VERIFIED 2021 WAEC Completed Physics Questions & Answers - EXPO
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Today’s WAEC English OBJ Answers: (2021 Answers)
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PHYSICS THOERY:
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(1)
Workdone = 1/2Fe
= 1/2 × 40 × 1/100
= 1/5 = 0.2J
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(2a)
(i)Navigation satellites e.g global position system - GPS
(ii) Communication satellites eg ANIK
(2b)
V = 2πR/T
For every second,
Speed, V = 2πR
V = 2 × 22/7 × 6300
V = 39,600km/s
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(3)
V=u cosθ -gt
V=40*cos30-10*1
V=34.64-10
V=24.64m/s
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(5a)
Wave particles duality is the concept in quantum mechanics that every particles or quantum entitle may be described as either a particle or wave.
(5b)
λ=h/MV
4.2*10^-¹¹=h/1.6*10^-²³
h=4.2*10^-¹¹ * 1.6*10^-²³
h=6.72*10^-³⁴
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(6)
(i)reflection
(ii)refraction
(iii)diffraction
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(7a)
Fiber optics work on the principle of total internal reflection.
(7bi)
Core: This is the light transmission area of the fiber, either glass or plastic.
(7bii)
cladding is made of a material with a slightly lower index of refraction than the core
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(8ai)
Hooke's law states that the extension e produced by an elastic material is directly proportional to the applied force provided the elastic unit is not exceeded.
(8aii)
(i) Given extension, e = I2 - I1 = 0.75 - 0.20 = 0.55m
Force applied, Fe = (1.95 - 0.30) × 10
= 1.65 × 10 = 16.5N
Force constant , K = F/R = 16.5/0.55 = 30N/m
(ii) Using F = K (I1 - Io)
m1g = K ( I1 - Io)
= 0.30 × 10 = 30(0.20 - Io)
= 0.10 = 0.20 - Io
Io = 0.20 - 0.10
Io = 0.10m
Length of spring when it was unloaded = 0.10m
(8bi)
Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration.
(8bii)
(i) Temperature
(ii) Size of particles
(8biii)
Rate of diffusion is inversely proportional to the square root of density under given conditions of temperature and pressure. ie R ∝ 1/√d
(8c)
T = circumference of the orbit/orbit velocity
T = 2πR/v
But V = √GM/R
T = 2π√R³/GM
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(9ai)
The dew point is the temperature to which air must be cooled to become saturated with water vapor.
(9aii)
Dew is a condensation phenomenon. The water vapour when cooled below its dew point and now comes in contact with a colder surface forms a dew. The metal part of the bicycle tends to be colder than rubber part of the bicycle because the metal surface get cold easily.
(9bi)
The specific heat capacity of copper is 400 JKg−¹K−¹ means that 400 J of heat energy is required to raise or lower the temperature of 1kg of a piece of copper by 1 kelvin.
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(11ci)
The capacitance of a capacitor is the ratio of the amount of charge on its plates to the potential difference between them. ie C = ∑/V
(11cii)
(i) Charge in both capacitors are the same
∑1 = Q2
ie C1V1 = C2V2 .........(1)
Total capacitance ,C = C1C2/C1+C2
C = C2/1+C2/C1 = C2/1+V/V2
C = C2V2/V1+V2 = C2V2/2
C= 1/2C2V2
(ii)Voltage across, V1 = (C2/C1+C2) × V
V1 = (C1/C1+C2) × V
But C1/C2 = d2/d1 = 5/2
V1 = (1/(5/2+1)) × 2 = 2/7 ×2 = 4/7Volts
Voltage across, C2 : V2 = 2-4/7 = 10/7volts
(11ai)
Root measurement value of an alternating current is the value of A.C current that has the same heating affect as a D.C current
(11aii)
Impedance is the affective resistance of an electric circuit arising from the combined affects of ohmic resistance and reactance .( ie resistors, indicators and capacitors)
(11b)
Current in circuit , I = 60/120 = 0.5A
Voltage across device , Vr = 120V
Voltage across capacitor , Vc = √V² - Vr
Vc = √240² - 120²
Vc = 207.846V
Ic = I = 0.5
Xc = Vc/Ic = 207.846/0.5 = 415.7Ω
Capacitance, C = Xc/2πf = 415.7/2*3.142*50 = 1.323F
(11ci)
The capacitance of a capacitor is the ratio of the amount of charge on its plates to the potential difference between them. ie C = ∑/V
(11cii)
(i) Charge in both capacitors are the same
∑1 = Q2
ie C1V1 = C2V2 .........(1)
Total capacitance ,C = C1C2/C1+C2
C = C2/1+C2/C1 = C2/1+V/V2
C = C2V2/V1+V2 = C2V2/2
C= 1/2C2V2
(ii)Voltage across, V1 = (C2/C1+C2) × V
V1 = (C1/C1+C2) × V
But C1/C2 = d2/d1 = 5/2
V1 = (1/(5/2+1)) × 2 = 2/7 ×2 = 4/7Volts
Voltage across, C2 : V2 = 2-4/7 = 10/7volts
WAEC physics Questions 2021 Theory Answers (Expo)
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