br Home » KCSE Physics marking schemes for exams of 1995-2014

KCSE Physics marking schemes for exams of 1995-2014

K.C.S.E 1995 PAPER 1 MARKING SCHEME

  1. Micrometer screw gauge
  1. Effort would reduce
  2. Flow from a to B
  3. 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    

  1. No change in flow OR the flow will still continue
  2. 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
  3. The liquid expand uniformly, expansion  is measurable ( large enough), thermal conductivity
  4. Rectilinear propagation/ light travels  in a straight line
  5. Water/ or glass are poor conductor of heat
  6. 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)
  7. Can be short – circuited without  being destroyed
    1. Longer life/ electrolyte never need attention
    1. Can stay discharged without being destroyed
    1. Can be charged with  large currents faster charging
    1. More rugged/ not damaged by rough condition of use/ robus
    1. Delivers large current, light
  8. Surface tension / adhesive forces supports water column or more capillarity in tube 2 than tube 1
    1. Surface tension is the same in both tubes and equal to the  weight  of water column supported
    1. Narrow tube has longer column to equate weight to wider tube
    1. Volume of water  in the tubes  is same hence narrower tube higher column
  1. – Length of conductor in the field

– Angle between conductor and fields

  1. All ferromagnetic  materials are attracted by magnets or any magnetic materials is attracted
  2. – increasing the tension

– Reducing the length 

  1. At equilibrium sum of clockwise moment = sum of anti – clockwise moments

Clockwise moments = P  x    X = QY

                                     Px = Qy

  1. h glass = V air / V glass                                  1.5 = 3 x 108 Ö g

Vg = 3 x 108 / 1.5                                = 2 x 108 ms-1

  1. V = f l sine V is constant reducing f to 1/3 Þ          l increases 3 fold
  2. 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
  1. A body at rest or in state of uniform motion tends to stay in that state unless an unbalanced force acts on it.
  1. Heat capacity  is quantity  of heat required to raise the temperature  of the  body by 1 k or 1 0C while, specific heat capacity is quantity of heat required to raise temperature of unit  mass of body by 1 k/ 10 C.
  1. (If x ≠z but both above y give  1 mk. Accept difference of 1.0 mark)
PHY 011

            hX = hZ > hY

  1. – Reducing                                          – Increasing
  1. 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
  1. E2 = E1 + h f  i             or E2 – E1= h = c/l

h= plank constant

c- Velocity of light

l- Wave length of light

  1. – Lead                         – Very dense/ has high atomatic mass
  2. Extrapolation on graph  ( line to touch frequency)

Reading on graph to (4.0 + – 0.2) x 1014Hz

  1. Lines parallel to the one shown but cutting  of axis further in
  2. Quality / Timbre
  3. X = 14
  4. The point where the weight of the body acts
  5. Temperature of source be the same

– Length of rods be the same / wax

– Amount of wax (detector) be the same

PHY 013

K.C.S.E 1995 PHYSICS PAPER 232/2 MARKING SCHEMES

PHY 020

1. (a)

(b) Constant Vel0                                Uniform vet                – zero accln

(c)        Ö4.5 = 118 – 50 = 15m/s         15.5 + -1.5 ( 14-17)

                        6.5-2

            Ö 6.5 = 112 – 70 = 6 m/s         (4=6)

                           7

            Average accln = ∆v = v – 11  = ( 6-15)

                                        t             t           2

                                                                                                = – 4.5 m/s2     

2.         l =   7   +  l  +  l

            RC   R1      R2   R3

            =          1    +   1   +   1

                        6    +   3         6

            = 1

                6

            RC = 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  Wm2

                                         L                        1.0                 m

                        = 3.0 x 10-5W m

3.         (a)

PHY 021
PHY 022

            (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

PHY 023

            Image correctly drawn

PHY 024

            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 MV2 > mg

                                                 r

            (ii) Frictional force F   = Centripetal force

                                                            MV2                = 1200 x (25)2

                                                              R                        150

                                                                        = 5.0 x 103N

5. (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

6. (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.

PHY 025

(ii)        Path differences equals to an odd number of half wavelengths or completely out of phase ( 1800)

(iii)       Photometer / photocell or thermometer with a bulb

7.         (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) p-n-p 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 (∆V0) is determined and a corresponding change input ( ∆V1) also attained change amplification.

PHY 026

(d) (i)

(ii) i2 = lC r lB

(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
  1. 15.00 + 0.30 = 15. 30 mm; or 1.53 / 1.53 x 102m
  2. Frequency: OR wavelength or energy
  3. Length of container/ height

Width of the base/ base area/ diameter/ radius of the base/ thickness

  1. hp p1 g = h2 p2g            Same as h1 p1= h2 p2

h1 = h2 p2g                   = 8 x 18

            pg                              08

                        = 18cm;

  1. (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

  1. Concrete mixture and steel have approximately the same linear expansively. The expand/ contract at the same rate;
  2. Radiation is at the electromagnetic waves Φ infrared while conduction involves particles, which move at lower  speed
  3. 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
  4. like charges repel unlike  charges attract
  5. Mass per unit length

Or (linear density/ thickness/ cross – sectional area/ diameter, radius

  1. Adhesion

Cohesion/ surface tension

  1. As the thermistor is heated its resistance reduces/ conductivity  increases hence drawing more current through it;  hence  less current flowing through B;
  2. (i) (ii)
  1. 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

  1. Turn anticlockwise  about O, OR Oscillate about O
  2.  
  1. The wavelength/ velocity of the water waves reduces; away from  the centre  because the pond  becomes shallower/ pond  deeper at centre
  2. Interferences  ( accept beat)
  3. Parallel resistor allow diversion of current; hence may not overheat; / current shared by parallel resistor
  4. Heat gained     5(80 – 40) = m(40-15) Heat gained MCD θ ( 80- 40)

5(40) = 25m                                        Heat post MCD θ =m (40 – 15) MC 40 – 15

                                                                        5(80-40) = 25 m

                                                                        25m = 200 = m = 8 kg

  1. Equal qualities of heated supplied;

MCWθW = MCPθP                                MCW (Qw –Q) = MCP (Qp – Q)

Since θP > θW                        or                     MCw > θ0 = MCP >QP

CW > θPCp

  1. Magnified, enlarged upright, virtual , image behind the mirror, negative  distance
  2. Apparent depth = Real Depth                                                12m = 0.9 m

Refractive indese of water      1.3

  1. Pressure is inversely proportional to the speed OR speed increases as pressure distance
  2. Maintaining a stable voltage during make and break/ storing charge during make and break and stops arcing sparking
  3. 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)
  4. A Polaroid absorbs/ cuts off light waves in all planes except in a particular plane of propagation                                                                                        ( 1mk)
  5. 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)
  6. 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)
  7. The weight of the  door and the force are  perpendicular to one  another     ( 1 mk)
  8. Eddy current                                                                                       ( 1 mk)
  9. Low negative voltage is applied on control grid, which control the number of electrons reaching the screen                                                                                    ( 1 mk)
  10. Low speed  / high charge / more massive/ size is large/ bigger`          ( 1 mk)
  11. n.p.n
  12. Limit the current through the base controls the current/ protect transistor from high current or voltage/ regulate reduce voltage.
  13. 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

  1. (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 at2

 (iii) Using t = v – u; in s = ut – ½ at2

                            a           

s = u (v-u) + ½ a (v-u)2 = V2 = u2 ÷ 2 as

            a                a

(b) u = 50 – v = 0 a =2

Using v2 = u2 – 2as;

Substitute 0 = 502 + 2 (-2) s;

S = 625m;

2.         (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)

                                                                  Rp = 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

                  = 1-0.5 V; = 0.5                                                                      ( 2 mks)

                  = 0 -0.5 V; + 0.5V                                                                  Total 13 mks)

  1. (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  Φ1 final temperature Φ2

            Current I voltage V;

            Heat given = heat gained by electrical heater the block

                        1 Vt = mc ( Φ2 – Φ1)

                        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 450C

            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 10-6J/Kg;

  1. (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

                        II. Central fringe is white; fringes on either side are colored;

  1. (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 = mbW2r

            Where mb = mass of the body

            Mbw2 r = Gradient of the graph = 1.167

            W2 = 1.167 = 11.67

                        0.1

            W = Ö 11.67

            = 3.42 rad s1

  1. ( a)
PHY 041

Close switch S

Vary pd until G deflects

  • l)
K (J) x 10-19 5 10 10 30 4
F = C/D ( HE) 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 hf0

      F0 is the x – intercept of graph

      F0 ( from graph) = 1.2 x 1015 HE

      Φ = 6.63 x 10-340.5 x 1.2 x 1015

      = 7.96 x 10-19 J

No. 6

PHY 042

(a)

KCSE 1997 PHYSICS PAPER 232/1 MARKING SCHEME

  1. Volume = 7.4 – 4.6 cm

2.8cm

Density = mass

                 Volume

= 11g

    2.8 cm3

= 3.9gcm-3

  1. F1 and F6
  1. Either altitude or latitude/ radius of earth changes/ acceleration due to gravity from place to place away from the earth
  1. Balance: meat + 0.5 kg on one side and 2 kg on the other:
  1. H1 P1g = h2 p2g

H2 = 1.36 x 104 x -64

            8 x 102

= 1088cm;/ 10.88m.

