K.C.S.E 1995 PAPER 1 MARKING SCHEME
 Effort would reduce
 Flow from a to B
 Pressure difference between liquids in A and B is P = egh where e is liquid, g = acceleration due to gravity and h is height
But force = P x cross section area of siphon, P = F/A
Thus F = egh A Since e.g. A are constants
Fα h
 No change in flow OR the flow will still continue
 Oil spread until it is one molecule thick or film taken as a perfect circle or oil drop has been taken as perfect sphere/ cylinder/ uniform thickness
 The liquid expand uniformly, expansion is measurable ( large enough), thermal conductivity
 Rectilinear propagation/ light travels in a straight line
 Water/ or glass are poor conductor of heat
 Each material is brought in turn to touch the cap. The conductor will discharge the electroscope while the insulator will not ( accept bring near conductor gauge)
 Can be short – circuited without being destroyed
 Longer life/ electrolyte never need attention
 Can stay discharged without being destroyed
 Can be charged with large currents faster charging
 More rugged/ not damaged by rough condition of use/ robus
 Delivers large current, light
 Surface tension / adhesive forces supports water column or more capillarity in tube 2 than tube 1
 Surface tension is the same in both tubes and equal to the weight of water column supported
 Narrow tube has longer column to equate weight to wider tube
 Volume of water in the tubes is same hence narrower tube higher column
 – Length of conductor in the field
– Angle between conductor and fields
 All ferromagnetic materials are attracted by magnets or any magnetic materials is attracted
 – increasing the tension
– Reducing the length
 At equilibrium sum of clockwise moment = sum of anti – clockwise moments
Clockwise moments = P x X = QY
Px = Qy
 h glass = V air / V glass 1.5 = 3 x 10^{8} Ö g
Vg = 3 x 10^{8} / 1.5 = 2 x 10^{8} ms^{1}
 V = f l sine V is constant reducing f to 1/3 Þ l increases 3 fold
 While light is composed of seven colour different/ many colour. For each colour glass had different value of refractive index/ different velocities of different l. So each colour is deviated differently causing dispersion
 A body at rest or in state of uniform motion tends to stay in that state unless an unbalanced force acts on it.
 Heat capacity is quantity of heat required to raise the temperature of the body by 1 k or 1 ^{0}C while, specific heat capacity is quantity of heat required to raise temperature of unit mass of body by 1 k/ 1^{0} C.
 (If x ≠z but both above y give 1 mk. Accept difference of 1.0 mark)
hX = hZ > hY
 – Reducing – Increasing
 Polarization
Type of radiation  Detector  Uses 
Ultra violet  Photographic paper fluorescence material  Cause ionization kills bacteria OR operating photosular cells photography 
Infrared  Phototransistor blackened thermometer  Warmth sensation 
Radio waves  Radio receiver or TV receiver  Communication 
 E_{2} = E_{1} + h f i or E_{2} – E_{1}= h = c/l
h= plank constant
c Velocity of light
l Wave length of light
 – Lead – Very dense/ has high atomatic mass
 Extrapolation on graph ( line to touch frequency)
Reading on graph to (4.0 + – 0.2) x 10^{14}Hz
 Lines parallel to the one shown but cutting of axis further in
 Quality / Timbre
 X = 14
 The point where the weight of the body acts
 Temperature of source be the same
– Length of rods be the same / wax
– Amount of wax (detector) be the same
K.C.S.E 1995 PHYSICS PAPER 232/2 MARKING SCHEMES
 (a)
(b) Constant Vel^{0} Uniform vet – zero accl^{n}
(c) Ö4.5 = 118 – 50 = 15m/s 15.5 + 1.5 ( 1417)
6.52
Ö 6.5 = 112 – 70 = 6 m/s (4=6)
7
Average accln = ∆v = v – 11 = ( 615)
t t 2
= – 4.5 m/s^{2}
 l = 7 + l + l
R_{C} R_{1} R_{2} R_{3}
= 1 + 1 + 1
6 + 3 6
= 1
6
R_{C} = 6 = 1.5 W
4
(b) Total resistance = 1.5 + 2.5 = 4 W
E = 1(YFR) Or l = V
R
2 = Ll
Current through xy l = 0.5 A
P.d across yz = 0.5 x 1.5 V
s= current through 3 W = 0.5 x 1.5 = 0.25 A
3
(c) R = /L A
I = RA = 6 x 5.0 x 10^{6} Wm^{2}
L 1.0 m
= 3.0 x 10^{5}W m
 (a)
(ii) Magnification = V Isign = 1.1 OR 1.75
u Osign 1.6 2.5 = 0.7 ± 0.05
 l = l + l l = 10
f u v u 60
l = l + l u = 6cm
10 u v
l = l + l Objects is 6 cm from the lens
U 10 15
4 (a) Lens symbol object between f & F 2 appropriate rays position of image
Image correctly drawn
The diagram in figure 3 shows a certain eye defect
(b) (i) Name of defect is long sightedness
(Refer to the diagram in the figure 3 above)
(c) (i) For water not to pour weight of the water must be less centrifugal force OR for water to pour out MV^{2} > mg
r
(ii) Frictional force F = Centripetal force
MV^{2} = 1200 x (25)^{2}
R 150
= 5.0 x 10^{3}N
 (a) (i) The magnitude of the induced e.m.f is directly proportional to the rate at which the conductor cuts the magnetic field lines
The induced current flows in such a direction as to oppose the changes producing it.
(ii) Plugging a magnetic into a coil
 in speed its g twins as straight of magnetic field
 Results in an increased in the induced e.m.f
(b) (i) Energy is neither created nor destroyed
Make power constant
VU = Joules ( ½ ) current = charge ( ½ )
Count time
P = IV
For large V, 1 must lower for power input to be equal to power output
(ii) Vs – Vp OR Vs – Na
Ns Vp Vp NP
Ns = Vs x Np = 9 x 480
Vp 240 Ns = 18
SECTION II
 (a ) Progressive wave Wave profile moves along with the speed of the wave
Stationary wave – wave profile appears static
Progressive wave – Phase of points adjacent to each other is different
Stationary wave – All points between successive node vibrate in phase
Progressive wave – Energy translation in the direction of the wave travels
Stationary wave No translation of energy but energy associated in the wave
(b) (i) A glass slide i.e. blackened with soot or paint lines are drawn close together using
a razor blade or pin.
(ii) Path differences equals to an odd number of half wavelengths or completely out of phase ( 180^{0})
(iii) Photometer / photocell or thermometer with a bulb
 (a) Common or sillen ( semiconductor) is doped with impurity atoms which trivalent ( e.g boron or indium) intensity in currency on pole group 4 doped with trivalent
(b) pnp emitter and carries made of p type material are of n type material for charge carries holes
 n – p – n – emitter and collector made of n type material are made of p type ( or charge carries electrons)
(c) At the middle of the reaction of a curve a tangent is drawn change on output (∆V_{0}) is determined and a corresponding change input ( ∆V_{1}) also attained change amplification.
(d) (i)
(ii) i_{2} = l_{C} r l_{B}
(e) Base – emitter – forward biased Base collector – reversed biased
PHYSICS PAPER 231/1A 1996 MARKING SCHEMES
 Correct full marks to be given
 Wrong units no marks given
 Wrong substitution no mark
 No units full mark
 15.00 + 0.30 = 15. 30 mm; or 1.53 / 1.53 x 10^{2}m
 Frequency: OR wavelength or energy
 Length of container/ height
Width of the base/ base area/ diameter/ radius of the base/ thickness
 h_{p} p_{1} g = h_{2} p_{2}g Same as h_{1} p_{1}= h_{2} p_{2}
h_{1} = h_{2} p_{2}g = 8 x 18
pg 08
= 18cm;
 (i) Rubber is elastic and when a nail pushed through it stretches and grips the nail firmly without allowing air leakage
(ii) Valve effect pressure from inside causes tyre rubber to press firmly on the nail
 Concrete mixture and steel have approximately the same linear expansively. The expand/ contract at the same rate;
 Radiation is at the electromagnetic waves Φ infrared while conduction involves particles, which move at lower speed
 There are three different sources of light of the different intensities; brighten/ dimmed / different direction/ amount quality. Similar sources/ at different distances from the object
 like charges repel unlike charges attract
 Mass per unit length
Or (linear density/ thickness/ cross – sectional area/ diameter, radius
 Adhesion
Cohesion/ surface tension
 As the thermistor is heated its resistance reduces/ conductivity increases hence drawing more current through it; hence less current flowing through B;
 (i) (ii)
 T< F or F> T
Moments of T and F about are equal; but the perpendicular distance from O to T perpendicular distance from O to F/ Resultant moment are zero
 Turn anticlockwise about O, OR Oscillate about O
 The wavelength/ velocity of the water waves reduces; away from the centre because the pond becomes shallower/ pond deeper at centre
 Interferences ( accept beat)
 Parallel resistor allow diversion of current; hence may not overheat; / current shared by parallel resistor
 Heat gained 5(80 – 40) = m(4015) Heat gained MCD θ ( 80 40)
5(40) = 25m Heat post MCD θ =m (40 – 15) MC 40 – 15
5(8040) = 25 m
25m = 200 = m = 8 kg
 Equal qualities of heated supplied;
MC_{W}θ_{W} = MC_{P}θ_{P} MC_{W} (Qw –Q) = MC_{P} (Q_{p} – Q)
Since θP > θW or MC_{w} > θ_{0} = MC_{P} >Q_{P}
C_{W} > θ_{P}C_{p}
 Magnified, enlarged upright, virtual , image behind the mirror, negative distance
 Apparent depth = Real Depth 12m = 0.9 m
Refractive indese of water 1.3
 Pressure is inversely proportional to the speed OR speed increases as pressure distance
 Maintaining a stable voltage during make and break/ storing charge during make and break and stops arcing sparking
 High temperature causes high – pressure build up in the cylinder, which causes the explosion; OR increases of KE of gas molecules which result to pressure, build up causing an explosion ( 2 mks)
 A Polaroid absorbs/ cuts off light waves in all planes except in a particular plane of propagation ( 1mk)
 A hears a constant frequency produced by the siren/ same roundness/ pitch B hears a frequency that increases as the vehicle approaches/ sound of increasing loudness/ higher sound ( 2 mk)
 Solid copper is denser than water hence the solid sphere sinks; weight is greater than upthrust. Hollow sphere experiences an upthrust equal to its weight so it will float/ density of hollow sphere is less than that of water ( 2 mks)
 The weight of the door and the force are perpendicular to one another ( 1 mk)
 Eddy current ( 1 mk)
 Low negative voltage is applied on control grid, which control the number of electrons reaching the screen ( 1 mk)
 Low speed / high charge / more massive/ size is large/ bigger` ( 1 mk)
 n.p.n
 Limit the current through the base controls the current/ protect transistor from high current or voltage/ regulate reduce voltage.
 Diode is forward biased; Base currents flows; hence collector current flows and lights the bulb/ current amplification ( 3 mks)
air molecule are in constant random motion; smoke particles collide with these air molecules hence their random motion
PHYSICS PAPER 232/1B MARKING SCHEMES 1996
 (a) (i) Acceleration a is rate of change of velocity
a = v – u
t
V = U + at
(ii) Distance is average velocity * time
S = (v + u)t;
2
Substitution for V with u + at;
S = ut + ½ at^{2}
(iii) Using t = v – u; in s = ut – ½ at^{2}
a
s = u (vu) + ½ a (vu)^{2} = V^{2} = u^{2} ÷ 2 as
a a
(b) u = 50 – v = 0 a =2
Using v^{2} = u^{2} – 2as;
Substitute 0 = 50^{2} + 2 (2) s;
S = 625m;
 (a) (i) Each bar is suspended at a time using the string;
The suspended bar is allowed to rest;
Its orientation is observed and recorded;
This is repeated several times for confirmation
(ii) The bar magnet settles in the N – S specific direction, due to its
Interaction (l) with magnetic field of the earth (l)
The iron bar settles in any direction; (l) because it does not have a magnetic field to the interact with that of the earth; (l)
(b) P and Q are magnetized to the same level, by applying two different (l) current lp and lq such that lq > lp (l)
Thus Q requires greater magnetizing power, (l) since its domains are more difficult to align; (l) P is easier to magnetize, since its (l) domain are more easily aligned: ( 1 mk)
(Total 14 mks)
3 (i) Series resistors 4 + 1 + 5W ( 1 mk)
Parallel resistors 2 + 3 + 5 W ( 1mk)
R_{p} = ^{5}/_{2} = 2.5
Total effective resistance 5.5 + 2.5 = 8.0 W ( 1 mk)
(ii) Current l = V; = 4.0; = 0.5A;
R 8.0
(iii) Current through each wing = 0.5 = 0.25 A; ( 1 mk)
2
Potential at Y = 0.5 x 4; 11; ( 2 mks)
Potential at Q = 0.5 x 2; = 0.51; ( 2 mks)
2
Potential difference between Y and Q
= 10.5 V; = 0.5 ( 2 mks)
= 0 0.5 V; + 0.5V Total 13 mks)
 (a) (i) The aluminium block is heated using the electric immersion heater for some time
t; The temperature changes (2) ∆ Φ of the block is recorded;
(ii) Mass of the block m
Time taken t
Initial temperature Φ_{1} final temperature Φ_{2}
Current I voltage V;
Heat given = heat gained by electrical heater the block
1 Vt = mc ( Φ_{2 – }Φ_{1})
C = 11.1
M (Φ – Φ)
(iii) Oiling the holes for better thermal; contact lagging
(b) Heat gained by calorimeter
= 60 x 10^{3} x 378 ( 45 – 25) J;
= 453.6 J
Heat gained by water
= 100 x 10^{3} x 4.200 ( 45 – 25J;
= 8.400J
Heat lost by condensing steam = m/
( 163.5 – 160 ) x 10^{3}/J
= 3.5 x 10^{3} x / J
Heat lost 3.5 g of ( condensed steam) water cooling to 45^{0}C
3.5 x 10^{3} ( 100 – 45) x 4,200;
= 808.5J
Heat given = heat gained
Hence:
3.5 / x 10^{3} + 808.5 J = 453 6J + 8,400J;
= 2.3 x 10^{6}J/Kg;
 (a) (i) Particles of the transmitting medium vibrate in the direction of the wave for a longitudinal wave, but at right angles for a transverse wave:
Sound requires medium but no medium required for electromagnetic wave; speed of sound lower than that of electromagnetic wave;
(b) (i) Speeds of sound;
2.5 x s = 400 x 2
S = 320 m/s;
(ii) 2 ( x – 400) = 2.5 + 2);
 = 1120m;
(c) (i) Double slit provides coherent sources;
(ii) Dark and bright fringes;
The central fringe is the brightest while the intensity of the other fringes reduces away from the central fringe;
(iii) I. The separation of fringes increases
 Central fringe is white; fringes on either side are colored;
 (a) Keep angular velocity Wl constant;
Centripetal force provided by mg;
Fix the mass m and measure of m;
Repeat for different values of m;
(b) (i) graph ( see on the next page
Axes labeled
Scale
Pts plot
Straight line
(ii) Gradient of the graph
= 0.625 – 0.1 = 1.167 N
0.525 – 0.075
Force F on the body = m_{b}W^{2}r
Where mb = mass of the body
M_{b}w^{2} r = Gradient of the graph = 1.167
W^{2} = 1.167 = 11.67
0.1
W = Ö 11.67
= 3.42 rad s^{1}
 ( a)
Close switch S
Vary pd until G deflects
 l)
K (J) x 10^{19}  5  10  10  30  4 
F = C/D ( H_{E}) x 10^{15}  1.89  2.64  4.11  5.55  6.5 
Finding f
See graph
Axes labeled
Scale
Pointed plotted
Straight line
(ii) Work function Φ is given by Φ hf_{0}
F_{0} is the x – intercept of graph
F_{0} ( from graph) = 1.2 x 10^{15} H_{E}
Φ = 6.63 x 10^{34}0.5 x 1.2 x 10^{15}
= 7.96 x 10^{19} J
No. 6
(a)
KCSE 1997 PHYSICS PAPER 232/1 MARKING SCHEME
 Volume = 7.4 – 4.6 cm
2.8cm
Density = mass
Volume
= 11g
2.8 cm^{3}
= 3.9gcm^{3}
 F_{1} and F_{6}
 Either altitude or latitude/ radius of earth changes/ acceleration due to gravity from place to place away from the earth
 Balance: meat + 0.5 kg on one side and 2 kg on the other:
 H_{1} P_{1}g = h_{2} p_{2}g
H_{2} = 1.36 x 10^{4} x 64
8 x 10^{2}
= 1088cm;/ 10.88m.
