PERATURAN PEMARKAHAN
BAB 6: GELOMBANG BAHAGIAN A
NO.1
ANSWER
MARK
(a)
The light which has one wavelength / one colour
1
(b)
Wavelength of red light is greater than wavelength of blue light
1
(c)
Distance between consecutive bright fringes for red light is more than that of blue light The distance between consecutive bright fringes for the same light are constant The longer the wavelength, the longer the distance between
1
1
(d)
1
consecutive bright fringes (e)(i)
The distance between two consecutive bright fringes will
1
decrease (e)(ii) (f)
a is inversely proportional proportional to x
1
Diffraction // Interference Interference of light
1 TOTAL
NO.2
ANSWER
8
MARK
(a)(i)
Pantulan
1
(a)(ii)
Pembiasan
1
(b)(i)
1.2 – 1.2 – 0.4 0.4
1
0.8 s
1
2T dengan 0.8 s // T dengan 0.4 s
1
2T = 1500×0.8 // T =1500×0.4
1
600 m
1
Tinggi
1
Dapat menembusi bumi dan minyak // kehilangan tenaga yang
1
(b)(ii) (b)(ii )
(c)(i)
rendah (c)(ii)
(d)
Tinggi
1
Mendapat isyarat yang lebih kuat
1
P
1 TOTAL
12
146
PERATURAN PEMARKAHAN
BAHAGIAN B
NO.3 (a) (b)
ANSWER Number of complete oscillation in one seconds
MARK 1 1
1
1
Length of pendulum X is shorter than the length pendulum metal bob Frequency of pendulum X is higher than the frequency of metal bob pendulum. Length and frequency of pendulum Y as same as that of metal bob pendulum. Amplitude of oscillation of pendulum Y is higher/bigger than amplitude of oscillation of pendulum X. Resonance
(c)
When the singer sings, it produces the high frequency of sound waves. The frequency of of sound waves vibrates the particles in the glass. When the frequency of sound waves same as the natural frequency of the glass, The particles of the glass will vibrate at maximum amplitude/ energy cause the glass to break.
1 1 1 1 1 1
(d) CHARACTERISTICS 1. Low density of string 3. High tension of string
2. 4. 6.
5. Nylon string/strong string/st rong 8.
7. Sound hole must be big
9. Fred must be widened
REASON Easy to carry / handle Can produce higher frequency/pitch More stiffer /can produced louder sound//not easy to break More air can be trapped//The coupled resonance of the front and back plates produces a resonance
Max= 10
10. Can differentiate different notes more
TOTAL
20
147
PERATURAN PEMARKAHAN
BAHAGIAN C NO.4 (a) (b)
ANSWER Distance between two successive points of the same phase in a wave
(c)(i)
(c)(ii) (d)
When the wave passing through shallow water// convex area, the wavelength and speed decrease After the wave passing through shallow water// convex area, the circular waves converge at focal point an then diverge from the focal point
OR Diagram λ = 633 nm = 633 x 10 -9 m a = 0.5 mm = 0.5 x 10 -3 m 633 × 10− × 4 = 0.5 × 10− X = 5.064 x 10 -3 mm
5. Low damping effect
D=4m
1+1
1 1 1
2. 4. 6. 8.
