WAEC 2018 PHYSICS PRACTICAL ANSWER EXPO

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(1a)

TABULATE:

S/N: 1, 2 , 3, 4, 5

L(cm): 90.00, 80.00, 70.00, 60.00, 50.00

T(s): 38.00, 36.00, 34.00, 31.00, 28.00

T = t/20(sqt): 1.90, 1.80, 1.70, 1.55, 1.40

√L(cm^1/2): 9.49, 8.94, 8.37, 7.73, 7.07

(1aix)

Slope(s) = Δvertical/Δhorizontal

=1.90 – 1.55/9.49 – 7.75

= 0.35/1.74

S = 0.201

(1ax)

g = 4π²/S²

= 4 ×(3.142)²/(0.201)²

= 977.4cm/s²

(1axi)

(i) I ensured the angle of oscillation is relatively small.

(ii) I ensured that no external force is added to the system of oscillation.

(iii) I ensured the oscillation is perfect

(2ai)

fo=15cm

2av)

a=60.00cm

b=20.00cm

Hence L=a/b=60.00/20.00

L=3

(2avi)

TABULATE

S/N:1,2,3,4,5

b(cm):20.00,25.00,35.00,40.00

a(cm):60.00,37.50,30.00,26.25,24.00

L=a/b:3.00,1.50,1.00,0.75,0.60

2avii)

Slope=Change in L/Change in a

=(3-0.25)/(60-18.6)

=2.75/41.4

=0.006642

2aviii)

S^-1=1/S

=(1/0.0066425cm)

S=15.05

S=15cm

(2aix)

PRECAUTIONS

-I ensured that all apparatus are in straight line

-I avoided error due to parallax when reading the metre rule

-I avoided zero error on the metre rule

(2bi)

u=10cm

f=15cm

Using 1/v+1/u=1/f

1/f-1/u=1/v

1/v=1/15-1/10

1/v=(2-3)/30

1/v=-1/30

v=-30cm

The characteristics of image formed are:

-It is virtual

-It is enlarged and magnified ie twice or two times as big as the object m=2

(2bii)

The concave mirror mounted in its holder is moved to and fro in front of the screen until a sharp image of the cross wire of the ray box is formed on the screen adjacent of the object.The distance between the mirror and the screen was measured as 30cm since the radius of curvature r=2fo then half is distance

(2bi)

The characteristics of imaged formed are :

i)It is virtual

ii) It is enlarged or magnified i.e. twice or two times bigger as the object(m=2)

(3a)

TABULATE:

x(cm):10,20,30,40,50,60

V(v):0.65,0.75,1.00,1.20,1.45,1.55

I(A):0.20,0.30,0.35,0.40,0.45,0.55

LogV(v):-0.187,-0.125,0.000,0.079,0.161,0.190

LogI(ohm):-0.699,-0.523,-0.456,-0.396,-0.391,-0.360

SLOPE(s):=( LogI2-LOgI1)/(LogV2-LogV1)

=-0.2-(-0.7)/0.3-(-0.2)

=0.5/0.5

=1AV^-1

(3axi)

(i) I ensured neat and tight terminals

(ii) I opened the key when reading was taken

(3bi)

(i)The brightness of the bulb increases

(ii) The voltage and current through the bulb increases

(3bii)

(i) diode

(ii) transistor

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