  1. Volume of 1 molecule = 18cm3

    6 x 1023

            Diameter of the molecule = 18cm3

                                                         6 x 1023

                               3   18cm3

                                    6 x 1023

                                    = 3.1 x  108 cm

                                    = 3.11 x 108                                                                

  1. Glass is a bad conductor of heart, the difference in temperature between the inside and the outside cause unequal expansion
  1. Adhesion of water to glass is greater than cohesion
  1. 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

  1. A ray from A                          A ray from B

Relative positions of A and B correctly drawn

  1. Solar cell ( photovoltaic) photocell/ photo electric cell
  1.             Soft magnetic materials loose their magnetism easily while hard magnetic

                  materials retain magnetism longer

  1.             Q = It                          Q = 0.5 x 4x x 60;                   = 120C
PHY 015
  1.             d= speed x t;               340 x 2;                       680m
  2.             At low speeds  the speed is streamline

At high speed the flow is turbulent

  1.             Vr =l

V    lr

240      = 30     lr = 0.75A;

 6              lr

  1.             mgh = ½ mv2   OR      V2 = U2 + 2 as;

h = ½                           S = V2 = 36

= 18m;                                        2as    2(10)

S = ut + ½ at2                                      = 1.8m;

  1.             V = f;

V= 3.0 x 108 ms-1        = 3.14m;

F    95.6 x 106S-1

  1.             6V
  1.             parallel             l   =     l    +     l    +   2

                         RP      400               400   400

            YZI     = V = 12 = 0.02A

                           R      60

            I           = V = 12 = 0.02 A

                           R     60

                        400 x 12 = 8V

                        600

  1.             ( No of irons) x 1000) = IV

Number = 13 x 240 = 3.12;

                        1000

  1.  Extra heat is required to change ice to water / latent heat of fusion
PHY 016
  1.  
  1.  A trolley slows down/ motion decreases since mass increases and the momentum is conserved, the velocity goes down
  1. CT = C1 – C2 = 1 = 1 + 1

   CT     CP        C3

= CT = CP   C3

            CP + C3

  1. 0C + 273 = -20 + 273 = 252K
  2. (a) Dark and bright fringes                 (b) Coloured fringes
  1. Small differences in frequencies
PHY 022
  1. By using laminated core
PHY 019
  1.  
  1. After 3 secs number decayed = ½ x 5.12 x 1020 = 2.56 x 1020

Next 3 secs. Number decayed = ½ x 2.56 x 1020 = 1.28 x 1020

Total number decayed                        = (1.28 + 2.56) x 1020

                                                            = 3.84 x 2020

PHYSICS PAPER 232/2 K.C.S.E 1997

MARKING SCHEME.

1.         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 q2

                Heat lost by water = 4200 x 0.2 ( 20- q2 )

            Heat lost by glass = 0.2 x 670 x (20 – q2 )

            Heat gained by ice = 0.04 x 334 x 103

                        Heat gained water = 0.04 x 4200 ( q2 – 0)

                        Heat lost = Heat gained.

                        4200 x 0.2 (20 – q2 ) + 0.2 x 670 x ( 20 – q2 ) = 0.04 x 334 x 103 + 0.04

                        X 4200 ( q2 – 0 )

                                                q2 = 5.36oC

PHY

2(a)      i)

            ii)         Extrapolation F4 – 0                           10.6m 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

PHY 001

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 at2  

            10 – 202 = 400 =20                                         =(20 x 2.2) + 1/2 + 10 x 2.22

            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 = 3.0 x 105 x 10 x 360 = 3.0 x 105wx

                                                60 x 60

Total = ( 3 + 2_ x 103 = 5.0 x 103kw 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   =  hfo + ½ mv2

           = (3.2 + 82 ) x 10-19 = 11.2 x 10-19                                f =     11.2 x 1019

                                                                                                           6.63 x 10-19

                       l= c = 3.0 x 108 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.

PHY 002

b)i)

  ii) IB = 0.5/100 x 2 =0.01 mA               IC = 2- 0.01 = 1/99MA

      IE = IC + Irs

iii)       IB = 0.5 x 4 = 0.02mA                        Ic = 3.98mA

                   100                                  r Ib = 0.02 – 0.01 = 0.01

           IC = 4 – 0.02 = 3.98mA          r Ic = 3.98 – 1.99 = 1.99

                   hFE = 3.98

                               0.02         = 1.99

           rIc = 3.98 – 1.79 = 1.99

           rIb = 0.02 – 0.01 = 0.01

            HFE = r Ic   = 1.99  = 1.99

                       r Ib = 0.01

                       r Ic  =  1.99   = 199  

                      r Ib         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.

PHY 003

 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 3/0.5 =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

1.         Accuracy of measuring tape is 10m or o.1 cm + 5cm or o.o5m.

2.         Length of post is 1.5 (1.50 x 1.55) Rangep = N3=

3.         Quantity of heat equation 20x (42-26)x C=103 x 15 x60

                                                C=2.8x 103JKg -1K     =          (2812.5 OR2813)

4.         Detecting imperfection in metal structures/block/flaws

5.         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.

PHY 017

6.

7.         Low contact pressure between tyre and earth/no sinking.

8.         IP =N3  =          Np= 20000×3= 2000

            Is=NP                                         30

9.         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.

11.       E=IR +h

            I=  E    =          2.0

            R+r                  2.0×0.5            =0.8A

12.       50        = (I)n n =3(half-lives)

            400          (2)n

            Half –life 72 = 24 min.

13.       High resistance voltmeter takes less current/low current recording low current.

14.       Domains/Dipoles initially organized are disorganized by mechanical forces.

15.       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.

16.       Length of the rod; diameter/cross sectional area of the rod/thickness nature/type of rod material/conductivity.

17.       R=P1/4 I = 2.0 x 106x0.5 = 2m OR = 2.041 or 2.0408

                                    4.9 x207

18        Some energy is lost due to friction/air friction acts on the pendulum/air dumping on the apparatus air resistance.

19.       In TV (CRT) deflection is by magnetic field, while in CRO deflection is by electric field.  X-Y 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.

20.       Heating/ cooking/communication/eye/photographic film or plate/LDR/photocell.

21.       Diode is forward-biased, no current flows

            Current flows when the switch is closed but when terminals are reversed, no current flows 

22.       Angle of inclination/nature of surface/length of inclination

            Height of inclination/frictioal force between the surface.

23.       layers of the crystal material are arranged according to faces/ plans/ flat surfaces.  Cleavage is only possible parallel to those faces/places/flat surfaces.

24.       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

PHY 018

25.

26        Addition of impurities with higher boiling points/presence of impurities.  Water heated under a higher pressure than atmospheric/below sea level.

27.       Moon covers the sun/obstruction of sun by the moon

            Both heat and light have same velocity/both are electromagnet waves.

28.       Overtones/harmonics

29.       Since F=MV2/V the sharper the corner (as B) the small the value of R hence the greater the F. (M& V constant).

30.       Gas through the nozzle gains velocity.  Hence its pressure reduces above the nozzle. The higher atmospheric pressure pushes air into the gas stream.

31.       When mercury is heated (during a fire); it expands and makes contact, completing the circuit to ring the bell.

32.       There will be no variation of intensity of light/ uniform intensity/no bands/one

33.       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.

34.       Component of weight down the slope =50 sin 300 =25N

            Total force parallel to slope= (29+25) N 54N.

PHYSICS PAPER 232/2 K.C.S.E 1998 MARKING SCHEME

1.         iii)        Scale, axes label, unit-plotting 8-10-2             5-7-1 Curve (smooth)

iv)        As the number of turns is increased, alignment of domain with field increases. After 35-36

            turns, all domains are aligned, so that magnet is saturated.

PHY 025

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.

PHY 026

2.                                                                    

            bi)        Volume of block = 4x4x16 = 256 cm3

                        Mass of block = 154 gm

                        D= m=154=0.6g/cm3 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

3.         a     i)   The bullet will land on the track   It has some horizontal (inertia) velocity

as the track.

            (ii)        (Use g = 10ms-2}

                        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

4.         a)         i)

            (ii)        Voltage, current, time

            (iii)       Q v/t                Rate= Q/t = v/tT (T=time taken for sun to heat)

PHY 028

b)         Fig. 4 shows a photocell.

ii)         When light rays strike cathode C surface electrons gain photon (energy) hence the cathode.

PHY 029

iii)        Draw a simple circuit including the photocell to show the direction of flow of current.

5          a)         i)

PHY 030

                        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 + 0.2A.

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.

1.         Reading on the vernier calipers

            0.5 + 0.01(5)               0.5 +0.05cm = 0.0055m/5.50mm.

