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QUESTION 1

Part A

You are provided with the following:

  • A uniform metre rule.
  • A spring balance.
  • Two 15 cm long string.
  • 2 complete stands.
  • 30 cm or half metre rule.
  • A set square.

 

Proceed as follows:

a) Set the apparatus has been set ready for use as shown in the figure below. The metre rule has

suspended at the 99.0 cm mark with a length of string securely tied to the clamp of the retort stand Q.

So not change this position throughout the experiment.

 

 b) Adjust the position of the clamp of the retort stand P so that the metre rule is suspended

  at 0.5 cm mark and is horizontal.

 c) Note and record the distance, x from A to B and also the tension, T, of the spring balance.

  Enter your results in the table below.

x (cm)

T (N)

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1 (m-1)

x

 

  

 (3 mks)

 d) Adjust the position of the clamp of the retort stand O so that the metre rule is suspended

  at the 10.0 cm mark and is horizontal. Note and record the distance x and T in the above table.

 e) Repeat part (d) of the experiment with the spring balance suspended at the 15, 20, 25

  and 30 cm marks. Enter your results in the table and complete the table.

 f) (i) On the grid provided, plot a graph of T (y –axis) against 1  (5 mks)

  (ii) Determine the slope, S, of the graph.  (2 mks)

  (iii) Given that M = S , find a value M, the mass of the metre rule (1 mk)

4.8

 B. a) (i) Place a candle and screen about 50 cm apart. Place a lens (convex) in between

  the screen and the candle. Move the lens from about 10 cm from the candle towards the   screen until a sharp image is focused on the screen. Mark this point U1. Move the lens until the second sharp image of a smaller size is focused on the screen. Mark this point U2.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 (ii) Measure the displacement of the lens i.e. distance between U1 and U2 and let it be X1. Let the

distance between the screen and the candle be Y1 i.e. Y = 50 cm.

 (iii) Repeat the procedure in (a) above by using a value of Y of 40 cm. Let it by Y2. Find the

  displacement X2

Find the value of x = x1 + x2, and value of Y = Y1 + Y2.  (2 mks)

2

 (iv) Given the equation  4f = X2 – Y2

  Y

  Calculate the focal length of the lens(f).  (3 mks)

2.  You are provided with the following apparatus. Two cells of 1.5v each, Nichrome wire gauge

30. An ammeter 0-5 A or 0-25 A range, cell holder, voltmeter 0.3v or 0.5 eight conductors at

least 4 with crocodile clips,, A switch and a metre rule.

Procedure

  1. Connect the circuit as shown in the diagram below.

 

 

 

 

 

 

 

 

 

 

 

  1. (i) Connect the ends A and point C where AC is 100 cm across the terminal as shown. Close

the switch and measure both current I and p.d. across the wire AC.

Current I =  (1 mk)

p.d.(v)  =  (1 mk)

(ii) Measure the e.m.f. of the cells E =  (1 mk)

(c) Reduce the length AC as shown 100, 70 cm, 60 cm, 50 cm, 40 cm and 20 cm. In each case

record the current (I) and the corresponding p.d.(v)

(d) Enter the length in table.

Length L (cm)

100

70

60

50

40

20

I (A)

      

V (Volts)

      

E – V (V)

      

Complete the table  (6 mks)

(e) (i) Plot a graph of (E-V) (V) on the y –axis against I(A) on the x – axis. (5 mks)

(ii) Determine the gradient of the graph.  (3 mks)

(iii) Given the equation E + V + Fr determine the internal resistance of each cell.  (3 mks)

 

 
 

SOTIK DISTRICT 2ND TERM PHYSICS PRACTICAL ANSWERS

 Q 1.  PART A:

 

x (cm)

 

T (m)

1 (m – 1

x

94

89

84

79

75

70

0.8

0.7

0.6

0.5

0.4

0.3

1.06

1.12

1.19

1.27

1.33

1.43

(3 mks)

 f(i)  Graph

– Labeling axes. 1

– Appropriate Scale. 1

– Plotting: 5 – 6 pts 2 1 1  (5 mks)

: 3 -4 pts 1

– Straight line  1

  (ii) Slope – 1 (2 mks)

  Evaluation 1  (2 mks)

  Answer  (0.48 + 0.05)

 

  (ii) Given M = S when S =0

4.8

M = 0.48

4.8 = 0.1 kg 1

 

QUESTION 1: PART B

 f (i) X1 = 7.5 cm 1 mk

  X2 = 50 cm 1 mk

  (ii) X2 = 6 cm 1

  Y2 = 40 cm1

X = 7.5 + 6

2  = 6.75 cm 1

  Y = 40 + 40

2

= 40 cm 1

  (iii) 4f = (6.75)2 – (40)2
1

  1. 1

f = 10 cm 1

9

QUESTION 2

b) (i) I = 0.12A + 0.01 1

V = 2.6V + 0.1 1

(ii) E = 3.0v (Max. range) 1

d) Table

L (cm)

100

70

60

50

40

20

I (A)

0.12

0.19

0.20

0.24

0.28

0.42

V (v)

2.60

2.50

2.40

2.35

2.30

2.00

E-V

0.40

0.50

0.60

0.65

0.70

1.00

(7 mks)

e) (i) Graph

 – Labeling axis 1 mk  – Appropriate scale 1 mk

 – Plotting: 5 – 6 pts 2 mks

: 3 – 4 pts 1 mk

 – Straight line 1

(ii) Slope – 1 mk  Evaluation – 1 mk

 Answer – 1 mk

(iii) Slope of the graph = r, internal /resistance 1 3 mks


 




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EcoleBooks | SOTIK DISTRICT 2ND TEXAM PHYSICS PRACTICAL QUESTIONS AND ANSWERS

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