7. Low frequency
1+1
1 1
Decrease CHARACTERISTICS 1. High wave amplitude 3. Longer wave length
MARK 1
REASON Louder Can diffract easily Can be heard clearly Low pitch// distinguished from other sound//can be heard by human
Max= 10
9. Q is chosen 10. Because.....1357//2468 TOTAL
20
148
PERATURAN PEMARKAHAN
KERTAS 3 (BAHAGIAN B)
NO. 5 (a) (b) (c)(i) (c)(ii)
(c)(iii)
ANSWER Rate of oscillations / frequency depends on the diameter/thickness of the string The thicker the diameter of the string, the lower the frequency// The thinner the diameter of the string, the higher the frequency To investigate the relationship between the thickness of a wire and the frequency of the oscillations MV: Thickness of wire, d RV: Frequency of oscillation, f CV: length of wire microphone, copper wires with different thickness or s.w.g, CRO, retort stand, connecting wires
MARK 1 1 1 1 1 1 1
(c)(iv)
(c)(v)
1. Set up the apparatus as shown. Tie up the copper wire to retort stand tightly 2. Start the experiment with the length, L=30 cm and s.w.g 16. 3. Pull the string in the middle so that it will vibrate. Use a microphone and record the frequency of the waveform on the screen of CRO. 4. Repeat by using different swg/thickness of copper wire. d/ cm
1 1
1 1
f/ s-1
(c)(vi)
1
(c)(vii)
TOTAL
12
149
PERATURAN PEMARKAHAN
BAB 7: KEELEKTRIKAN BAHAGIAN A
NO.1 (a)
ANSWER The power of device is the rate of of which it transfers energy// =
(b)
Electrical energy to light energy + heat energy
(c)(i)
Heat energy
(c)(ii)
Filament bulb: =
MARK 1
1+1 1
11.25 × 100% 75
1 1
= 15 % Energy saver bulb: =
11 20
1
× 100%
1
50 % (c)(iii)
(d)
Energy saver bulb
1
High efficiency// less wasted energy
1
Refrigerator // any other appliances
1
Frost-free refrigerator// not allow frost to accumulate too thick in the
1
freezer// close the door tightly TOTAL
12
150
PERATURAN PEMARKAHAN
NO.2 (a)
ANSWER 12 J of energy per second is released when it is connected to a 6 V
MARK 1
supply (b)
(c)
Series
1
Parallel
1
Voltage for each bulb in diagram 8.2 is more than 8.1
1
Total resistance in Diagram 8.2 is less than 8.1// current flow in
1
each bulb in Diagram 8.2 is more than in 8.1 (d)(i)
R1 = 4 + 4 + 4 = 12 Ω =
=
.
1
= 0.5 A (d)(ii)
1
=
1 4 4
+
1 1
1 3 + = 4 4 4
= = 1.33 3 =
6.0 1.33
= .
1
I flow each bulb =
(e)(i) (e)(ii)
4.5 3
1
= .
Diagram 8.2
1
If one bulb blow, another bulb can still function
1
Less effective resistance// more current flow
1 TOTAL
12
151
PERATURAN PEMARKAHAN
BAHAGIAN B
NO.3 (a) (b)
ANSWER
MARK 1
Diagram 10.1 connected in series and Diagram 10.2 connected
1 1 1 1 1 1
The rate of charge flows.
in parallel.
The reading of ammeter in Diagram 10.2 is greater than in Diagram 10.1.
(c)
The reading of voltmeter in Diagram 10.1 > Diagram 10.2.
The effective resistance in Diagram 10.2 < Diagram 10.1.
Effective resistance increases, the current flows decreases.
Circuit connected in parallel, the effective resistance decreases.
The ammeter reading increased
The voltmeter reading decreased.
Effective resistance in the circuit decreased.
1 1 1
(d) CHARACTERISTICS Attach one fuse to the live wire in the consumer unit/ fuse box. Using the insulating wires // thicker wires Attach switch for each lamp. Connect the metal fitting lamp to the earth wire/cable. Using only 240 V light bulb.
Max= 10
REASON To break/switch off the circuit when large current before the wire become hotter and produce fire. To prevent short circuit // To reduce resistance, improve efficiency. To allows each lamp to be switched on and off independently. To flows electron (extra) to earth to avoid lethal shock. To ensure the bulbs light up with normal brightness.