2.         Third force F3 acting on the ruler is either upwards or downwards. 

PHY 001

                                                                                    No: My must be at the centre.

3.         Center of gravity rises when the body is tilted slightly and lowers when released / returns to original  position.

PHY 002

4.         Y must be below x

            Reason: P water is greater than  paraffin = height of water required is therefore less than that of paraffin.

5.         Cohesion between Hg molecules is greater than adhesion between Hg and glass molecules/cohesion force or adhesion. Force.

PHY 003

6.         (NB: with or without labeling one mark.)

7.         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

8.         Angle of rotation of reflected ray=2(angle of rotation of mirror)

                                                                =2 x 30 = 600

9.         Charge concentrate at sharp point causing heavy discharge/ ionization neutralization, leaf falls off.

10.       V = IR   Þ I = V/R   I = 3/! = 3A

            1/R= 1/R1 + 1/R2= 2/2

            1/R = 1=R=1

11.       4mm=20N

            1.5 =?                    F =    Ke

            1.5x 20            K = F = 20   = 5 x 103 N

            4                             e    4x 10-3

            = 7.5 N            F=5x 103 x 1.5 x 10-3

                                                            =7.5N

12.       -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.

PHY 004

14.       Moment of couple = Force x distance between forces.

            =10 x 2 = 20NM.

15.       F = Ma                        = 70 x 0.5        F  35N

            35N = 20a       a = 35              = 1.75M/s2

                                          20

16.       P = force x velocity     Power = Fd/t = 20 x 10x 20

            Mg x h/t = 20x 10 x 20/40                               40

                        = 100w                                      =  100j/I

17.       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

18.       Beat frequency f = f2 – f1                              F = f2 – f1

                                        = 258 – 256                         256 – 258

                                         = 2Hz                                 =/-2/ = 2

19.       P = V1 = 15000 = V x 2         W = QV but Q = It     e = I2Rt

                   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

20.       Heat lost by substance = heat gained by water

            MsCsrq1 = MwCwrq2

            2 x 400 x 60 = Mw x 4200 x 1

            Mw = 2 x 400 x 60 = 30 = 11.4kg

                            4200            7

21.       V = I(R +r)

             5 = 10 (R + 50)500 Þ R + 50 Þ R = 500 – 50 = 450W

                  1000         

22.       Apparent depth = 30 – 10 = 20cm real depth = 30 = 1.5

                                                                                      Apparent depth 20

23.       Kinetic energy ray / heat energy.

24.       – Horizontal acceleration is zero because g component horizontally is 0

            -Horizontal velocity remains constant

            – Resultant horizontal force is zero                             – resultant force is Zero.

25.       V2 is smaller than V1                                       V1 is larger than V2

PHY 005

26.

27.       P1 = 1.03 x 105            T1 = 20:C = 393K       V1 = V

            P2 =?                            V2 = 1/8V or v/8

            P1V1 = P2V2                1.03 x 105 – P2/8          = p2 = 3.24 x 105N/M2

28.       Radio waves, infrared, x-rays, Gamma rays.

29.       Up thrust = PV x 10 = 10 PV

30.       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.

31.       Tension = centripetal force.

            T = Mv2/r         but v = wr                   2 = 0.1 x w2 x 0.33

            T = Mw2r         t = 0.2 x 10 = 2N        2N = Mw2r                  2 = 0.1 x w2 x 0.03

            -w2 = 2/0.003  w Ö2000/3       w = Ö666.7      = 25.82 rads/s

PHY 006

32.       Object should be between F and lens.

33.       Downwards into the paper.

34.       A-earth wire                B – live wire                C neutral wire

35.       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 103

              F = 1/T = 50 Hz.                   1/20 x 10-3

PHY 013

ii)

2.a)

3i)     Average velocity at intervals AB and CD.

         T = 1/50 x 56               VAB = 1.5cm/0.1s        VCD = 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

PHY 014

    ci)

4a)       Figure 5 represents a simple voltage amplifier circuit.

 b  i)     Base current.

                                    Current gain =  Collector current        p2 = 1a/Ib

                                                                                  Base current

                                                            62.5 = 2.5 x 10-3

                                                                        Ib

                                                                Ib = 2.5 x 10-3 = 40uA                       (4×10-5)A

                                                            62.5

ii)         Load resistance, RL                                                      IcRL = Vcc = 5.5

            P.d across RL                                                                   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 = V2 = (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

                        Mw       = 1000                                     R.D = Wt. in air = 2.0 = 0.8

                        Vw                                                                                          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

  1. 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.
  2. Beaker becomes more stable because the position of C.O.G is lowered on melting or water is denser than ice.
  3. 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.
  4. 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 105 x 4 x 104 x 15 = W x 60

                                    W = 20N

  1. Particles of gases are relatively far apart  while those of liquids and liquids are closely parked
  2. 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
  3. 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
  1. Can withstand rough treatment

Do not deteriorate when not in use

  1. Struts are DE,  DC, AD, BD                          Ties are BC; AB
  2. The keepers become magnetized thus neutralizing the pole, this reduces repulsion at the poles, thus helping in retention of magnetism
PHY 033
  1.  

Force F2 at the ends perpendicular and turning to opposite to F1

  1. VR = 4;
  2. Efficiency of the  system

Efficiency       = M.A x 100               = 100  x 1 x 100 = 89.3%

                           V.R                               20     4

                                                = 89%

  1. Sound waves
  2. Let A’s represent current through the Anometers using Kirchoffs Law

A1 + A2 = A3

But                  A1 =  A2

So                    A1 = A2 = ½ A3

Similarly          A4 + A5 = A3

So that             A4 = A5 = ½ A3

So                    A1 = A2 = A4=A5

  1.             P = V2;                                    40 = 2402                     R = 1440W

       R                                      R

PHY 034
  1. Wire expands becoming longer (reduces tension) this lowers frequency hence pitch.
  1. Boiling point of spirit is lower than that of water. Specific heat capacity is lower than that of water.
  1. Fig 12 shows a  ray of light incident  on a convex mirror
  1. 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
  1. The air above paper travels faster than below causing lower pressure  above. Excess pressure causes paper to be raised.
  1. Combined capacitance                        = 1.5 μ F

= CV = 1.5 x 3            = 4.5 μC.

PHY 037
  1.  
  1. Microwave / cooker/ telephone/ radar etc
  1. U.V removes electrons from zinc surface so leaf will not only collapse if electroscope was negatively charged.
  1. Number of turns/ strength of magnetic field
  1. To reduce eddy currents in the armature
  1. Difference  in energy of the state/ nature  of atoms
  1.  X – rays produces                              – Hard X – rays  are produced
  1. 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

  1. (a)        (i) Convex mirror – driving mirror/ supermarkets mirrors
PHY 039

     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/f-1          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       

PHY 040

2. (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 L1is greater than that of L2 (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

3. (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 – 2730C) since tube is uniform L α T.

(iii)       The water bathy allows the air to be heated uniformly.

(c)        P1V1 = P2 V2               = 1.5.x 105 x 1.6 = 1.0 x 105 x V2

            T1         T2                                 285                  273

                                    = V2 = 23m3

4.         (a) (i) Easily magnetized and demagnetized

(ii)  Vp = Np                 240 = 500

       Vs   Ns                  Vs        50

                                    Vs = 24;           V= VPR

                                    VQP = 1/3 ;       VPR = 8 V

(b) Volume of A displaced = 6.0 x 12 cmcm3 or P = G * g

                                    Mass                            = 12 x 106 x 800          F = PXA

                                                                        = 0.0096 kg                 ans = 0.09N

                                    Weight = mg = 0.096N

(ii)        Volume of B displaced = 6.0 x 3        = 18 cm3

            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 10-6m3

            =657 kgm-3 can also be in g/cm3

5.   (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= Mw2r = 0.4

                                    W2 =  0.4                                 or F = Mw2r

  • x 0.08                                0.4 0.1 w2 x 0.08

W= 7.07rad/s                          W  = 7.07 rad/s

                        II         F= Mw2r = 0.1 x 7.072 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

6. (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 LL2 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) L1 A – L2 A = l

            From the figure L1A = 18.5cm + 0.1

            L2 A = 18 cm + 0.1

            L2A = L1 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.

7. (a)    Pure semi- conductors doped  with impurity  of group 3, combination creates a hole ( positive), this accepts electrons.

PHY 041

(b i)

(i)         At Ve E = 0

            Vcc = Ic R  L

            Lc = 9/1.8 K W l­c = 10

            VeE = V­cc = 9

(ii)        ∆lc = (see graph) = 3.5 – 1.2 = 2.3 mA

                                    B =      ∆lc

                                                ∆lc

                                    2.40A

                                    40 μ A

                                                = 60.

PHYSICS PAPER 232 /1 K.C.S.E 2001 MARKING SCHEME

1.         Volume removed = 11.5cm3

            Density =   mass    =   22   1.9cm-3

                             Volume      11.5

2.         Weight on side A has bigger volume when water is added.

3.         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.

