END OF TERM 3
FORM 3 PHYSICS PP2
SECTION A : (25 MARKS)
Answer all the questions in this section in the spaces provided.
- Figure 1 below shows an object infront of plane mirror.
Figure 1
Sketch image of object using mirror shown. (1mk) - Figure 2 below shows an object infront of concave mirror and it’s image.
Figure 2
Locate position of its principal focus. (2mks) - State the use of Manganese (IV) oxide in dry cell. (1mk)
- Use figure 3 below to answer following questions.
figure 3
Determine- Total resistance. (3mks)
- Potential difference across 4Ω resistor. (3mks)
- Figure 4 shows conductor carrying current in magnetic field and moves in direction shown.
Figure 4
Identify polarities X and Y. (2mks) - A man standing between two parallel walls fires a gun. He hears an echo after 1.5 seconds and another one after 2.5 seconds and yet another one after 4 seconds. Determine the separation of the walls. (Take velocity of sound 340 m/s)
- A student shouts and hears an echo after 0.6 seconds. If the velocity of sound is 330m/s. Calculate the distance between student and reflecting surface. (3mks)
- Figure 6 shows water waves moving towards barrier. Show the emergence of the reflected waves
Figure 6 -
- Define refractive index. (1mk)
- The critical angle of a material is 43.20. Determine the refractive index of that material. (2mks)
- A battery of emf E drives a current of 0.25A when connected to 5.5W resistor. When the 5.5W resistor is replaced with 2.5W resistor, the current flowing becomes 0.5A. Find the emf E and the internal resistance r of the battery. (3 marks)
- Define the term sulphation as applied to lead acid cells. (1 mark)
SECTION: B(55MARKS)
ANSWER ALL QUESTIONS IN THIS SECTION
-
- An uncharged metal rod brought close but not touching the cap of a charged electroscope causes a decrease in the divergence of the leaf. Explain the observation. (1mark)
- In experiment to investigate factors affecting capacitance of a capacitor, a student increased the area of the plates and decreased the separation of the plates. Explain the effect on the capacitance when
- the area of plates increased (1 mark)
- the distance of the separation of the plates decreased (1 mark)
- Figure 7 illustrates a method of charging a metal sphere.
Fig. 7.- Name the method of charging shown in fig 8.(1mar(ii) Indicate the final charge on the sphere in fig 8. (1 mark)
- Figure 9 shows an arrangement of capacitors connected to a 10V d.c supply.
Determine- the combined capacitance (2 marks)
- the total charge in the circuit (1 mark)
- the total energy stored in the circuit. (2marks)
-
- Distinguish between e.mf. and terminal voltage of a battery. (2 marks)
- The graph in figure 8 shows the variation of potential difference V against current I for a cell when current is drawn from it.
From the graph determine- The e.m.f of the cell. (2 marks)
- The internal resistance of the cell. (4marks)
- On the space provided below, draw a circuit that could be used to obtain the results represented by the graph. (2 marks)
-
- Figure 9 is an illustration of a wave pattern.
- State with reason the type of wave shown. (2 marks)
- Determine the wavelength of the wave. (1 mark)
- Calculate the frequency of the wave given that the speed of the wave is 9m/s. (3 marks)
- Figure 10 show s monochromatic source of light L behind a barrier with a single slit S placedbehind another barrier with two identical slits S1 and S2. A screen PQ is placed in position as shown.
- Explain what is observed on screen PQ. (2 marks)
- What is the significance of S1 and S2 ? (1 mark)
- Figure 9 is an illustration of a wave pattern.
- Figure 11 shows an electromagnetic relay being used to switch an electric motor on and off. The electromagnet consists of a coil of wire wrapped around a core. The motor in figure is switched off.
- Suggest suitable material for the core. (1mark)
- What happens to the core when switch S is closed? (2marks)
- Why do the contacts A and B close when the switch S is closed. (2marks)
- When the switch S is opened, what will happen to;
- The core (1mark)
- Soft iron armature. (1mark)
- Give one other application of an electromagnet. (1mark)
- State two ways in which an electromagnet could be made more powerful. (2marks)
- Figure 12 below shows a narrow beam of white light onto a glass prism.
- What is the name of the phenomenon represented in the diagram? (1mk)
- Name the colour at X and Y. Give a reason. (3mks)
- What is the purpose of the slit? (1mk)
- The figure 4 shows a circuit with a coil used to warm oil in a beaker.
