Physics Paper 1 Form 3 End Term 2 Exams 2021 with Marking Schemes
INSTRUCTIONS:
 Answer all the questions in the spaces provided after each question.
 All numerical answers should be expressed in decimal notations.
 You may use electronic calculators and tables.
 The figure below shows a spherical ball placed between 2 wooden blocks and a meter rule.
What is the volume of the ball? (3 mks)  A solid weighs 16.5N on the surface of the moon. The force of gravity on the moon is 1.7N/kg.
Determine the mass of the solid. (2 mks)  30cm^{3} of a liquid X was added to 70cm^{3} of water and the resulting mixture had a volume slightly less than 100cm^{3}, explain the observation. (2 mks)
 Explain how heat loss by;
 Radiation is minimised in a vacuum flask. (1 mk)
 Conduction is minimized in a vacuum flask. (1 mk)
 The figure below shows part of a scale of vernier caliper. Given that the device has a zero error of –0.02 and has been used to measure the diameter of a ball.
What is the radius of the ball? (2 mks)  A pipe of radius 6mm is connected to another pipe of radius 9mm. If water flows in the wider pipe at 2m/s, what is the speed in the narrower pipe? (3 mks)
 The springs below are identical and have negligible weight. The extension produced on the system of springs is 20cm.
Determine the constant of each spring. (4 mks)  An air bubble of volume 0.5cm^{3} when released from the bottom of a lake rises to the surface of the lake.

 Explain why the bubble rises. (2 mks)
 Calculate the volume of the bubble at the surface of the lake given that the lake is 92.7m deep and the atmospheric pressure is equivalent to 10.3m of water pressure. (4 mks)
 What assumption have you made in arriving at your answer? (2 mks)

 A fixed mass of gas at constant pressure has a volume of 600cm^{3} at 0^{o}C. At what temperature will its volume be 1099 cm^{3}? (4 mks)

 State three uses of magnets. (3 mks)
 Define the following terms as used in Physics:
 Magnetic materials. (2 mks)
 neutral point. (2 mks)
 State three conditions for a body to be in equilibrium. (3 mks)
 State four practical applications of friction. (4 mks)
 Use simple sketches of a cone to illustrate the three states of equilibrium and name. (6 mks)

 Give a reason why water is not suitable as a barometric liquid. (3 mks)
 Explain the application of (a) above. (3 mks)
 Use domain theory of magnetism to explain how a magnet may lose its magnetism on heating and hammering. (4 mks)
 Explain the following observations:
 A boy jumping from a high table tends to spread his legs. (1 mk)
 Convex mirrors are not preferred for use as driving mirrors. (1 mk)
 Why convex mirrors are used as driving mirrors and in supermarkets. (1 mk)

 State three practical applications of c.o.g. (3 mks)
 Name two factors that affect the c.o.g of a body giving a reason for each. (4 mks)
 A car travelling at a speed of 72kmh^{1} is uniformly retarded by application of brakes and comes to rest after 8 seconds. If the car with its occupants has a mass of 1250kg, calculate
 breaking force. (2mks)
 Work done by bringing it to rest (2mks)
 A block and tackle system is used to lift a mass of 200kg. If this machine has a velocity ratio of 5 and an efficiency of 80%;
 Sketch a possible arrangement of the pulleys, showing how the rope is wound. (2 mks)
 Calculate the effort applied. (Take g = 10N/kg) (2mks)
Marking Scheme
 5.8 – 4.5 = 1.3
1.3 = 0.65
2
V = ^{4}/_{3} x^{ 22}/_{7} x 0.652
= 1.151cm^{3}. Ans  M = W
g
M = ^{16.5}/_{1.7}= 9.706 Kg Ans  This is because some of the particles of liquid X are able to fit in between water particles.

 Silvered inner walls
 Presence of a double wall with a vacuum in between.
 Diameter = 7.44 + 0.02
= 7.46
Radius = 3.73cm  A_{1}V_{1} = A_{2}V_{2}
36 × V_{1} = 81 x 2
V1 = 4.5 m/s
3  For parallel springs extension
40N
2k
For single spring, extension
= 20N
K
Total extension = 40N + 20N
2k k
= 20cm
(40 + 40N) = 20cm
2k
80 = 20cm
2k = ?
K = 80N
20 x 2cm
= 2N/cm Or 0.02N/m 

 The bubble rises up since the density of air is less than that of water.
  Vol of air bubble @ bottom = V_{1} and @ top = V_{2} Pressure acting on bubble at the surface
= 10.3m of water pressure
 Pressure acting on bubble @ bottom
= P_{1} = 10.3 + 92.7
= 103.0m of water pressure
From Boyle’s law
P_{1}V_{1} = P_{2}V_{2}103.0 x 0.5 = 10.3 V_{2}V_{2} = (103 x 0.5)
10.3
= 5cm^{3} Ans
  The temperature of the lake and that of the air bubble is constant.

 V_{1} = 600 cm^{3} T_{1} = 0^{o}C = 273K
V_{2} = 1099 cm^{3 }T_{2} = ?
V_{1} x T_{2} = V_{2} x T_{1}600T_{2} = 1099 x 273k
T_{2} = 1099 x 273k
600cm^{3}∴ T_{2} = 500 K 
  Are those that can gain magnetic qualities and acts as magnets e.g iron.
  Are materials that cannot be magnetized e.g wood
  This is a point in a magnet where the magnetic power of attraction or repulsion is not experienced.
  Making compass (any 3)
 In radio speakers, amplifiers, video tapes etc
 In hospitals to remove objects from eye/body
 Used in TVs generators, telephone receivers
  Area of base
 Position of centre of gravity
 Forces on both side of pivot should be same   movement of bodies
 braking system
 lighting match stick
 writing or erasing blackboard
 skidding
 Lubrications (any four) 
  Due to its anomalous expansion
 Has high b.p and low m.p
 It is uncompressible due to its state and delocalized electrons.   sea breeze
 ice bergs
 land breeze
  Due to its anomalous expansion
  On heating a management, dipoles in their domains face one directs
 They disorient on heating further
 Hammering while in EW direction distorts alignment of magnetic atoms
 Magnets are kept facing in NS directions. 
  10 maintain the centre of gravity.
  Forms diminished images
  They have a wide field of view


 Tripod stand
 Luggage buses
 Racing cars   Area of base – the greater the area the more the stable a body is.
 Position of c.o.g/the lower the c.o.g, the stable the body.


 F = ma and acceleration a = v – u
t
but u = 72km/h
= 20m/s
∴ a = 0 – 20
8
= 2.5 m/s
Hence 1250 x 2.5
= 3125 N  W.d = ke lost by the car
= ½ mv^{2} – ½ mu^{2}
= ½ x 1250 x 0^{2} – ½ x 1250 x 20 x 20
= 2.5 x 10^{5}J
 F = ma and acceleration a = v – u

 Efficiency = M.A x 100%
V.R
= 80 = M.A.
100 5
M.A. = 80 × 5 = 4
100
But M.A = L and L = Mg
E
K = 200 x 10
Hence, 4 = 200 x 10
4
= 500N
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