TOTAL
20
152
PERATURAN PEMARKAHAN
BAHAGIAN C
NO.4 (a) (b)
ANSWER
MARK
Electric field is a region around a charged object which any other charged body experience a force
Charges on the disc neutralize the negative charges on the ping pong ball Likes charges on the disc and the ball repelled each other The ball attracted by the positively charged disc The ball oscillates between the two plates
1 1 1 1 1
(c) CHARACTERISTICS 1. Fuse for each circuit 3. Bulbs and power sockets arrange in parallel
REASON 2. Cut off current if overloaded 4. Other bulb or power socket still working when one of it blows out
5. Circuit for lighting is in parallel with power circuit
6. Supply different value of current
7. Kilowatt-hour meter installed near main fuse
8. Record total power usage and prevent overloading of current
Max= 10
9. Circuit L 10. Because ...... 1357// 2468
(d)(i) (d)(ii)
10 – 13 A V = IR = 9 x 26.7 = 240.3 V
1 1 1
P = IV = 9 x 240.3 = 2162.7 W TOTAL
1 1 20
153
PERATURAN PEMARKAHAN
KERTAS 3 (BAHAGIAN B)
NO. 5 (a) (b) (c)(i) (c)(ii)
(c)(iii)
ANSWER Resistance is depends to the diameter/ cross-sectional area of wire The resistance of the wire decreases as its diameter increases// when the diameter of wire increase, the resistance is decrease To investigate the relationship between diameter of wire and resistance MV: diameter// cross-sectional area RV: resistance CV: length// temperature Constantan wire, meter rule, connector wire, batteries, rheostat, ammeter, voltmeter and switch
MARK 1 1 1 1 1 1 1
(c)(iv)
(c)(v)
1. The circuit is set up as shown in figure with the 20 cm of constantan wire s.w.g= 20 across the point X and Y. 2. The switch is on, and the rheostat is adjusted until the ammeter shows a reading. 3. The ammeter and voltmeter reading are recorded. 4. The resistance of the constantan wire is calculated using a formula =
1 1
5. The experiment is repeated with a constantan wire as s.w.g= 22, s.w.g= 24, s.w.g=28 and s.w.g= 30
1 1
(c)(vi)
diameter/s.w.g (cm) 20 22 24 28 30
R (Ω)
1 (c)(vii)
TOTAL
12
154
PERATURAN PEMARKAHAN
BAB 8: KEELEKTROMAGNETAN BAHAGIAN A
NO.1 (a)
ANSWER The current that is induced by electromagnetic inductions when the circuit is complete//The current that can be produced without any electrical supply/source
MARK 1
1
(b)(i)
X= N Y= S
(b)(ii)
Diagram 6.1 bar magnet towards the solenoid Diagram 6.2 bar magnet away from the solenoid
(c)(i)
Repulsive Attractive
(c)(ii)
Lenz
1
Increase the speed of bar magnet/increase the number of turns of the solenoid/use a stronger magnet TOTAL
1
(d)
NO.2 (a) (b)(i) (b)(ii) (b)(iii) (c) (d)
(e)
1 1 1 1 1
ANSWER An electromagnet is a magnet in which a magnetic field is produced by the flow of electric current. Increase the current. The strength of an electromagnet increases Soft-Iron core Easily magnetised and demagnetised
9
MARK 1 1 1 1 1 1 1 1 1 1
Increase the number of turns. The strength of an electromagnet increases L Increase // More The poles for a U-shaped electromagnet are closer together P= mgh t = 250 x 10 x 4 5 = 2000 W
1 1 TOTAL
12
155
PERATURAN PEMARKAHAN
BAHAGIAN B
NO.3 (a) (b)(i) (b)(ii) (b)(iii) (b)(iv) (c)
ANSWER The current produced when the magnetic flux is cut across by a conductor// Changing of flux at conductor 10.2 - no relative motion between the magnet and the coil // 10.3 - there is relative motion Number of turns in 10.3 is less than number of turns in 10.4 Number of turns increases, the change in magnetic flux increases Induced current increases Faraday’s Law
MARK 1 1
1 1 1 1 1 1
1
Magnetic flux is cut , induced current is produced When ϴ= 90o maximum current produced // ϴ= 0o minimum current produced Commutator is used to ensure the direction of the current that flows through the external circuit is in one direction