- State the Ohm’s Law. (1mk)
-
- Explain how heat is produced in the coil. (2mks)
- Given that the reading of the ammeter is 2.5A, determine the resistance of the coil. (3mks)
- How much heat is produced in the coil in a minute? (3mks)
- Give two changes that can be made in the set-up in order to produce more heat per minute. (2mks)
- How much heat is produced in the coil in a minute? (3marks)
MARKING SCHEME
SECTION A
-
-
rays with arrowsposition of F - Depolarizer
-
- 4 + 3 = 7Ω
1 + 1 = 12
7 5 35
= 35/12= 2.916Ω - I = 12 + 12 = 144 = 4.1143A
35 35
C(4Ω ) = 5/12 x 4.1143 x 4 = 6.857V
- 4 + 3 = 7Ω
-
- X - North
- Y - South
-
Alternative: 2d = V
t
Time for second echo = 5/2 = 2.5 seconds
1.5 + 2.5 = 40 seconds
D= 5 x t = 330 x 4 = 1320m -
- S = vt =330 x 0.6 = 99m
-
- E = IR + Ir
E = 0.25 × 5.5 + 0.25 r
E = 0.5 × 2.5+ 0.5 × 5
E = 1.35 + 0.25r
E = 1.25 + 0.5 r
0 = 0.10 - 0.25 r
0.25r = 0.10
0.1 = 0.4Ω
0.25
E= 1.35 + 0.25 x 0.4
= 1.45
- E = IR + Ir
-
-
- Ration of sine of angle of incidence to sine of angle of refraction for a given pair of media.
- n = 1 = 1 = 1.461
Sin C Sin C 43.20
- Hardening of lead sulphate on the lead plates
SECTION B 55 MARKS
-
- A metal rod is a good conductor of charge hence the electroscope
-
- Increase
- Increases
-
- Induction
- Electrons from the earth to the sphere to neutralize the repelled positive charges
- Negative charges.
-
- Series = (3 x 3)/(3+2) = 9/6 = 1.5μF
in parallel with 2μF
CT = 1.5 + 2 = 3.5μF
In series with 1μF
CT = (3.5 x 1)/(3.5+1) = 3.5μF/4.5μF
= 0.78μF
= 7.8 x 10-7F - 7.8 x 10-7 x 10
= 7.8 x 10-6 C - E = ½CV2
= ½ x 7.8 x 10-6 x 10
= 3.9 x 10-5J
- Series = (3 x 3)/(3+2) = 9/6 = 1.5μF
-
- E.m.f is the p.d. across the battery in the ü1 open circuit while terminal voltage is the p.d. across the cell is a closed circuit.
-
- E.m.f = 1.625V ü1 (show unit)
- E = IR + Ir
V = E - Ir
V = Ir + Slope =
-r = 1.5 - 1.0 = 0.5 = 0.78
0.16 - 0.8 -0.64
r= 0.78Ω
-
-
- Transverse wave
- The particles vibrate at right angle to the direction of motion of the wave
- The wavelength = 50cm or 0.5m
- v = f
f = 9
0.5
= 18Hz
- Transverse wave
-
-
- Bright and dark fringes are formed
- Bright fringes are formed where constructive interference occurs while dark fringes are formed were destructive interference occurs
- Produce coherent sources of light.
-
-
- Soft iron
- The current flows through the solenoid; it is magnetized and attracts the soft iron armature.;
- The magnetized core attracts the soft iron armature. The pivot armature pushes the springy metal strip which joins contact B and A.;;
-
- The core (1mark)
- It loses its magnetism;
- Soft iron armature. (1mark)
Soft iron goes back to its original position thus switching off the current in the circuit.;
- The core (1mark)
- Give one other application of an electromagnet. (1mark)
- Electric bell, telephone receiver, moving coil loudspeaker and circuit breaker.
- State two ways in which an electromagnet could be made more powerful. (2marks)
- Using a soft iron core, increasing the current and
- Increasing the number of turns;;
-
-
- Dispersion of light.
- X – Red
Y – Violet
- Red has the lowest frequency/longest wavelength hence least deviated while violet has the highest frequency/shortest wavelength hence most deviated.
- Act as point source of light.
-
-
- The current flowing through the conductor is directly proportional to the potential difference across its ends provided temperature and other physical conditions are kept constant.
-
- When the switch is closed the current flows through the coil which offer resistance hence dissipating heat.
- V = IR
R= V= 12 = 4.8Ω
I 25
- Heat = Vlt
= 12 x 2.5 x 60
= 1800J -
- Increasing the number of coils.
- Increasing the current.
- The readings will decrease because the resistance is decreased. ¹
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