# KCSE Past Papers 2013 Physics paper 1 (232/1) 5.4.1 Physics Paper 1 (232/1)

1. 5.32 cm (1 mark)

2. – magnitude of the force

– The perpendicular distance between the force and the pivot. (1 mark)

3. Patmosphere = Pmercury + pair enclosed;

Pair = 760 – 600:

= 160 mm Hg;

(3 marks)

4. (a) F = Ke;

20 = 0.5 K;

K = 40 Ncm” (2 marks)

(b) F = 40 x 0.86 = = 34.4 N;

(1 mark)

5. – Weight of object in air

– Weight of object when fully immersed in ﬂuids (2 marks)

6. Upthrust = weight in air – weight of object in ﬂuid. (1 mark)

7. Wood is a poor conductor of heat; hence heat is used to bum paper, while most heat is conducted away by copper: hence paper takes long to bum. (2 marks)

8. Clockwise moments = anticlockwise moments;

0.l8x = l(50-x)+0.12(l00-x)

0.l8x = 50-x+l2-l2x

O.l8x = 62- 1.l2x

7.30x = 62

x = 47.69 cm; (3 marks)

9. Air is compressible; so the transmitted pressure is reduced; (2 marks)

10. The high velocity of the gas causes a low pressure region;

Atmospheric pressure is higher;

Pressure difference draws air into the region; (3 marks)

11. Water molecules have a high adhesion forces; With glass molecules and hence rise up the tube while mercury molecules have greater cohesion;

Forces within than adhesion with glass hence do not rise up. (2 marks)

12. Allow for expansion;

Water expands on cooling between 4° C and 0° C;

13 Diffusion of the ink molecules;

SECTION B

14 (a) – increasing the angular velocity;

– Reducing the radius of the path;

(b) (i) Tension in the string;

(ii) Arrow to centre of circle;

(iii) Direction of motion of object changes and causes the velocity to change with time;

(iv) F = mv/r

0.5 * 8*2/2

= l6N;

(c) (i) V2 = uz + 2as;

O€u’—2>< 10>< 100

u= ‘/ 2000

44.72 ms”;

(ii) V = u + at ;

O = 44.72 – 10 X t

t= 4.472

Total time = 2 X 4.472

= 8.94s ;

temperature;

(b) (i) E=Pt;

=6O><5 X60;

= 13000];

(ii) Mass of water = 190 – 130 = 60g;

mlf = Pt.

60 = .

——1OO0l, 60><60><5,

1,= 3 >< 105 J/Kg;

15 (a) Quantity of heat required to convert 1 kg of ice at 0° C to water without change in temperatures

(iii) Heat from the surrounding melts the ice; (1 mark)

16 a) F = Ma;

F = 2 >< 5

= ION;

friction force = 12 – 10

= 2N; (3 marks)

(b) (i) OA – the ball bearing decelerates; as the upthrust increases to a maximum; (2 marks)

AB – ball attains terminal velocity; when upthrust = weight; (2 marks)

(c) (i) VR = 2 (1 mark)

(ii) To change direction of effort; (1 mark)

(iii) Efﬁciency =—1‘”% >< 100;

so=%><1o0%

MA=l.6;

..1.6_5O0

” L=500X 1.6

=800 N; (3 marks)

17 (a) (i) F = mg

= 10 X 10

= 100 N ;

Additional pressure = = 1 Ncm ’

new reading = 10 + 1 = ll N; (4 marks)

(ii) Pressure has increased; because, when the volume reduces, the collisions between the gas molecules and walls of the container increases; (2 marks)

(b) (i) Pressure = ll Ncm’2 _ (1 mark)

(11) Q = i;

Tr T1

L = L-

300 T1 ‘

T1=i3°‘)l:)‘ 11 = 330k;

T2 = 57° C (4 marks)

18 (a) (i) (I) – Reading decreases on spring balance;

(II) – Reading on weighing balance increases.

(ii) As the block is lowered, upthust increases;

and hence it apparently weighs less;

(b) (i) Upthrust – weight in air – weight in water

= 2.7 – 2.46

= 0.24 N;

Reading in weighing balance = 2.8 + 0.24

= 3.04 N;

(ii) Relative density = weight in air;

upthrust

= 11

0.24

= 11.25 ;

Density = R.d X density of water

= 11.25 >< 1000

= 11250 kgm‘;

c) The hydrometer sinks more;

The density of(t.he water is reduced;

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