O Level Pure Physic TYS Physics 1999 to 2008 Answers, for REFERENCE only
42
1 Nov 99 PHYSICS 1 A 2 A 3 C 4 B 5 B 6 A 7 D 8 C 9 D 10 B 11 B 12 A 13 A 14 C 15 A 16 A 17 B 18 D 19 D 20 B 21 B 22 B 23 D 24 D 25 C 26 A 27 B 28 B 29 A 30 B 31 C 32 C 33 B 34 A 35 A 36 C 37 B 38 C 39 A 40 C Paper 2 1ai,ii) bi) the car decelerates and the speed of the car decreases. ii) the acceleration decreases. ci) work done = force x distance moved = 0.11 x 0.90 = 0.099 J ii) KE at bottom = loss in pe + KE at top – work done against friction = 0.50 + 0.30 –0.099 = 0.701 J 2a) solid : C liquid: A gas : B b) The separation between molecules in a gas is much greater than the separation between molecules in a solid. c) i) molecules in a solid vibrates about a fixed position. ii) molecules in a liquid are mobile. iii) molecules in a gas move freely and randomly in all direction. 3a) b) The rod moves into the paper. Using Fleming’s left hand rule, the force (thumb) is perpendicular to both the direction of magnetic field( first finger) and the direction of the current in rod (second finger). ci)the force on the rod is increased, it has a higher acceleration.
Transcript
1
Nov 99
PHYSICS
1 A 2 A 3 C 4 B 5 B 6 A 7 D 8 C 9 D 10 B
11 B 12 A 13 A 14 C 15 A 16 A 17 B 18 D 19 D 20 B
21 B 22 B 23 D 24 D 25 C 26 A 27 B 28 B 29 A 30 B
31 C 32 C 33 B 34 A 35 A 36 C 37 B 38 C 39 A 40 C
Paper 2
1ai,ii)
bi) the car decelerates and the speed of the car decreases.
ii) the acceleration decreases.
ci) work done = force x distance moved = 0.11 x 0.90 = 0.099 J
ii) KE at bottom = loss in pe + KE at top – work done against friction
= 0.50 + 0.30 –0.099
= 0.701 J
2a) solid : C
liquid: A
gas : B
b) The separation between molecules in a gas is much greater than the separation between
molecules in a solid.
c) i) molecules in a solid vibrates about a fixed position.
ii) molecules in a liquid are mobile.
iii) molecules in a gas move freely and randomly in all direction.
3a)
b) The rod moves into the paper. Using Fleming’s left hand rule, the force (thumb) is
perpendicular to both the direction of magnetic field( first finger) and the direction of the
current in rod (second finger).
ci)the force on the rod is increased, it has a higher acceleration.
2
ii) it moves in the opposite direction.
4ai) On the diagram, join two ends of Y to the buzzer.
ii) low high
high low
OR
bi) A :AND gate B: NAND gate
ii) output of AND gate is 0 whereas the output of NAND is 1.
The output from the NAND gate is always the opposite of the output from the AND
gate.
5a) Electrons are transferred from the cloth to the metal.
bi)
+ + + + + + + + +
----------------------
ii) The negative charges on the leg neutralise with the positive charges on the
aluminium foil leaving behind negative charges on the foil.
6. ai) The fuse is to prevent too large current flowing throught the bulbs. When the
current is larger than the normal operating current, the fuse will blow.
ii) So that when the fuse blows, the bulbs are electrically isolated from the power supply.
b) P =IV, I = P/V = 60 / 230 =0.26 A
c) no of lamps = 5 /0.26 = 19.2.
max no = 19
7a) Geiger-Muller tube
b) electron
c) some of the radiation is absorbed by the carton.
d) i) All the alpha –particles would be completely absorbed by the carton.
3
ii) It is highly penetrating, it would pass through the carton and not stopped by it.
8ai) The constant temperature at which a substance is changed from the solid to the
Place the block on a piece of paper. The ray box shines a ray of light through the
block and the ray is visible on the paper. The incident ray and the emergent ray are then
traced on the paper.
b) i = 34.5o , r = 22.5
o , n = sin i / sin r = 1.48
c) Graph
di) when x = 90o, y= critical angle = 43
o .
ii) The largest possible value for x is 90, and the corresponding y is 43, therefore it
is impossible to obtain y=50.
d) Similarity: as x increases, y also increases.
Difference: both graphs give different values of critical angle.
10. a) As the volume of air decreases, the separation between the air molecules
decreases. The frequency of collision between the air molecules and the walls of the
pipe increases , causing pressure to increase.
b) i) P1 V1 = P2 V2
P2 = 1.012 x 105
Pa
ii) different in pressure = 1.012 x 105 -
1.000 x 10
5
= 0.012 x 105 Pa
4
iii) P = h ρ g
h = 0.12 m
ci) The water level in the left side rises,
and the level in the right falls until both levels are the same.
Diagram
ii) use a liquid with density lower than that of water.
11.ai) ii) radio wave, infrared visible, ultraviolet.
iii) radiowave: use for transmission of radio and TV signals.
infrared : intruder alarm, remote control.
ultra violet : detect counterfeit notes; cause tanning of skin and skin cancer.
iv) can pass through vacuum,
same speed 3 x 108 m/s in vacuum/ air
transverse wave (any 1)
bi) Very high frequency alternating voltage is applied across a piezoelectric crystal in a
transducer. When the natural frequency of the vibration of crystal matches the frequency
of the applied frequency, resonance occurs and a strong beam of ultrasonic waves is
emitted, and then transmitted into the body in the form of longitudinal wave where it is
reflected or absorbed by the internal organ.
ii) f = 100 x 20000 =2 x 106
Hz
λ = v/f = 1500 / 2 x 106
= 7.5 x 10-4
m
5
Nov 2000
PHYSICS
1 D 2 A 3 C 4 B 5 B 6 D 7 A 8 C 9 D 10 D
11 C 12 C 13 C 14 B 15 A 16 D 17 D 18 B 19 C 20 B
21 B 22 D 23 -B 24 B 25 A 26 C 27 C 28 B 29 B 30 B
31 C 32 A 33 B 34 A 35 B 36 B 37 D 38 C 39 D 40 C
Paper 2
1 ai) C
ii) At A it has max KE. As it travels upwards, potential energy increases and kinetic
energy decreases. Therefore it travels slowest at C.