 Volume of 1 molecule = 18cm^{3}
6 x 10^{23}
Diameter of the molecule = 18cm^{3}
6 x 10^{23}
3 18cm^{3}
6 x 10^{23}
= 3.1 x 10^{8} cm
= 3.11 x 10^{8}
 Glass is a bad conductor of heart, the difference in temperature between the inside and the outside cause unequal expansion
 Adhesion of water to glass is greater than cohesion
 The rate of cooling depends on the rate of evaporation
Rate of evaporation depends on the surface area
Surface area A, < surface area B for evaporation
 A ray from A A ray from B
Relative positions of A and B correctly drawn
 Solar cell ( photovoltaic) photocell/ photo electric cell
 Soft magnetic materials loose their magnetism easily while hard magnetic
materials retain magnetism longer
 Q = It Q = 0.5 x 4x x 60; = 120C
 d= speed x t; 340 x 2; 680m
 At low speeds the speed is streamline
At high speed the flow is turbulent
 V_{r} =l
V l_{r}
240 = 30 l_{r} = 0.75A;
6 l_{r}
 mgh = ½ mv^{2} OR V^{2} = U^{2} + 2 as;
h = ½ S = V^{2} = 36
= 18m; 2as 2(10)
S = ut + ½ at^{2} = 1.8m;
 V = f;
V= 3.0 x 10^{8} ms^{1} = 3.14m;
F 95.6 x 10^{6}S^{1}
 6V
 parallel l = l + l + 2
R_{P} 400 400 400
YZI = V = 12 = 0.02A
R 60
I = V = 12 = 0.02 A
R 60
400 x 12 = 8V
600
 ( No of irons) x 1000) = IV
Number = 13 x 240 = 3.12;
1000
 Extra heat is required to change ice to water / latent heat of fusion
 A trolley slows down/ motion decreases since mass increases and the momentum is conserved, the velocity goes down
 C_{T} = C_{1} – C_{2} = 1 = 1 + 1
C_{T} C_{P} C_{3}
= C_{T} = C_{P }C_{3}
C_{P} + C_{3}
 ^{0}C + 273 = 20 + 273 = 252K
 (a) Dark and bright fringes (b) Coloured fringes
 Small differences in frequencies
 By using laminated core
 After 3 secs number decayed = ½ x 5.12 x 10^{20} = 2.56 x 10^{20}
Next 3 secs. Number decayed = ½ x 2.56 x 10^{20} = 1.28 x 10^{20}
Total number decayed = (1.28 + 2.56) x 10^{20}
= 3.84 x 20^{20}
^{ }
PHYSICS PAPER 232/2 K.C.S.E 1997
MARKING SCHEME.
 i) To make and beak contact / circuit
– It bends and straightens or the metals expand differently.
 ii) Current flows, heating takes place, temperature rises, strip is heated and bends way from contact ; disconnects heater; temperature; drops reconnected heater or completes circuit.
 b) Let final temperature be q_{2}
_{ }Heat lost by water = 4200 x 0.2 ( 20 q_{2} )
Heat lost by glass = 0.2 x 670 x (20 – q_{2} )
Heat gained by ice = 0.04 x 334 x 10^{3}
Heat gained water = 0.04 x 4200 ( q_{2} – 0)
Heat lost = Heat gained.
4200 x 0.2 (20 – q_{2} ) + 0.2 x 670 x ( 20 – q_{2} ) = 0.04 x 334 x 103 + 0.04
X 4200 ( q_{2} – 0 )
q_{2} = 5.36^{o}C
2(a) i)
 ii) Extrapolation F4 – 0 6m force is zero
Leading x axis = 10.6 + 0.2 10.6 8
Intercept 10.6
10.6 – 8 = 2.6 = 2.6m away from B
 b) 10w + ( 10×60) = 2.0 x 40 Þ 10w + 6x = 80 w =^{ x}/_{10} =2N
3a)
 b) i) V = u+ at Deceleration = u – v
0 = 20 + 2a OR t
a = – 10ms^{2} = 20 – 0
2
=10ms^{2}
 ii) Stopping time = 2.2s Total time stop = 2.2 sec
Before stopping = 0.2 x 20 = 4m S = ut +^{1}/_{2} at^{2 }
10 – 202 = 400 =20 =(20 x 2.2) +^{ 1}/_{2} ^{+ }10 x 2.2^{2}
2(10 ) 20
20 + 4 = 24m = 19.8m
4a) AB: (2000 x 20) + (600 x 200) + ½ x 10 x 4000) + ( ½ x 30 x 4000)
40000 + 120000 + 60000
Total 200000J = 200KJ
 b) 6000 x 0.6 = 3600w
 c) Power Input = 0 x 10^{5} x 10 x 360 = 3.0 x 10^{5}wx
60 x 60
Total = ( 3 + 2_ x 10^{3 }= 5.0 x 10^{3}kw Eff. ^{3}/_{5} x 100 = 60
5a) Amount of current No of coils / shape of core / X – core
 b) i) End of coil facing up becomes a south pole and the metre rule is pulled down / attraction occurs. Or Rule tips; core magnetized; top of core becomes south pole; attracts magnet.
 ii) The metre rule to have appointer attached to read zero when switch S is open. Use rheostat to vary current to maximum and calibrate accordingly.
c)HF = hf_{o} + ½ mv^{2}
= (3.2 + 82 ) x 10^{19} = 11.2 x 10^{19} f = 11.2 x 10_{–}^{19}
6.63 x 10^{19}
l= c = 3.0 x 10^{8} x 6.63 x 10^{34} = 1.76 x 10m
F 11.2 x 10^{9}
SECTION 2
6ai) Semiconductors – conducting is by holes Conductors – conducting is by electrons
 ii) Semiconductors – silicon, germanium Conductors – copper , tin iron.
b)i)
 ii) I_{B }= _{5/100 }x 2 =0.01 mA I_{C} = 2 0.01 = 1/99MA
I_{E }= I_{C} + I_{rs}
iii) I_{B} = 0.5 x 4 = 0.02mA I_{c} = 3.98mA
100 r I_{b} = 0.02 – 0.01 = 0.01
I_{C} = 4 – 0.02 = 3.98mA r I_{c} = 3.98 – 1.99 = 1.99
h_{FE }= 3.98
0.02 = 1.99
rI_{c} = 3.98 – 1.79 = 1.99
rI_{b} = 0.02 – 0.01 = 0.01
HFE = r Ic = 1.99 = 1.99
r Ib = 0.01
r I_{c} = 1.99 = 199
r I_{b } 0.01
7a(i) Transverse – particles in the wave perpendicular to the direction of the wave.
Longitudinal – particles move in the same direction as the wave.
b)i)
 ii) Velocity decreases since the frequency remains the same. No loss of energy therefore amplitude does not change.
 c) a) Frequency = ^{30}/_{60} = 0.5 Hz
 b) Speed = ^{6}_{/2} = 3m/s l = V/f ^{3}/_{5 }=6m
 d) A long AA’ – loud and soft sound (constant)
a long OO’ – loud and solid.
PHYSICS PAPER 232/1 K.C.S.E 1998 MARKING SCHEME
 Accuracy of measuring tape is 10m or o.1 cm + 5cm or o.o5m.
 Length of post is 1.5 (1.50 x 1.55) Rangep = N3=
 Quantity of heat equation 20x (4226)x C=10^{3} x 15 x60
C=2.8x 103JKg ^{1}K = (2812.5 OR2813)
 Detecting imperfection in metal structures/block/flaws
 addition of soap solution to pure water reduces the strength of the skin total was holding pin from sinking and so it sinks. Surface tension supports the pin. Addition of soap reduces tension/weakens/broken.
6.
 Low contact pressure between tyre and earth/no sinking.
 I_{P} =N_{3} = Np= 20000×3= 2000
I_{s}=N_{P }30
 surface area of water . Nature of surface of the container/colour/texture /material/ (ambient temperatures).
10 Evaporation and cell reaction cause loss of water. Distilled water does not introduce impurities to the cell.
 E=IR +h
I= E = 2.0
R+r 2.0×0.5 =0.8A
 50 = (I)^{n }n =3(halflives)
400 (2)^{n}
Half –life 72 = 24 min.
 High resistance voltmeter takes less current/low current recording low current.
 Domains/Dipoles initially organized are disorganized by mechanical forces.
 As the rod approaches the cap, negative charges/electrons on the cap are repelled towards the rod. The leaf collapses since the positive charges on it are neutralized attraction. As the rod gets even closer to the cap moved more negative charges/electrons charges are repelled to the leaf, causing it to diverge.
 Length of the rod; diameter/cross sectional area of the rod/thickness nature/type of rod material/conductivity.
 R=P^{1/4} I = 2.0 x 10^{6}x0.5 = 2m OR = 2.041 or 2.0408
4.9 x20^{7}
18 Some energy is lost due to friction/air friction acts on the pendulum/air dumping on the apparatus air resistance.
 In TV (CRT) deflection is by magnetic field, while in CRO deflection is by electric field. XY plates.
ATV (CRT)has two time bases while a CRO has only one.
In CRT it produced 625 lines per second while CRO is 25 lines per second.
 Heating/ cooking/communication/eye/photographic film or plate/LDR/photocell.
 Diode is forwardbiased, no current flows
Current flows when the switch is closed but when terminals are reversed, no current flows
 Angle of inclination/nature of surface/length of inclination
Height of inclination/frictioal force between the surface.
 layers of the crystal material are arranged according to faces/ plans/ flat surfaces. Cleavage is only possible parallel to those faces/places/flat surfaces.
 Principles of moment.
200 x1.5 R x 0.5, 0.5f=1x20x10or 0.5,R=600. R=F +200 = 400N take moments about O
F=600 200 =400N
F=400N
25.
26 Addition of impurities with higher boiling points/presence of impurities. Water heated under a higher pressure than atmospheric/below sea level.
 Moon covers the sun/obstruction of sun by the moon
Both heat and light have same velocity/both are electromagnet waves.
 Overtones/harmonics
 Since F=MV2/V the sharper the corner (as B) the small the value of R hence the greater the F. (M& V constant).
 Gas through the nozzle gains velocity. Hence its pressure reduces above the nozzle. The higher atmospheric pressure pushes air into the gas stream.
 When mercury is heated (during a fire); it expands and makes contact, completing the circuit to ring the bell.
 There will be no variation of intensity of light/ uniform intensity/no bands/one
 Is the one which cannot form on a screen Is formed by rays which are not real
Is formed by extending rays. Formed by apparent rays.
 Component of weight down the slope =50 sin 30^{0 =}25N
Total force parallel to slope= (29+25) N 54N.
PHYSICS PAPER 232/2 K.C.S.E 1998 MARKING SCHEME
 iii) Scale, axes label, unitplotting 8102 571 Curve (smooth)
 iv) As the number of turns is increased, alignment of domain with field increases. After 3536
turns, all domains are aligned, so that magnet is saturated.
Sketch – curve above 1 to some saturation, and from origin.
 b) When switch is closed electromagnet attracts soft iron. This causes T to close and so circuit 2 is put on.
 bi) Volume of block = 4x4x16 = 256 cm3
Mass of block = 154 gm
D= m=154=0.6g/cm^{3 }deny ½ mk if not to d.p
V 256
 ii) Volume of liquid ¾ of 256 = 192 cm3
Density of liquid = 154 = 0.8g/cm3
192
 a i) The bullet will land on the track It has some horizontal (inertia) velocity
as the track.
(ii) (Use g = 10ms2}
S = ut + ½ at2
For freefall u = 0 t=Ö2h/g T= 6sec
Horizontal distance = vxt = 6×50 = 300m
V2= U2 + 2as OR v= 2U + at OR ½ Mu2 = mgh
From above u = 30m/s
S=ut+ ½ at2
T=ut + ½ at2
T= 6 D= vxt = 50×6 =300cm
(bi) Measure pressure with Bourdon gauge
Measure the length of air (reg volume at tone).
(ii) Tabulation values of p and length of air column (volume )
Plot graph of I/V vs P OR L vs I/P
Graph is a straight line. Hence pa I/v
Tabulate P and V (I) Calculate PV or PL
PV (1) = PL Hence Pa ^{1}/_{v}
_{ }
 a) i)
(ii) Voltage, current, time
(iii) Q v/t Rate= Q/t = v/tT (T=time taken for sun to heat)
 b) 4 shows a photocell.
 ii) When light rays strike cathode C surface electrons gain photon (energy) hence the cathode.
iii) Draw a simple circuit including the photocell to show the direction of flow of current.
5 a) i)
 ii) Since sin i is common and r < re then sin rv < sin re
 b) n Sin C=1 OR Sin C ^{1}/_{n}
_{ }sin C= ^{1}/_{1.4} C= 45.600 (45.58) or 45.35 min/45.36
SECTION II
6 a) When T and Y are connected C is charged by E, until C achieves same
p.d. across it as for E C max p.d is achieved when T and Y are connected after first process. C acts, as source of e.m.f and discharges through r unit no more current flow or current is zero.
 b) Current = dQ draw target at 30. Substitution I =3.6mA +2A.
7a) 2 complete rays, 2 with arrow at one end image (inverted real) (continuous tie) locating F size 2.4 +0cm
 b)
U (cm)  20  25  30  40  50  70 
V(cm)  20  16.7  15  13.3  12.5  11.6 
1 V(cm^{1})  0.50  0.040  0.033  0.025  0.020  0.014 
1 V(cm^{1})  0.50  0.060  0.067  0.075  0.080  0.086 
 ii) ^{1}/_{f} = ^{1}/_{u} +^{1}/_{v }Intercept ^{1/}_{f}
0.1 = 1/f .^{.}. f = 10cm
PHYSICS PAPER 232/1 K.C.S.E 1999 MARKING SCHEME.
 Reading on the vernier calipers
0.5 + 0.01(5) 0.5 +0.05cm = 0.0055m/5.50mm.
 Third force F3 acting on the ruler is either upwards or downwards.
No: My must be at the centre.
 Center of gravity rises when the body is tilted slightly and lowers when released / returns to original position.
 Y must be below x
Reason: P water is greater than paraffin = height of water required is therefore less than that of paraffin.
 Cohesion between Hg molecules is greater than adhesion between Hg and glass molecules/cohesion force or adhesion. Force.
 (NB: with or without labeling one mark.)
 aParticles are + vely charged, if majority deflected most Þatom is empty.
Deflection Þ existence of a +vely charged nucleus.
Few deflected Þ nucleus is small/mass is concentrated at the centre
 Angle of rotation of reflected ray=2(angle of rotation of mirror)
=2 x 30 = 60^{0}
 Charge concentrate at sharp point causing heavy discharge/ ionization neutralization, leaf falls off.
 V = IR Þ I = ^{V}/_{R} I = 3/! = 3A
^{1}/_{R}= 1/R1 + 1/R2= 2/2
^{1}/_{R} = 1=R=1
 4mm=20N
1.5 =? F = Ke
1.5x 20 K = F = 20 = 5 x 10^{3} N
4 e 4x 10^{3}
= 7.5 N F=5x 10^{3} x 1.5 x 10^{3}
=7.5N
 Dipping a magnet into a container with iron fillings, most of them will cling at the poles Þ
– Use of plotting compass to trace.
13.
 Moment of couple = Force x distance between forces.
=10 x 2 = 20NM.
