KCSE 2000 Marking Scheme

The documents below is a collection of KCSE marking scheme for the year 2000. Download by clicking the links shared:

KCSE 2000 BIOLOGY P1 ANSWERS

KCSE 2000 BIOLOGY P2 ANSWER

KCSE 2000 CHEMISTRY PAPER 1 ANSWERS

KCSE 2000 CHEMISTRY PAPER 2 ANSWERS

KCSE 2000 CRE PAPER 1 & PAPER 2 ANSWER

KCSE 2000 Geography Paper 1 answers

KCSE 2000 Geography Paper 2 Answers

KCSE 2000 KISWAHILI PAPER 1 ANSWER

KCSE 2000 MATHEMATICS P1 ANSWERS

2.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.

3.Beaker becomes more stable because the position of C.O.G is lowered on melting or water is denser than ice.

4.On earthing negative charges flow to the leaves from earth to neutralize positive charges when the rod is withdrawn the leaves are left with net negative charge.

5.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

 

6.Particles of gases are relatively far apart  while those of liquids and liquids are closely parked

7.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

8.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

9.

 

10. Can withstand rough treatment

Do not deteriorate when not in use

11. Struts are DE,  DC, AD, BD                          Ties are BC; AB

12.The keepers become magnetized thus neutralizing the pole, this reduces repulsion at the poles, thus helping in retention of magnetism

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

15. VR = 4;

16. Efficiency of the  system

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

V.R                               20     4

= 89%

17. Sound waves

18. 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

 

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

R                                      R