( If the object is moving vertically upwards, velocity is zero at the highest point.)
bi) weight = mg = 0.20 x 10 = 2.0 N
ii) KE = ½ mv2
v = 5.0 m/s
2.a) scale : 1cm : 5.0 N
Diagram
resultant force = 34.5 N
b) WD = F x d = 16 x 0.20 = 3.2 J
3.a) The air molecules are in continuous, random motion. They collide with smoke
particle, causing it to move in a erratic manner.
b)
c) all the points of light would drift in one particular direction.
4 a) Diagram
6
b) diminished, inverted.
c) camera
5a) amount of energy required to change 1 kg of water to steam at constant temperature.
b) the steam condenses to form water.
c) heat required = mc ∆θ = 500 x 4.2 x (100 –20) = 168000J
d) final mass = 500 + 168000/2250 = 575 g
6.ai) arrows are in clockwise direction.
Diagram
ii) region nearest to the wire,ie along the surface of the wire.
b) refer to Physics Essay, similar to plotting magnetic field around a magnet.
7a) Helium nucleus
b) beta has weaker ionising power.
c) The electric field produces forces on the ions. The positive ions are attracted to the
negative plate and the electrons are attracted to the positive plate. The moving charges
result in a current.
d) 241 Am 237 Np + 4 He
95 93 2
8a) R = V /I = 3.8 /0.25 = 15 ohms
b) As current increases, temp rises, the gradient of the graph decreases showing an
increase in resistance as R = 1/ gradient.
Section B
9a) i) work done in bringing a unit charge between the two points.
ii) I = P /V = 36 ? 6 = 6 A
iii) The lamp would be over heated and blow.
7
bi) V = 4V, I = 6 A
ii) 1. Diagram
no of turns in primary = 20
no of turns in sec = 12
2. The alternating current in the primary coil produces changing magnetic field . The
iron core links this changing magnetic field to the sec coil, emf is induced in the sec coil.
iv) B has the better solution. There is very little power wastage in using transformer.
In A, there is power wastage as heat in R.
10a) vibration of air molecules
direction of wave motion
vibration of air molecules is parallel to the direction of wave motion.
8
DEC 2001 PHYSICS 5054/1
1 C 2 A 3 D 4 A 5 D 6 D 7 A 8 B 9 A 10 B
11 C 12 C 13 C 14 B 15 B 16 C 17 D 18- 19 D 20 A
21A 22 A 23 B 24 C 25 A 26 D 27 B 28 A 29 D 30 A
31 A 32 B 33 B 34 D 35 D 36 A 37 A 38 A 39 B 40 A
Paper 2
1a) velocity and acceleration
bi) Both forces are in directly opposite direction.
ii) both forces are in the same direction.
iii)
5N 5N
resultant= 5N
equilateral triangle
2a) 00 C
b) t= (lt – l0 ) / ( l100 – l 0 ) x 100 = 75 0 C
c) energy = C ∆€ = 2.4 x 100 = 240 J
d) resistance of platinum wire.
3a) i)label the part where the coils are closer to each other.
ii) compression-a region where the separation of the two nearest coils is the
smallest.
iii) both the particles and the coil oscillate in a direction along the direction of the
wave.
bi) measure from one compression to another compression , 3.0cm
ii) f = v/wavelength = 75/ 3 =25 Hz
4) a) 1.8/10 =0.18 cm
b)The light rays undergo refraction since the speed of light decreases as the light rays
enter the plastic disc and the refracted rays bend towards the normal.
9
ci) draw rays that bend towards normal as they enter the plastic and bend away from the
normal as they travel out of plastic.
ii) 0.7/10 =0.07cm ( approximately)
d) move the lens vertically upwards.
5a) P=IV
I= 10.4 A
b)i) 1.25mm
ii) the wire may overheat and lead to fire.
C) live and neutral wire comes into electrical contact when the insulation of the wiring
becomes damaged.
6) a)
bi)negative charge from drop 1 neutralised with the positive charge from drop 2.
ii) drop 1: positive
drop 2 : negative
c) energy required in bringing a unit charge from one point to another.
7)a) iron can be easily and strongly magnetized. It does not retain its magnetism.
Steel is less easily and less strongly magnetized but can retain much of its magnetism.
b)i) there is a change of magnetic flux in the coil.
ii) deflect in opposite direction.
8)a) Unstable nuclei emit alpha, beta and gamma rays in a random manner.
b) strong penetrating power, able to penetrate through the even packaging.
Not deflected by magnetic and electric field.
c) radiation can cause cancer, radiation burns and damages to our body.
9) ai)The energy can be transferred through the vibration of molecules. The molecules
of the heated end pass on the vibration to the neighbouring molecules.
10
ii) KE no change ( no change in temp)
PE increases ( there is a change of state)
iii) When the steam molecules collide with the walls, the steam molecules exert a
force on the walls, Since P = F/A, a pressure is produced.
Bi) F = PA =60 N
ii) weight of the mass M required = 20 N
( use F1x d1 =F2 x d2 )
M= W/g = 2.0 kg
iii) Move the mass M nearer towards the pivot.
Increase the surface area of valve and hole.
10) ai) As temp rises, resistance of thermistor decreases. The total resistance of the
circuit decreases and the current increases.
ii) 1. I = V/R = 12/504=0.024 A
2. V=IR=0.024 x 4.0= 0.096 V
bi) When the temp rises, the resistance of the thermistor drops. The current through the
relay coil increases and closes the relay switch, current now flows through the switch.
ii) 1.pd across the thermistor = 10 V
2. R = V/I = 10 /0.1 =100 ohms
c) Fig 10.2, pd across motor is always 12V,it can operate at its max capacity.
11)ai) distance = speed x time = 15 x 20 = 300 m
ii) work done = F x d = 1200 x 300 = 360000 J
iii) Power = WD /t = 360000/ 20 = 18000W
bi) KE = ½ mv2
= 90000J
ii) At the beginning of braking period, the speed was high. The loss in kinetic energy at
the beginning is greater than the loss in KE towards the end.
