RSS

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) 2.00 2.50 3.00 3.50 4.00 4.50
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)

Well done. Sometime you will feel like craving for something to keep you going like…
Affiliate Program ”Get Money from your Website”

-end-

 

9 responses to “2012 KCSE Physics P1-20120918Q

  1. BENARD KIRUI

    April 12, 2013 at 11:08 pm

    Actualy me personally I am very motivated by the very challenging subject,physics,which is now made easier through getting revision questions from the internet and acquiring answers for my preparation of 2013 KCSE

     
  2. Dickson Kiprono kigen

    July 16, 2013 at 2:33 pm

    I like the questions, its very helpful and i thankfuly appreciate!!!

     
  3. Mwanza Stephen

    August 10, 2013 at 6:51 pm

    this is my library. thank you

     
  4. shalom owen

    June 6, 2014 at 8:07 pm

    yahhhhhhhhhh! we gonna get it through this rev quizes.thank u.God bless………

     
  5. s.k

    July 6, 2014 at 7:46 pm

    keep up wth the goodwork

     
  6. rcnderitu@gmail.com

    September 23, 2014 at 8:21 pm

    Awesome site

     
  7. Memusi Fred

    August 10, 2016 at 7:25 pm

    Lucky of me, getting here as early as this. counting one year to sit for the knec exams. Will I not be benefiting for this revisn surely? Thanks in advance

     

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

 
%d bloggers like this: