Waec 2019 Physics Practical Answers – May/June Expo

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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

Under T = t/20(s)
1.8305, 1.6830, 1.5660, 1.4310, 1.2965

Under T²(s²)
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

Intercept, C = 4.0

(1axi)
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.

(1 GRAPH)
CLICK HERE FOR THE IMAGE

(1bi)
Draw the diagram
(α) K.E is maximum
(β) acceleration is maximum

(1bii)
Period, T = 2π√m/k
Given : weight, W = 120N
Mass = w/g = 120/10 = 12
T = 45
π = 3.142
4 = 2(3.142)√12/k
0.6365 = √12/k
Square both sides
0.405 = 12
K = 12/0.405 = 29.6

=====================
(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

Under V(v)
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

(3aix)
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

C = -(Ro + β)
= -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.

(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

Power = I²ab Rab
=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

(3)
CLICK HERE FOR THE IMAGE

(3x)
(i) I avoided parallax error when taking readings from the voltmeter
(ii) I ensured that the key was opened when reading is not been taking

COMPLETED!

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