11
NOVEMBER 2002
Paper 1
1. D 2. D 3. C 4. C 5. B
6. C 7. D 8. B 9. A 10. A
11. C 12. C 13.'B 14. B 15. A
16. A 17. A 18. C 19. D 20. D
21. D 22. B 23. B 24. D 25. C
26. C 27. A 28. D 29. D 30. D
31. C 32. A 33. C 34. C 35. B
36. D 37. C 38. C 39. A 40. C
Paper 2
1.
(a)
(b) see txt
(c) (i) 180 N cm
(ii) F=4.5N
(d) the cg is lower so it will only topple at larger angle of tilt/ it will take a larger angle of tilt
before line of action of weight falls outside area of base.
2.
(a) (i) TIR occurs, light is from denser to less dense and i > c
(ii) refraction occurs, theres change in speed, faster in air.
(b) Angle x = 23°
(c) can transfer more data at any time/ less interference/less data loss/less energy loss/ligther
and cheaper
3.
(a) (i) Radiowaves
(ii) v = fλ since v is constant, when λ increases, f will decreaser
(b) (i) tall bldgs will block the wave
(ii) there is energy loss so repeater will increase the energy again
4.(a) 2.0, 0.2, 0.9, 1.2, 3.0
(b) Air-conditioner
(c) Meter reading = 6357.8 kWh
(d) heater carries larger current, so to reduce heat loss/power loss/overheating R is kept low by
increasing area of wire (P=I2R)
5.
(a) (i) The hot gases will expand thus will have a lower density
12
(ii) when heated molecules will gain KE and vibrate faster. They will knock against neighbouring
molecules and transfer the KE to them..so heat is transferred from hot to cold region.
(b) when gases pass through –ve wires, they will become –vely charged. So they be attracted to
the +vely charged plate as unlike charges attract.
6.
(a) liquid
(b) It is the amount of energy required to raise the temperature of unit mass by 1 °C
c) use 1 g of liquid
heat = 1.4 x1x15 = 21 J/g so less than 200 J/g
7.
increase ( Vol increase)
same (same as Patm)
increase ( temp increase, KE increase, vel increase)
decrease ( since P is same and vel increase, frequency of collision must decrease)
8.
(a) as turbine rotate, there will be change in magnetic flux in the coil, so there will be induced emf.
As N pole approach coil, N pole will be induced in coil to oppose it. When the N pole leaves, a S
pole will be induced on coil. So emf will be alternating.
(b) double peak (V = 0.1 V), ½ period (T = 10mms) and time to second wave is also halved
9. Not in syllabus
10.
(a)(i) Pressure = 15 N/cm2
(ii) 15 N/cm2
(iii) Force = 6000 N
(b)(i) Volume = 100 cm3
(ii) Distance = 0.25 cm
(c) valve A - opens
valve B - closes
piston Q - remains at the same place
(d) Air is compressible whereas oil is not.
11. Not in syllabus
13
12.
(a)(i)
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0 1 2 3 4 5
(ii) grad 330 m/s
(b)(i) Time interval t = 5 ms
(ii) Distance d = 1.65 m
(c) distance too small/time taken too small/ human reaction time will make time recorded for
sound very inaccurate/ too many echoes
(d) faster speed as sound travel faster in water. Water particles are closer together compared to
air particles.
12 Or
(a)(i) PE = mgh
(ii) 130 m
(b)(i) energy cannot be created or destroyed but can transformed from one form to another
(ii) Total energy = KE + PE = 1118000
So KE = 1118000 – PE (from column B)
(c) Velocity of the car = 51.0 m/s
14
Nov 2003
Paper I 1. B 2. A 3. C 4. D 5. C 6. A 7. A
8. B 9. C 10. D II. C 12. B 13. A 14. A
15. C 16. A 17. D 18. B 19. B 20. D 21. D
22. C ' 23. C 24. D 25. g . 26. D 27. B 28. B
29. D 30. B 31. C 32. D 33. B 34. A 35. D
36. C 37. A 38. D 39. B 40. A
Paper 2
1
(b) Gravitational field is the region where gravitational force acts on any object in that
region.
(c) (i) 195 N
(ii) 305 N
(iii) 305 / 65 = 4.69
2. (a) Position C, since the perpendicular distance from line of action is greatest.
(b) F= (120 x 34) / 80 = 51 N.
3. (a) virtual, upright, magnified
4. (a) The air molecules oscillate back and forth about their individual equilibrium
positions in the direction parallel to the direction of propagation of sound.
(b) (i) Frequency of wave is the number of oscillations in one second.
(ii) λ =v/f=320/2000 =0.16m
15
distance =3λ = 0.48m
(c) T = 0.002s; frequency = I /T = 500 Hz
5. (a) The randomly and rapidly expanding moving air molecules hit the wall of the bottle
and exert a pressure on it. Pressure is the force exerted by the air molecules on a unit area
on the wall.
(b) When temperature of the air falls, the air molecules travel at lower velocities, thus the
air pressure in the bottle decreases and the air pressure outside the bottle is higher than
the one in the bottle, causing the bottle to become partially crushed.
6. (a) (i) Compass needle rotates anti-clockwise and points to the West.
(ii) Compass needle rotates clockwise and points to the East.
(b) Compass needle vibrates around the North direction at 50 times a second.
7. (a) (i) The force acts upward, perpendicular to side AB.
(ii) Using Fleming's left hand rule, the direction of the force is indicated by the thumb
which points upward.
(b) (i) Upward
(ii) When the coil makes a half turn, a counter moment that opposes the original rotation
is produced in the coil. This prevents the coil from rotating continuously.
(c) The insulation prevents current flow during the half turns. Thus the moment that
opposes the original rotation would not be produced at every half turn.
8. (a)
1. Increasing the number of turns of secondary coil.
2. Decreasing the number of turns of the primary coil.
(b) Iron can be demagnetized more easily than steel.
(c) (i) Power = Voltage x Current
36W=24VXI
I=36/24= 1.5 A
(ii)
Ip(240 V) = 36 W
Ip = 36/240 = 0.15 A
9. (a) (i) The resistance across soil contacts is very high when soil is dry. Thus input B is
low.
(ii) The resistance of LDR is high when it is dark . this leads to input A becoming low
when it is dark. Conversely, when it is bright, the resistance in LDR decreases. Hence
input A becomes higher.