4.         No air on moon surface / no air pressure / no atmosphere.

5.         When the permanganate dissolves / or breaks up into particles (molecules) these diffuse through the water molecules

6.         When rises up the tube into the flask or water is sucked into the tube or bubbles are seen momentally.

7.         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.

PHY 020

8.

9.         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.

10.       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.

PHY 021

11.

12.       F2 F3 or F1 and F4

13.       Moment of a couple = one force x distance between the two forces.

            Distance between F1 and F4 = 0.8sin 30o.       Moment = 0.8sin 30o x 100 =10NM

            Alternative (F2 and F3)                        Moment =  f x 1M = 60N x1M = 60nM(or J)

14.       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/s2

15.       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.

PHY 022

16.

17.       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.

18.       Current in heater =  p  =    3000  =  12.5A

                                            V         240

            Fuse not suitable since current exceed the fuse value.

19.       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.

PHY 023

21.

22.

PHY 024

23.

24.       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.

PHY 025

25.

26.       Fh = f1 – f2                 OR Fh = f1 – f2

            Fh = 6 – 4                    = 6.25Hz – 4Hz

            Fh = 2                          =2.25Hz.

27.       Longer radio waves are easily diffracted around hills/ radio waves undergo diffraction easily.

28.       Tension in A = 1.05N – 1.0N = 0.05N

            Tension in B = tension due to A + Tension due to B

PHY 026

                                     0.05 _+ 0.05 = 0.10N

29.

PHY 027

30.

31.

32.       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

33.       Pd across Anode – cathode                Or anode potential (voltage)

34.       r – b (Beta)or ie                       B = 82             C = 206

PHYSICS PAPER 232/2 K.C.S.E 2001. MARKING SCHEME

1.         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 (20-T) J

            Heat gained = Heat lost

            13600J + 168 TJ = 1680 (20-T)J                                             T= 10.80C

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)         Np  =  Vp                                 Vs =    240  x 20 =  12v

                        Ns  =   Vs                                         400

                                                            Vs = V/R = 12/50        =0.24 A

                                                            Is Peak = 0.24A x 2

                                                                        =0.34A

3.         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

1-pr

= 100x 10 x 10 x 2×2 x1o-4                             =  0.4kg

=   4N                                                              = 0.4 x 10     =4N

4.         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 = Ef – Ei = 5.44 x 10-19j = 4.08-19j

                                                            rE = hf          =          h C

                                                                                                   l      

                                                            l = 6.63 x 10-34 x 3.0 x 108m

                                                                             4.08 x 10-19

                                                            = 4.88 x 10-7 m (4.87 – 4.90)

PHY 030

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)

V2(M2/s2) 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 106 kgs-1

            OR      S = ½ at2

                        T = 100 x 2

                        T = 20 sec                                Momentum p = Ft

                        F = ma

                            – 200 x 1000 x 0.5 = 106

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(m3) 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 103 paM3

                        = 4.94 ± 0.65

             Slope = 4.94 ± 0.65

              Slope = 2RT

                        R = 4.87 x 103

                                   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 105 (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

1.         11.72/11.72 CM/0.01172M

PHY 015

2.

3.         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.

4.         R = V = 0.35 = 0.5W

                    I      0.70

            P = RA = 0.5 x 8 10-3 = 8 x 10-3W m.

                   C              0.5

5.         p          = F                               P = F

                        = 2500                                A

                          425,000pg                 Total press =                2500                =2,000N/m2

                        =250,000PG                                                    0.025              

6.         -Low temperature reduces K.E / velocity of molecules

            – Hence lower rate of collision / less collision -Reduction in pressure

7.         Can B                                                              Good absorber of radiation.

PHY 016

8.

19.       (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

20.       Mattress increases stopping time/time of collision increased this reduces the rate of change of momentum.

21.       C = C1 + C2                 Q = CV

            CT = 3×2         5mF      V = Q  V=1 x 10-4  = 20V

PHY 017

                                                       C

22.

23.

24.       V         =fl

            l          = v       =          330/30 = 11m

                            F

25.       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 103

            V = 1.838 x 10-5 m3 = 18.4cm3

                    1.839 x 10-5m3

PHY 018

28.

29.

30.       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

31.       107 – 42 = 65

32.       Penetrating power

33.       Downwards

34.       Work function of metal / min energy required to eject e-1 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 108 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         = 38.80– 38.48

                                                    38.7 – 38.42

c)         Sin q = 1.596

            sin 21.1

            Sin q = n

            Sin 21.1

            q = 35.250 – 35.151

            35.350– 35.211

2.         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)

PHY 028

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 100                                             = 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/S2

(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 40C. 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 103+ 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)         1st resonance l I1fe                 l          = I12-I2                        OR V= 2F (I2-I1)

                                                4                      2                        f=        

                                                                                                                        V                 

                                                                                                            2(I2-I1) 129-77

                                                                        l= 129-77

                                                                        2

            2nd resonance 3l = I2+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 ToC is plotted; graph is straight line cutting T at – 273oC (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 104 pa

ii)         p1 = 1.15 x 105 pa                   q1 = 52.0oC     

            p2 = 1.25 x 105 pa                   q2 = 80.0oC

                                    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 105

                                           40

                        = 14000/40         350pac ()

            KT       =          Constant

            C         =          9.6 X 104

            350 T=          9.67 x 104

            to         =          274.3 (266-284)

5.         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 1015

                                    H = 6.7 x 10-34 fs

                        At Kemax = q hf = 0

                        Extrapolation shown or

                        Read off fo = 1.07 x 1015 Hz

                        Q =1.07x 1015x 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 1014

                                    Or        q  =      2.5 x 1.6 x 10-19

PHYSICS PAPER 232/1 K.C.S.E 2003 MARKING SCHEME.

PHY 016

1.

2.         30.0 + 0.5 = 30.5         (No mark if working not shown)

3.         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 104 x 0.7

                = 9.52 x 104 or 95200 Nm-2   Allow g = 9.8m/s2 (follow through working)

4.         Air molecules are in continuous random motion. They bombard / knock / collide with smoke particles

5.         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.

7.         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.

PHY 017

8.        

            R = I = 90 = 40o

                        Or

            R = 180 – 100 – 80 = 40o

                     2                  2

9.         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)

11.       Temperature

12.

PHY 019

13.       It does not retain magnetism / Iron is easily magnetized / demagnetized / Iron enhances / strengthens magnetism.

14.       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.)

PHY 020

15.       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.

17.       P = V2 / R                                P = VI = I2R

            75 = 240 x 240            or         Do not accept p = VI alone without I2R

                           R                              R = p/12

            = 768 W                                   R = 75 x 240 / 75 x 240 / 75 = 168 W

18.       Beta particle b (Do not a ward for beta ray)  Beta radiation Beta emission

19.       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.

20.       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.

21.       Speed = distance / time           speed = 600m/s

            = 1200                         1.75

                  3.5                         =343 m/s

            = 343m/s                     (Range 342.8  – 343 m/s)

PHY 021

22.

23.       Circuit A

            Current draw from each cell is less than in B / In A there is les internal resistance.

PHY 022

24.      

25.       To with stand high temperature / high melting point.

PHY 023

26.

27        Fringes will be closer together / more fringes of violet light has a shorter wavelength Red light has longer wavelength.

28.       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.

PHY 024

29.

30.       F = ma             a2 = a1                         Accept F = ma for formula mark

            F = 2 ma2                 2                          a2 = a1m

            2ma2 = maf                                                   2m

31.       Radio waves, Infrared, visible light, U.V light, X-rays (accept correct order)

32.       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.

PHY 025

33.       Maximum deflection of G will be double; flux linkage doubles when the turns are doubled.

34.

35.       Q = hf0 =Wo or & = hfco

                          = 6.63 x 10-34 x 9.06 x 1014j

                          = 6.01 x 10-19 J or 6.0061 x 10-10 or 6.0 x 10-19 if working is shown.

36.       Fast air causes low / – reduced pressure at the top. So there is net force upwards on pith ball / pressure difference pushes pith ball upwards.

37.       Parallel C = (1.3 + 0.7 0)mF = 2.0 mF or 2 x 10-6F

            Series 1 = 1 + 1 =1

                        CT 2.0 / 2.0

                        CT = 1.0 mF // 1.0 x 10-6 F.

PHYSICS PAPER 232/2 K.C.S.E 2003 MARKING SCHEME.

PHY 035

            (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 t2                                  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;           = 30ms-1                                 (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   = mv2

                                                                                    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 106 J kg -1

4.         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;

5.         a)         e.m.f is total work done in transferring unit charge from on terminal of battery to the other;

PHY 036

            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

                                                1.0-0.5

                                                = 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)        

PHY 038

            (i)         C is intercept and C= O;

                        K is gradient given by

                        Gradient = 15.2 x 10-4 x 10

                                                400-105

                                          = 11.2 x 103

                                                   295

                                          = 37.97 pak-1

           (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

7. 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

1.         15.5 + 0.33 = 15.83mm/1.583cm

2.         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.

