2025 WAEC PHYSICS: 2025 WAEC MAY/JUNE PHYSICS ANSWERS (7230)

Friday 28th August 2020
Physics 2 (Essay) – 9:30am –11:00am
Physics 1 (objective) – 11:00am –12:15pm

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PHYSICS OBJ
1-  10: DCCAACDCCC
11-20: BDBABBACAC
21-30: ACBACCCBAD
31-40: BDDDDACCCA
41-50: CAAABBCBBC

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==COMPLETED====

==No. 12  completed==

Welcome to official 2025 Physics WAEC answer page. We provide 2025 Physics WAEC Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct WAEC Phy Expo. WAEC Physics Questions and Answers 2025. WAEC Phy Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2025 WAEC MAY/JUNE PHYSICS ANSWERS. 2025 WAEC EXAM Physics Questions and Answers

(1)
Original length Lo = 3m
extension e = 3Lo = 3×3 = 9m
F ∝e
F = ke where k = 982.3Nm‐²
F = 982.3×9 = 8840.7N
The force that produces the extension F = 8840.7N

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(2a)
Metal flat plate: It is a collector of heat that converts the radiant solar energy from the sun into heat energy by the process of greenhouse effect.

(2b)
Thermal Insulator: It prevents loss of accumulated heat in the solar collector and thereby improves its absorption efficiency.

(2c)
Tubes: They are used to capture almost all the sun's energy. It is the most efficient solar collector

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(3a)
Plastic

(3b)
The refractive index of the core must be greater than that of the cladding.

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(4)

V ∝ T^x U^y
LT‐¹ ∝(MLT^‐²)^x(ML^‐1)^y
LT‐¹ ∝ M^xL^xT^‐²x × M^yL^-y
LT‐¹ ∝ M^(x+y) L^(x-y) T^‐²x
Comparing
X-y = 1 -----(1)
X+y = 0 -----(2)
-2x = -1 ----(3)
From equation 3,
X = 1/2
Putting X = 1/2 into eqn 2
y = -1/2
Equation is:
V ∝ T½ U‐½
V = √T/U
V = K√T/U
Where k is constant

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(5)
Work done in a gravitational field = force × distance
= f × r
= GMeM/R² × R
= GMeM/R
Work done is also kinetic energy, K.E = 1/2MVe²
Equating both
GMeM/R = 1/2MVe²
Ve² = 2GMe
Ve = √2GMe/R -----(1)
Also,Force f = GMeM/R² = mg
Me = gR²/G -----(2)
Putting (2) into (1)
Ve = √2GgR²/GR = √2gR

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(6)
θ = 30°, g = 10m/s²
Time of flight, T = 25seconds
Using,
Time of flight, T = 2usinθ/g
25 = 2usin30°/10
2u(0.5) = 250
u = 250
Velocity of projection, u = 250m/s

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(7)
(i)Coherence
(ii)Directionality
(iii)Monochromatic
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(8ai)
A torque is defined as a measure of the force that can cause an object to rotate about an axis

(8aii)
Factors that determine torque are;
- The magnitude of the force
- The radius
- The angle between the force and the lever arm

(8bi)
Freefall simply means the act of which are objects fall freely under the influence of gravitational force. An object under free fall does not encounter air resistance

(8bii)
Height of tower H = 40.0m
Velocity v = 10.0m/s, time t= ? , g=10.0m/s

Using the relation
S= ut + 1/2gt^2
H= ut + 1/2gt^2
40= 10t + 1/2*10t^2
4= t + t^2/2
= t^2 + 2t - 8=0 ( solving quadratically)
= (t+4)(t-2)=0
t+4=0 or t-2=0
t= -4 or t=2

Therefore the time to fall on the ground t=2s

(8c)
Side of the cube of wood = 8.0cm
2.0cm = 1/4 of the side immersed in water
R.density of water Rw = 1.00
R.density of Oil = Ro = 0.72
Volume of the wood V = L³
= 8×8×8
V = 512cm³
Volume immersed in water = 1/4×512 = 128cm³
Weight of water displaced = 128g × Rwater = 128×l
By applying the principle of flotation
Mass of wood = 128×Rw
=128 × l = 128g

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(10ai) 
The critical angle is defined as the angle of incidence in the denser medium when the angle of refraction in the dense medium is 90°

(10aii) 
The conditions necessary for total internal reflection to occur 
1. The light must be travelling from an optically more dense to an optically less dense medium. 
2. The angle of incidence in the denser medium must be greater than the critical angle. 

(10aiii)
Three practical applications of total internal reflection:
[Choose three] 
(i) Mirages
(ii) Field of view of fish underwater.
(iii) Refractometer. 
(iv) Optical fibre.
(v) Binoculars. 

(10b) 
Two effects of refraction: 
(i) The apparent depth of a swimming pool.
(ii) Apparent bending of a stick partly immersed in water.

(10ci) 
Progressive wave is defined as a wave that transfers energy in travelling or moving away from the source of the disturbance.

(10cii) 
y = 0.5sin(1000πt - 100πx)/17 = y = Asin(2πx/λ - wt)

(∝) at = 1000πt
===》 w = 1000
But w = 2πf
2πf = 1000π
f = 1000/2
f = 500Hz

(β) 2πx/λ = 100π/17 x
2/λ = 100/17
λ = 2×17/100
=0.34m

(γ) V = fλ
= 500 × 0.34
=170m/s

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(12)

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