 F = Ma = 70 x 0.5 F 35N
35N = 20a a = 35 = 1.75M/s^{2}
20
 P = force x velocity Power = Fd/t = 20 x 10x 20
Mg x h/t = 20x 10 x ^{20}/_{40} 40
= 100w = 100j/I
 F = I/T =^{ 1}/_{0.5} = ^{10}/_{5} = 2HZ
OR
F = No. of waves made in 1 second = 2 Hz
OR
F = No of waves
Time = 2/1 = 2.5 / 1.25 = 2Hz
 Beat frequency f = f2 – f1 F = f2 – f1
= 258 – 256 256 – 258
= 2Hz =/2/ = 2
 P = V1 = 15000 = V x 2 W = QV but Q = It e = I^{2}Rt
10 x 60 =V = W 15000 1500 = 2 x 2 x R x 60 x 10
^{W}/t = VI = V = 1500 Q 60 x 10 x 2 60 x 10 x 2150 = 24R
10 x 60 x 2 V = 12.5v 25 = 4R
150 V = 25 x 2
12 4
12.5V V = 12.5V
 Heat lost by substance = heat gained by water
M_{s}C_{s}rq_{1} = M_{w}C_{w}rq2
2 x 400 x 60 = M_{w }x 4200 x 1
M_{w }= 2 x 400 x 60 = 30 = 11.4kg
4200 7
 V = I(R +r)
5 = 10 (R + 50)500 Þ R + 50 Þ R = 500 – 50 = 450W
1000
 Apparent depth = 30 – 10 = 20cm real depth = 30 = 1.5
Apparent depth 20
 Kinetic energy ray / heat energy.
 – Horizontal acceleration is zero because g component horizontally is 0
Horizontal velocity remains constant
– Resultant horizontal force is zero – resultant force is Zero.
 V_{2 }is smaller than V_{1} V_{1} is larger than V_{2}
26.
 P_{1} = 1.03 x 10^{5} T1 = 20:C = 393K V1 = V
P_{2} =? V2 =^{ 1}/_{8}V or_{ v}/_{8}
P_{1}V_{1} = P_{2}V_{2} 1.03 x 10^{5} – P^{2}/_{8} = p^{2} = 3.24 x 10^{5}N/M^{2}
 Radio waves, infrared, xrays, Gamma rays.
 Up thrust = PV x 10 = 10 PV
 Ultra violet releases electrons from zinc plate by thermal emission.
On removal of electrons, zinc becomes +vely charged.
Positive charge on zinc discharges/ neutralizes the charged on the electroscope.
 Tension = centripetal force.
T = Mv^{2}/r but v = wr 2 = 0.1 x w^{2} x 0.33
T = Mw^{2}r t = 0.2 x 10 = 2N 2N = Mw^{2}r 2 = 0.1 x w2 x 0.03
w^{2} = 2/0.003 w Ö2000/3 w = Ö666.7 = 25.82 rads/s
 Object should be between F and lens.
 Downwards into the paper.
 Aearth wire B – live wire C neutral wire
 Z _{Y} Z b Z_{+1 }+^{o} – 1e
Or Atomic number charges by / New is a head of the old or Z + 1
PHYSICS PAPER 232/2 K.C.S.E 1999. MARKING SCHEME
1a) Longitudinal waves – direction of the disturbance while ½ .Transverse waves – direction of propagation is perpendicular to that of the disturbances.
b i) YP – XP = 2l
 ii) Dark fringes; crests and troughs arrive at the same time OK destructive interferences Bright fringes; crests arrive together at the same time OR constructive interference.
iii) No interference pattern because no diffraction takes place.
C i) T = (2.5 – 5) x 10 – 3
= 20 x 10 – 3s 10^{3}
F = 1/T = 50 Hz. 1/20 x 10^{3}
ii)
2.a)
3i) Average velocity at intervals AB and CD.
T = 1/50 x 56 V_{AB }= 1.5cm/0.1s V_{CD} = 3.2cm/0.1s
= 0.1s 15cm/s 32cm/s
 ii) Average acceleration of the trolley.
(b) V2 = U2 + 2gh mgh = 1/2MV2
V = Ö 2gh V = Ö 2gh
ci)
4a) Figure 5 represents a simple voltage amplifier circuit.
b i) Base current.
Current gain = Collector current p2 = 1_{a}/I_{b}
_{ } Base current
62.5 = 2.5 x 103
I_{b}
_{ }Ib = 2.5 x 103 = 40uA (4×105)A
62.5
 ii) Load resistance, R_{L }IcRL = Vcc = 5.5
P.d across R_{L }RL = 5.5 = 2.2kW
2.5 x 10^{3}
10 – 4.5 = 5.5 ICRL = 5.5
RL = 5.5
2.5 x 10^{3}
5a) Ammeter reading decreases.
The resistance of metals decreases with increase in temperature.
 i) P = V^{2} = (240)^{2} P = 576w
R 100
 ii) P = VI
I= P = 576 = 2.4A
V 240
SECTION II
6a) Benzene sinks in liquid benzene.
Water increases in volume on solidifying while benzene reduces in volume; ice is less dense that liquid water. Solid benzene is denser that liquid benzene.
b i) Weigh the metal block in air and in water
Fill the overflow can in water and place on a bench / diagram
Collect the overflow in the beaker and weigh
Compare difference in weight of metal block and weight of overflow
Repeat
Up thrust = tension + weight
= (0.5 + 2.0) = 2.5N alternative
Weight of H2O) = 2.5N Up thrust = 2.5N
M_{w} = 1000 R.D = Wt. in air = 2.0 = 0.8
V_{w }Upthrust 2.5
Vw = 0.25 volume of wood €wood
1000 €wood
Density of wood = 0.2 €wood
0.25/100
0.2 x 1000
25
800kg/m3
c i) Time taken for half of the radio acute material to disintegrate.
 ii) Correct readings for 60 and 30 time 25 + 2 minutes
PHYSICS PAPER 232/1 K.C.S.E 2000 MARKING SCHEME
 Acceleration of gravity on Jupiter is higher than that of earth, so a bag of saw dust must be less massive if the greater acceleration on earth is to produce the same pull as sugar bag on earth.
 Beaker becomes more stable because the position of C.O.G is lowered on melting or water is denser than ice.
 On earthing negative charges flow to the leaves from earth to neutralize positive charges when the rod is withdrawn the leaves are left with net negative charge.
 Since the system is in equilibrium let A be the area of piston and P the pressure of steam
P x A x 15 = W (15 + 45)
2.0 x 10^{5} x 4 x 10^{4} x 15 = W x 60
W = 20N
 Particles of gases are relatively far apart while those of liquids and liquids are closely parked
 Since the strip is bimetallic when temperature rises the outer metal expands more than the inner metal; causing the strip to try and fold more; this causes the pointer to move as shows
 This is because metal is a good conductor, so that heat is conducted from outer parts to the point touched; while wood is a poor conductor
 Can withstand rough treatment
Do not deteriorate when not in use
 Struts are DE, DC, AD, BD Ties are BC; AB
 The keepers become magnetized thus neutralizing the pole, this reduces repulsion at the poles, thus helping in retention of magnetism
Force F_{2} at the ends perpendicular and turning to opposite to F_{1}
 VR = 4;
 Efficiency of the system
Efficiency = M.A x 100 = 100 x 1 x 100 = 89.3%
V.R 20 4
= 89%
 Sound waves
 Let A’s represent current through the Anometers using Kirchoffs Law
A_{1} + A_{2} = A_{3}
But A_{1} = A_{2}
So A_{1} = A_{2} = ½ A_{3}
Similarly A_{4} + A_{5} = A_{3}
So that A_{4} = A_{5} = ½ A_{3}
So A_{1} = A_{2} = A_{4}=A_{5}
 P = V^{2}; 40 = 240^{2} R = 1440W
R R
 Wire expands becoming longer (reduces tension) this lowers frequency hence pitch.
 Boiling point of spirit is lower than that of water. Specific heat capacity is lower than that of water.
 Fig 12 shows a ray of light incident on a convex mirror
 Fig 13 shows a semicircular glass block placed on a bench. A ray of light is incident at point O as shown. The angle of incidence, i is just greater than the critical angle of glass
 The air above paper travels faster than below causing lower pressure above. Excess pressure causes paper to be raised.
 Combined capacitance = 1.5 μ F
= CV = 1.5 x 3 = 4.5 μC.
 Microwave / cooker/ telephone/ radar etc
 U.V removes electrons from zinc surface so leaf will not only collapse if electroscope was negatively charged.
 Number of turns/ strength of magnetic field
 To reduce eddy currents in the armature
 Difference in energy of the state/ nature of atoms
 X – rays produces – Hard X – rays are produced
 From 300 – 150 = 74 S 200 – 100 = 76 S
Average = 75 ± 1 other values on the graph could be used
Donor impurity is the atom introduced into the semiconductor(doping) to provide an extra electron for conduction.
PHYSICS PAPER 231/2 K.C.S.E 2000 MARKING SCHEME
 (a) (i) Convex mirror – driving mirror/ supermarkets mirrors
Parabolic mirror solar heater reflector, reflector, torch reflector etc.
(ii)
(b) (i) V= 45 M = 3.5 ( from graph) m = v/u Þ 3.5 = 45/u
U = 12.9 cm ± 0.4
(ii) Choosing convenient value of ‘m’
M = I, V = 20 =u M= v/f1 M = v/f 1/f= 1/45+ 1/12.9
1/f = 1/20 + 1/20 v= 45m = 3.5 m= 0 = f = v
f= 10cm f = 9.8 – 10.3 f= 10 cm f = 10cm
 (a) Initially the balls accelerates through the liquid because terminal viscosity is greater than viscous and upward forces after sometimes the vicious forces equals mg and the balls move at constant velocity. The difference due to the fact that the viscosity L_{1}is greater than that of L_{2} (coefficient of viscosity)
(b)
(ii) (I) A. plot the graph of acceleration against the mass m
See graph paper
Graph 5 marks
Plot 2 marks
Axes 1 mark
Scale 1 mark
Line 1 mark
(II) Intercept = μg
Intercept = 2.80 ± 0.2 (from graph)
Μ = 2.80 ± 0.2
10
Μ = 0.28 ± 0.02
 (a) When temperature rises, K.E/speed of molecules of the gas increases. Since volume is constant this increases the rate of collision, with the walls of the container, and increase in collision increases pressure.
(b)
(i) Length of column of dry air Temperature
Length/ height of the head Volume of air
(ii) Temperature is varied and values of L and T. Measured and recorded; a graph of L versus T. (A) is plotted. This is a straight line cutting T axis at O (A) (or – 273^{0}C) since tube is uniform L α T.
(iii) The water bathy allows the air to be heated uniformly.
(c) P_{1}V_{1} = P_{2} V_{2} = 1.5.x 10^{5} x 1.6 = 1.0 x 10^{5} x V_{2}
T_{1} T_{2} 285 273
= V_{2} = 23m^{3}
 (a) (i) Easily magnetized and demagnetized
(ii) V_{p} = N_{p} 240 = 500
V_{s} N_{s} V_{s} 50
V_{s} = 24; V= V_{PR}
V_{QP} = 1/3 ; V_{PR} = 8 V
(b) Volume of A displaced = 6.0 x 12 cmcm^{3} or P = G * g
Mass = 12 x 10^{6} x 800 F = PXA
= 0.0096 kg ans = 0.09N
Weight = mg = 0.096N
(ii) Volume of B displaced = 6.0 x 3 = 18 cm^{3}
Weight = 18 x 106 x 1000 x 10 = 0.18N
(iii) Weight of block = weight of third displaced
0.096 + 0.18 = 0.276
Mass = 0.027 kg
Volume = 0.0276 kg
42 x 10^{6}m^{3}
=657 kgm^{3} can also be in g/cm^{3}
 (a) When whirled in air centripetal force is provided by bottom of container because of the holes, there is no centripetal force on water on the water, so it escapes through holes leaving clothes dry.
(b) (i) I Centripetal force equals force of friction
F= Mw^{2}r = 0.4
W^{2} = 0.4 or F = Mw^{2}r
 x 0.08 4 0.1 w^{2} x 0.08
W= 7.07rad/s W = 7.07 rad/s
II F= Mw^{2}r = 0.1 x 7.07^{2} x 0.12
= 0.60N
Force required = 0.60 – 0.40
0.20N
(ii) The block will slide this is because although the frictional force is greater centripetal force would be needed to hold it in place.
SECTION II
 (a) Conditions of interference: Waves must equal frequency and wavelength; to be in phase or have constant phase relationship ( comparable amplitude)
(b) Walking along PQ creates path difference between waves from L_{1 }L_{2} when the path difference is such that the waves are in phase of full of wavelength loud sound is heard, when the path difference is such that the waves are out of phase. (½ of odd ½ l) low sound is heard.
(ii) L_{1} A – L_{2} A = l
From the figure L_{1}A = 18.5cm + 0.1
L_{2} A = 18 cm + 0.1
L_{2}A = L_{1} A = 0.5 cm + 0.2
Using scale given l = 0.5 x 200
= 100cm
V= f l = 350 x 1
350m^{1}
(iii) The points interferences are closer; higher frequency Þshorter wavelength; so if takes shorter distance along PQ to cause inference.
 (a) Pure semi conductors doped with impurity of group 3, combination creates a hole ( positive), this accepts electrons.
(b i)
(i) At V_{e} E = 0
V_{cc} = I_{c} R_{ L}
L_{c} = 9/1.8 K W l_{c} = 10
V_{e}E = V_{cc} = 9
(ii) ∆l_{c} = (see graph) = 3.5 – 1.2 = 2.3 mA
B = ∆l_{c}
∆l_{c}
2.40A
40 μ A
= 60.
PHYSICS PAPER 232 /1 K.C.S.E 2001 MARKING SCHEME
 Volume removed = 11.5cm3
Density = mass = 22 1.9cm^{3}
Volume 11.5
 Weight on side A has bigger volume when water is added.
 Centre of gravity of A is at (geometric) centre while that of B is lower when rolled. Centre of gravity of A stays in one position while that of B tends to be raised resisting motion as it resists; thus slowing down B. OR B there is friction force between the surfaces which resists motion.
 No air on moon surface / no air pressure / no atmosphere.
 When the permanganate dissolves / or breaks up into particles (molecules) these diffuse through the water molecules
 When rises up the tube into the flask or water is sucked into the tube or bubbles are seen momentally.
 Cold water causes air in the flask to contract // reduces pressure inside flask or when cold water is poured it causes a decrease in volume of air the flask or pressure increases in the flask // volume of the flask decreases.
8.
 Point action takes place at sharp points (A , B, C, D ), charge concentrates at sharp points causing high pd, this causes air the surrounding to be ionized. The positive ions are repelled causing points to move in opposite direction.
 By forming hydrogen layer / cover or hydrogen atoms or molecules which insulate the copper plate OR forming it cells between hydrogen and zinc which opposes the zinc copper cell or by forming a hydrogen layer / cover which increases internal resistance.
11.
 F_{2} F_{3 }or F_{1} and F_{4}
 Moment of a couple = one force x distance between the two forces.
Distance between F_{1} and F_{4 }= 0.8sin 30^{o}. Moment = 0.8sin 30^{o} x 100 =10NM
Alternative (F_{2} and F_{3}) Moment = f x 1M = 60N x1M = 60nM(or J)
 V2 – U2 = 2aS OR S = v+u t
1502 – 3002 = 2a (0.5) 2
a= 67, 500ms^{2} 0.5 = V = 150m/s u = 300m/s s = 0.5
or deceleration = 67,500ms^{2} 300 + 150/t t = 1/450s
2
a = v – u = 150 – 300
t 1/450 = 667,500m/s^{2}
 Efficiency = work done by machine x 100 E = work out x 100
Work done on machine Work input
; Work done on machine (work input) = 550,000j.
16.
 R = V/I = 1.5 / 0.1 = 15’W
R = 15’W – 12’W = 3’W
OR E = 1(R +r)
1.5 = 0.1 ( 12 + r) = 1.5 =1.2 + 0.1r
0.3 = 0.1e = r = 0.3/0.1’W
R = 3’W.