4.         Density of water is low/It will result to a very log barometer/ very long tube

PHY

5.

            NB at 40 c graph must be curved

            – 40 must be marked

            – If drawn using a ruler N0 mk

            – If 200 c is marked, it must be higher than 00c          

6.         Wooden Block

            Wooden block is a poor conductor of heat all the heat goes in melting the wax.

7.         NB- Check correct rays with arrows.

                 – at least one angle on each reflecting surfaces must be marked.

PHY 001

8.

9.         To depolarize/ oxidizer/ reduces polarization/oxidizes H2 to H2 to H2 0/Changes H2 to H20/ removes H2 (any give 1 mark)

10.       Adding detergent/Impurities/increasing temp/heating (Any give 1mk)

11.

PHY 002

Check

-correct pattern

– correct direction

NB- at least 4 lines of forces must be shown

  • Lines of forces must start at the poles.
PHY 003

12.

NB forces must be straight

Lines must touch a conduct

13.       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)

14.       Sum of clockwise moment=sum of anticlockwise moments

            Wx20= 30×5

            2w=15

            Higher, reducing the current.

16.       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)

17.       Distance= Area under graph

                          = 2x 1/2 x2x 20

                           = 40m

                        Or s = ut+ 1/2at2

                             S=2(20) + ½(-10)4

                              S= 20

                              S=2×20

                              40m

18.       W= Fd

                   Mg sin q

                 = 60x 10×0.5×4

                 = 1200J

19.       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

20.       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.

21.       E=pt t=450- 150 =300s

            E= 50×300

            1=  150,000J

22.       Q=ml

            15000=0.1×1

            1= 150,000J/kg

PHY 004

23.

24.       -Correct rays must be refracted to the eye and should be diverging.

-Dotted lines should show image position.  (-should not have arrows-must intersect within container)

PHY 005

25.       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.

26.       Either on closing on closing s1 while s2 open

            Q= CV = 3C

            When s1 is open s2 closed charge is shared between the two capacitors

CT=C+C=2q

Since q is the same equal to 3C1 the new pd=V1

Q=CTV1 =3C

V1           =1.5V

Or

S1 ­closed S2 open lower capacitor charges to 3V

S1 open S2 closed lower capacitor charges the upper to same charge (p.d)

Final pld = 3/2 V = 1.5V

Or

Q=CV=3C

S2 closed charge is shared

CV= Q/2

V= QC/2C = 3C/2C= 1.5V

27.       Either V1/T1=V2/T2

            200/293=V2/353

                        V2=241ml

Or V= KT

200=293K

K=0.6828

V2=0.6828 x 353

V2= 240.96 ml

The other answers for

V2 240.9/240.94ml

28.                   X-rays                                                             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)

29.       T=Mv2/r                                  or T sin q -mv/r            or  tanq =V2/rg

            81=MV2/r                               q=86.46                       V2=0.499×86.4

            81=5V2/0.5                                 r=0.488                        V280.63

            V=9m/s           V2=0.499×81 x 0.9981/0.5                  V=8.979 m/s

                                    V2=80.70

                                    V=8.983m/s

PHY 006

30

NB.     At least three ware forms must be drawn.

            Ware length (spacing) must be maintained

PHY 007

31.

32.

            Check-at least Three complete troughs/Crest

  • Amplitude range 6.5 squares

   8.5 squares

33.                   X-rays (Hard)                                                  Soft-rays

                        -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

34.       hf0=Wc=q

Fo        = Wc/h=32×16 x10/6.62×10-34

                        = 7.73 x 1014 H2 or 7.732 x 1014 H2 or 7.734 x 1014 H2       

=          7.73 x 1014H2 or 7.732 x 1014H2 or 7.734 x 1014H2   

PHYSICS PAPER 2 2004 MARKING SCHEME

PHY 008

1.         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.  

PHY 009

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

                                                = 27.9cm3

Density = Mass/Volume         = 4.9/27.9g/cm3 = 1.760kg/m3

2.         a)         i)          Mass m1 of melted ice/mass of water. Time t1 take

                        ii)         Q=m1              Vit=ml

                                    P=ml/t               p=Vi=ml/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=4/3Wr3 =Wr2h(r=radius of drop)

                        Volume of patch =Wr2h (h=Thickness of molecule)

                        4/3 Wr3=Wr3/Wr2h (equating)

                        H=4/3Wr3/Wr2+2=4x(0.25)3/3x(100)2 2.1×10-6mm

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 4/3Pr3 = 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.

 3. 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)

PHY 010

     ii) Delayed 1×1020___________ ½ x 1020 – ________ ¼ x 1020 _______ 1/8 x 1020

                        = 0.125 x 1020 = 1.25 x 1020

                Left   1 x 1020 ———– 1/8 = 0.875 x 1020

            (Subtraction )              = 8.75 x 1019 Atoms.

4. a) i) 0.30cm

        ii) 0.65-0.25=0.4 Sec.

        iii) f=1/T = 1/0.4 2.5 HZ

        iv) V= fx=V/x =200/25 =80cm = 0.8m

b.         i)         

PHY 011

                        ii)         m= ht of Image = distance of image

                                          ht of object     distance of object

                                    h0/200  = 25/5 h0= 200×25/5  = 100m

5.         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 (H3-1) 3.91 3.5 2.9 2.3 2.1 2..0

                        12-11 0.65 0.57 0.48 0.39 0.34 0.32

                        ii)         Slope (Gradient) = V/2 = (0.67-0.10)m/4.0-0.75)x 10-3H3-1

                                                                                    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.

7.         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 = 1.28-0.10/(7.7-4.8) x 10 14

                        h= Slope x e

                          = 1.18 x 1.6x 10-19/29 x1014

                          = 6.51 x 10-34JS

                        (5.82 – 6.66) x 10-34 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 2014

Work function Æ = 6.51 x 10 -34 x 4.55 x 10 14

                        = 2.96x 10-19J

Range (2.56-3.06) x 10 -19J

c)         ½ mv2 max = hf-Æ

            hf= 6.51 x 10x 3×1015

            KE max = 1.953 x 10-18 – 6.4 x 10 -19

                          = 1.31 x 10-18

            Range (1.12 – 1.31) x 10 -18J

PHYSICS PAPER 232/1 K.C.S.E 2005 MARKING SCHEME

1.         Volume of 55 drops                =8ml                accept cm3

            Or Volume of one drop          =8/55

                                                            = 0.1454/0.1455/0.145/0.15cm3

PHY 019

2.

3.         Water in A expands reducing/lowers density

            This reduces/lowers up-thrust on block causing tipping to side A

4.         There is extra/ more/higher/ increased pressure in (b) due to the wooden block increasing distance d2

PHY 020

5.         Reduce/ minimize the transfer of heat by radiation OR Reduce the loss of heat OR gain of heat by radiation.

6.         2 sec of rays with arrows labeling of umbra (totally dark) and partly dark (Penumbra)

7.         A or tube with air

PHY 021

            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

8.        

9.         A-Positive

            B-Negative

10.       C- Ammonium jelly/chloride /paste/solution/NH4Cl

            D-Mixture of carbon and manganese (iv) oxide/MnO2

11.       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.

PHY 022

12.      

13.       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.

14.       S=nt+1/2 st2 where t is the time to reach the ground

            15= 0 + ½ St2 since the initial velocity is zero and t= 3 = 1.732

            Horizontal distance= Horizontal speed x t = 300x 3 o 519.62m

15.       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

16.       =4cm or 0.04m from the graph

       V   = fl= 5 x 0.04

            = 0.2ms-1 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.

PHY 023

18.                                                                   Accept cells in parallel and other symbols of rheostats

19      (i)

                                                            = V2/-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 I2  R

            R=  240 x 240

                    2500

             R= 23.04R

20. The liquid is boiling

PHY 024

21.

PHY 025

22.                                                       2 Rays: Each correct ray 1 mark

                                                            Image position 2.2 +0.1 cm from mirror.

23.  C=470-100 =370 + 7=370

24.    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.

Text Box: Velocity
Text Box: momentum

25.

26.      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.

           2.  Provided a finish is non-viscous, incompressible and its flow steamline and increase in its velocity produces a corresponding decrease in pressure

           3.          When the speed of a fluid increases, the pressure in the fluid decreases and vice versa.

27.                   273+ -281.3 = 8.3K (accept – 8.15 was use.)

28.

PHY 027

                        (i)                                                         (ii)

29.                   (i)         F = MV2/r

                                                4800 = 800 x V2

                                                                20

                                                V = 10.95m (allow 10.09 of a slide is used)

                        Alternatives.

                        (ii)        Vmax = √Mrg but         Fr = Mμg

                                                                        M = Fr             =          4800   

                                                                                Mg                      800×10

                                                                                                = 0.6

                        (iii)       F = Ma

                                    4800 – 800 x a,           a = 6m/s2        

                                    A = v2/r

                                                OR

                                    6 = V2/20

                                    V = 10.95

                        (iv)       F = MR, M = F/R =    4800   

                                                                        800 = 0.6

                                    Tan ө = 0.6

                                    V2 = rg tan ө

                                                OR

                                    V2 = 20 x 10 x 0.6

                                    V = 10.95

30.                   Image changes from real to virtual

                        Image changes from inverted to upright

                        Image changes from behind lens to the same side as object.