(b) 1,0,1,0
(c) (i) input A - 0, input B - 0, input C - 1
(ii) light level - dark; soil condition - dry
(d) The light level will be lower in order to switch on the pump. This is because when
resistance of RI is increased, the voltage across R, or input A is higher. Hence resistance
in LDR can be higher - which means light level to sum on the pump can be lowered.
10. (a) (i) The denser cool air in the ice box sinks and displaces the warmer, less dense air.
This sets up convection currents which eventually keep the whole food compartment cool.
(ii) The fins are black as black is a better radiator of heat than lighter colours. Heat can be
lost more quickly to the surrounding.
(b) (i) Energy required = mxlf
=(20gx16) x330)
16
=10560J
(ii) There is no change in temperature during the freezing. Thus no heat is absorbed by
the plastic tray. As a result, the heat capacity of the plastic tray does not affect the answer.
(iii) Heat energy removed = Power x time
10560=30xt
T =352 s
11. (a) This is to ensure that light signals will arrive at the output end given that the
transmission of light is not 100% efficient.
(b) A cladding B Glass core
The speed of light in the cladding is higher than the speed of light in the glass core. Thus
the refractive index in the cladding ik Ahan that of the glass core. As a result when light
strikes the boundary surface (line Ab) with angle of incidence that is greater than the
critical angle, total internal reflection occurs.
(c) (i) 40 x 100 V 4000 V
(ii) p.d. = I x R = 0.8 A x(0.7 x 7500)!Q = 4200 V
(iii) p.d. = 4000 V + 4200 V = 8200 V
(iv) thermal energy lost per second per km = I2 x R = 0.8 z x 0.7 = 0.448 J OR
(a) First of all, the background radiation is measured by recording the count rate as
measured using a GM tube. Then the count rate from the radioactive source is measured
at regular fixed intervals of 30 minutes over a period of 6 hours. The background count
rate is subtracted from each measurement and the actual count rate from the source is
measured as shown in the table below. A graph of the count rate of the radioactive source
against time is plotted. From the graph, the time taken for the count rate to fall by half is
measured. A number of measurements are made and an average value is calculated. The
average value is the half-life of the radioactive source. The results can be tabulated as
shown below.
Result
(ii) Alpha radiation can cause ionisation of atoms in human cells. Hence when the source
causes ionisation of atones In the t)NA, the cell will do something very different from
what it is supposed to do and in tum may become cancerous. Gamma rays have less
ionisation effect,
Plot a graph of
17
18
Nov 2004
1 B 21 B
2 C 22 A
3 B 23 B
4 D 24 C
5 D 25 A
6 D 26 D
7 A 27 D
8 A 28 D
9 C 29 C
10 A 30 B
11 B 31 A
12 C 32 D
13 A 33 A
14 D 34 C
15 A 35 C
16 B 36 B
17 B 37 B
18 B 38 A
19 C 39 C
20 B 40 C
Paper 2
1a) P has a greater gradient as compared to Q. gradient of dist-time graph = speed
b) speed = dist / time= 120 km / 1.5 h =80 km /h
c) draw a line starts from 30 mins , parallel to line P.
2.a) larger
b) difference in height = 30mm
Pressure difference = h d g = 30/1000 x 1000 x 10 = 300 Pa
c) i) same diff in water level ; 30mm
ii) double in difference in height , ie 60mm
3 a) He reaches max speed when forward force= resistive force
forward force F = 320N, therefore resistive force = 320N,
From graph, v=8.4 m/s
19
b)i) KE = ½ mv2= ½ x 60 x 8.42 =2100 J ( only 2 to 3 sf accepted)
ii) mgh = 2120
60 x 10 x h = 2120
h = 3.5 m (2sf)
4. a)smoke particles are collided by air molecules at different angles, different times or
unevenly resulting eratic motion of smoke particles.
b) air molecules are in continuous, random motion in all direction .
5. Many candidates did not realise that ray 1 was totally internally reflected and drew a
refracted ray
inside the air bubble. A significant proportion of candidates incorrectly drew the
continuation of ray
2 as a refracted ray bending downwards (towards, rather than away from, the normal) as
it enters
the air bubble but almost all candidates drew ray 3 correctly with no deviation on
entering the air
bubble.
ai) draw a line: (normal: pass through centre of circle)
ii) mark angle between ray and normal.
b)draw lines: bend away from normal, then bend towards the normal
ci) When light travels from Vacuum(air ) to a medium, sini / sin r = n
ii) n = speed in air/ speed in medium = 3.0 / 2.2 =1.4 (2sf)
6ai)
ii) up and down 4 times in 1 sec
b) v = f λ
20
0.80= 4 x λ
λ = 0.2 m
c)use f= 2 Hz
7a)i) Due to repulsion of like charges ,electrons are repelled to the right end of the
sphere, leaving positive charge on the left side of the sphere.
+ -
+ -
+ _
b)
+
+
+
21
c) positive charges evenly distributed.
+
+ +
+ + +
d) plastic
8)a) to step down the voltage
b)i) P= IV , when high V is used, I can be minimised.
Less energy will be wasted as heat in the cable due to joule heating. ( Power loss = I2 R)
ii) Thicker cable has lower resistance. Power loss = I2 R . Lower resistance will cause
smaller thermal energy loss.
c) i) Np / Ns = Vp / Vs
ii) Np / 48000 = 20k / 275 k
Np= 3490 turns
Section B
Question 9
(a)(i) Conduction: When outer layer of the copper pipe is heated, the molecules vibrate
more. The molecules pass on the vibration to the neighbouring molecules. In this way,
energy is transferred from molecule to molecule.
(ii) Boiling requires energy. When the alcohol absorbs enough heat energy , it starts to
boil. When the energy is transferred to the cold water, it condenses as condensation
releases this energy.
(iii) the movement of the alcohol vapour, molecules or liquid is fast down the tube, as the
large pressure difference is likely to cause a rapid movement from the high to the low
pressure ends of the tube.
(b) specific latent heat is the energy needed to change the state of 1 kg of the liquid to gas
or gas to liquid without any change in temp.
(b)(ii) ml = mc x change in temp
25 x 840 = 500 x 4.2 x change in temp
22
change in temp= 10 0C.
