam= 231.2g 
      Temperature of the cold water = 180C 
      Final temperature of the water=290 C
      1. Determine the amount of steam that condensed. (1 mark)
      2. Calculate the amount of heat lost by the condensed steam. (specific heat capacity of water=4200J/Kg/K) (3 marks)
      3. Calculate the amount of heat absorbed by water and the calorimeter (specific heat capacity of copper = 390J/kg/K) (3marks)
      4. Calculate the specific latent heat of vaporization Lv of water. (2 marks)
      5. Explain why cooling the water used in the calorimeter to below room temperature could have led to more accurate result. (1 mark)
  3.        
    1. Distinguish between elastic collision and inelastic collision. (1 mark)
    2. A van of mass 1500 kg travelling at a constant velocity of 72 km/h collides with a stationary car of mass 900kg. The impact takes 2 seconds before they move together at a constant velocity for 20 seconds. Calculate:
      1. Their common velocity (3marks)
      2. The distance moved after the impact.(2marks) 
      3. The impulsive force (3marks) 
      4. The change in kinetic energy (3marks)
      5. Why is the kinetic energy not conserved in this collision (1 mark)
  4.          
    1. Define angular velocity. (1 mark).
    2. Figure 4 below shows a mass 500g moving in vertical circle having a radius of 35cm at a constant velocity. It makes 2 revolutions in one second.
      1phyf4mt221p1q18b
      1. Indicate on the diagram the direction of centripetal force. (1 mark)
      2. Calculate the linear velocity of the mass. (3marks)
      3. Calculate the centripetal acceleration of the object. (2marks)
      4. Determine the centripetal force. (3marks)
      5. Giving a reason, state the point at which the string is likely to snap. (2mark)
  5.            
    1. State the law of conservation of energy. (1 mark)
    2. Figures4 below shows a ball of mass of 5kg rolling along a frictionless path as shown.
      1phyf4mt221p1q19b
      1. Calculate the potential energy of the ball at point O. (2 marks)
      2. Determine the velocity of the ball at point A (2 marks)
      3. If the ball rolls back when it reaches point B, state the energy changes that takes place O to B. (1 mark)
      4. It is observed that the efficiency of the machine increases when it is used to lift large loads. Give a reason for this (1 mark)

MARKING SCHEME

  1. Length= 5.35cm
    Volume= 5.35 x 5.35 x 5.35 = 28.6225cm3
    p= m/v = 1533/28.6225 = 53.5593g/cm3

  2. Total pressure = atm + ρgh 
    = 760mmHg+20mmHg
    = 780mmHg
  3. This breaks the surface tension of water running over the canvas tent hence increasing adhesive force so that the tent leaks.
  4. Cohesive forces are stronger in solids than in liquids
  5.    
    • In case of fire outbreak, the temperature increases, brass expands more than iron. 
    • The bimetallic strip bends towards the iron side and makes the contact. 
    • This completes the circuit causing the electric bell to ring.
  6. The metallic chair is a good conductor of heat and gains heat faster than the wooden bench, which is a poor conductor of heat
  7. For a system in equilibrium, the sum of clockwise moments is equal to the sum of anticlockwise moments
  8. (xx24) = (100-x)16+(x-50)10
    24x=1600 - 16x+10x-500
    24x=1100-6x
    x=36.67cm
  9. F=ke
    50=k x 0.2
    K= 500/2 = 250N/M
    K=250 n/m
  10.    
    1phyf4mt221p1qa10
  11. The length of the spring reduces as the masses are added until it cannot reduce anymore
  12. For a fluid that is non-viscous incompressible and the flow is streamline, then an increase in velocity causes a corresponding decrease in pressure it exerts. 
  13.    
    1.    
      1phyf4mt221p1qa13    
    2. Water level in manometer B is lower than the levels in manometer A and C. Water level in manometer A is the highest hence high pressure compared to B and C.
  14.      
    1. B will have a higher temperature than the water in A. Dull surfaces are good obsorbers of heat
    2.      
      1phyf4mt221p1qa14b     
  15.      
    1. For a fixed mass of gas volume is directly propotional to absolute temperature provided pressure is kept constant. 
    2.    
      1.    
        • Length of the air column trapped / volume of air (L) 
        • Temperature of the water bath
      2.      
        • Temperature is varied and corresponding value of L and T are recorded
        • A grapgh of L vs absolute temperature T is then plotted
        • Straight line is obtained cutting the x-axis at OK(-273ºC)
      3.        
        • Sulphuric acid index acts as a pointer in the volume of the gas on the scale
      4. Pressure of the trapped air is the same as the atmospheric pressure pluse pressure due to acid index which remains constant through the experiment
      5. Vi=1200cm3 , T=27+273=3000K
        p=1.2atm
        p1V1 = p2v2 → 1.2 x 1200 = 3 x 600
         T1         T2           300               t2
        T2 → 375K or 102ºC
  16.       
    1. A quantity of heat required to change the temperature of a given mass of a substance by 1Kelvin
    2.           
      1. 231.2-228.8=2.4g
      2. Steam condensed to water at 29ºC
        Steam 100ºC → water 100ºC → Water 29ºC
        Q=MLV+MCO
        =(0.003 x lv+894.6)
      3. Heat absorbed = mc cc o + mw cw o
        =0.105 x 390 x (29-18)+0.1236x x 42.0x(29-18)
        =480.45+5710 32
        =6190.77j
      4. Heat lost - heat gained
        0.003Lv + 894.6=6190.77
        0.003Lv=Ω96.17
        Lv= 1,765390 0r 1.765 x 106 j/kg
      5. This balances the heat exchange between the calorimeter with its contents and the surrounding.
  17.         
    1.         
      • Elastic collision- both momentum and kinetic energy of colliding bodies is conserved 
      • Inelastic collision- only momentum is conserved but not kinetic energy
    2. Moment before collision = momentum after collision
      1. (1500 x 20)+0=2400 x v
        12.5m/s=v
      2. s=ut=12.5m/s x 0=250m
      3. ft=mv -mu
        fx2=1500(20-12.5)
        fx2=11250→f= 5625
        or ft=mv-mu
        fx2=900x12.5
        f-5625N
      4. Initial KE=1/2 mu2
        =1/x 1000x 202= 300,000
        Final KE=1/2(m1m2)v2
        =1/x 2400 x 12.52=187,500J
        KE change= 300000-187500= 112500J
      5. This is the energy converted to sound and heat or energy used in doing work
  18.        
    1. Rate of change of angular displacement
    2.       
      1.   
        1phyf4mt221p1qa18b
      2. f=2Hz
        ω= 2πf
        = 2 x 3.142x2
        = 12.56
        V= r ω
        = 0.35 x 12.56
        = 4.3982m/s
      3. a=v2/r=(4.3982)2
                       0.35
        =55.270m/s2
      4. f=mv2/r=0.5 x 55.270
        =27.635N
      5. At point R. This is where tension is maximum in the string
  19.          
    1. The sum of kinetic energy and potential energy of a system is constant
      1. PE=mgh=5x10x20=1000J
      2. PE=KE
        mgh=1/2 mv2
        v2=2gh
        v2=2x10x20
        v2=400
        v=20m/s
      3. PE →KE → PE + heat energy + sound
      4. Efficiency increase because as the load increases energy used in overcoming frictionand lifitng the weight of the machine parts becomes insignificant.
        • MA advatage increase because friction and weight of the machine parts remain constant.
    2.       
      1phyf4mt221p1qa19b

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