31.                   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.

PHY 028

32.                  

33.                   To withstand the high temperature (immerse heat) prevent the target from melting due to high temperature or immense heat.

34.                   Methylated spirit evaporates faster/highly volatile than water taking latent heat away faster from the hand.

PHY 029

35.                  

36.                   m- Alpha (µ) particle/ radiation/decay

                        n- Beta (b)

                         x- Polonium (Po)                       

37.                   When the switch is closed and nails attracted.

                        When the switch is opened, the nail on the iron end drops first.

38.                   Conductor allows charge to be distributed/movement/spread.

PHY 030

PHYSICS PAPER 232/2 K.C.S.E 2005 MARKING SCHEME

PHY 006

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

PHY 007

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

PHY 008

(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)

2. (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

3.         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 10-6C

4. (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 D2 and D3 to be forward 

biased, so current flows  through  D2 R and  D3 to  B.

In the second  half – cycle, B is positive making  D4 and D1  forward  biased. The  current flows through D4 R  and  D1 to A

PHY 009

(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

              ∆IB

     120 = ∆Ic

              20B/A

Therefore ∆Ic = 120  x 20 MA = 2.4mA

Output p.d charge       = RL x ∆IC

                                    1000R x 2.4 mA

                                    = 2.3v

5.         (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 10-4m2

                                    = 5.8 x 104 Pa

NB:  can work with N/cm2

Accept 5.6 – 5.8) x 104 pa

            (iii) Strain        = Ext                           = 5       = 0.025

                                    Original  length           200

(c) ED and DC

6.   Angular velocity is the ratio of angle covered (angular displacement) to the time interval

or W = θ2 – θ1

                               t2 – t1

(b) w = 300 – 170 = 10 radis-1

                         13

                        10t = 170

PHY 010

                        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

PHY 011

7. (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

PHY 012

K.C.S.E 2006. MARKING SCHEME

PHYSICS PAPER 1

1.               Volume = 68cm3

                  Mass    = 567g

                  Density = m = 567

                                   V      68

                  = 8.34 gcm-3                                                                (3 marks)

PHY

2.

3. 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)

  1. 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)

5.   Lower spring extend by 15 cm;

      Upper springs extended by 10 cm;

      Total = 15 + 10 = 25 cm                                                                     (3 marks)

PHY 001

6.

7. 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)

8. 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.

PHY 002

10. At F, radius of curve is smallest and so greatest centripetal force is required to

keep luggage on carrier; ( F =mv2)                                                      ( 2 marks)

                                                       R

11. A1V1 = A2V2;

      π x 62 x V1 = π x 92 x 2;

                  = 4.5 ms-1                                                                                ( 3 marks)

12. 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)

13. Ft = ∆ mv;

      720 x 0.1 = 0.6 x v;

                  = 120ms-1                                                                                ( 3 marks)

14. (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 r3

                  = 4/3 π x 0.0253

                  = 6.54 x 10-5 cm3                                                                     ( 2 marks)

      (ii)        Area = π r2

                  = π x 102

                  = 314 cm2                                                                                ( 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)

15. (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 ± 0.1
  • (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)

16.       (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 105

                        ( 3.5 – 0) x 106

                 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

                        V1        = V2

                        T1            T2

                        T1 = 273 + 37 = 310k

                        T2 = 273 + 67 = 340k

                        4000    = V2

                        310      340

                        V2 = 4387 litres                                               ( 4 marks)

17.       (a) A body fully or partially immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced                                                         ( 1 mark)

PHY 003

(b) (i)

(ii)                    100g:   Uw = 0.12N     Us = 0.09N

                        150g:   Uw = 0.18N     Us = 0.14N

                        200g:   UW = 0.24N     Us=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)

18.       (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 ( T-0)

                  = 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 ( 60-T)

                  6680 + 84T = 50400 – 84OT + 4320 – 72T

                  996T = 48040

                  T = 48.20C                                                                   ( 4 marks)

K.C.S.E 2006: MARKING SCHEME

PHYSICS PAPER 2

PHY 004

1.

2.   Magnification  =                                 

                              Im age dist = ht of image

                              Object dist     height of object

                              10        = 16

                              600          h

PHY 005

                              H = 9.6 m                                                        (3 marks)

3.

4. To allow escape of gases ( H2 and O2) from battery

5.    (i) Longitudinal wave

PHY 006

      (ii) Length of the spring, from one point to  a similar point of vibration

PHY 007

6.

PHY 008

7.  

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)

8.         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)

10.       Work function of a metal is the minimum energy required to set free (release) an electron from the surface of the metal                                             (1 mark)

11.       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)

PHY 009

12.       Straight beam from gun to screen OR no gravitational effect on the beam. ( 1 mark)

13.

14.       Resulting X- rays have shorter wave length/ hard/ high frequency because electrons have higher K.E                                                        ( 2 marks)

15.       a = 234 + 4 = 238

PHY 010

            b = 92 – 2 = 90                                                                        ( 2 marks)

16.

17.       (a)        Charge Q, on C1 is given by

                        Charge Q1 = C1 V;

                        = 0.3 μ F x 4.5;

                        1.35μC;                                                                       ( 3 marks)

            (b)        CT = C1 + C2;

                        = (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 C1 is distributed to C2. Since values of C1 and C2 remain constant, when Q on C1 reduces, then Q = C1V implies V must reduce also, hence voltmeter reading reduced.

PHY 011

18.       (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)

  1. (a) Refr. Index n = sin i   velocity in air

      Sin r   velocity in substance

OR

n = Real depth

      Apparent depth                                         ( 1 mark)

PHY 012

            (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 = 38.70 = critical angle         ( 3 marks)

                        Sin θ                   16

20.       (a) (i)   P = slip rings

                        Q = Brushes                                                                (2 marks)

                 (ii) 0-90 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 = Ns; Þ €s = 240 x  60 = 12 volts                             ( 2 marks)

                      €p     Np                      1200

                  (ii) Pp = Ps (power) or ls Vs = lp Vp

                        IS = Ip Vp = 0.5x 240; = 10A;

                                    Vs            12                                                    ( 3 marks)

21.       (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/mm3                   (0.550)3cm      substitution = 10.5g/cm3                         answer            – allow transfer of error 10500kg/m3 3 mks
3. V2V4 V1 V3 ( 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 PA ghA = PaghB                     formula         or correct PAg x 24 = 1200 g x 16         substitute     substitution Pa = 800  kgm-3                     answer         answer 3 mks
6. Radiation 1 mk
7. X2 is made greater than X1 / X1 is made shon X2 X2 is made larger than X1 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.                             F1 d1 = f2 d2 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/S2 17 x 0.2 (12 – 14.5) m/s2 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   V2 – u2 = 2as V2 – 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 = mlv + mc ∆θ         = formula Heat lost by steam = 0.003 x 2.26 x 106 = substitution Heat lost  by steam water = 0.003 x 4200 ( 100-T) 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 ( 100-T); Allow  transfer of error 1680T – 16800 = 6780 + 1260 – 12.6T 1692 .6 T = 24840 T = 14.70C           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 = MV2            V=rw For thread to cut         r                   w = 3.049 F= 5.6  N               5.6 = 0.2 x v2             0.15 (l) = 13.7 radius      V2 = 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

PHY 031

1.

2.         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)          

3.         X is north        (both correct)

            Y is north                                                                                            (1 mark)

PHY 032

4.

5. T = 0.007S              (T)

             3

    F = l/T = 3/0.007 ( f)

            = 429Hz 428.57 – 434. 80H2                                                               (3 marks)

PHY 033

6.

7.

PHY 034

8.         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)

9. R1 = V2                    R2 = V2;

              P                            8P

     R1 = V2 x 8P

    R2      P     V2

            = 8                                                                                                       (3 marks)

10.  The process of the eye lens being adjusted to focus objects at various distances

(1 mark)

PHY 035

11.

12.       The higher the intensity implies greater number of electrons and hence higher saturation current                                                                            (1 mark)

13.       a = 234

            b= 82

14.

PHY 036

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

PHY 037

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

17.       (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) Ip = 600/400 = 1.5A                                                                      ( 2 marks)

(iii) Ps = Pp = 600W

            ls = 600/40 = 15A                                                                      ( 1 mark)

18.       (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

= 6.25Hz

PHY 038

(iii)

K.C.S.E 2008 EXAMINATIONS

PHYSICS PAPER 1

MARKING SCHEME.