Question 10
(a) (i)
1 0 ( becos OFF) OFF
0 1 ON
(ii)Component P is a thermistor,.
(b)(i) the resistance of the thermistor decreases as
temperature rises,
(ii)current increases
(iii) pd across Q increases.
(iv) the p.d. across P decreases
(c)The advantage of a variable resistor is that it
allows an operator to change the temperature that the buzzer comes on at.
(d) 0 input to the NOT gate, output would be 1. This occurs when the temperature is low
and then there is a greater p.d.
across component P.
Question 11
EITHER
(a)(i) When there is in fact no current at all in the coil since there is a short circuit.
In (ii) the brushes may not be in contact with the rings and thus no current can flow.
(b)(i) moment as the product of the force and the perpendicular distance from the line of
action of the force to the pivot.
(ii) total moment = Fx d + F x d= 3x 0.065/2 x2 = 0.195 Nm.
(iii) the perpendicular distance is the greatest when the coil is horizontal.
23
(iv)
moment
time
1 rev
OR
(a)
beta gamma
1. negatively charges No charge
2. is an electron Electromagnetic wave
3. stopped by a few mm aluminium Stopped by a few cm lead
4. can be deflected by electric and magnetic
field
not deflected by electric and magnetic field
(b)- it takes time for the radioactive iodine to travel through the bloodstream,
-decay is random.
(b)(ii) (40+38+36+40) / 4 = 38.5 per second,
(iii) 38.5 : 2 cm3 = 144000: x
X = 7480 cm3
(3sf)
8days 8 days
(iv) 40 20 10
Ans: 10 per second.
(c) Use tongs/ forceps to handle the material.
The most often quoted precaution was that the doctor should wear a lead-lined suit or use
lead-lined gloves. A “radioactive suit” is not sensible. Many candidates gave suggestions
such as
“wear a mask or coat” but these were not considered sensible suggestions for a Physics
examination.
24
2005 PHYSICS 5052
1 C 21 C
2 C 22 C
3 D 23 C
4 C 24 B
5 B 25 B
6 A 26 C
7 B 27 D
8 D 28 B
9 A 29 B
10 A 30 D
11 C 31 C
12 C 32 D
13 C 33 A
14 A 34 D
15 B 35 C
16 D 36 C
17 D 37 B
18 B 38 C
19 C 39 B
20 D 40 D
Paper 2
Section A
1 (a) Mass is the amount of matter in a body.
(b)(i) The names given to the two forces : tension and weight. It was expected that the
two forces should be drawn along the central line of the diagram.
tension
Weight
(ii) Weight = m g = 0.40 x 10 = 4.0 N ( must give 2 s f)
Tension = weight = 4.0 N (2sf)
(c) the mass moves upwards because there is a resultant force upwards.
The tension had increased ,the tension is greater than the weight, thus producing this
resultant force upwards.
2 (a) Pressure is due to molecules colliding on the walls of the container, producing a
force on the wall. [1]
As there are more impacts on the larger area, but that the number of impacts per unit area
was the same, therefore the pressure is the same.[1]
(b)(i) Work Done = Force x distance= 50 N x 0.015 m = 0.75 J
25
(ii) Boyles’ Law: PV = constant
P1V1 =P2 V2
1.0 x105 x V = P2 x 0.8 V
P2 = 1.25 x105 Pa
Answers: (b)(i) 0.75J, (ii) 1.25 x105 Pa.
3 (a)
, r = 19.8°
(b) (i) wavefront - the line joining points having the same phase on the wave, e.g. the line
joining crests along a wave.
(ii) 1. the wavelength decreases ( as seen in the diagram)
2. the speed of the wave decreases.
3. frequency is constant
4 (a) x –ray , ultraviolet , infrared, microwave
(b) The uses of ultra-violet radiation
: 1)production of a suntan, 2)sterilization, 3)vitamin D,
4)the testing of banknotes ( checking of counterfeit notes)
(c) Properties :
1) transverse waves,
2)able to reflect or carry energy,
3) having the same speed in vacuum or air
4) able to travel in a vacuum
5 (a) The diagram showing increased loudness and pitch refer to the board.
(ii) louder sound has larger amplitude and higher pitch has
higher frequency or shorter wavelength.
(b) (i) electrical energy changes into chemical energy when a battery is charged.
(ii) chemical energy changing to electrical energy and then to sound. (two stage energy
process )
6(a) In (i), Due to the repulsion of like charges(electron in metal discharge ball repelled
by the electrons in metal dome), the negative electrons move to the right. The electrons
move down the rod making the metal discharge ball positive.
(ii)
X
(b) Q = It ,1mC = 10-3
C,Answer: (b) 0.00133A.
s i n 6 01 .4 7
s i n 3 6n
ο
ο= =
sin30
sinn
r
ο
=
26
7 (a) The iron rods were magnetised with the ends of each rod next to each other have the
same magnetic pole. Therefore they repel each other.
(b) There should be no movement when one of the iron rods is replaced with a copper rod.
Copper is a non-magnetic material.
8 (a) alpha and beta particles are stopped by the lead or inner container and that some of
the gamma rays can escape from the box.
(b) 1) handle the source using tongs or tweezers
2) wear gloves and special suits
(c) The Geiger-Muller tube.
The number of counts in any time period, or the reading of a rate meter, varies at
different times shows the random nature of radioactivity.
Section B
Question 9
(a) the resistance increases as the p.d. increases, because the gradient of the I - V graph
decreases as resistance = 1/ gradient.
(b)(i) 1.5 v + 12 v = 13.5V,
(ii) V = IR , R = 1.5 V / 0.8 A= 1.88 Ω;
(c)(i) 0.8A ( from graph) ,
(ii) I through L is 2A, total I = 2.0 + 0.8 = 2.8A,
(iii) R = V / I = 12 V / 0.8 A = 15Ω,
(iv) 1 / R = 1/ 15 + 1 / 1.87 , R = 4.3Ω.
10(a)(i) When the card falls, it first block off the light from the light beam, less light falls
on the LDR , its resistance increases.
When the light passes through the square hole, more light falls on the LDR , its resistance
decreases.
Then the card block off the light again as it falls , less light falls on the LDR , its
resistance increases.