 Current in heater = p = 3000 = 12.5A
V 240
Fuse not suitable since current exceed the fuse value.
 Heat loss will be higher in A
Methylated spirit will boil faster / evaporates / more volatile causing loss of heat through latent heat of vaporization.
20.
21.
22.
23.
 Since masses are the same, there are more hydrogen molecules than oxygen molecules/more collision in B than in A and hence more pressure in B. Collision in B is higher than in A.
25.
 Fh = f1 – f2 OR Fh = f1 – f2
Fh = 6 – 4 = 6.25Hz – 4Hz
Fh = 2 =2.25Hz.
 Longer radio waves are easily diffracted around hills/ radio waves undergo diffraction easily.
 Tension in A = 1.05N – 1.0N = 0.05N
Tension in B = tension due to A + Tension due to B
0.05 _+ 0.05 = 0.10N
29.
30.
31.
 E = pt = 60 x 30 x 60 x 60J E = 60/1000 kW x 36hrs
In kWh = 60 x 36 +60 x 60 J E = 0.06 x 36
1000 x 60 x 60
= 2.16 Wh E = 2.16kWh
 Pd across Anode – cathode Or anode potential (voltage)
 r – b (Beta)or ie B = 82 C = 206
PHYSICS PAPER 232/2 K.C.S.E 2001. MARKING SCHEME
 Let final temperature be T
Heating gained by melted ice MCT = 0.040 x 340,000J
Heat lost by water. = MCq 0.040 x 4200 x (20T) J
Heat gained = Heat lost
13600J + 168 TJ = 1680 (20T)J T= 10.8^{0}C
2 a i) So as to have opposite polarity on the poles.
 ii) since the current is varying with time; it causes the current in the solenoid to vary, with time causing the diaphragm to vibrate this vibration is at the frequency of speech; hence reproducing speech.
iii) No vibration/receiver does not work, steel core pieces would become permanent magnet/so force of attraction would not be affected by variation in speech current.
 b) N_{p} = V_{p }^{ }V_{s }= 240 x 20 = 12v
N_{s } = V_{s } 400
V_{s }= V/R = 12/50 =0.24 A
I_{s} Peak = 0.24A x 2
=0.34A
 a) Fill tray with water to the brim and level on bench; sprinkle lycopodium
powder on the water surface either pick an oil drop with kinked wire; and measure the volume of a drop; put one drop at centre of the tray let oil spread and measure maximum diameter d of the patch; hence reproducing speech.
 b) Hydrogen since its less dense it diffuses faster.
 c) p= pgh; Or mass = D x V
= 1000 x 2×10 = 1000x 2x/1000
1pr
= 100x 10 x 10 x 2×2 x1o^{4} = 0.4kg
= 4N = 0.4 x 10 =4N
 i) Filament heats up cathodes; causing electrons to boil off the cathode.
 ii) Grid controls brightness of spot since it is negatively charged it repels the electrons reducing number of electrons
iii) A vertical line would appear/spot oscillates vertically
 iv) Deflection in TV is by magnetic fields.
 v) Magnetic field produces greater deflection on electrons beam allowing wider screen.
 b) Energy released rE = E_{f} – E_{i }= 5.44 x 10^{19}j = 4.08^{19}j
rE = hf = h C
l
l = 6.63 x 10^{34} x 3.0 x 10^{8}m
4.08 x 10^{19}
= 4.88 x 107 m (4.87 – 4.90)
5a)
 bi) IE = IC + IB
100 + 0.5
= 100.5mA
(ii) b = Ic / IB = 100 = 200
0.5
SECTION II.
6 a i) A body at rest or in motion at constant velocity stays in that state unless acted on by an unbalanced force; the rate of change of momentum of a body is directly proportional to the force acting on the body(F = ma) for every action, there is and equal and opposite reaction: any one for;
(ii)
V^{2}(M^{2}/s^{2})  0.04  0.16  0.36  0.64  1.00  1.44 
Graph – see graph papers Axis – labels
Scale Plot – 5.56 point
Line – 4 point Slope = 1.24 – 0.100 = 5.88 + 0.27
0.210 – 0.016
 b) V2 + u2 = 2as
When m = 0
V2 = 2 x 0.5 x 100
Momentum = mv = 200 x 1000 x ( 2x 0.5 x 100)
2.0 x 10^{6} kgs^{1}
OR S = ½ at^{2}
T = 100 x 2
T = 20 sec Momentum p = Ft
F = ma
– 200 x 1000 x 0.5 = 10^{6}
7 a i) The pressure of a fixed mass of an ideal gas is directly proportional to the absolute temperature provided the volume is held constant.
ii)
I/V(m^{3})  40.0  5  58.8  71.4  83.3  90.9 
Graph – see graph paper Axis – labels
Scale Plot – 5 – 6 points
Line – 4 points
Slope 4.24 – 2.00 x 105
86 – 40
= 4.87 x 10^{3} paM^{3}
= 4.94 ± 0.65
Slope = 4.94 ± 0.65
Slope = 2RT
R = 4.87 x 10^{3}
2 x 300
= 8.12NM/K or JK
= 8.23 ± 0.11
 b) P1 = P2
T1 = T2
T1 = 12 + 272 = 285
T2 = 88 + 273 = 361
P2 = 1.0 x 105 x 361
285
I/P x 10^{5 }(pa 1)  0.5  0.40  0.33  0.29  0.25  0.22 
Y = intercept = 3.8 Log 600R
600r = 6309.57
R = 10.5 + 5.0
PHYSICS PAPER 232/1 K.C.S.E. 2002 MARKING SCHEME
 11.72/11.72 CM/0.01172M
2.
 g moves / shifts to the right / C.O.M. moves/ shifts/ more weight or mass of he right/ weight will have a clockwise movement about O/causing greater moment of force towards right than left.
 R = V = 0.35 = 0.5W
I 0.70
P = RA = 0.5 x 8 10^{3} = 8 x 10^{3}W m.
C 0.5
 p = F P = F
= 2500 A
425,000pg Total press = 2500 =2,000N/m^{2}
=250,000PG 0.025
 Low temperature reduces K.E / velocity of molecules
– Hence lower rate of collision / less collision Reduction in pressure
 Can B Good absorber of radiation.
8.
 (Assume no heat losses)
Heat gained = heat lost E = pt = mcrq
2 x c x (30 – 20) = 90 x 15 x 60 90 x 15 x 60 = 2 x c 10
C = 90 x 15 x 60 4050j / kgk = c
20
C = 4050j/kgk
 Mattress increases stopping time/time of collision increased this reduces the rate of change of momentum.
 C = C_{1} + C_{2} Q = CV
CT = 3×2 5mF V = Q V=1 x 10^{4} = 20V
C
22.
23.
 V =fl
l = v = ^{330/}_{30 }= 11m
F
 Law of floatation – a floating body displaces its own weight
Weight of block = weight of mercury displaced
0.250 x g = 13.6g
0.25 = v
13.6 x 10^{3}
V = 1.838 x 105 m^{3} = 18.4cm^{3}
1.839 x 105m^{3}
28.
29.
 p = VI
Kettle Iron box TV
I = p/n =^{ 2500}/_{250 }= 8A 750/250 = 3A ^{300}/_{250 }= 1.2A
Total = 8 + 3 + 1.2 = 12.2A = Appropriate fuse = 15A
 107 – 42 = 65
 Penetrating power
 Downwards
 Work function of metal / min energy required to eject e1 for excess energy work function.
PHYSICS PAPER 232/2 K.C.S.E 2002 MARKING SCHEME
1a) (speed of light in vacuum e = 3.0 x 10^{8} ms^{1})
Refractive index = speed of light in vacuum
=3.0 x 102 m/s
1.88×102 m/s
= 1.596 = 1.60
 b) sin C = 1
n
1
1,596
C = 38.8^{0}– 38.48
38.7 – 38.42
 c) Sin q = 1.596
sin 21.1
Sin q = n
Sin 21.1
q = 35.25_{0} – 35.15^{1}
35.35_{0}– 35.21^{1}
 b – beta radiation
Force is of the circle implying negatively charged (Fleming’s left hand rule)
(bi) K= alpha (ii) X= 88 Y= 288
(ci) Increase in thickness
(ii) Increase in thickness reduces the radiation reaching the Geiger tube
(iii) Increase in pressure
(iv) Increase roller pressure squeezes metal sheet (possess more) reducing the thickness of foil coming out of them.
(v) Alpha particles have little penetration very few or none pass though foil.
(vi)
3.
a i) R to pass through the c.o.g Forces not labeled. A ward half for each
(ii) = mg Sinq = 30.0 x 10 sin 10^{0} = 52.1 N (accept 52.08, 52.08, 52.09)
(ii) A = F Net force down = Mg sin q – friction = 52.1 20
= 32.1
M = 32.1
3.0 = 1.07M/S^{2}
(iii) Acceleration increases with the increase in angle
4 a i) A ice absorbs latent heat without in temperature (or ice melting no change of temperature heat goes to latent heat fusion)
B Water molecules gain K.E (increase in K.E.)
C heat is used to change water into vapour.
 ii) Water has anomalous expansion, where we have maximum density at 4^{0} Anomalous behaviour/explain.
iii) Frozen seawater has a lower temperature than frozen fresh water boiling point of sea water is higher than fresh water.
(b) (heat gained = ML + MCq
= 3 x 336 x 10^{3}+ 3 x 4200×5
= 1.07 x 106J
5 a i) Transverse waves (accept elliptical)
 ii) As waves move in the medium, the particles of medium do not move: they vibrate in positions so cork does not move.
iii) Period of wave T= 0.205
f= 1 = 5Hz
T
V = fx
X = 0.30 = 0.60M
5
 iv) Velocity decreases when depth decreases hence the x decreases (since frequency is constant wavelength decreases)
 b) 1^{st} resonance l I_{1}fe l = I1_{2}I_{2} OR V= 2F (I_{2}I_{1})
4 2 f=
V
2(I_{2}I_{1}) 12977
l= 12977
2
2^{nd} resonance 3l = I_{2}+C l= 104 cm =340
V=fl
340=fx 1.04 = 326.9 Hz.
F= 327 Hz (326.9)
 a) Charles law: for a fixed mass of a gas at a constant pressure the volume is directly proportional to the absolute temperature Kelvin thermodynamics.
 bi) Volume of gas trapped by drop of cone sulphuric acid, water in heated (in both) and volume (height) of gas: in tube increase as temperature rises; values of height H and T are tabulated; a graph of volume V versus temperature T^{o}C is plotted; graph is straight line cutting T at – 273^{o}C (absolute Zero); so volume is directly proportional to absolute temperature.
 ii) Short temperature range – Keeping pressure constant is difficult
 ci) When q – qT – 273k Extrapolation on graph show:
Pressure read off b = 9.7 x 10^{4} pa
 ii) p1 = 1.15 x 10^{5 }pa q_{1} = 52.0^{o}C
p2 = 1.25 x 10^{5} pa q_{2} = 80.0^{o}C
p1 p2
To + q1 To + q2
1.115 x 105 1.25 x 105
To + 52 To + 80.0
To 270
– Rise in volume height – Rise in temperature
Recording of tabulation – Graph
Analysis of graph Conclusion
Alternatives
P = mx + c
P = kq + kto when K gradient.
K = Dv = (1.14 – 1) x 105
Dx 50 – 10
= 0.14 x 10^{5}
40
= ^{14000}/_{40} 350pac ()
KT = Constant
C = 9.6 X 10^{4}
350 T_{o }= 9.67 x 10^{4}
to = 274.3 (266284)
 ai) mV light removes electrons on zinc plate. This lowers the excess charge
constant (negative) on leaf leading to collapse/ becomes less negative (more positive)
 ii) Since mv light removes electrons positive charge re attracts the electrons thus keeps the charge constant and so leaf does not collapse.
 bi) Frequency of incident light / energy of proton / energy of light work function of surface
 ii) From Kemax = hf – q
h is slope of graph
Slope = (10 – 20) x 10^{19}
(2.6 – 1.4) x 10^{15}
H = 6.7 x 10^{34} fs
At Kemax = q hf = 0
Extrapolation shown or
Read off f_{o} = 1.07 x 10^{15} Hz
Q =1.07x 10^{15}x 6.67 x 10^{34}
= 7.4 x 10^{19}
 c) Kemax = hf q
= 6.67 x 1034 x 5.5 x 1014
1.6 x 10^{19}
= 2.29 eV
Since hf< q no photo elective effect
E = hf = 6.67 x 10^{34} x 5.5 x 10^{14}
Or q = 2.5 x 1.6 x 10^{19}
PHYSICS PAPER 232/1 K.C.S.E 2003 MARKING SCHEME.
1.
 30.0 + 0.5 = 30.5 (No mark if working not shown)
 Low density / weight / mass lowers Cog Lower Cog increases stability. Or higher mass / weight / density raises Cog. Higher Cog. reduces stability.
P = ò hg / p = dhg
= 1.36 x 10^{4} x 0.7
= 9.52 x 10^{4} or 95200 Nm^{2} Allow g = 9.8m/s^{2 }(follow through working)
 Air molecules are in continuous random motion. They bombard / knock / collide with smoke particles
 Glass flask initially expands / Heating increases the volume of the flask; hence the lignin level drops. Eventually water expands more than glass, leading to the level rising.
 Initially the wire gauze conducts heat away so that the gas above does not reach the ignition temp/point. Finally the wire gauze becomes not raising the temp of the gas above ignition point.
R = I = 90 = 40o
Or
R = 180 – 100 – 80 = 40^{o}
2 2
 The negative charges on the rod initially neutralize the positive charges on the leaf and the plate / A the road is moved towards the cap electrons are repelled to the leaf, making it to fall.
As the road is brought nearer, the excess negative charges on the leaf and the plate.
Current for a longer (Do not accept cheaper)
 Temperature
12.
 It does not retain magnetism / Iron is easily magnetized / demagnetized / Iron enhances / strengthens magnetism.
 Clock wise moments about pivot = Anticlockwise moments about pivot.
F x 2.5 Sin 30 = 2.5 x 20 F = 40N
Acc. F cos 60o = 20.
F = 20
Cos 60 = 40N (Do not accept symbols for principle.)
 Light travels from optically an optically denser to a less dense / rarer medium / the incident ray is inside the optically denser of the two media.
16.
Rays marked independently: Correctly if in the right direction with arrows. Object distance is 9.1cm + 0.2 (8.9 – 9.3). No arrow on the virtual. Any through optical centre.
Other rays to principal axis and dotted through F.
 P = V^{2} / R P = VI = I^{2}R
75 = 240 x 240 or Do not accept p = VI alone without I^{2}R
R R =^{ p}/_{1}^{2}
= 768 W R = 75 x^{ 240} / _{75} x^{ 240} / _{75} = 168 W
 Beta particle b (Do not a ward for beta ray) Beta radiation Beta emission
 Dope with group III element (e.g. Boron, Al, Ga). Three silicon electrons pair up with impurity atom electrons. One electron of silicon has no electron to pair up; hence a hole is created(For correct structure without explanation but showing a group three element.
 Piece of metal does not displace own weight but the two together displace their own weight/ weight of water displaced is less than the weight of metal while weight of water displaced equals the weight of the tow/up thrust equal to combined weight.
 Speed = distance / time speed = 600m/s
= 1200 1.75
3.5 =343 m/s
= 343m/s (Range 342.8 – 343 m/s)
22.
 Circuit A
Current draw from each cell is less than in B / In A there is les internal resistance.
 To with stand high temperature / high melting point.
26.
27 Fringes will be closer together / more fringes of violet light has a shorter wavelength Red light has longer wavelength.
 Do not accept: Heat loss = heat gain
Pt = mcq or VIt = mcrq
2500t = 3.0 x 4200 x 50
T = 252s / 4.2min / 4 min 12s.
29.