1.         Water              V= Mw           or MW = ML              RD =   ML  = P

                                            I                                P                             ML

2.         For liquid        V= ML                        P = ML            P = ML

                                            P                                 MW                 MW

            P = ML

                   MW

P 013

3.         (a)

            b)         R – Increases   OR      R – Approaches W

                        F – Reduces                F – Reduces

4.         – Atmospheric pressure is higher than normal/ standard or boiling was below

– Pressure of impurities

5.         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)

6.         Heat conductivity/ rates of conduction/ thermal conductivity (NB: If heat conduction no mark)

7.         X sectional area/diameter/thickness/radius

8.         P1 = 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 104 pa

            Total pressure

            =          8.58 x 104 pa

                        (85800pa)

9.         – Intermolecular distances are longer/ bigger/ in gas than in liquids

            – Forces of attraction in liquids are stronger/ higher/ greater/ bigger/ than in gases

10.       (In the diagram)

P 014

11.       Stable equilibrium

            When it is tilted slightly Q rises/ c.o.g is raised when released it turns to its original position

12.       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.

13.       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

14.       Heated water has lower density, hence lower up thrust

15.       (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 at2

                                    49 = 0 + ½ x a x 72

                                    a = 2M/S2

                        (ii)        V = u + at        or         v2 = u2 + 2 as

                                    = 0 + 2 x 7 = 14m/s     v2 = 02 + 2 + 2 x 2 x 49

                                                                        V2 = 14m/s

            (c)        (i)         S= ut + ½ gt2               either V2 = u2 + 2gs

                                    1.2 = 0 + ½ x 10 x t2              v = u + gt

                                                                        V2 = 02 + 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 M1

                        Final mass of water & calorimeter, M2

                        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 103 (34-25) = 3780J

                        (iii)       MmCMDT                  or sum of (i) and (ii)

                                    = 150 x 103 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

17.       (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 m3 /40 x 10-6m3 / 40cm3

(ii)        -2750C – 2800C

  • a real gas

Liquefies/ solidifies

            (d)       P1 V1 = P2 V2 but V1 = V2      If  P = P2 is used max marks 3

                            T1         T2                                T1   T2

                        P2 = P1T2 = 9.5 x 104 x 283    P2 = P1 T2

                                    T1                        298               T1

                                = 9.02 x 104pa             = 9.5 x 104 x   283

                                                                                    298

                        = (90200pa)                             (90200 pa)

                           (90.2 x 103 pa)                      (90.2 x 103pa)

18.       (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 103

                                      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%

19.       (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 = 180cm3                 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

P 015

                                                                        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

1.         BC       – Total absence of light; umbra, completely dark

                        – Total darkness

                        Rays are completed blocked from this region by the object

2.         Leaf in A falls a bit while leaf in B rises a bit

            The two leaf electroscope share the charge

P 016

            Correct circuit.

3.

4.         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.

P 017

5.         Needs not be dotted

6.         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)

7.         Movement equals 1.75 oscillations

            T          = 0.7/ 1.75

                        = 0.4 sec

                        F = I/T

                        = 1/0.4 = 2.5 HZ.

P 018

8.

9.         (i) V =                         O volts

            Reason                        No current

            (ii) V = 3 volts

            Current flows in the resistors

10.       P = V2/R            P = 220Ù2/240Ù2/100

            R = 2402

                   100

            = 84 J/S

11.       Short sightedness/ myopia

            Extended eyeball/ lens has short focal length/ eye ball too long any two

12.       Spot moves up and down

13.       Frequency increases

            Accept             Becomes hard

                                    Wavelength decreases

                                    Strength / quality

14.       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

15.       (a)        Temperature

                        Density

  • Graph
    • 46.5 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

16.       (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 = 49.80

                                    Angle < 49.80

  • X = 900 – θ

= 40.20

  • Sin θ/ sin X = 1.31

Sin θ = 1.31 sin 40.20

= 0.8460

= θ= 57.80

P 019

17.       (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

18.       (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 = IsVs

                                    Vs           Ns     Is = 800

                                                                     40

                                    400 = 200

                                    Vs      200

                                    Vs = 40 Volts  = 20A

                        (ii)        Pp         Ps

                                    800 = 400 Ip

                                    Ip = 800

                                           400

                                    = 2A

19.       (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 10-34xf

                        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 1018Hz

                        Accept ev = Gf

                        F = ev/g

K.C.S.E PHYSICS YEAR 2009

1.         Volume run out= 46.6 cm3

            Density = m/v = 54.5 / 46.6 = 1.16953

            = 1.17g/ cm3

2.         T2 = 4 Π 2L/g

            = 1.72 = 4 Π2 x 0.705

                                    g

            g= 9.63m/s2

3.         Needle floats due to the surface tension force

            Detergents reduces surface tension, so the needle sinks

4.         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.

5.         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.

6.         Prevents/ holds, traps breaks mercury thread/ stops return of mercury to bulb  when thermometer is removed from a particular body of the surrounding

7.         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)

8.         Heat travels from container to test tube by radiation so the dull surface P, gives more heat to the test tube.

9.         Center of gravity located at the intersection of diagonals

10.       Parallel

            F= 2 ke

            40= 2 x ke

            E1 = 40/2k = 20/k

            Single = f= ke2

                        20 = ke2

                        E2 = 20/k

            ET = e1 + e2

            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

11.       Air between balloon is faster that than outside so there is pressure reduction between.

12.

Displacement  
0  

Time

13.       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

14.       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)

15.       (a)        – Pressure

                        – Dissolved impurities

            (b)

            (i)         BPt = 780C

            (ii)        (I) ∆t = 4.5 min

                        Q = pt = 50 x 4.5 x 60J

                        = 13500J

            (II) Q = 70  – 16 = 540C                      (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 105J/kg

                    0.18

16.       (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%

17.       (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 0.2m3

                        Weight = mg = 0.3 x 0.2 x 0.2 x 10500 kg/m3 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 cm3 Density m/v = 0.008/50 x 106 m3

18.       (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 105 Pa

                        (II) I/v = 1/1.15 = 0.87 cm-3

            (iii) ∆ P = (1.88 – 0.8) x 105 pa = 1.08 x 105 Pa

                        ∆P =  ℓgh = ℓ x 0.80 x 10

                        P = 1.08 x 105 kg/m3

                                    0.80 x 10

                        = 13500 kg/m3

            (iv)       Pressure at top = atmospheric

                        0.8 x 105 pa

c.         p1v1/T1 = p2v2/T2              = 2.7 x 105 x 3800 = 2.5 x 105 x v2

                                                       298                          288

            250C = 298 k                           = 3966 cm3

            150c = 288k

19.       (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) v2 = u2 + 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

                                             0. 14

PHYSICS PAPER 2 YEAR 2009

SECTION A

1. Infinite ( very many, uncountable, several

2.

3. Negative change

4. Allow gassing/ release of gases

            OR, release H2 and O2 produced at the electrodes

5. Increase the magnitude of l

            Increase the number of turns per unit length

            Use of U shaped iron core

6.         F = 0.5 sec

            F = 1/T

            = 1/0.5

            = 2 Hz

7.         1.33 = 3/v x 105

            V= 3 x 105

                  1.33

                        = 2.26 x 108 m/s

8.         T = lA

9.         (L-q) cm

10.       (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).

11.       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

12.       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

13.       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)

14.       (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 QB = 5/8 x 32

              = 32 x 106                              = 20 PC

                  8 x 106                               VB = 20 x 10-6

                = 4 V                                            5 x 10-6

                                                            = 4V

15.       (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

16.       (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

17.       (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.

18.       (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 = lf0

            F0 =     3.0 x 108

                        8.0 x 10-7

            = 3.75 x 1014 Hz

            (ii) W = hf0

            = 6.63 x 10-34 x 3.75 x 1014

            = 2.49 x 10-19J = 1.55 e V

  • x 10-19

            (iii) KEMAX = hf – hf0

            = h (8.5 – 3.75) x 1014

            = 6.63 x 4.75 x 1014

            = 3.149 x 10-19 joules

                = 1.96828 e

19.       (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

PHYSICS

 PAPER 1 YEAR- 2010

MARKING SCHEME

  1. 1.62cm / 1.62
  2. 2.53 + 0.5 = 3.03s (working must be done)
  3. Air (Molecules ) expand by heating

