WAEC 2018 PHYSICS PRACTICAL ANSWER
EXPO
Posted by Emmylish on Tuesday, April 3,
2018
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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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WAEC 2018 PHYSICS PRACTICAL ANSWERCreated at 2018-04-07 13:57:46
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PHYSICS PRACTICAL ANSWER
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