 F = ma a2 = a_{1 } Accept F = ma for formula mark
F = 2 ma_{2} 2 a2 = a1m
2ma_{2} = maf 2m
 Radio waves, Infrared, visible light, U.V light, Xrays (accept correct order)
 Galvanometer deflects; Changing flux produced in p is linked to Q causing an e.m.f to be induced / by mutual inductance an emf / current is induced in Q.
 Maximum deflection of G will be double; flux linkage doubles when the turns are doubled.
34.
 Q = hf0 =W_{o} or & = hfco
= 6.63 x 10^{34} x 9.06 x 10^{14}j
= 6.01 x 10^{19} J or 6.0061 x 10^{10} or 6.0 x 10^{19} if working is shown.
 Fast air causes low / – reduced pressure at the top. So there is net force upwards on pith ball / pressure difference pushes pith ball upwards.
 Parallel C = (1.3 + 0.7 0)mF = 2.0 mF or 2 x 10^{6}F
Series 1 = 1 + 1 =1
C_{T} 2.0 / 2.0
C_{T} = 1.0 mF // 1.0 x 106 F.
PHYSICS PAPER 232/2 K.C.S.E 2003 MARKING SCHEME.
(i) Velocity equal zero; (ii) body is uniformly accelerated;
(iii) Body is uniformly decelerated to origin
(b i) S = ½ at 2 a = 10 ms 2
45 = ½ x 10 x t^{2} t= 3 s; (3mks)
(ii) the initial horizontal velocity of the ball.
S –V at; 50 Va x 3; VA = 16.7 ms 1
(iii) V = U + at;
V = O + 10 x 3; = 30ms1 (total 13 marks)
2ai) work= force x distance;
= 2000 x 3.0 x 10;
6000J;
 ii) Power = work done
time
= 60000
6
= 10000w;
iii) 12.5 kW
% efficiency = work output = power output
= work input = power input
= 12.5 x 103
iii) Force is centripetal = mv^{2}
^{ r}
= 20 x 4.24
4
= 89.9V Total 14 marks
3 a) Specific latent heat of vaporization is the quantity of heat required to change 1 kg of a liquid at boiling point completely to vapour at the same temperature and atmospheric pressure
B i) I Mass of condensed steam = 123 120 =3g;
II Heat gained by water
= 0.070 x 4200 x 25J;
Heat gained by calorimeter
= 0.05mx 390 x 25; = 487.5J;
= 7837.5J;
 ii) Q = mL;
II Q= 0.003 x L
0.003 x L = 7837.5;
L= 2.61 x 10^{6} J kg 1
 a i) I 4cm; II A= 2cm;
 ii) I 0 to A 9cm containing 2 ¼ waves
time for 1 wave = 0.04 s
f=^{1}/_{7}; = ^{1}/_{0.04} f= 25Hz;
II V = f; 15 x 0.04 = 1 ms^{1}
Ai to allow all radiations to penetrate;
(ii) On entry radiation ionizes argon gas
Avalanche of ions flows between terminal causing condition;
Pulse of current flows; Pulse registered as particle;
iii) Quenching the tube;
 a) e.m.f is total work done in transferring unit charge from on terminal of battery to the other;
b)
(ii) i) E = V + Ir;
(iii) From the graph determine the; current (A)
I internal resistance = slope of graph
Slope = 1.2= 0.90
1.00.5
= 0.3
0.53
= 0.6W
(c) Current through shunt = 3.0 0.03 = 2.97a;
Pd across g= Pd across shunt = 10x 0.03; 4 marks
Resistance of shunt Ir = 10 x 0.03
= 2.97 x r= 10 x 0.03
R = 0.101 W
SECTION II
6 a) Water is heated and gently stirred;
Values f pressures and temperature are recorded to intervals;
Temperature is converted to K and atmospheric pressure p added to P;
Graph of pressure p against (K)
Plotted giving straight line;
 b)
(i) C is intercept and C= O;
K is gradient given by
Gradient = 15.2 x 104 x 10
400105
= 11.2 x 103
295
= 37.97 pak1
(ii) Gas would liquidify;
(c) 270C = 300k
3270C = 600k
P1=p2
T1=T2
2.1 x 105m= p2
300 600
P2 = 4.2 x 105 Pa
 a) i) The candle is placed at a distance u from lens and screen position adjusted until
sharp image is obtained; the distance v between lens and screen is measure; Process is repeated for other values of V;
For each set of u, v, f is found 1/f = 1/u + 1/v; average f determined;
(ii) Image is virtual and so not formed on screen
 c) m = v= 2
^{v}/_{15} + ^{1}/_{30};
=^{1}/_{f} = ^{1}/_{15} + ^{1}/_{30}
F= 10cm
PHYSICS PAPER 1 2004 MARKING SCHEME
 15.5 + 0.33 = 15.83mm/1.583cm
 Air in the balloon expands/volume of balloon increases; displaces more air raising the up thrust of air;
3 i) Stability reduced/Lower /less stable
Upper section heavier/hollow section becomes heavy/more massive top
– Raising the c.og of the block.
 Density of water is low/It will result to a very log barometer/ very long tube
5.
NB at 4^{0 }c graph must be curved
– 4^{0} must be marked
– If drawn using a ruler N0 mk
– If 20^{0} c is marked, it must be higher than 0^{0}c
 Wooden Block
Wooden block is a poor conductor of heat all the heat goes in melting the wax.
 NB Check correct rays with arrows.
– at least one angle on each reflecting surfaces must be marked.
8.
 To depolarize/ oxidizer/ reduces polarization/oxidizes H_{2 }to H_{2} to H_{2 }0/Changes H_{2} to H_{2}0/ removes H_{2 }(any give 1 mark)
 Adding detergent/Impurities/increasing temp/heating (Any give 1mk)
11.
Check
correct pattern
– correct direction
NB at least 4 lines of forces must be shown
 Lines of forces must start at the poles.
12.
NB forces must be straight
Lines must touch a conduct
 Increasing current/increasing no. of turns or length of coils/ increase strength of field same as moving magnet close to core & using U shaped winding coil on soft iron core/increasing the angle between conductor and the field. (give any 2mks)
 Sum of clockwise moment=sum of anticlockwise moments
Wx20= 30×5
2w=15
Higher, reducing the current.
 Either in (10b)current from each cell is less than in (10 a)
Or
Power supplied in 10(b) is less than in 10(a)
 Distance= Area under graph
= 2x ^{1}/_{2 }x2x 20
= 40m
Or s = ut+ 1/2at^{2}
S=2(20) + ½(10)4
S= 20
S=2×20
40m
 W= Fd
Mg sin q
= 60x 10×0.5×4
= 1200J
 Electromagnetic Mechanical
can travel through vacuum – Cannot travel through a vacuum
Travel at speed of light – Travel at varying speeds
are faster – are slower
– Does not necessarily
Refuse a material media – Refuse a material media
 Either p=VI = V2/r
When V reduces power reduces
So rate of heating reduces
Or V=IR
P=I2 R (reducing IR reduces power so rate of heating reduces.
 E=pt t=450 150 =300s
E= 50×300
1= 150,000J
 Q=ml
15000=0.1×1
1= 150,000J/kg
23.
 Correct rays must be refracted to the eye and should be diverging.
Dotted lines should show image position. (should not have arrowsmust intersect within container)
 Plasticine increases mass of body since momentum is conserve or weight of trolley/normal reaction increases so fiction forces increases or Mass of trolley increases, the driving force being constant.
 Either on closing on closing s_{1} while s_{2 }open
Q= CV = 3C
When s_{1} is open s_{2} closed charge is shared between the two capacitors
CT=C+C=2q
Since q is the same equal to 3C_{1} the new pd=V_{1}
Q=CTV_{1} =3C
V_{1} =1.5V
Or
S_{1 }closed S_{2} open lower capacitor charges to 3V
S_{1} open S_{2} closed lower capacitor charges the upper to same charge (p.d)
Final pld = ^{3}/_{2} V = 1.5V
Or
Q=CV=3C
S_{2} closed charge is shared
CV= Q/2
V= ^{QC}/_{2C} = ^{3C}/_{2C}= 1.5V
 Either V1/T1=V2/T2
^{200}/_{293}=V2/353
V2=241ml
Or V= KT
200=293K
K=0.6828
V_{2}=0.6828 x 353
V_{2}= 240.96 ml
The other answers for
V_{2 }240.9/240.94ml
 Xrays Gama Rays
produced by fast moving electrons As a result of disintegration of nucleus
Produced due to energy changes in
Level of atoms due to energy changes with nucleus
Of atoms
Produced when energy changes in
Electronic structure of atoms produced due to change in nucleus
Of atoms.
(Any one comparison give 1mk)
 T=Mv2/r or T sin q mv/r or tanq =^{V2}/_{rg}
81=MV2/r q=86.46 V^{2}=0.499×86.4
81=^{5V2}/_{0.5} r=0.488 V^{2}80.63
V=9m/s V^{2}=0.499×81 x 0.9981/0.5 V=8.979 m/s
V^{2}=80.70
V=8.983m/s
30
 At least three ware forms must be drawn.
Ware length (spacing) must be maintained
31.
32.
Checkat least Three complete troughs/Crest
 Amplitude range 6.5 squares
8.5 squares
 Xrays (Hard) Softrays
Shorter Wavelength Longer wavelength
More energetic Less penetrating
High Frequency Low frequency
Produced by high voltage Produced by low voltage.
Produced by fast moving electrons Produced by slow moving e
electrons
 hf0=Wc=q
Fo = Wc/h=32×16 x10/6.62×1034
= 7.73 x 10^{14} H2 or 7.732 x 10^{14} H2 or 7.734 x 10^{14 }H2
= 7.73 x 10^{14}H2 or 7.732 x 10^{14}H2 or 7.734 x 10^{14}H_{2}
PHYSICS PAPER 2 2004 MARKING SCHEME
 a) – Put in water and mark
Put in liquid and Mark
– Space between the 2mks which represent the reciprocals of densities is divided into equal parts.
b)
 i) Up thrust=0.49N
 ii) Up thrust=weight of liquid displaced (Archimedes Principle) = 0.4N
Mass of Liquid =0.049kg=49g(converting m to kg or g)
Volume of liquid = 6.2 x 4.5
= 27.9cm^{3}
Density = Mass/Volume = 4.9/27.9g/cm^{3} = 1.760kg/m^{3}
 a) i) Mass m_{1} of melted ice/mass of water. Time t_{1} take
 ii) Q=m1 Vit=ml
P=^{ml}_{/t} p=Vi=^{m}l/_{t}
iii) Part of heat produced by heater is wasted temperature of ice may be lower than zero.
 b) i) When oil drop is placed at the centre of tray, oil spreads on water until it is one molecule thick producing patch (monolayer)
 ii) Volume of drop=4/3Wr^{3} =Wr^{2}h(r=radius of drop)
Volume of patch =Wr^{2}h (h=Thickness of molecule)
4/3 Wr^{3}=Wr^{3}/Wr^{2}h (equating)
H=4/3Wr3/Wr^{2}+2=4x(0.25)^{3}/3x(100)^{2} 2.1×10^{6}mm
Because oil does not necessary spread to a monolayer/ one molecule thick or Big errors in radius of oil drop and patch or errors in measurement of diameter/radius.
iii) Put oil in a burette and read level, let 100 drops fall and read new level, obtain radius using ^{4}/_{3}Pr^{3} = Volume
or
Obtain thin wire and make Kink; deep in oil and let drop form on kink use a milimetre scale to measure diameter of drop.
 a) i) Produce alcohol vapour
Cools alcohol vapour below condensation temperature or cools air so that alcohol vapour condenses.
 ii) Radiation from source ionizes air along its path; alcohol condenses
around these ions; forming droplets or traces; nature of traces
identifies radiation.
iii) Can detect,__ While electroscope on , can identify nature of
radiations, is more sensitive.)
b)i)
 ii) Delayed 1×10^{20}___________ ½ x 10^{20} – ________ ¼ x 10^{20} _______ 1/8 x 10^{20}
= 0.125 x 10^{20} = 1.25 x 10^{20}
^{ }Left 1 x 1020 ———– 1/8 = 0.875 x 1020
(Subtraction ) = 8.75 x 1019 Atoms.
 a) i) 0.30cm
 ii) 0.650.25=0.4 Sec.
iii) f=^{1}/_{T} = ^{1}/_{0.4 2.5} HZ
 iv) V= fx=^{V}/_{x} =^{200}/_{25} =80cm = 0.8m
 i)
 ii) m= ht of Image = distance of image
ht of object distance of object
^{h0}/_{200 } = 25/5 h0= 200×25/5 = 100m
 a) i) Increasing me of turns/coils
Increasing speed (rate) of rotation
 b) In a motion produces Eddy currents. These cause force to act on plate causing damping in B Eddy currents are reduced by slots
 c) Rms = V peak/2
V peak = 12×14142=16.97v=17v
6 a) One turning fork is loaded with a small amount of plasticine sounding together again one can produce detectable beats.
 b) i) ^{1}/_{f} x10^{3} (H_{3}^{1}) 3.91 3.5 2.9 2.3 2.1 2..0
1211 0.65 0.57 0.48 0.39 0.34 0.32
 ii) Slope (Gradient) = ^{V}/_{2 }= (0.670.10)m/4.00.75)x 10^{3}H3^{1}
V=340 ^{10m}/_{s}
iii) Sound waves entering tube is reflected at water surface forming standing wares with incoming wares, when an antinode is at the mouth loud sound is heard. By adjusting length of air column this can be achieved.
 i) Photoelectric effect is the emission of electrons from a surface when radiated with radiations of sufficient frequency.
Correct circuit must work i.e cathode connected to (ve) Emphasize on mA cell connected and v in parallel
 ii) Slope = 1.280.10/(7.74.8) x 10 14
h= Slope x e
= 1.18 x 1.6x 1019/29 x1014
= 6.51 x 1034JS
(5.82 – 6.66) x 1034 JSAlt – Selecting 2 pts from graph
– Substitution in simultaneous equs
Value of h
Value of Æ
Fs (Threshold Frequency) = 4.55 x1014 (where graph cuts the axis)
Range (4.4 – 4.6) x 20^{14}
Work function Æ = 6.51 x 10 ^{34} x 4.55 x 10 ^{14}
= 2.96x 10^{19}J
Range (2.563.06) x 10 ^{19}J
 c) ½ mv2 max = hfÆ
hf= 6.51 x 10x 3×10^{15}
KE max = 1.953 x 10^{18} – 6.4 x 10 ^{19}
= 1.31 x 10^{18}
Range (1.12 – 1.31) x 10 ^{18}J
PHYSICS PAPER 232/1 K.C.S.E 2005 MARKING SCHEME
 Volume of 55 drops =8ml accept cm^{3}
Or Volume of one drop =8/55
= 0.1454/0.1455/0.145/0.15cm^{3}
2.
 Water in A expands reducing/lowers density
This reduces/lowers upthrust on block causing tipping to side A
 There is extra/ more/higher/ increased pressure in (b) due to the wooden block increasing distance d_{2}
 Reduce/ minimize the transfer of heat by radiation OR Reduce the loss of heat OR gain of heat by radiation.
 2 sec of rays with arrows labeling of umbra (totally dark) and partly dark (Penumbra)
 A or tube with air
Gas molecules move faster/quicker than water molecules OR Diffusion of gases is faster/more than in water/Grahams law the density of air is less than that of water
Figure 6
 APositive
BNegative
 C Ammonium jelly/chloride /paste/solution/NH_{4}Cl
DMixture of carbon and manganese (iv) oxide/MnO_{2}
 In (a) cohesive forces between water molecules are greater than adhesive forces between water and wax while in (b) adhesive forces between water and glass molecules are greater than cohesive forces between water molecules.
 to make the rotation continuous by changing the direction in the coil every half cycle/turn also accept changing direction of the current every half cycle/turn/maintaining the direction of current in field.