Pressure inside is less than atmospheric pressure

  • Flame heats air which expands /becomes less dense /and more upwards

This will push the blade upwards /creates convention currents hence rotate

  • Flask which is in contact with heat expands first

Liquid expands more than glass

  • Clockwise moments = anticlockwise moments

Or W1d1 = W2 d2 / F1d1 = F2 d2

W x 0.2 = 2 x 0.25                             Either

W = 2.5N

  • Water flows faster in Y than X hence pressure is lower at Y than X (ie 1st  mark – compare velocity 2nd mark – compare pressure
  • a). 8N

b). 14 – 8 = 30a or F – Ma

      a = 6/30 = 0.2m/s2

  • Patch is one molecule thick or monolayer
  1. a). u = 5.0 – 4.04 (working must be shown)

     u = 0.96N

b). Weight of liquid displaced = 0.90N

Mass of liquid displaced = 0.96kg

V = M = 0.096 = 1.2 X 104m

        P     800      1.2 X 102cm2

                           120cm3

  1. Volume decreases, so more collusions per second
  1. F = mw2 = mg.           or        F = Mv2/v  but V = wr

0.2 x 1x w2 – 0.5 x 10          w2 = f   =  0.5 x 10

        W2 =   5                                  mr      0.2 x 1

                  0.2                              w = 5rad/s

W = √25 = 5 Rad/sec

  1. Newton per metre
  2. Increase the base area or lowers the centre of gravity

SECTION B

  1. a) Potential energy               Kinetic energy            heat + sound (sound not a must)

b).i).  Work done by force = Fd = 200 x 22.5 = 4500J

   ii) = mgh = 30 x 10 x 7.5 = 2250J

iii)  Work done by force – work done on mass = 4500 – 2250

                                                                                 = 2250J

iv). Eff = work output x 100% or eff = work output

                work input                                 work input

2250 x 100% = 50%                                           = 2250 = 0.5

4500                                                               4500

                                                            EFF – MA x 100%    1.5 x 100% = 50%

                                                                        VR                  22.5

                                                                                                7.5

c). Reduce friction by use of rollers/smoothening (polishing /oiling surface)

method of reducing friction must be stated

  1. a). = 66.4 – 43.2 = 23.2g

b). = 23.2g = 23.2cm3

1g/cm3  (NB: Working must be shown)

c). 23.2cm3

d) (67.5 – 43.2)g: 24.2g (working must be shown)

e). 82.3 – 67.5 = 14.8g (working must be shown)

f). Volume of sand = volume of bottle – volume of added water

= 23.2 – 14.8

=  8.4cm3         

g)  P = M        = 24.3g = 2.983g/cm3

     V                 8.4cm2

                                                                                NB at least 2 dec .places

  1. a). At high attitudes pressure is low so boiling point is low s o pressure cooker increase pressure inside it which raises boiling  point

or  pressure inside the cooker is higher raising the boiling point)

b) i). Q = MCDE or MCO or MCDT

          = 3 x 4200 x 80 = 100800J

    ii). Q = CO/CDO/CDT = 450 x 80

                                        = 3600J

iii). PL = MCDO + CDA   L = 34.85

       3000t = 1008000 + 36000

        3000t = 1044.000

iv). Either

                                                                        or

      Mlv = Pt                                                   MlV = Pt

3 x 2.3 x 106 = 3000t                                  3 x 2.3 x 10-3 = 3000t

Therefore t = 2300s                                                 t = 2.3 x 10-6S

(38.3 minutes)

  1. a). V = M                                                       or V =  4

             P                                                                     3000

                                                            V = 1.33 x 103m3

                                                            At least 2.    d. p

b). Upthrust = weight of liquid displaced = Vpg

                        = 800 x 1.33 x 10-3 x 10

                        = 10.64N

                                    OR

      Upthrust = weight of liquid displaced = Vpg.

           = 1000 x 1.33 x 10-3 x 10

          = 13.33 N

c). Weight of stone in all = 4 x 10 = 40N

   Reading of spring balance = 40 – 10.64 = 29.36N

                                    OR

            40 – 13.33 = 26.67N

d) 85 – 10.64 = 74.36N                               OR 85 – 13.33 = 71.67N

19.a) i).  Body moves with constant acceleration/increasing velocity or velocity

               increasing uniformly with time either

      ii).  Body moving with decreasing/reducing acceleration

     iii).  Constant (uniform) velocity / zero acceleration

b. i)     V = u + at

            V = 10 – 2.5 x 1.5

            V = 6.25m/s

   ii).    S = ut + ½ at2

            S = 10 (1.5) – ½ (2.5) 1.5)2 = 12.1875m

                                                            =  12.19m

PHYSICS YEAR 2010 PAPER 2

  1.  
  2. Reflected ray rotates 2 x 20 = 200
  3. Find deviation = (800 + 200) = 1000
  4.  
  5. Any slight deviation of the N – Pole to the right

3.

(2marks)

4. Initially attracted because of the charge Vi(+ve or –ve)

Then neutralized and charged positive and hence repel

Charging by contact and law of electrostatics

5. Distance = 2f = 2 x 25 = 50cm

Alternative

Just 50cm

OR

2 x 25 = 50cm

6. Implies low current. So reduces heat losses/power loss of I2R loss reduced

P = I2R should be accompanied by power loss

Nb: Heat/losses/power loss

7. More practice /relationship between and f and t

8. V1 = Ft1      or   η = Tt

                                T1

V2 = Ft2          or    η = 18  = 1.25

                                  14.4

Accept all expression

9. 20g         10g – 5   5g  2.5  5         1.2 g

Mass remaining

10. IO – initial current

                                         12 = 7IO

P = I2R = I2OR                 P = (7IO)2R = 49I2OR

Power is 49 times the initial value

Apply the power formula 

11. Motion out paper/moves upwards or increases in p d increases heating effect.

12. Increasing the accelerating voltage OR increase the p d between anode and cathode.

13.  F = V = C

              λ      λ

    = 3 x 108     13.0 x 105Hz

         1000

14. Look for biasing only. (any other device that does not affect the working should be ignored) e.g diodes resists

15 a) i). current falls off to zero 1√ / falling to zero/deflects to max. Then zero. Reducing gradually or after sometime.

ii). Current flows when the capacitor is charging 1√ 

When fully charged current stops (No current) and p . is equal to charging voltage 1√ /

b). Vc = 5V1√

c). Touch both axis, award for no labeled axis

d). i). 1 = 1 + 1 = 5 + 4 =  9

         Cs   4   5        20      20

Cs = 201√

         9

C1 = 20 + 31√   = 5.22µ

          9

ii) Change on series section = Q = CV1√

    = 20 x 101√ µC

        9

  =  22.2µ C

OR

Q series = Q – Q3µF1√

= (5.22 – 3) x 101√ µ C1

 = 22.2µ C1

Charge is the same on series section hence on 5.0µ F is 22.2µ  c1√

16.i).

ii). i)500mm 1√   ±5mm

     ii) M = V  = h11√   = 50 = 2mm ± 0.2

                   h    h0       25

iii) Move the object towards F but not beyond 1√ move object away from lens

iv) No answer.

b). 1 = 1 + 11√  

     F    u    v

1   = 1 + 11√  

50    80   v

V = 400/31√  

17. a).i)  L2     ii) Brighter

         iii) Total resistance is less/ red reduced √1

b). i) 1.5V 1√  

       ii). Ir = 1.5 – 1.2 = 0.31√  

             0.4r = 0.3

              V = 0.75R1√  

     p. d and E.M.F / more practice and practical approach

c). R = 3 + 0.75 + R1√             0.15(R + 3.75) = 1.51√                 

    R = + 3.75                          R + 3.75 = 1.5 = 10

                                                                0.15

E = IR                                                R = 10 – 3.75

1.5 = 1 (R+ 3.75)                  = 6.25Ώ1√  

OR

R = E = 1.5 = 101√  

       T    0.15

R = R – (V + 3) 1√      + 3.75

R = 6.25Ώ1√  

1.5 – 0.75 x 0.15 = 1 (3 + R) 1√  

1.5 – 0.1125 = 0.15 = 1 (13 + R) 1√  

1.3875 = 3 + R1√  

0.15

R = 6.25Ώ1√  

18.a) i). Deflected towards +ve plate (N) 1√  

        ii). Deflection will be greater1√  

        iii). I). Sport moves back and forth1√  

               II). There will be a horizontal line1√  

b). Electrons are given off as a result of heat produced by current so that electrons gain energy to break off from the surface1√  

Thermionic emission

c). Increasing the filament current 1√    p.d across filament is increased1√   

so that more electrons are released

d. P = IV √1

      = 100 x 1.5 x 10-3

      = 1.5J/S1√  

19 a) intensity or radiation1√  

b) i) (Min p.d)

negative potential sufficient to just stop the movement of electrons.

ii) I). Gradient = h1√  

                          e

h =          30 -) √1                                 =               3

        (12 – 4.4) x 7.6 x 10-19                             7.6 x 1014

Gradient = 0.3947 x 10-14

      h = 0.3947x 10-14  1.6 x 10-19

         = 0.6316 x 10-33

          = 6.316 x 10-34

II).    w  = 1.751√  

          e

Wo = Y intercept  x e

       =  1.75 x 1.6 x 10-19√1

                1.6 x 10-19

               = 1.7 ev  √1

     Alternative

   Wo = hfo √1

         = 6.32 x 44 x 10-14 x 10-34  √1

                1.6 x 10-19

Or

Range 1.7           1.8ev    Wo  = Y intercept

                                        e

Wo = 1.75                  -Wo = -175 or Wo = 1.75

                                      e                       e

Wo = 1.75ev             OR    Wo = 1.75v x e

                                                 = 1.75ev

OR

-Wo = 1.75ev  

WO = 1.75√2 ½                           Reject 1.75v = Wo

Penalize –ve and units in                         Wo = Y intercept = -1.