Waec 2021 Physics Answer

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WAEC PHYSICS ANSWERS
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THEORY-ANSWERS

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PHYSICS OBJ:
01-10: AAABACCDDB
11-20: ACCDCBCBAC
21-30: CCCBBDDBBB
31-40: ACBBCCCAAD
41-50: BBADCCBAAC

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PHYSICS THEORY
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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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(1, 2, 3 & 5)

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(8)


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(9)

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(11)

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

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
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(12ai)
A radioactive nucleus is said to be relatively stable if it does not further disintegrate or form any other atomic nucleus

(12aii)
Reason; The more the atomic mass of a nucleus, the more it’s radioactive tendency an the lesser it’s stability

(12bi)
Ionization potential is the amount of energy eV, which helps to remove an electron completely from an atom. It is represented with V.

(12bii)
E = eV
IeV = 1.6×10-¹⁹

V = hv/λ where h is the planks constant

h = 6.626 ×10-³⁴ , c = 3×10⁸m , λ = 1.02 × 10-⁷m

E = (1.6×10-¹⁹) × (6.626 ×10-³⁴) × (3×10⁸)/1.02 × 10-⁷

E = 31.805 × 10-⁴⁵/1.02 × 10-⁷
E = 31.2 × 10-³⁸ volts

 

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