 S=nt+^{1}/_{2} st^{2} where t is the time to reach the ground
15= 0 + ½ St^{2} since the initial velocity is zero and t= 3 = 1.732
Horizontal distance= Horizontal speed x t = 300x 3 o 519.62m
 Efficiency = Ma/VR OR Ma/VRx 100%
0.75 = 600/400
V.R
V.R = 2
ACT
M.A ^{600}/_{400}= 1.5
^{1.5}/_{V.R}= 0.75
V.R=2
 =4cm or 0.04m from the graph
V = fl= 5 x 0.04
= 0.2ms1 or 20cm/s
17 The pitch decreases as the siren falls
The higher the speed away from the observer, the lower the frequency heard and so the lower the pitch hard.
 Accept cells in parallel and other symbols of rheostats
19 (i)
= V^{2}/R
2500= 240 2/R
R=23.04 or (23.03)
(ii) P=IV
I P/V = 2500/240 =10.417A
V = V/I= 240/2500
2500
= 23.04R (23.03)
(iii) P= IV and V=IR or I^{2} R
R= 240 x 240
2500
R= 23.04R
 The liquid is boiling
21.
 2 Rays: Each correct ray 1 mark
Image position 2.2 +0.1 cm from mirror.
 C=47^{0}10^{0} =37^{0} + 7=37^{0}
 n = i
Sin C
n = I = 1
Sin 37 0.6018
n =1.66/1.551/1.662 Allow TE from question 23 and allow all the marks.
25.
 1. At steady rate, the sum of pressure, the potential energy per unit volume and kinetic energy per Unit volume in fluid in a constant.
 Provided a finish is nonviscous, incompressible and its flow steamline and increase in its velocity produces a corresponding decrease in pressure
 When the speed of a fluid increases, the pressure in the fluid decreases and vice versa.
 273+ 281.3 = 8.3K (accept – 8.15 was use.)
28.
(i) (ii)
 (i) F = MV^{2}/r
4800 = 800 x V^{2}
20
V = 10.95m (allow 10.09 of a slide is used)
Alternatives.
(ii) V_{max} = √Mrg but Fr = Mμg
M = Fr = 4800
Mg 800×10
= 0.6
(iii) F = Ma
4800 – 800 x a, a = 6m/s^{2}
A = v^{2}/r
OR
6 = V^{2}/20
V = 10.95
(iv) F = MR, M = F/R = 4800
800 = 0.6
Tan ө = 0.6
V^{2} = rg tan ө
OR
V^{2} = 20 x 10 x 0.6
V = 10.95
 Image changes from real to virtual
Image changes from inverted to upright
Image changes from behind lens to the same side as object.
 In excited state the electron is in a higher (outer) energy level. As it falls back it releases energy and may fall in steps releasing different energies (radiations) (proton) packets energy.
 To withstand the high temperature (immerse heat) prevent the target from melting due to high temperature or immense heat.
 Methylated spirit evaporates faster/highly volatile than water taking latent heat away faster from the hand.
 m Alpha (µ) particle/ radiation/decay
n Beta (b)
x Polonium (P_{o})
 When the switch is closed and nails attracted.
When the switch is opened, the nail on the iron end drops first.
 Conductor allows charge to be distributed/movement/spread.
PHYSICS PAPER 232/2 K.C.S.E 2005 MARKING SCHEME
1.
With distance between lens and object being greater than facal length f;
 Adjust the lens distance until a sharp image of object is formed besides object
 Distance between the lens and the object is measured and repeated several times
 The average of the distance is the focal length of the lens
Alt Method: No parallax method is also marked
The pin is adjusted until there is no parallax between the object pin and the pin image. The distance between the lens and pins is the focal length of the lens
(b) On the graph paper
NB: position = 5.2 x 4 cm
= 20.8 cm
= 21 ± 1 cm
(c) (i) Long sightedness/ hypermetropia/ presbiopia
(ii)
 (i) Distance traveled by the effort in one revolution = 2πR
Distance traveled by load = 2πr
Velocity ratio (V.R) = effort distance = 2πR = R
Load distance = 2πr r
Therefore V.R = R
r
(ii) V.R = R = 8cm = 1.6
R 5 cm
Efficiency = M.A = 80
V.R 100
But M.A = Load = 20N
Effort E
Therefore 20N ÷ 1.6 = 0.8
E
20N x 1 = 0.8
E 1.6
Effort E = 20N
1.6 x 0.8 = 15.6 (3) N
= 15.6N
(iii) When the load is large, the effect of friction and weight of the moving
parts is negligible
NB friction and weight of moving parts to be mentioned
 Total resistance R = 6 W + 5 W + 1 W = 12 W
Total current 1 = ^{V}/_{R} Check correct substitution
(ii) P.d across each capacitor = 1R
= 0.25 x 11
= 2.75v
Charge = CV = 1.4 x 2.75 x 10^{6}
= 3.85 x 10^{6}C
 (a) (i) Pure Silicon or germanium is doped with prevalent impurity i.e.
phosphorous.
(ii) Four of the fire valence are paired with semi conductor electrons
(iii) The fifth electron is left unpaired and so conducts
NB; Doping pairing and conducting must be mentioned
(b) (i) In the first half – cycle A is a positive making D_{2} and D_{3} to be forward
biased, so current flows through D_{2} R and D_{3} to B.
In the second half – cycle, B is positive making D_{4} and D_{1} forward biased. The current flows through D_{4} R and D_{1} to A
(ii)
(iiii) The capacitor is charged when p.d is rising and stores charge
It discharges through the resistor when p.d is falling
This makes output smooth i.e reduces humps
(c) hfe = ∆Ic
∆I_{B}
120 = ∆Ic
20B/A
Therefore ∆Ic = 120 x 20 MA = 2.4mA
Output p.d charge = R_{L} x ∆IC
1000R x 2.4 mA
= 2.3v
 (a) Extension is directly proportional to the extending force provided the
elastic limit is not exceeded.
(b) (i) 3.2 N or 3.3 N
(ii) At 5 cm F = 1.45N
Stress = F/A = 1.45
0.25 x 10^{4}m^{2}
= 5.8 x 10^{4} Pa
NB: can work with N/cm^{2}
Accept 5.6 – 5.8) x 104 pa
(iii) Strain = Ext = 5 = 0.025
Original length 200
(c) ED and DC
 Angular velocity is the ratio of angle covered (angular displacement) to the time interval
or W = θ_{2} – θ_{1}
t_{2} – t_{1}
(b) w = 300 – 170 = 10 radis^{1}
13
10t = 170
T = 17 sec
(c) (i)
(ii) T = mco2r – C slope = mr = 1.5 – 0.25 = 0.061
28.5 – 8.0
M= 0.061 = 0.203 Kg (0.2 kg)
30 x 10^{2}
iii) Extent graph (calculate) C= 0.2
It represents frictions between table and body
 (a) Radioactivity is the spontaneous disintegration of unstable nuclei so as to stabilize
When radiation enters via mica windows, the argon gas is ionized; the electrons going to the anode and positive ions going to cathode; thus a discharge is suddenly obtained ( PULSE) between anode and cathode and registered as a particle by counter. The discharge persists for a short time due to the quenching effect of halogen vapour.
(c) Half life average t ½ = 24.5 min (error transfer)
12 12 12
(d) t(min 40 28 16 4
Activity 480 960 1920 3840
3 half – lives
t = 4 min
K.C.S.E 2006. MARKING SCHEME
PHYSICS PAPER 1
 Volume = 68cm^{3}
Mass = 567g
Density = m = 567
V 68
= 8.34 gcm^{3} (3 marks)
2.
 Pressure at a point in a fluid is transmitted equally to all points of the fluid and to the walls of the container. (1 mark)
 On heating, the bimetallic strip bends; This causes the position of the centre of
gravity of the section to the left to shift to the right causing imbalance and so tips to the right (2 marks)
 Lower spring extend by 15 cm;
Upper springs extended by 10 cm;
Total = 15 + 10 = 25 cm (3 marks)
6.
 Effect of weight of second pulley reduces efficiency of A. Load in B is larger and
so effect of friction is less in B increasing efficiency. ( 1 mark)
 In B some of the heat is used up in melting the ice, while in A all the heat goes to
raise the temperature of the water to reach boiling point ( 2 marks)
9.
 At F, radius of curve is smallest and so greatest centripetal force is required to
keep luggage on carrier; ( F =mv^{2}) ( 2 marks)
R
 A_{1}V_{1} = A_{2}V_{2};
π x 6^{2} x V_{1} = π x 9^{2} x 2;
= 4.5 ms^{1} ( 3 marks)
 As the temperature changes the volumes of the gases in the balloons change
differently. The change in volume and hence the change in upthrust will differ. ( 2 marks)
 Ft = ∆ mv;
720 x 0.1 = 0.6 x v;
= 120ms^{1} ( 3 marks)
 (a) In solids the molecules are held in position by intermolecular forces that are
very large. In liquids the molecules are able to roll over one another since the forces are smaller ( 1 mark)
(b) (i) Volume = 4/3 π r^{3}
= 4/3 π x 0.025^{3}
= 6.54 x 10^{5} cm^{3} ( 2 marks)
(ii) Area = π r^{2}
= π x 10^{2}
= 314 cm^{2} ( 2 marks)
(iii) A x diameter of molecule = volume;
314 x d = 6.54 x 10^{5}
d = 2.1 x 10^{7} cm ( 3 marks)
(c) (i) The oil is assumed to have spread to thickness of one molecule ( 1 mark)
(ii) Sources of errors:
 Getting the right oil
 Measuring drop diameter
 Measuring diameter of patch
 Getting drop of a right size ( any 2 x 1 = 2 marks)
 (a)
 Make diameter of springs different
 Make number of turns per unit length different
 Make lengths of springs different ( any 2 x 1 = 2 marks)
 (i) 2.2 N ; 2.2 ±1
 (ii) Spring constant = gradient
= 2.1
4.1 x 10^{2}
= 5/Nm^{1}
For each spring k= 102 Nm^{1} ( 1 mark)
(iii) Work = Area under graph
= 0.75 + 1.65 x 1.7 x 10^{2}
2
= 2.04 x 10^{2} J ( 3 marks)
 (a) A gas that obeys the gas laws perfectly ( 1 mark)
(b) (i) By changing pressure very slowly or by allowing gas to go to original temperature after each change ( 1 mark)
(ii) k is slope of graph
K = ( 2.9 0) x 10^{5}
( 3.5 – 0) x 10^{6}
K = 0.083 NM
(iii) Work done on the gas ( 4 marks)
(iv) Use dry gas ( 1 mark)
Make very small changes in pressure ( any 1 x 1 = marks)
(c) Since pressure is constant
V_{1} = V_{2}
T_{1} T_{2}
T_{1} = 273 + 37 = 310k
T_{2} = 273 + 67 = 340k
4000 = V_{2}
310 340
V_{2} = 4387 litres ( 4 marks)
 (a) A body fully or partially immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced ( 1 mark)
(b) (i)
(ii) 100g: U_{w} = 0.12N U_{s} = 0.09N
150g: U_{w} = 0.18N U_{s} = 0.14N
200g: U_{W} = 0.24N U_{s}=0.18N ( 2 marks)
(ii) Relative density = upthrust in spirit
Upthrust in water
= average 0.09 0.14 0.18
0.12, 0.18, 0.24
= 0.76 ( 3 marks)
(c) Weight of air displaced = ρVg
1.25 x 1.2 x 10N
=15N;
= upthrust
Weight of helium = ρVg
0.18 x 1.2 x 10N
= 2.18N;
Weight of fabric = 3N
Forces downwards = 2.16 + 3 = 5.16N;
Tension = 15 – 5.16
= 9.84 N ( 4 marks)
 (a) Specific latent heat of fusion of a substance is the quantity of heat required to melt completely one kilogram of the substance ( at its normal melting point) to liquid without change of temperature. ( 1 mark)
(b) (i) Q = ml
= 0.02 x 334000J
= 6680J ( 2 marks)
(ii) Q = mcθ
= 0.02 x 4200 ( T0)
= 84 TJ ( 2 marks)
(iii) Heat lost by warm water
= mcθ
= 0.2 x 4200 ( 60 T)
Heat lost by calorimeter = mcθ
0.08 x 900 ( 600 – T) ( 2 marks)
(iv) Heat gained = Heat lost
6680 + 84T = 0.2 x 4200 ( 60 –T) + 0.08 x 900 ( 60T)
6680 + 84T = 50400 – 84OT + 4320 – 72T
996T = 48040
T = 48.2^{0}C ( 4 marks)
K.C.S.E 2006: MARKING SCHEME
PHYSICS PAPER 2
1.
 Magnification =
Im age dist = ht of image
Object dist height of object
10 = 16
600 h
H = 9.6 m (3 marks)
3.
 To allow escape of gases ( H_{2} and O_{2}) from battery
 (i) Longitudinal wave
(ii) Length of the spring, from one point to a similar point of vibration
6.
Reflected waves are curved. Either converging circular reflected waves. Converging to F; OR two perpendicular lines from the surface of one of the curves meeting at F. (2 marks)
 Distance moved by sound waves = 2x;
2x = speed x time
X = 330 x 1.8
2
= 297m ( 3 marks)
9.
 Constant temperature
 No mechanical strain ( 1 mark)
 Work function of a metal is the minimum energy required to set free (release) an electron from the surface of the metal (1 mark)
 Threshold frequency K.E of electron = 0 hence velocity of the electron would be zero; (No motion) thus photo electric effect cannot be observed ( 2 marks)
 Straight beam from gun to screen OR no gravitational effect on the beam. ( 1 mark)
13.
 Resulting X rays have shorter wave length/ hard/ high frequency because electrons have higher K.E ( 2 marks)
 a = 234 + 4 = 238
b = 92 – 2 = 90 ( 2 marks)
16.
 (a) Charge Q, on C_{1} is given by
Charge Q_{1} = C_{1} V;
= 0.3 μ F x 4.5;
1.35μC; ( 3 marks)
(b) C_{T} = C_{1} + C_{2};
= (0.3 + 0.5) μ F
= 0.8 μ F ( 2 marks)
(c) (i) 4.5v ( 1 mark)
(ii) Observed on voltmeter p.d drops to less than 4.5 (1 mark)
(iii) The drop of p.d in C (ii) is because the charge on C_{1} is distributed to C_{2}. Since values of C_{1} and C_{2} remain constant, when Q on C_{1} reduces, then Q = C_{1}V implies V must reduce also, hence voltmeter reading reduced.
 (a) (i)
(ii) Image at 10cm from mirror (using scale) (2 marks)
(iii) Magnification
Size of image _{= }4.0 cm = 2
Size of object 2.0 cm
OR
Image distance _{=} 2.0 cm = 2
Object distance 1.0 cm
(b) ( i) I Image distance
I = I + I
f v u
I = 1 – I = 3
v 5 20 20
v = 20 = 6.67 cm
3
II Magnification
= v = 6.67 = 0.33; ( 2 marks)
u 20
(ii) Image characteristics: real, inverted, diminished, less bright
( 2 marks)
 (a) Refr. Index n = sin i velocity in air
Sin r velocity in substance
OR
n = Real depth
Apparent depth ( 1 mark)
(b) (i)
(ii) Slope of graph = ^{16}/_{24} = 2/3
Refr. Index n = Real = I
Apparent slope
= 3 = 1.5 ( 4 marks)
2
(c) n= sin 90; Þ sin θ = 1; Þ θ = 38.7^{0} = critical angle ( 3 marks)
Sin θ 16
 (a) (i) P = slip rings
Q = Brushes (2 marks)
(ii) 090 magnetic flux cut changes from high to low. (decreasing);
90 – 180 magnetic flux change from low to high. (increasing)
At each peak 0 – 180 magnetic flux change is maximum though in different directions, (position of coil). ( 3 marks)
(b) (i) €_{s} = N_{s}; Þ €_{s} = 240 x 60 = 12 volts ( 2 marks)
€_{p} N_{p} 1200
(ii) P_{p} = P_{s} (power) or l_{s} V_{s} = l_{p} V_{p}
_{ }
I_{S} = I_{p} V_{p} = 0.5x 240; = 10A;
V_{s} 12 ( 3 marks)
 (a) (i) P = Ring circuit ( 1 mark)
X = Neutral ( point or terminal)
Y = Live ( point or terminal) ( 2 marks)
(ii) I Purpose of R – or fuse; is a safety element in a circuit
against excess current
II R is connected to Y but not X to ensure that when it breaks
a circuit any gadget/ appliance connected does not remain live. ( 1 mark)
(iii) Earthing is necessary in such a circuit to guard against electric
shocks.
