# KCSE 2010 Physics Paper 2 Questions with Marking Scheme

## QUESTIONS

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

1. Figure 1 shows a ray of light indecent on a plane mirror at O. The mirror is then rotated anticlockwise about O from position M1 to position M2 through an angle of 100. The final reflected ray is OC.

Determine the angle BOC. (2 marks)

2. Figure 2(a) shows the magnetic compass placed under the horizontal wire XY

A large current is passed from X to Y. Draw the final position of the magnetic compass needle in figure 2(b). (1 mark)

3. Figure 3 shows a diagram of current carrying wire wound on a U- shaped soft iron.

Draw the magnetic field pattern around P and Q. (2 marks)

4. A positively charged sphere is suspended by an insulating thread. A negatively charged conductor is suspended near it. The conductor is first attracted, after touching the sphere is repelled. Explain this observation. (2 marks)

5. Figure 4 shows a bright electric lamp placed behind a screen which has a hole covered with a wire gauze. A concave mirror of focal length 25cm is placed in front of the screen. The position of the mirror is adjusted until a sharp image on the gauze is formed on the screen.

Determine the distance between the mirror and the screen. (2 marks)

6. Explain why electric power is transmitted over long distances at high voltages. (2 marks)

7. Figure 5, shows the displacement of a point varies with time as a wave passes it.

On the same diagram, draw a wave which passes the point with a half the amplitude and twice the frequency as the one shown. (2 marks)

8. A water wave of wavelength 18 mm is incident on a boundary of shallow water at right angles. If the wavelength in the shallow end is 14.4 mm, determine the refractive index of water for a wave moving from the deep to the shallow end. (3 marks)

9. The initial mass of a radioactive substance is 20g. The substance has a half-life of 5 years. Determine the mass remaining after 20 years. (2 marks)

10. A certain I flowing through a wire of resistance R was increased seven times. Determine the factor by which the rate of heat production was increased. (3 marks)

11. Figure 6, shows a horizontal conductor in a magnetic field parallel to the plane of the paper.

State the direction in which the wire may be moved so that the induced current in the direction shown by the arrow. (1 mark)

12. An x-ray tube produces soft x-rays. State the adjustment that may be made so that the tube produces hard x-rays. (1 mark)

13. The wavelength of a radio wave is 1 km. Determine its frequency. (2 marks)
(Take the speed of light as 3.0×108 ms -1. )

14. Figure 7 shows a block diagram of p-n junction diode.

On the diagram, show how the battery may be connected so that the diode is reverse biased. (1 mark)

SECTION B (55 marks)
Answer ALL the questions in this section in the spaces provided.

1. Figure 8, shows a circuit that may have been used to charge a capacitor.

1. State the observation on the milliameter when the circuit is switched on. (1 mark)
2. Explain the observation in (a) above. (2 marks)

2. The circuit in figure 8 is left on for some time. State the value of the p.d. across:
1. the resistor R: (1 mark)
2. the capacitor C. (1 mark)

3. Sketch the graph of potential difference (V) across R against time. (1 mark)

4. Figure 9 shows three capacitors connected to a 10V battery.

Calculate:
1. The combined capacitance of the three capacitors; (3 marks)
2. The charge of the 5.0 μF capacitor. (3 marks)

1. Figure 10 shows an object placed in front of a converging lens of focal length 50mm.

1. On the same figure, draw a ray diagram showing the location of the image. (3 marks)
2. Use the ray diagram to determine:
1. Image distance; (1 mark)
2. Magnification. (2 marks)

3. State the adjustment that should be done to obtain a larger virtual image using the same lens. (1 mark)
4. State one application of the arrangement in figure 10. (1 mark)

2. Figure 11 shows a pin 60 mm long placed along a principal axis of the lens used in part (a). The near end of the pin is 80 mm from the lens.

Determine the length of the image. (5 marks)

1. Figure 12 shows an electrical circuit including S1, S2, S3, and three identical lamps L1, L2, L3. A constant potential difference is applied across X and Y.

1. Other than L1, state the lamp that will light when S1 and S2 are closed. (1 mark)
2. How does the brightness of L1 in (i) above compare to its brightness when all the switches are closed? (1 mark)
3. Explain the observation in (ii) above. (1 mark)

2. Figure 13 shows a cell in series with a 3Ω resistor and a switch. A high resistance voltmeter is connected across the cell.

The voltmeter reads 1.5V with the switch open and 1.2V with the switch closed.
1. State the electromotive force of the cell. (1 mark)
2. Determine the current through the 3Ω resistor when the switch is closed. (2 marks)
3. Determine the internal resistance of the cell. (2 marks)

1. Another resistor R is connected in series with the 3Ω resistor so that a current of 0.15A flows when the switch is closed. Determine the resistance of R. (3 marks)

15. Figure 14a, is a diagram of a cathode ray tube, M and N are parallel vertical plates.

1. When switch S is open, a spot is seen at the centre of the screen as shown in figure 14(b).
1. State what happens to the spot when S is closed. (1 mark)
2. State what would happen to the spot if the potential difference across MN is increased. (1 mark)
3. State what would be seen on the screen if the battery is replaced with an alternating emf of:
1. A low frequency about 1 Hz: (1 mark)
2. A high frequency about 50Hz. (1 mark)

2. Explain the process by which electrons are produced at F. (2 marks)
3. State with a reason how the brightness of the spot can be increased. (2 marks)
4. The accelerating voltage of the tube is 1000V and the electron current in the beam is 1.5mA. Determine the energy conveyed to the screen per second. (2 marks)

1. State the property of radiation that determines the number of electrons emitted when a radiation falls on a metal surface. (I mark)