1
(d) CHARACTERISTICS
1. Step down transformer 3. Ns: Np = 240: 6 = 40 : 1 5. Use diode 7. Use capacitor 9. Use laminated soft iron core
Max= 10
REASON
2. To reduce voltage 4. To reduce 240V to 6V 6. To change AC to DC 8. To smooth the output current 10. To reduce heat loss due to eddy current
TOTAL
20
156
PERATURAN PEMARKAHAN
BAHAGIAN C
NO.4 (a) (b)
ANSWER
MARK
Current that flows in one direction
The (magnadur) magnets produce a magnetic field / diagram The current in the wire produces a magnetic field / diagram The two magnetic fields interact/combine to form a resultant / catapult field / diagram
1 1 1 1
1 S
N
(c) CHARACTERISTICS
1. Low density material
3. High number of turns
5. High strength magnets 7. More segments
REASON
2. Coil has smaller mass / lighter 4. Larger force acting on the coil / higher electromagnetic field strength 6. Larger force acting on the coil 8. Coil rotates smoothly
Max= 10
9. Motor R is chosen 10. Because.......1357//2468 (d)(i) (d)(ii)
3 x 8 // 24 (J) 24 ()
1 1
5 4.8 W
(d)(iii)
1 1
4.8 ()
× 100 % 12 40 % // 0.4
TOTAL
1 20
157
PERATURAN PEMARKAHAN
KERTAS 3 (BAHAGIAN B) NO. 5 (a) (b) (c)(i) (c)(ii)
(c)(iii)
ANSWER The brightness of the light depends on the speed of rotation of the wheel. The greater the velocity of the magnet in the coil wire, the greater magnitude of the induced current. To investigate the relationship between the velocity of the magnet and the magnitude of the induced current. //When the height of the magnet higher, the velocity increases MV: velocity of the magnet RV: magnitude of the induced current CV: number of turns of the coil Bar magnet, a coil of cooper wire, miliammeter, meter ruler, connecting wires
MARK 1 1 1
1 1 1 1
(c)(iv)
(c)(v)
1. The apparatus is arranged as shown in diagram. The height of bar magnet is adjusted at h = 20 cm. 2. The bar magnet is dropped into the coil of wire. Reading of miliammeter for current is recorded. 3. The steps are repeated for h = 30 cm, h = 40 cm, h = 50 cm and h = 60 cm
1 1 1 1
(c)(vi)
h/ cm 20 30 40 50 60
I/ mA
1
(c)(vii)
TOTAL
12
158
PERATURAN PEMARKAHAN
BAB 9: ELEKTRONIK BAHAGIAN A NO.1 (a)(i) (a)(ii)
(a)(iii)
ANSWER Diagram 6.1 – anode of the diode is connected to positive terminal of dry cell. Diagram 6.2 – anode of diode is connected to the negative terminal of dry cell. Bulb in diagram 6.1 lights up, bulb in Diagram 6.2 does not lights up. The bulb does not light up when the positive diode is connected to negative battery // reversed biased OR The bulb light up when the positive diode is connected to positive battery// forward biased
MARK 1
1 1
1 (a)(iv) 1 1
- AC power supply - arrangement of diodes (output & input)
1 (a)(v) (a)(vi)
Capacitor TOTAL
NO.2 (a)
ANSWER
MARK 1 1 1
Convert AC to DC Four diode
(b)(i) Full wave rectifcation With capacitor
1 1
(b)(ii) (c) (d) (e)(i)
1 8
Smoother the current Circuit R Allows current to flow in only one direction T = 4 x 0.02 = 0.08 s f = 1/0.08
1 1 1 1 1
(e)(ii) = 12.5 Hz
1 1
(e)(iii) No change in amplitude, half period TOTAL
12
159
PERATURAN PEMARKAHAN
BAHAGIAN B
NO.3 (a) (b)(i) (b)(ii) (c)(i) (c)(ii) (d)
ANSWER Cathode ray is a narrow beam of a fast electrons moving in a vacuum. Negative/ (-) Voltage of EHT in diagram 10.2 is greater than 10.1 The deflection in diagram 10.2 is greater than 10.1 Voltage of EHT increases, the strength of electric field increases // directly proportional The strength of electric field increases , the deflection of the cathode ray increases// directly proportional
MARK 1 1 1 1 1 1
The cathode is heated emits electrons The electron/cathode ray accelerated Cathode rays travel in a straight line. Cathode rays is blocked by maltase cross Cathode rays carry kinetic energy and converts to light energy when they hit the screen.