(b) Cost of electricity
1.5 kw x 30h x 8 Kshs = Kshs 360/=
KCSE 2007 PHYSICS MARKING SCHEME
PAPER 1
1.  0.562 – 0.012 = 0.550cm 5.62 – 0.12 = 0.55 cm 5.5 mm  Or 5.62 – 0.12 5.5  1 mk 
2.  Density p= m/r D = m/v = 1.75g formula – accept g/mm^{3} (0.550)^{3}cm substitution = 10.5g/cm^{3} answer – allow transfer of error 10500kg/m^{3}  3 mks  
3.  V_{2}V_{4} V_{1} V_{3} ( correct order)  1 mk  
4.  Sucking air reduces pressure inside the tube; so that atmosphere pressure forces the liquid up the tube  1 mk  
5.  Look for symbols P_{A} gh_{A} = Pagh_{B} formula or correct P_{A}g x 24 = 1200 g x 16 substitute substitution P_{a} = 800 kgm^{3} answer answer  3 mks  
6.  Radiation  1 mk  
7.  X_{2} is made greater than X_{1} / X_{1} is made shon X_{2} X_{2} is made larger than X_{1} Since B receives radiation at a higher rate, it must be moved Further from sources for rates to be equal: since A receives radiation at a lower rate than B. F_{1} d_{1} = f_{2 }d_{2}  2 mks  
8.  Taking moments and equating clockwise movements = anticlock movements 0.6 N x 7cm = mg N x 30cm; W = mg = 1.4 N:  3 mks  
9.  Distance = area under curve between 0 and 3. 0 second; = 120 x 3 x 0.2 = 72M: Trapezium Rule (3 trapeziua) Mid – ordinateral = 70.5  
10.  Acceleration = slope of graph at t = 4.0 s Or a= ∆ V or trapezium rule (6 trapezia) ∆ t = 72m = 16 x 3 = 14.11 m/S^{2} 17 x 0.2 (12 – 14.5) m/s^{2} or trapezium (1) or 1 triangle = 76.5m  2 mks  
11.  Pressure, impurities::  2 mks  
12.  Kelvin ( K) in words ( one triangle used follow)  2 mks  
13.  The pressure of a fixed mass of a gas is directly proportional to its absolute ( Kelvin) temperature provided the volume is kept constant P & T volume constant  1 mk  
14.  Since the quantity of water A is smaller, heat produces grater change of temperature in A; This causes greater expansion causing the cork of temperature in A; this cause greater expansion causing the cork to sink further. Per unit volume/ greater decrease in density/ lower density in A  
SECTION B  
15 (a)  Smoke particles Show the behavior or movement of air molecule Smoke particles are larger than air molecules/ visible and light enough to move when bombarded by air molecules Lens Focuses the light from the lamp on the smoke particle; causing them to be observable Microscope Enlarge the smoke particle So that they are visible/ magnifies smoke particles  2 mks)
2 mks)
2 mks)  
(b)  Smoke particle move randomly / zigzag / haphazardly Air molecules bombard the smoke particles/ knock, hit Air molecules are in random motion  3 mks  
(c)  The speed of motion of smoke particles will be observed to be higher smocking particles move faster, speed increases, increased random motion  1 mk  
16(a)
(b) (i)
ii
iii  A body at rest or motion at uniform velocity tends to stay in that state unless acted on by an unbalanced force/ compelled by some external force to act otherwise. S = ∆u Nd or 98. 75 – 0 ( m/s)^{2} 16 – 0 = 6.17ms^{2}
20k = s = 6.09 depend on (i) K = 6.09 20 = 0. 304 Increase in roughness increases k and vice versa Uniform speed in a straight line – uniform velocity  1 mk
3 mks
2 mks
1 mk  
(c)  Applying equation
V^{2} – u^{2} = 2as V^{2} – 0 = 2 x 1.2 x 400 Momentum p = mv
= 800 x 2 x 1.2 x 400
= 24787.07 = 24790  4 mks  
17.(a)  Quantity of heat required to change completely into vapour 1 kg of a substance as its normal boiling point without change of temperature; Quantity of heat required to change a unit mass of a substance from liquid to vapour without change in temp  1 mk  
(b) (i)  So that it vaporizes readily/ easily  1 mk  
(ii)  In the freezing compartment the pressure in the volatile liquid lowered suddenly by increasing the diameter of the tube causing vaporization in the cooling finns, the pressure is increased by the compression pump and heat lost to the outside causing condensation. Acquires heat of the surrounding causing the liquid to vaporize  
(iii)  When the volatile liquid evaporates, it takes away heat of vaporization to form the freezing compartment, reducing the temperature of the latter. This heat is carried away and disputed at the cooling finns where the vapour is compressed to condensation giving up heat of vaporization  
(iv)  Reduces rate of heat transfer to or from outside ( insulates) Reduces / minimizes, rate Minimizes conduction/ convertion of heat transfer  1 mk  
(c) (i)  Heat lost = ml_{v} + mc ∆θ = formula Heat lost by steam = 0.003 x 2.26 x 106 = substitution Heat lost by steam water = 0.003 x 4200 ( 100T) Total = 6780 + 126 ( 100 – T) = 8040 – 12.6T  3 mks  
(ii)  Heat gained by water = MC θ = 0.4 x 4200 ( T 10) Or = 1680 T – 16800  1 mk  
(iii)  Heat lost = heat gained OR correct substitute 1680 (T – 10) = 6780 12.6 ( 100T); Allow transfer of error 1680T – 16800 = 6780 + 1260 – 12.6T 1692 .6 T = 24840 T = 14.7^{0}C 14.68  1 mk
15 mks  
18.(a)  Rate of change of velocity towards the centre Acceleration directed towards the centre of the motion Acceleration towards the centre of orbit/ nature of surface  2 mks  
(b) (i)  Roughness / smoothness of surface. Radius of path/ angular velocity/ speed (Any two)  2 mks  
(ii)  II) _{A}> (l)_{B} (l)_{C} ( correct order)  1 mk  
(c)  F = m(l)^{2} r F = MV^{2} V=rw For thread to cut r w = 3.049 F= 5.6 N 5.6 = 0.2 x v^{2} 0.15 (l) = 13.7 radius V^{2} = 4.2 = 13.66 13.66 v = 2.0494  4 mks  
19 (a)  A floating body displaces its own weight of the fluid on which it floats  
(b)(i)  To enable the hydrometer float upright / vertically  1 mk  
(ii)  Making the stem thinner/ narrower ( reject bulb)  1 mk  
(iii)  Float hydrometer on water and on liquid of known density in turn and marks levels; divide proportionally and extend on either side/ equal parts  2 mks  
(c)i)  Tension; upthrust; weight  3 mks  
(ii)  As water is added, upthrust and tension increase; reaching maximum when cork is covered and staying constant then after weight remains unchanged as water is added  3 mks
11mks 
K.C.S.E 2007 PHYSICS MARKING SCHEME
PAPER 2
1.
 Alkaline cell lasts longer than lead acid cell/ remain unchanged longer
Alkaline cell is more rugged than lead acid cell/ robust/ can withstand rough handling
Alkaline cell is lighter than lead – acid cell (any one (1 mark)
 X is north (both correct)
Y is north (1 mark)
4.
 T = 0.007S (T)
3
F = ^{l}/_{T} = 3/0.007 ( f)
= 429H_{z} 428.57 – 434. 80H_{2} (3 marks)
6.
7.
 l = 1.5 : or l = E
R + r R + r
0.13 = 1.5
10 + r
R + 1.5W;
R =1.5 W (3 marks)
 R_{1} = V^{2} R_{2} = V_{2;}
P 8P
R_{1} = V^{2} x 8P
R_{2} P V^{2}
= 8 (3 marks)
 The process of the eye lens being adjusted to focus objects at various distances
(1 mark)
11.
 The higher the intensity implies greater number of electrons and hence higher saturation current (1 mark)
 a = 234
b= 82
14.
SECTION B
15 (a) The ratio of the pd across the ends of a metal conductor to the current passing through it is a constant (conditions must be given)
Also ^{V}/ _{l} = R
(b) (i) It does not obey Ohm’s law; because the current – voltage graph is not linear through line origin / directly proportionate
 Resistance = ^{V}/_{l }= inverse of slope ; gradient = ∆ l
∆V
= (0.74 – 0.70) V
(80 – 50) mA
= 0.4V
30 x 10^{3} A
= 1.33W
1.20 – 1.45 W (range) ( 3 marks)
(iii) From the graph current flowing when pd is 0.70 is 60.MA
Pd across R = 6.0 – 0.7 = 5.3v
R = 5.3 V
36mA
= 147W
= 139.5 – 151. 4W ( 3 marks)
(c) Parallel circuit 1/30 + 1/20 = 5/60 or 60/50
R = 12 W
Total resistance = 10 + 12 = 22W ( 2 marks)
(ii) l = ^{V}/_{R} = ^{2.1}/_{22} = 0.095A ( 1 mark)]
(iii) V = lR = 10 x 2.1
22
= 0.95
16.
Diverging effects should be seen ( 2 marks)
(b) (i) A diaphragm
B Film ( 2 marks)
(ii) The distance between the lens and the film / object is adjusted; so that the image is formed on the film
Adjust the shutter space/ adjust the aperture ( 2 marks)
(iii) Shutter – opens for some given time to allow rays from the object to fall on the film creating the image impression/ exposure time is varied
A (diaphragm) controls intensity of light entering the camera (3mks)
B (film) – coated with light sensitive components which react with ight to crate the impression register/ recorded or where image is formed.
(c) (i) magnification = v/u = 3
Since v + u = 80
U = 80 – v
v = 3
80 – v
V= 240 – 3v
V= 60cm ( 3 marks)
(ii) From above u = 20cm
^{l}/_{f }= l/v + l/u = l/60 + l/20 ( 2 marks)
F = 15cm ( 15 marks
 (a) The induced current flows in such a direction that its magnetic effect
oppose the change producing it.
(b) As the diaphragm vibrates, it causes the oil to move back and forth in the magnetic cutting the filed lines, this causing a varying e.m.f to be induced in the coil which causes a varying current to flow. ( 1 mark)
(ii) Increasing number of turns in the coil – increasing of the coil
Increasing the strength of the magnet ( any two correct) ( 2 marks)
Vp = Np
Vs Ns
400 = 1200
Vs 120
Vs = 40V
(ii) I_{p} = 600/400 = 1.5A ( 2 marks)
(iii) Ps_{ = }P_{p} = 600W
l_{s} = ^{600}/_{40} = 15A ( 1 mark)
 (a) (i) A Grid
B Filament ( 2 marks)
(ii) Filament heats cathode
Electron boil off cathode ( theremionic emission) ( 2 marks)
(iii) Accelerating ( 1 mark)
Focusing
(iv) Across X – plates ( 1 mark)
(v) To reduce collisions with air molecules that could lead to ionization
(b) Height = 4 cm
Peak value = 4 x 5
= 20V
(ii) 2 wavelength = 16 cm
T = 8 x 20 x 10^{3}
= 0.16S
f = ^{l}/_{T} = ^{1}/_{0.16}
= 6.25H_{z}
(iii)
K.C.S.E 2008 EXAMINATIONS
PHYSICS PAPER 1
MARKING SCHEME.
 Water V= Mw or MW = ML RD = ML = P
I P ML
 For liquid V= ML P = ML P = ML
P MW MW
P = ML
MW
 (a)
 b) R – Increases OR R – Approaches W
F – Reduces F – Reduces
 – Atmospheric pressure is higher than normal/ standard or boiling was below
– Pressure of impurities
 When flask is cooled it contracts/ its volume reduces but due to poor conductivity of the glass/ materials of the flask water falls as it contraction is greater than that of glass. (3 mks are independent unless there is contradiction)
 Heat conductivity/ rates of conduction/ thermal conductivity (NB: If heat conduction no mark)
 X sectional area/diameter/thickness/radius
 P_{1} = pgh or Pr = PA + heg
= 1200 x 10 x 15 x 10^{2} = 8 x 10^{4} + 15 x 1200 x 10^{2} x 10
= 1800 pa = 8.58 x 10^{4} pa
Total pressure
= 8.58 x 10^{4} pa
(85800pa)
 – Intermolecular distances are longer/ bigger/ in gas than in liquids
– Forces of attraction in liquids are stronger/ higher/ greater/ bigger/ than in gases
 (In the diagram)
 Stable equilibrium
When it is tilted slightly Q rises/ c.o.g is raised when released it turns to its original position
 This reduces air pressure inside the tube, pressure from outside is greater than inside/ hence pressure difference between inside and outside causes it to collapse.
 Diameter coils different/ wires have different thickness/ No. of turns per unit length different/ length of spring different.
(x Larger diameter than Y
Or in one coils are closer than in the other
 Heated water has lower density, hence lower up thrust
 (a) Rate of change of momentum of a body is proportional to the applied
force and takes in the direction of force.
(b) (i) S= ut + ½ at^{2}
49 = 0 + ½ x a x 7^{2}
a = 2M/S^{2}
(ii) V = u + at or v^{2} = u^{2} + 2 as
= 0 + 2 x 7 = 14m/s v^{2} = 02 + 2 + 2 x 2 x 49
V2 = 14m/s
(c) (i) S= ut + ½ gt^{2} either V^{2} = u^{2} + 2gs
1.2 = 0 + ½ x 10 x t2 v = u + gt
V^{2} = 0^{2} + 2 x 10 x 1.2
T = 1.2 = v = 24 = 4.899
5
4.899 = 0 + 10t
= 0.49s T = 0.4899s
(ii) s = ut
u = 8 = 2.5 = 5.10215.103m/s
t 0.49
Heat energy required to raise the temperature of a body by 1 degree
Celsius/ centigrade of Kelvin
Measurements or
Initial mass of water and calorimeter M_{1}
Final mass of water & calorimeter, M_{2}
Time taken to evaporate (M1 – M2), t
Heat given out by heater = heat of evaporation= ML
Pt = (m1 – m2)1
L= pt
M1 – M2
(c) (i) = CDT
= 40 x (34 – 25) = 40 x 9 = 360J
(ii) MWCWDT
100 x 10^{2} x 4.2 x 10^{3} (3425) = 3780J
(iii) MmCMDT or sum of (i) and (ii)
= 150 x 10^{3} x cm 6 360 + 3780
= 9.9 cmJ = 4140J
(iv) 150 x 10^{3} x cm x 66 = 4140 heat lost = heat gained + heat
by water gained
by
cm = 4140 9.9 cm = 360 + 3780
150 x 10^{3} x 60 cm = 4140
0.15 x 60
418J/Kgk 418J/Kgk
 (a) Lowest temperature theoretically possible or temperature at which/
volume of a gas/ pressure of gas/K.E (velocity) of a gas is assumed to be zero
 Mass/ mass of a gas
Pressure / pressure of a gas/ pressure of surrounding
 (i) 4 x 10^{5} m^{3} /40 x 10^{6}m^{3} / 40cm^{3}
(ii) 275^{0}C – 280^{0}C
 a real gas
Liquefies/ solidifies
(d) P_{1} V_{1} = P_{2} V_{2} but V_{1} = V_{2} If P = P_{2} is used max marks 3
T_{1} T_{2} T_{1} T_{2}
P_{2} = P_{1}T_{2} = 9.5 x 104 x 283 P2 = P_{1} T_{2}
T_{1} 298 T_{1}
_{ }= 9.02 x 10^{4}pa = 9.5 x 10^{4} x 283
298
= (90200pa) (90200 pa)
(90.2 x 10^{3} pa) (90.2 x 10^{3}pa)
 (a) VR = Effort distance
Load distance
(b) (i) Pressure in liquid is transmitted equally through out the liquid
NB; if term fluid is used term in compressive must be staled
Work done at RAM = work done on the plunger
(ii) P x A x d = P x a x d or vol of oil at plunger = at RAM
A x D = a x d a x d = A x D
d = A d = A
D a D a
VR = A VR = A
a a
(c) (i) MA = load
Effort
4.5 x 10^{3}
135
= 33. 3 (33 ^{1}/_{3})
(ii) Efficiency = MA x 100% OR efficiency = MA = 33.3
VR VR
= 33.3 x 100%
45
= 74%
= 0.74
(iii) % work wasted = 100% – 74%
= 26%
 (a) When an object is in equilibrium sum of anticlockwise moments about
any point is equal to the sum of clockwise moments about that point
(b) (i) V= 100 x 3 x 0.6 = 180cm^{3} W = Mg
M = VP OR = Pvg
180 x 2.7 = 486 g = 2.7 x 3 x 0.6 x 100 x 10
100
W = Mg
486 x 10 = 4.86N
1000
= 4.86 N
(ii) Taking moments about F pivot; 20F = 15 x 4.86
F = 15 x 4.86 = 3.645
20
Or
F = taking moments about W, 15R = 35F – (i)
F + W = F = R – 4.86 – (ii) substitute
F = R – 4. 86 — 1
F = 3.645N
OR
Taking moments about F = 20R = 4.86 x 35
R = 8.51 and F = R – W
F= 8.51 4.86 = 3.645N
(iii)
(iv) As x increase/ anticlockwise moments reduces/ moments to the left reduces/ distance between F and pivot reduces F has to increase to maintain equilibrium
K.C.S.E 2008 MARKING SCHEME
PHYSICS PAPER 2
 BC – Total absence of light; umbra, completely dark
– Total darkness
Rays are completed blocked from this region by the object
 Leaf in A falls a bit while leaf in B rises a bit
The two leaf electroscope share the charge
Correct circuit.