2. Figure 15 is a graph of the supporting potential Vs against frequency in an experiment on photoelectric effect.

1. What is meant by stopping potential? (1 mark)
2. Given that the stopping potential Vs is related to the frequency by the equation.

where e is the charge of an electron, (e = 1.6 x 10-19C)

Determine from the graph:
1. Plank’s constant, h; (4 marks)
2. The work function ωo for the metal in electron volts (eV). (3 marks)

## MARKING SCHEME

1.
• Reflected ray rotates 2 x 20 = 20°
• Find deviation = (80° +20") = 100°
2. Any slight deviation of the N-pole to the right
3. Correct poles  √1 Correct direction + pattern√1
4. Initially attracted because of opposite charge √1
(+ve or -ve)
Then neutralised and charged positive and hence repel √1
Charging by contact and law of electrostatics. √1
5. Distance = 2f = 2 x 25 √1/2= 50cm√1/2
Alternative Just 50cm√1
2 x 25 = 50cm√1/2
Or
6. implies low current √1 So reduces √1 heat losses/ power loss Or
I2R loss reduced
P = I2R should be accompanied by power loss
NB Heat losses/ Power Loss
7. Mare practicel relationship between f and t
Displacement
8. V1 =  FT1   or     η = T0
T1
V2 = FT2            η = 18
14.4
=1.25
Accept all expressions
9. 20g → 10g → 5g → 2.5g → 1.25g
Mass remaining
10. lo - Initial current          I2 = 710
P= I2R = I20R              PE (710)2R = 4912OR
Power is 49 times the initial value
Apply the power formula
11. Motion out of paperi moves upwards.  Or Increases in p.d increases heating effect
12. Increasing the accelerating voltage OR Increase the P.d between a anode and cathode.
Accept extra high tension increased
13. f = = c
λ     λ
= 3.0 x 108         3.0 x M05Hz
1000
14. Look for biasing only. (any other device that does not affect the working should be ignored eg. diodes/ resists.
15.
1.
1. Current fails off to zero falling to zero/ deflects to max. then zero
Reducing gradually or after sometime
2. current flows when the capacitor is charging
When fully charged capacitor stops (No current) and P.dis equal to the charging voltage
2. Vc = 5V
3. Touch both axis, Award for no labelled axis
4.
1. 1 = 1  + 1  = 5 + 4  = 9
cs   4     5         20     20
Cs = 20
9
C120 +3  = 5.22μF
9
Accept 5.22μF only
2. Change on series section = Q = Cv
=20  ×  10 μc
9
=22.2μc or
Q series = QT - Q3μF
= (5.22-3) x 10   μC1
= 22.2 μC1
Charge is the same on series section hence charge on 5.0μF is 22.246641
Accept 22.2μC only
16.
1.
1. Each rayamk indipendenih imkon dotted extrapolation
1mk for datted image
2.
1. 50mm                 ±5mm
2. M = v = h1 = 50 = 2mm  ±0.2
h    ho    25
3. Move the object towards F but not beyond
Move obiect away from lens
4. no answer
2. 1 1 + 1
f      u    v
11 + 1
50   80   v
V = 400/3
17.
1.
1. L2
2. Brighter
3. Total resistance is less/reduced
2.
1. 1.5V
2. Ir = 1.5 - 1.2 = 0.3
0.4r = 0.3
V = 0.75R
P.d and E.M.FI more practice and practical approach
3. Rr = 3 +0.75 + R                0.15(R+3.75) = 1.5
RT = R+3.75                      R +3.75 = 1.5 =  10
0.15
E=IR                                 R = 10 -3.75
1.5 = I(R+ 3.75) = 6.25Ω
Or
R = E = 1.5 = 10
f     0.15
R = RT - (V  + 3)  + 3.75
R = 6.25Ω
1.5 - 0.75 x 0.15 = I(3 + R)
1.5 -0.1125 = 0.15(3 + R)
1.3875  = 3 + R
0.15
R = 6.25Ω
18.
1.
1. Deflected towards +ve plate (N)
2. Deflection will be greater
3.
1. Spot moves back and forth
To and fro (Not along across)
2. There will be a horizontal line 1
2. Electrons are given off as a result of heat produced by current
So that electrons gain energy to break off from the surface
Thermionic emission
3. Increasing the filament current so that more electrons are released p.d across filament is increased so that more electrons are released
4. P= IV
= 100 x 1.5 x 10-3
= 1.5J/s
Accept J, W, J/S
19.
1. Intensity of radiation
2.
1. (Min p.d)
Negative potential sufficient to just stop the movement of electrons.
2.
1. Gradient =
e
h =            3.0 - 0                  =     3
(12-14) x 76 x1014             7.6 x 1014
Gradient = 0.3947 x 10-14
h = 0.3947 x 10-14 x 1.6 x 10-19
= 0.6316 x 10-33 = 6.316 x 10-34
2. =w=1.751.
e
W0 = y-intercept x e
= 1.75 x 1.6 x 10-19
1.6 x 10-19
=1.75ev
Alternative
= W0 = hf0
=6.32 x 0.4 x 1014 x 10-34
1.6 x10-19
=1.74ev
OR
Range 1.7 →1.8ev
wo = y intercept =  1.75
e
Wo =  - 1.75  or we  = 1.75
a                       e
Wo =  1.75ev     OR   Wo = 1.75v  x  e
= 1.75ev
OR
-Wo = 1.75 -1.75ev
e
OR
Wo = 1.75  →  rej: 1.75v = Wo
penalise -ve and units in
Wo =  Y intercept
= - 1.75

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Read 3393 times Last modified on Monday, 22 November 2021 09:15

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