1 1 1 1 1
(e) CHARACTERISTICS 1. Filament 3. Cathode
5. Control Grid
7. Focusing anode
9. Accelerating anode
11. Y-plates
13. X-plates
REASON 2. To heat up the cathode 4. Emits electrons 6. Controls the number of electrons// control the brightness of the image on the screen 8. Focuses the electrons into a beam 10. To accelerate electrons to towards the screen 12. To deflect the electron beam vertically 14. To deflect the electron beam horizontally
Max= 10
TOTAL
Max=20
160
PERATURAN PEMARKAHAN
BAHAGIAN C
NO.4 (a) (b)
ANSWER Semiconductor is a material with electrical conductivity better than insulator but weaker than a conductor Doping process/Silicon is doped with pentavalent atoms/Phosphorus/ Antimony To produce covalent bond Increase the free electron inside the semiconductor Majority charge-carriers is negative electron
MARK 1 1 1 1 1
1
Max=4 Free
(c)(i) (c)(ii) (c)(iii)
V V
X-Z =
6V X-Y = 6 – 1 = 5 V
V M
electron
1 1
R M 6V R R N M
R R 1000 6V
5
M
M
5R M+ 5 000 = 6 R M 5 000 Ω R M=
1 1 1
161
PERATURAN PEMARKAHAN
(d) CHARACTERISTICS 2. 1. LDR is connected at base circuit
3. Terminal positive of batteries is connected to collector
4.
6. 5. Bulbs are arranged in parallel circuit 8.
7. Relay switch is used
REASON When intensity of light is low / dark, resistance of LDR increases / so V base is large / transistor switched on So that the transistor is forward biased All bulbs are connected to voltage supply of 95V So that the secondary circuit will switch on // So that the electromagnet will switch on the secondary circuit
Max= 10
9. R is chosen 10. Because.....1357//2468 TOTAL
20
162
PERATURAN PEMARKAHAN
BAB 10: RADIOAKTIF BAHAGIAN A NO.1
ANSWER
MARK
Time for the activity of the radioactive substance to become half of (a)
1
the original activity // Time taken for half of the mass of radioactive substance to decay
(b)(i)
(b)(ii)
(c) (d)
X : 10 minutes
1
Y : 05 minutes
1
X : 50 %
1
Y : 25 %
1
The decay rates for substance X is slower//vice versa
1
The shorter the half life, the higher the decay rate
1
Background radiation
1 TOTAL
NO.2 (a)(i)
(a)(ii)
(a)(iii)
ANSWER
8
MARK
8 hours
1
Shorter time taken
1
Strong
1
Gamma ray
1
Liquid
1
Easier to dissolved
1 TOTAL
6
163
PERATURAN PEMARKAHAN
BAHAGIAN B
NO.3 (a) (b)(i)
(b)(ii)
(b)(iii) (c)
ANSWER Unstable nucleus Exponential graph The time taken for the activity to become half of its initial value is constant The time taken for the activity of P to be half its initial value is constant //5 hrs The time taken for the activity of Q to be half its initial value is constant //100 s Half-life
Put the radioactive source opposite the detector Detector is connected to the thickness indicator Detector detect the reading of the changes in counts Thickness is measured with the thickness indicator If the reading of the detector is less than the specified value, the thickness of the paper is too tick/ vice versa
MARK 1 1 1 1
1 1 1 1 1 1 1 Max= 4
(d) CHARACTERISTICS Uses thick lead box
Max= 10
REASON Radioactive rays cannot pass through
Packed into concrete drum and buried underground
Prevent the radioactive waste discharged to the environment
Use forceps/ robotic hand
Avoid direct touching
Use siren
Faster warning when leakage
Wearing photographic films
Detect the exposure radiations
TOTAL
Max=20
164
PERATURAN PEMARKAHAN
BAHAGIAN C
NO.4 (a)(i) (a)(ii)
ANSWER Unstable isotope
Beta particles penetrates through the paper and detected by the detector If the detector detect lower reading the paper is too thick // If the detector detect higher reading the paper is thin The roller has to compress harder if the paper is thick // Vice versa
MARK 1 1 1 1 1
(b) CHARACTERISTICS 2. 1. Graphite 4. 3. Boron / Cadmium 6. 5. Heavy water 8. 7. Thick
REASON To slow down the fast neutrons produced by the fission. To absorb some of the neutrons // reduce the rate of the fission reaction. To absorb heat from the nuclear reaction// have high specific heat capacity To prevent leakage of radiation from the reactor core
Max= 10
9. R is chosen 10. Because.... 1357//2468 (c)(i)
(c)(ii)
= 0.19585 × 1.66 × 10− = = 0.19585 × 1.66 × 10− × ( 3 × 1 0 ) = 2.92599 × 10− = 2.92599 × 10− = 5 × 1 0− = 5.8512 × 10−
1 1 1
1
TOTAL
1 20
165