3.
 Hammering causes the domains or dipoles to vibrate when setting, some domains
themselves in the N S – direction due to the earth’s magnetic field causing magnetisatioa.
 Needs not be dotted
 When the switch is closed, 1 flows the iron core in the solenoid is magnetized attracting the flat spring this causes a break in contact disconnecting current.
Magnetism is lost releasing the spring
– Process is repeated (make and break circuit)
 Movement equals 1.75 oscillations
T = ^{0.7}/ _{1.75}
= 0.4 sec
F = ^{I}/_{T}
= ^{1}/_{0.4} = 2.5 HZ.
8.
 (i) V = O volts
Reason No current
(ii) V = 3 volts
Current flows in the resistors
 P = ^{V2}/_{R} P = ^{220}^{Ù2}/_{240}_{Ù2/100}
R = 240^{2}
100
= 84 J/S
 Short sightedness/ myopia
Extended eyeball/ lens has short focal length/ eye ball too long any two
 Spot moves up and down
 Frequency increases
Accept Becomes hard
Wavelength decreases
Strength / quality
 Beta particle
Gain of an electron OR
Mass number has not changed but atomic number has increased by 1
Atomic number has increased by one
Nature will not affect the speed
 (a) Temperature
Density
 Graph
 5 m accept 46 m to 47 m
 T = 4 x
V
V=4x or slope = 4
t v
= 0.51 ^{1}
43
= V = 43 x^{ 4}/_{0.51} = 337 m/s
 For max internal observer is at one end and so the distance = 2L
337 x 4.7 = 2L
L= 792 M
(c) (i) Distance moved by sound from sea bed = 98 x 2 m
V= 98 x 2
0.14
= 1400M/S
(ii) Distance = v x t
1400 x 0.10/2
= 70m
 (a) Light must travel from dense to less dense medium
Critical angle must be exceeded (< i > c)
 1 n 2 = Sin i = Sin I
Sin r Sin r
= Sin 90 OR = Sin θ
Sin θ Sin 90
= I I
Sin θ n
= ^{1}/_{sin θ}
 (i) At greatest angle θ, the angle must be equal to critical θ angle of the medium
Sin θ = sin c
= ½
= 1/1.31 = 0.763 θ = 49.8^{0}
Angle < 49.8^{0}
 X = 90^{0} – θ
= 40.2^{0}
 Sin θ/ sin X = 1.31
Sin θ = 1.31 sin 40.2^{0}
= 0.846^{0}
= θ= 57.8^{0}
 (a) (i)
(ii)
(b) (i) Open circuit p.d = 2.1 v
 Different in p.d = p.d across
2.1 – 0.8 = 0.1 r
0.3 = 0.1 r
r= 0.3
0.1
= 3n
 When I is being drawn from the cell, the p.d across the external circuit is the one measured
01 x R = 18
R = ^{1.8}/_{0.1}
= 18 n
 (a) Flux growing/ linking
No flux change
Flux collapsing
Switch closed: Flux in the coil grows and links the other coil inducing an
E.M.F
Current steady: No flux change hence induced E.M.F
Switch opened: Flux collapses in the R.H.S coil inducing current in opposite direction
 (i) Reduces losses due to hystesis ( or magnetic losses)
Because the domain in soft iron respond quickly to change in magnetic (or have low reluctance) i.e easily magnetized and demagnetized.
 Reduces losses due to eddy current
Because laminating cuts off the loops of each current
Reducing them considerably
(c) (i) VP = NP P = I_{s}V_{s}
V_{s} N_{s} I_{s} = 800
40
400 = 200
Vs 200
Vs = 40 Volts = 20A
(ii) P_{p} P_{s}
800 = 400 I_{p}
I_{p} = 800
400
= 2A
 (a) (i) Hard X – Rays
(ii) They are more penetrating or energetic
(b) (i) A cathode rays/ electrons/ electron beam
B Anode/ copper Anode
 Change in P.d across PQ cause change in filament current
OR temperature of cathode increases
This changes the number of electrons released by the cathode hence intensity of X rays
 Most of K.E is converted to heat
 High density
 Energy of electrons is = QV= ev
= 1.6 x 10^{19} x 12000
Energy of X rays = Hf
= 6.62 x 10^{34}xf
6.62 x 10^{34} x f = 1.6 x 10^{19 }x 12000
F = 1.6 x 10 19 x 12000
6.02 x 10^{3f}
= 2.9 x 10^{18}Hz
Accept ev = Gf
F = ^{ev}/_{g}
K.C.S.E PHYSICS YEAR 2009
 Volume run out= 46.6 cm^{3}
Density = ^{m}/_{v} = 54.5 / 46.6 = 1.16953
= 1.17g/ cm^{3}
 T^{2} = 4 Π ^{2L}/g
= 1.7^{2} = 4 Π^{2} x 0.705
g
g= 9.63m/s^{2}
 Needle floats due to the surface tension force
Detergents reduces surface tension, so the needle sinks
 When equal forces applied, pressure on B is greater than on A due to smaller area./ pressure differences is transmitted through to liquid causing rise upward. Force on A is greater than hence upward tension.
 Molecules inside warm water move faster than in cold water. For Kinetic energy in warm water is higher than in cold water/ move with greater speed/ molecules vibrate faster in warm water.
 Prevents/ holds, traps breaks mercury thread/ stops return of mercury to bulb when thermometer is removed from a particular body of the surrounding
 Dull surface radiate faster than bright surface
P Looses more of the heat supplied by burner than Q OR
Q shinny surface is a poorer radiator/ emitter of heat thus retains more heat absorbed Or
P Dull surface is a better radiator/ emitter i.e. retains less of the heat absorbed. ( there must be a comparison between P & Q)
 Heat travels from container to test tube by radiation so the dull surface P, gives more heat to the test tube.
 Center of gravity located at the intersection of diagonals
 Parallel
F= 2 ke
40= 2 x ke
E_{1} = ^{40}/_{2k} = ^{20}/_{k}
Single = f= ke_{2}
20 = ke_{2}
E_{2} = 20/k
E_{T} = e_{1} + e_{2}
20 = 20 /k + 20/k
20 k = 40
K= ^{40}/_{20} = 2N/cm
OR Extension of each spring = 10
K = 20N/ 10 cm
– 2N/ cm
 Air between balloon is faster that than outside so there is pressure reduction between.
12.


Time
 The lowest temperature possible/ Temp at which ideal gas has zero volume ( Zero pressure) or molecules have zero / minimum energy OR
Temperature at which a gas has min internal energy/ zero volume
 V= r x 21 OR T = 1/33 = 0.030303
= 0.08 x 21 V 33m/s T = 2V / w =
= 16.6m/s w = 2v/0.0303 = 207.525
V= rw
0.08 x 207. 5292
= 16.5876m/s
SECTION B (55 MARKS)
 (a) – Pressure
– Dissolved impurities
(b)
(i) BPt = 78^{0}C
(ii) (I) ∆t = 4.5 min
Q = pt = 50 x 4.5 x 60J
= 13500J
(II) Q = 70 – 16 = 54^{0}C (accept 54 alone or from correct working)
(III) Q = MC ∆θ
C= 13500J
0.1kg x 54k
= 2500J/ kj
(iii) ∆ t = ( 7.3 – 6.8) min = 30s
Q = pt = ml = 30x 50J
L= 30 x 50 = 83.33 x 10^{5}J/kg
0.18
 (a) Efficiency = work output x 100% ( equivalent)
Work input
OR Ratio of work output to work input expressed as a percentage
(b) (i) work effort = F x S
= 420 N x 5.2 N
2184J
(ii) Distance raised = 5.2 sin 25 = 2.2 m (2.1976)
Work done = 900N x 2.2 m
= 1980J
(iii) Efficiency = work output x 100% = 1980 x 100
Work input 2184
= 90.7%
 (a) A floating body displaces its own weight of the fluid on which it floats
(b) (l) w = T + U
(ii) Vol = 0.3 x 0.2 x 0.2m^{3}
Weight = mg = 0.3 x 0.2 x 0.2 x 10500 kg/m^{3} x 10
= 1260N
(iii) Vol of liquid = vol of block
Weight of liquid displaced = Vpg
0.3 x 0.2 x 0.2 x 1200 x 10N
= 144N
(iv) T = w – u
1260 – 144N
1116N
(c) Weight of solid = weight of kerosene displaced
= 800 x 10 x 10^{6} x 10 = 0.08 N
Mass = 0.008 kg
Vol = 50 cm^{3} Density ^{m}/_{v} = ^{0.008}/_{50 x 106 m}^{3}
 (a) The pressure of a fixed mass of an ideal gas is directly proportional to the
Absolute temperature if the volume is kept constant.
(b)
(i) Volume increases as bubble rises because the pressure due to liquid column is lowered; therefore the pressure inside bubbles exceeds that of outside thus expansion.
(ii) (I) Corresponding pressure = 1.88 x 10^{5} Pa
(II) I/v = 1/1.15 = 0.87 cm^{3}
(iii) ∆ P = (1.88 – 0.8) x 10^{5} pa = 1.08 x 105 Pa
∆P = ℓgh = ℓ x 0.80 x 10
P = 1.08 x 10^{5} kg/m^{3}
0.80 x 10
= 13500 kg/m^{3}
(iv) Pressure at top = atmospheric
0.8 x 10^{5} pa
 ^{p1v1}/_{T1} = ^{p2v2}/_{T2} = 2.7 x 10^{5} x 3800 = 2.5 x 10^{5} x v_{2}
298 288
25^{0}C = 298 k = 3966 cm^{3}
15^{0}c = 288k
 (a) Rate of change of angular displacement with time
Acc. Without (rate)
(b)
(i) Mass, friction, radius ( any two)
(ii) Oil will reduce friction since frictions provide centripetal force; the frequency
for sliding off is lowered.
(c) v^{2} = u^{2} + 2 as
= 0 + 2 (0.28)h
V = √ 0.56 x 1.26
= rw
= 0.84 = 0.14 x w = 0.84 = 6 rad s
 14
PHYSICS PAPER 2 YEAR 2009
SECTION A
 Infinite ( very many, uncountable, several
2.
 Negative change
 Allow gassing/ release of gases
OR, release H_{2} and O_{2} produced at the electrodes
 Increase the magnitude of l
Increase the number of turns per unit length
Use of U shaped iron core
 F = 0.5 sec
F = 1/T
= 1/0.5
= 2 Hz
 1.33 = 3/v x 10^{5}
V= 3 x 10^{5}
1.33
= 2.26 x 10^{8} m/s
 T = lA
 (Lq) cm
 (i) Movement of magnet causes flux linkage to change
E.M.F is produced in the cell.
(ii) When 1 flow from Q to P, a N. pole is created which opposes the approaching pole (long’s law).
 Increases in P d increases 1 in filament OR. Increase in P d increases heating effect this produces more electrons by Thermionic Emission.
Hence results on more intense x – rays
 ^{2d}/_{05} = ^{2d}/_{0.6 }+ 34 OR V = ^{d}/_{t}
D = 17/0.2 = 85 m = 17 x 2
0.1
Speed = 2 x 86 = 340 m/s
0.5
= 340m/s
 Diode in (a) is forward biased while in 6 (b) is reversed biased Or Battery in 6 (a) enhances flow of e. across the barriers while in 6 (b) barriers potential is increased.
SECTION B (55 MARKS)
 (a) Capacitances decreases
Area of the overlap decreases
(b)
(i) Parallel, Cp = 5 + 3 = 8 pf
Whole circuit ¼ + 1/8
C = ^{32}/ _{12} = 2.6 + Pf
(ii) Q = CV
= 8/3 x 12 PC
= 32 PC
(iii) B = Q/C OR Q_{B} = ^{5}/_{8} x 32
= 32 x 10^{6} = 20 PC
8 x 10^{6} V_{B} = 20 x 10^{6}
= 4 V 5 x 10^{6}
= 4V
 (a) Increase in 1 causes rise in temp
Rise in temp causes rise in R
(b) R = v/l
2.5
1.2
= 2.1 W
(c) Read off P d across Y = P.O.V from graph
(d) Power P = IV
= 0.8 x 3
2.4 watts
 (a) (i)
(ii) Highest reading near red light
Red light has more heat than violet OR
Red light is close to ultra red which has more heat energy
(b) Depth = 11.5 – 3.5 = 8.0 cm
= 11. 5 = 1.4375
8
 (a) β = particle
(b) (i) Ionizes attracted towards electrodes
Collusions with other molecules cause avalanche of ions which on attraction to the electrodes causes the discharge.
(ii) are attracted towards electrodes
Collusion with other molecules causes avalanche are of ions which on attraction to the electrodes causes
(c) (i) x = 36
Y = 92
(ii) Small, decreases in mass
Loss of mass
Mass defec
(iii) Each of the neutrons produced at each collision further collision with Uranium atom causing chain reaction.
 (a) (l) Electrons are emitted from Zn plate
Reduced of charge on the leaf
(ii) Any electron emitted is attracted back to the electroscope
(iii) Photons of infra red have to lower f than U – V have energy to eject to the electrons.
(b) (i) Number of electrons emitted will increases
(ii) Max K.E of the emitted electrons will increase
(c) (i) V = lf_{0}
F_{0} = 3.0 x 10^{8}
8.0 x 10^{7}
= 3.75 x 10^{14} Hz
(ii) W = hf_{0}
= 6.63 x 10^{34} x 3.75 x 10^{14}
= 2.49 x 10^{19}J = 1.55 e V
 x 10^{19}
(iii) KE_{MAX} = hf – hf_{0}
= h (8.5 – 3.75) x 10^{14}
= 6.63 x 4.75 x 10^{14}
= 3.149 x 10^{19 }joules
^{ }= 1.96828 e
 (a)
(i) Attach two identical dippers to the same vibrator, switch on and the circular waves produced OR
Use one straight vibrator with two identical slits to produce coherent waves.
(ii) Constructive – Bright
Destructive – Dar
(b) C I –Two waves arrive at a point in phase
DI – Crest meets a trough and gives a zero intensity
– Path diff is ½ odd number of l