(ii) When the card falls, it first block off the light from the light beam, less light falls on
the LDR , its resistance increases.
For a potential divider, the fixed p.d. 6V was shared in varying amounts across the two
resistors, as resistance of LDR increases, the pd across the fixed resistor decreases and
pd across LDR increases.
V across LDR = ( RLDR / ( Rr + RLDR ) x 6 V
(b) NOT gate: a HIGH output (1) is obtained from a LOW (0) input, which refers to low
p.d.across LDR [1mark]( LDR has low R when there is light falls on it)
The LED was off when the card blocks the beam of light.[1]
(c) i) Distance = v x t = 1.0 x 0.03 sec = 0.03 m for pulse 1
27
For pulse 2 : average speed = dist / time = 0.03 / 0.015 = 2.0 m/s
ii) acceleration = V- U / t
(2.0 – 1.0 ) / 0.103 = 9.7m/s2
11. EITHER
(a) 1. Measure the mass of water boiled away.
( mass before boiling m2- mass after the boiling stops m1)
2. Measure the time taken for the mass of water to be boiled away (m2-m1)with
the time measured from the start of boiling.
Power x time = mass x lv
Power can be calculated .
(b)(i) E = P x t = 2000 x 6 x 60 = 720 000J,
(ii) E = 2 kW x 6 / 60 = 0.2 kWh,
(iii) 720 000 / 0.2 = 3 600 000J.
(c)When the liquid evaporates, latent heat is required to break the bonds between the
molecules or to separate them.Therefore , only the more energetic molecules can escape,
leaving behind the less energetic ones, therefore the liquid cools.
OR
(a) (i) 6x 60 / 0.00462 = 77 900 ( 3sf)
(ii) 0.00012 J x 77900 = 9.35 J ( 3sf)
(iii) 9.35 J = mc x temp rise = 50 g x 4.2 x ∆ θ
∆ θ = 0.04450C
b) the brain is not made from 50 g of water and the specific heat capacity of the brain and
water are likely to be different.
(c) i) The definition of power : energy change or work done per unit time
ii) 9.35/72 = 0.13
( 0.20 W x 6 x 60 s = 72 J)
Answers: (a)(i) 77 900, (ii) 9.35J, (iii) 0.0445°C; (c)(ii) 0.13.
28
PHYSICS 5052
Nov 2006 Answers
1 D 21 D
2 D 22 B
3 A 23 A
4 C 24 C
5 C 25 B
6 D 26 D
7 D 27 B
8 C 28 B
9 B 29 A
10 B 30 D
11 C 31 D
12 B 32 B
13 C 33 C
14 A 34 A
15 C 35 C
16 A 36 D
17 D 37 C
18 B 38 C
19 C 39 C
20 B 40 B
1
a distance = 12 x (28 – 12) = 192 m
b a = (v-u)/t = (28-12)/34-28)= 2.7 m/s2
c distance = 1/2 (30)(10) = 150m
therefore car did not reach end of town as 150m < 192m
2.
a 10 m/s2
b(i) weight of the paper is equal to the air resistance acting on it in the opposite
direction. So resultant force is 0. acceleration is 0. speed is constant.
(ii) weight is greater than the air resistant acting on the coin. So resultant force is not
zero. Coin will accelerate.
C both will fall together at the same speed
Fall faster
Time to reach bottom is shorter
Acceleration is constant at 10 m/s2
3
a(i) means both are at the same temperature and there is no net heat gain or loss
between the two teapots.
(ii) they are both at the same temperature – room temperature
29
b(i) teapot is at a higher temperature than the surrounding. Heat will be lost to the
surrounding through radiation. Heat will be be conducted to the outside of the pot and
then futher lost to the surrounding through convection of the air.
(ii) black. It is a better radiator of thermal heat/infra-red radiation.
4
a(i)
(ii) n = sin 40/sin 25 = 1.5 (2sf)
b(i) it cannot occur because the light is traveling from less dense to denser material
(ii) it is defined as the angle of incidence in the optically denser medium for which
the angle of refraction in the less dense medium is 90 degrees.
5.
a
b S N S N
c(i) test it with both poles of a magnet. It there is repulsion then it is still magnetized.
(ii) insert in solenoid with ac current. Slowly withdraw to a few metres away.
6
a potential energy is converted to KE as the water drops.
b. (2000-1200)/2000 x 100 = 40%
c sound
KE of the water that flows away from wheel
PE of water that fall midway of wheel
Friction as wheel turns
7
a electromagnetic induction
b deflect
25º
40º 40º
30
c no deflection. The wire does not cut the magnetic flux so there will be no induced
emf
d using fleming right hand rule. Index finger point from north to south. Thumbs
point downward. The middle finger will give the direction of the current.
8.
a 16 V
b(i) R = 8
V = RI
I = 16/8 = 2 A
(ii) V = RI
= 6 x 2 = 12 V
(iii) V = RI
9 = 6I
I = 1.5 A
V = RI
16- 9 = R (1.5)
R = 4.7
31
9
a(i) ti is the turning effect of the force about the pivot and is given by the
multiplication of the force with the perpendicular distance from the line of action of the
force to the pivot.
(ii) M = 0.1 x 2.6 cm = 0.26 Ncm
b(i)
(ii) the weigh is now vertically below the pivot. The perpendicular distance from the
weight to the pivot is now zero. Thus the moment is zero.
c
The CG is lower
The area of the base is bigger
10
a(i) earth
(ii) the lamp outer part is an insulator. So even when the live wire touches it there
will not be any current flow.
b(i) means it produces 100 J or energy in 1 s
(ii) P= IV
100 = Ix 230
I = 0.43 A
Fuse = 0.5A or 1A
(iii) E = Pt = 100 x 30 x 60 = 180 000 J
11
A
B
32
a potential energy is converted to Kinetic energy when the gas and dust fall inwards.
The KE is then converted to thermal/heat energy
b High temperature means high KE. High KE means high velocity.
C same no of protons but different no of neutrons. H has 1 p and 1 n. But H has 1 p
and 2 n.
OR
a NOT and AN
b(i)
b(ii) When sensor is hot, the resistance of the thermistor will be low. Thus the pd
across the thermistor will be low and the resistance across the fixed resistor will be high.
