WAEC PHYSICS ANSWERS 2021

WAEC PHYSICS ANSWERS 2021

Here is the waec physics answers for 2021.

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Physics Answers

(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

8ai

Hooke’s law states that the strain of the material is proportional to the applied stress within the elastic limit of that material.

8b

Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration down the concentration gradient.”

(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

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= 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

9a
The dew point is the temperature to which air must be cooled to become saturated with water vapor. It is assumed that air pressure and water content is constant. When cooled further, the airborne water vapor will condense to form liquid water.

,(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.

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(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.

11c

The capacitance of a capacitor is the ability of a capacitor to store an electric charge per unit of voltage across its plates of a capacitor. Capacitance is found by dividing electric charge with voltage by the formula C=Q/V. Its unit is Farad.

(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)

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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

We gat you all always

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