Free 2020/2021 WAEC Physics Practical Alternative To Essay/Obj/theory runz/expo May/June
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2019 PHYSICS PRACTICAL ANSWERS
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EXAM STARTS BT 9;30AM.
ANSWERS WILL BE READY BY 9AM DOWN
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EXAM STARTS BT 9;30AM.
ANSWERS WILL BE READY BY 9AM DOWN
STAY WITH YOUR PHONE TO MAKE USE OF THE RIGHT ANSWERS
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(1aviii)
In a tabular form
Under y(cm)
20.0, 30.0, 40.0, 50.0, 60.0
Under t(s)
36.61, 33.66, 31.32, 28.62, 25.93
36.61, 33.66, 31.32, 28.62, 25.93
Under T = t/20(s)
1.8305, 1.6830, 1.5660, 1.4310, 1.2965
1.8305, 1.6830, 1.5660, 1.4310, 1.2965
Under T²(s²)
3.3507, 2.8325, 2.4524, 2.0478, 1.6809
3.3507, 2.8325, 2.4524, 2.0478, 1.6809
(1ax)
Slope s = ΔT²/Δy
S = 4 – 2.5/0 – 39
S = 1.5/-39
S = -0.0385
Slope s = ΔT²/Δy
S = 4 – 2.5/0 – 39
S = 1.5/-39
S = -0.0385
Intercept, C = 4.0
(1axi)
SR = c
R = c/s = 4/-0.0385 = -103.9
SR = c
R = c/s = 4/-0.0385 = -103.9
(1axii)
(i) I ensured that the retort stand was well clamped.
(ii) I avoided error due to parallax when reading the metre rule.
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=======================================
(3ai)
E.m.f, E = 3.00v
(i) I ensured that the retort stand was well clamped.
(ii) I avoided error due to parallax when reading the metre rule.
===================================



=======================================
(3ai)
E.m.f, E = 3.00v
(3avii)
In a tabular form
Under R(ohms)
2.0, 5.0, 10.0, 12.0, 15.0, 20.0
In a tabular form
Under R(ohms)
2.0, 5.0, 10.0, 12.0, 15.0, 20.0
Under V(v)
0.86, 0.62, 0.43, 0.38, 0.32, 0.25
0.86, 0.62, 0.43, 0.38, 0.32, 0.25
Under V-¹(v-¹)
1.1628, 1.6129, 2.3256, 2.6316, 3.1250, 4.0000
1.1628, 1.6129, 2.3256, 2.6316, 3.1250, 4.0000
(3aix)
Slope s = ΔR/Δv-¹
=17.5 – 4.25/3.5 – 1.5
=13.25/2
=6.625
Slope s = ΔR/Δv-¹
=17.5 – 4.25/3.5 – 1.5
=13.25/2
=6.625
Intercept, C = -5.75
(3ax)
S = Roα
6.625 = 0.85(α)
α = 6.625/0.85 = 7.79
S = Roα
6.625 = 0.85(α)
α = 6.625/0.85 = 7.79
C = -(Ro + β)
= -5.75 = -(0.85 + β)
β = 5.75 – 0.85
β = 4.9
= -5.75 = -(0.85 + β)
β = 5.75 – 0.85
β = 4.9
(3axi)
(i) I ensured tight connections.
(ii) I avoided error due to parallax when reading the voltmeter.
(i) I ensured tight connections.
(ii) I avoided error due to parallax when reading the voltmeter.
(3bi)
Vab = 12 – 8 = 4v
Rab = 4×5/4+5 = 20/9ohms
Iab = Vab/Rab = 4/20/9 = 4 ×9/20 = 9/5
=1.8A
Vab = 12 – 8 = 4v
Rab = 4×5/4+5 = 20/9ohms
Iab = Vab/Rab = 4/20/9 = 4 ×9/20 = 9/5
=1.8A
Power = I²ab Rab
=1.8² * 20/9
= 7.2watts
=1.8² * 20/9
= 7.2watts
(3bii)
Current in line = power/voltage = 3600/240
=15A
Circuit breaker remain closed because current in line is less than 20A
NUMBER 3 PICTURES

Current in line = power/voltage = 3600/240
=15A
Circuit breaker remain closed because current in line is less than 20A
NUMBER 3 PICTURES

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1. 2019/2020 WAEC PHYSICS PRACTICAL ALTERNATIVE TO EXPO ANS RUNZ
2. 2019/2020 WAEC PHYSICS PRACTICAL QUESTION ANSWERS EXPO
3. 2019/2020 WAEC PHYSICS PRACTICAL ALTERNATIVE QUESTION ANSWERS RUNZ/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 = 4Ï€²/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
(3aviii)
Slope=Change in logI^-1/Change in d(cm)
(Y2-Y1)/(X2-X1)=(0.23-0.06)/(80.0-13.0)
=0.17/67=0.0025
S=0.0025
3viiii)
slope Dvertical/ Dhorizontal
=1 . 41- 1 . 00 / 80 - 25
=0 . 4 / 55
S =0 . 00727
3x)
from the graph C =32cm
3xi)
K =c/ 2 =33 / 2 =16 cm
3xii)
- precautions -
take any two
i) I removed key when reading is not taking
ii) I ensured tight connections at the terminals
iii) I avoided parallax error when taking reading from the ammeter .
3bi)
At I =1 . 5 A ,1 / I =1 / 1 . 5 =0 . 67A ^ - 1
Log ( I ^ - 1 ) = Log 0 . 67 =0 . 17
3bii)
i) length - longer wires have greater resistance.
ii) thickness - smaller diameter wires have greater resistance.
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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 = 4Ï€²/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
(3aviii)
Slope=Change in logI^-1/Change in d(cm)
(Y2-Y1)/(X2-X1)=(0.23-0.06)/(80.0-13.0)
=0.17/67=0.0025
S=0.0025
3viiii)
slope Dvertical/ Dhorizontal
=1 . 41- 1 . 00 / 80 - 25
=0 . 4 / 55
S =0 . 00727
3x)
from the graph C =32cm
3xi)
K =c/ 2 =33 / 2 =16 cm
3xii)
- precautions -
take any two
i) I removed key when reading is not taking
ii) I ensured tight connections at the terminals
iii) I avoided parallax error when taking reading from the ammeter .
3bi)
At I =1 . 5 A ,1 / I =1 / 1 . 5 =0 . 67A ^ - 1
Log ( I ^ - 1 ) = Log 0 . 67 =0 . 17
3bii)
i) length - longer wires have greater resistance.
ii) thickness - smaller diameter wires have greater resistance.
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