So the output is high.
c(i) NOT (light sensor) - high
NOT ( heat sensor) - high
AND – high
(ii) NOT (light sensor) - low
NOT ( heat sensor) - high
AND – low
d heater that turns on when it is night time
Output
from
Heater
33
5052 Physics November 2007
PHYSICS Paper 5052/01 Multiple Choice Question Number Key Question Number Key 1 C 21 C 2 B 22 A 3 B 23 C 4 C 24 D 5 D 25 B 6 C 26 D 7 A 27 B 8 D 28 B 9 B 29 C 10 B 30 C 11 C 31 B 12 C 32 A 13 A 33 D 14 A 34 D 15 D 35 C 16 A 36 C 17 B 37 C 18 A 38 B 19 D 39 B 20 D 40 A Difficult Q Question 9, 20 and Question 28 Section A
Question 1 (a) (i) 0-2 and 16-24 s. For both intervals, net F is zero.
(ii) 0 – 2 s, the car was initially at rest. 16- 24 s the car moves at constant speed
(b) (i) Use driving F = 0 or 2500N , both can be accepted. F = ma.
a= 0 or 2500/ 850 = 2.94 m/s2
or 0 m/s2
2 were (the formula had to be shown in both cases) (ii) Net F= ma = 850 x 2= 1700 N
Net F = Driving F - Resistive force 1700 = 2500 – Resistive F Resistive F = 2500 – 1700 = 800 N From the graph, t = 6.4 seconds.
Question 2
34
(a) X-rays are able to penetrate the body, are absorbed by bone, cause little harm or low ionisation in the body, or are able to affect a photographic film or cause
fluorescence. (b) The medical uses most often chosen were cancer treatment and sterilisation of medical equipment. gamma rays are able to penetrate through the body to the cancer or travel deep into the materials being sterilised, that they cause high ionisation in killing the cancer or bacteria, or even that the energy of the gamma ray is high. Question 3 (a) Vernier calipers are able to read to 0.1 mm.
(b) (i) W = mg 1.80 N / 0.1768 kg = 10.2 N/kg (ii) Density = mass / volume = 176.8 g / ( 4.01x 2.04 x 1.12) = 19.297g/cm
3
(iii) The calculated value of the density of the bar was 19.297 g/cm3 and the density of pure gold
was given as 19.281 g/cm3. It was intended that candidates should realise that the readings used to calculate the density were only given to 3 significant figures and that some error is possible. Thus it is entirely possible that the bar is made of pure gold. However most candidates stated definitely that the bar was not made of pure gold as its density was higher. Some strong candidates did correctly evaluate the situation and explain the possibilities. Question 4 (a) PV = constant, 150 x 10 x 1x10
5 = P x 1200
P = 1.25 x10
5 Pa
. (b) the pressure increases because more molecules per second or per square metre hit the
tyre walls. A reference to collision of molecules with the walls of the tyre was crucial. Force on the wall increases, P = F/A, therefore P increases. Question 5 (a)(i) energy needed to change the state of 1 kg of liquid to gas without any change in temperature. (ii) P xt = m l 1.2 x 1000 x t = 400 x 2250 t= 750 sec (b)The jet of steam produces more energy because the steam condenses to water and liberates latent heat. Or there was more internal energy in the steam or the increased energy was caused by breaking of bonds as the water boiled. Internal Energy = PE ( depending on state of matter, liquid) + KE ( depending on temp)
35
Question 6
(a) (i) The ray is not refracted as it enters the glass block because its angle of incidence is 0°.
(ii) n = speed of light in air / speed of light in glass
1.5 = 3.0 x 108 / v
V = 2.0 x 10
8 m/s
(iii) Both the speed and wavelength increase as the light leaves the glass block. b) Draw a refracted ray between normal and RS, bends away from the normal.
(c)The majority of candidates suggested that the ray of light is totally internally reflected at R. Since the diagram was not drawn to scale and the refractive index, as calculated from angles in the diagram, was not exactly 1.5, other answers were accepted by the Examiners, such as “refracted away from the normal”, “refracted along the surface” or even just “reflected”.
Question 7 (a) (i) power = potential difference x current (ii) fuse ratings of integral value between 9 and 20 A were accepted
P = 2000 W, P = I V , 2000 = I x 230 , I = 8.7 A (iii) 0.2 kW x 15/60 h = 0.05 kWh
(b) Fuse X “blows” when P and Q are connected and fuse Y is unaffected (or the current through fuse Y becomes zero).
“both air-con and the refrigerator are switch off, do not work or are unharmed”.
Question 8 (a) electrons move from the plate to the cloth.
(b)(i) the charge on the plate is neutralised.
(ii) The diagram was usually drawn correctly with some attempt to show negative at the bottom and positive at the top of the particle. (iii) Most candidates scored one mark for a statement that negative charge on the particle was attracted to positive charge on the plate. One mark was awarded either for realising that the repulsion between positive charges on the plate and particle was weaker than the attractive force or that the negative charge on the particle cannot flow onto the plate since the plate is an insulator. Section B
36
Question 9 (a) The ammeter reading decreases
The numerical values required were either that the reading of voltmeter 1 decreases from 12 V or that the reading of voltmeter 2 increases from zero. (b)(i) draw a curve current- voltage graph of a lamp.
(ii) As current increases, the lamp gets heated up, resistance increases, therefore the slope decreases ( slope = 1/ R) for I-V graph
(iii) Q = I t , From the graph, I = 0.15 A , Q = 0.15 x 5 x 60 = 45 C . Question 10 (a)KE = ½ m v
2 = ½ x 700 x 40
2 =5.6 x 10
5 J
(b) The marks in this question were reserved for an increase in gravitational potential energy ( due to slope)and for one other form into which the energy is transferred, such as heat .
(c)(i) PE = mgh = 700 x 10 x 3.0 = 21000 J
(ii) Total Energy = gain in PE + work done against friction 560000 – 21 000 = 539000 J
Work done against friction, F x d = 539000, F = 539000 / 40 = 13500 N
(d) Many possible factors were allowed, such as the roughness or type of surface or tyre, mass or weight of the car, the size, evenness or distribution of the stones and even the wind speed ordirection. Some candidates did not appear to understand what was meant by a “factor that affects the value of the frictional force” and merely stated “air resistance”, “friction” or even “distance”. Air resistance was not an acceptable answer as this would depend on the car’s speed. Question 11 EITHER Not in the syllabus OR (a) (i) Y-shift control was adjusted to move the trace down, that the Y-gain was adjusted to produce a larger trace vertically and that the time base was adjusted to give a larger distance between the pulses on the screen.
