# 2012 KCSE Physics P1-20120918Q

2012 KCSE Physics paper 1 20120918Q- Marking scheme

PHYSICS 232/1

Theory

Paper 1

Time:  2 hours

SECTION A – (25 MARKS) –  ANSWERS

1.      Viscous force     (1mk)

2.      Increasing the temperature increases the rate of collision of molecules with the walls (1mk) .  Therefore increasing pressure. (1mk)

3.      Increasing the temperature at point x on the water surface reduces the surface Tension on point x. (1mk)  there is high surface Tension on y pulling the box towards y.   (1mk)

4.      S = ut + ½ at2     (1mk)

9  = ½ xa  x 9     (1mk)

a = 2ms -1           (1mk)

5.a

Fig5.1

b.      Clockwise moment = Anticlockwise moment

F x 10cm = 100N x 5cm       (1mk)

F = 50N     (1mk)

6.      The glass flask expand first before the liquid making the level to fall (1mk)

The liquid then expand raising the level.   (1mk)

7.      The air is moving at high velocity inside the paper tube reducing the pressure inside.  (1mk) .    The high atmospheric pressure from outside make the tube to collapse.    (1mk)

8.a.   Upthrust  =  5.0  –  4.04  =  0.96N     (1mk)

b.      Upthrust = weight of liquid displaced   (1mk)

Mass of liquid displaced  =  0.096 kg   (1mk)

Volume of liquid  =  volume of solid

=  (0.096/ 800)  =  1.2  x 10-4 m3  (1mk)

c.       Density of the solid =  m/v.

=     0.5kg /0.00012m3     (1mk)

=  4166.67 kg/m3   (1mk)

9.      It is the temperature of a body when it has no energy.  (1mk)

10.              V = rw                                     (1mk)

=  0.12m  x 33 x 2π            (1mk)

=  24.884 m/s                       (1mk)

11.    Potential energy – kinetic energy – heat                   (1mk)

i.I.      Work done  =  Force  x  distance

= 400N x 25m              (1mk)

= 10000 Joules            (1mk)

II.       Work done  =  weight x height

= 600N  x 10m

= 6000 Joules

III.      Work done to overcome friction = (10000 – 6000)J = 4000 J  (1mk)

ii.       Efficiency =  Work output  x 100%         (1mk)

Work input

= 6000  x 100  = 60%  (1mk)

10000

iii.      Reducing friction between the mass and incline plane.     (1mk)

Reducing the angle of inclination.

12.a.i.To show the behavior of the air particles since they are invisible.

(2mk)

ii.       To focus light to the smoke particles in the smoke cell.   (2mk)

iii.      To magnify the smoke particles so that they can be seen clearly.(2mk)

b.      The smoke particles are seen to be moving in a continuous random motion.          (1mk)

The smoke particles are bombarded by the invisible air particles. (1mk)

c.       The smoke particles will move at a faster rate in random motion.

(1mk)

13.a. It is the quantity of heat required to change the state of a unit mass of a substance from liquid to solid or solid to liquid without change in temperature.             (1mk)

b.i.     Q  =  MLf                       (1mk)

=  0.02 x 334000

=  6680J

ii.       Q  = MCT                    (1mk)

=  0.02 x 4200 x  T

=  84T                      (1mk)

iii.      Q  =  0.2  x 4200  x (60 – T)  +  0.08  x 900  x (60  –  T)    (1mk)

=  840(60 – T) +  72  (60 – T)

= 54720  –  912T                                                               (1mk)

iv.      Heat lost by water  +  Heat lost by calorimeter

=  Heat gained by ice to melt   +  Heat gained by ice water  (1mk)

54720  –  912 T  =  6680  +  84T   (1mk)

996T  =  48040                         (1mk)

T   =  48.23°C                      (1mk)

v.       No heat lost to the environment              (1mk)

No heat is received by the thermometer   (1mk)

14.a.i. For a fixed mass of a gas the pressure is directly proportional to the

absolute temperature when the volume is kept constant.          (1mk)

b.i.

 Pressure P (x105pa) 2 2.5 3 3.5 4 4.5 Volume, V(m3) 0.025 0.02 0.017 0.014 0.012 0.011 1/V (m -3) 40 50 58.82 71.43 83.3 90.91

(2 mks)

On the graph paper              Naming the axis            (1mk)

Plotting                (2mks)

Curve                  (1mk)

Scale                             (1mk)

ii.       PV    =  2RT

R    = ( PV/2T)                     Where PV = gradient

T  = 300K

gradient =   Δ  P/ Δ(1/V)    =  PV

=  (( 2.5  –  1.5)  x 105 pa)/( (50 – 30) cm-3)(1mk)

=  5.0 x 103         (1mk)

R =  (5.0  x 10)/(2  x 300) =  8.33             (1mk)

15.a. In solids the intermolecular forces are higher than in liquids.  (1mk)

b.      Volume  =  43 πr3    (1mk)

43  x 3.142  x (0.025)3 cm3

6.546 x 10-5 cm3         (1mk)

ii.       Area  =  π r 2

=3.142 x 10 x 10cm2   (1mk)

=314.2cm2                                    (1mk)

iii.      Diameter of oil molecule

= Volume of one drop/ Area of the patch        (1mk)

= (6.546 x 10-5 cm3) /314.2 cm2 (1mk)

= 2.08 x 10-7cm           (1mk)

c.i.        Oil drop is spherical

Area formed is a uniform circle

Molecule are forming a monolayer

(any one given 1mk)

ii.       The volume of the oil drops not the same.

The diameter of the oil patch not constant.

Molecules may not be forming a monolayer.

(Any two 1mk each)

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-end-

### 10 responses to “2012 KCSE Physics P1-20120918Q”

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