37
(ii) 3.5 x 5 = 17.5 mV Time: 7 x 0.1 = 0.7 s (b) (i)the magnetic field or flux within the coil changes or that magnetic field lines cut the coil. Emf is induced .(In some answers it was not clear that the field within the coil itself was changing or that the lines actually cut the coil.) In (ii) the right-hand grip rule was correctly stated as showing that the direction of the current was correct as it predicts a north pole at the coil of the magnet which repels the incoming magnet.
38
NOVEMBER 2008
Paper 1 1.D 2.B 3.B 4.C 5.D
6.B 7.D 8.A 9.A 10 D
11.B 12.B 13.C 14.B 15.B
16.A 17.C 18.D 19.C 20.D
21.A 22.B 23.A 24.D 25.B
26.D 27.B 28.B 29.C 30.D
31.B 32.B 33.A 34.C 35.A
36.A 37.B 38.D 39.A 40.A
Paper 2
1. (a) Vector has direction and magnitude, scalar has magnitude only.
(b) Acceleration/ forces/velocity/displacement
(c)
Sin 30 = TC/200 Tc = 100N
Cos 30 = TB/200 TB = 173 N
2.
(a) GPE decreases, KE increases then decreases, elastic PE increases
(b) 80x10x56 = 44800 J
(c) ½(80)v2 = 23000
v = 24 m/s
200N
TC
TB
30°
39
3. (a)
(b) the voltage produced is proportional to the temperature difference between the cold
and hot junctions. thermocouple can be calibreated by measuring the voltage when the 2
junctions are placed in ice and steam.
(c) The metals have high melting points, hence it can be used to measure high
temperature/Suitable to measure quickly changing temperature/can measure temperature
at precise location or location that is remote.
4. (a) (i) 10-9
m
(ii) v = fλ
3 x 108 = 10
-9 x f
f = 3 x 1017 Hz
(b) Addition or removal of electrons from neutral atoms/molecules
(c) (i) X-ray radiation may cause damage to the molecules and cells
(ii) Microwaves when absorb can cause heating up
5. (a)
(b) n = sin 42/sin 30 = 1/sin c
c = 48.4°
(c)
milli
voltmeter
hot junction
in engine
cold
junction/ice
point
metal 1 metal 1
metal 2
40
(ii) Inside of optical fibre is optically denser than the outer part of the fibre. At each
point of incidence, the angle of incidence is greater than the critical angle. Hence total
internal reflection takes place each time.
6. (a)
(ii) Refraction of light takes place at the two surfaces of the lens. light ray changes speed
as it enters the lens and again as it leaves the lens. angle of incidence are not equal to
zero.
(b) Object is placed between the lens and the principal focus.
7. (a) (i) I remains zero.
(ii) Until a certain value of V about 0.6V I remains zero. When V varies above this value,
I increases slowly at first and then steeply.
(b) V = IR
0.8 = 8 x 10 -3
x R
R = 100Ω
(c) No, I don't agree. When V is negative, the current is zero. Resistance will be very high.
8. (a) When the a.c. passes in the primary coil, the direction of resultant magnetic flux
alternates. The iron core links this changing magnetic flux to the secondary coil. This
change of magnetic flux passing in the secondary coil, induces an output voltage in it.
(b) ratio = 25000:400000 = 1:16
(c) Current = 12000/16 = 750 A
(d) I will be small thus energy loss in the cables is low.
optical fibre light
ray
41
9. (a) (i)F = ma but the total mass is not double as it is mass of car + load, so a is not
halved.
(ii) as car travel faster, friction and air resistance increase. thus the resultant force acting
on car ( angine force – friction ) will reduce. When the friction/air resistance have the
same value as the engine force,resultant force = 0 so a = 0, car will not accelerate and v
wil not increase anymore.
(b) (i) 10.9 = 49 000/t
t = 4500 s
(ii) 4.24 x 10 3 x 4500 = 1.9 x 10
7 J
(iii) E = PVt
48 x 95 x t = 1.9 x 107
t= 4180 s
(iv) All energy supplied during the charging is stored in the battery. There is no energy
loss.
(c) There is no 'smoke (carbon particles or gases) emitted into the air. Lesser pollution.
10. (a) (i) 1. Polystyrene is a poor conductor of heat. it also traps air which is also a poor
conductor of heat so heat form outside cannot be conducted into the refrigerator.
(ii) cool air contracts and density increases, it sinks. warmer air, less dense, rises.
Convection current is set up and whole refrigerator is cooled.
(iii) Smooth white metal surface is a poor radiator of thermal energy.
(b) (i) 100 x 4.2 x 4 = 1680 J
(ii) 3 x 4.2 x 46 = 580 J
(iii) 1680 – 580 = 1100 J
(iv) 3 x lf = 1100, lf = 367 J/g
11. Either
(a) (i) The wire which is kept at high potential.
(ii) The wire which is kept at 0 V
(b)(i) the 5A fuse is supposed to limit the current to less than 5 A, if the fuse is changed
to 30 A, current larger than 5 A can still flow. this may cause overheating and damage to
appliances.
(ii) if there is a fault and live wire touches outer casing, the outer casing will become
“live” thus a person touching it may get electric shock. The earth wire would have
provided an a path for the current to flow to the earth and cause the fuse to blow, thus
appliance will be isolated form “live” wire.
(c) The circuit breaker can be reset, while the fuse wire has to be change
42
Set up circuit as shown
set rheostat to max and record ammeter reading
observe if fuse blows
repeat exp by slowly decrease rheostat resistance and I increases to 1A, 2A etc until 5A
observe if fuse blows
11 Or
(a) as temp ↑ R for thermistor ↓ so total R ↓so I↑
V = IR so V also ↑
(b) V = 36 V
1.6/(1.6+R) = 36
R = 1070 Ω
(c)
note the V for thermisto and 1.6 kΩ resistor must add up to 6 V
