WAEC GCE 2025 PHYSICS PRACTICAL ANSWER
(1a)
Precaution taken for mechanics:
(i) I would ensure slight displacement of the bob
(ii) I would avoid draught
(iii) I would avoid conical oscillation
(iv) I would ensure firm support
(1ax)
(i) I would ensure slight displacement of the bob
(ii) I would avoid draught
(1bi)
The statement simply means at the equator, the force of gravity on a body causes its velocity to increase at the rate of 10 m/s per second.
or
This means that at the equator, 10 N is the force of gravity acting on 1 kg mass moving towards the centre of the earth.
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(2b)
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(3)
(3v)
(3b)
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WAEC GCE 2025 PHYSICS THEORY AND OBJ ANSWER
NOTE:
Pls kindly trace it from your system. The options are being reshuffled. You might see number 1 as any number, so trace and be fast. You might also see option A as option C, B or D. Just use the answers in words and and trace properly. Kindly Trace it carefully to avoid errors!!!!
Physics -Obj
1; (B) coulomb per second.
2; (B) Gravity
3; (C) W
4; (B) weight and upthrust.
5; (C) II and III only.
6; (C) 20 m/s^-¹
7; (A) 10cm
8; (B) 2.0m/s^-¹
9; (B) 0.75m/s^-²
10; (B) Solidification
11; (C) 13.33m/s^-¹
12; (D) 325.3v
13; (B) below the image.
14; (B) An atom emits radiation when its electrons move from a higher to a lower energy level
15; (B) 63mg (rounded)
16; (C) 11 m/s^-¹ (rounded)
17; (C) 116.7k (rounded)
18; (B) W/I²R
19; (A) Wavelength
20 ; (C) 75%
21; (C) 0.30m
22; (B) 45°
23; (D) The diagram showing the ray reflecting back into the original medium at the boundary.
24; (B) Generator
25: (C) tungsten
26; (B) amplitude
27; (C) 2.0g cm³
28; (A) 2
29; (A) I(R¹+R²+R³)
30; (B) 20m
31; (C) 2x
32; (B) 20ms^-
33; (C) kg m^²s^-²
34; (A) choose the one facing down
35; (A) F×d²
36; (B) Displacement is a vector quantity.
37; (B) has dimension M^-¹L³T^-²
38; (D) I,II and IV only
39; (B) 0.02m³
40; (A) total time of flight of a projectile.
41; (C) 300N
42; (B) 42°
43; (D)
44; (A) follow a circular path.
45; (D) proportional to the square root of the tension in it.
46; (D) 1/16
47; (A) 4.83*10¹? Hz
48; (A) Parabolic mirror
49; (D) 2.12N
50; (D) latent heat does not involve any change in temperature.
(1a)
TABULATE
Under; Ferromagnetic Materials;
(i)Cobalt
(ii)Iron
(iii)Nickel
Under; Paramagnetic Materials:
(i)Aluminum
(ii)Platinum
(iii)Sodium
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(2a)
Half-wave rectification.
(2b)
Direct current.
(2c)
Diode.
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(3)
An absorption spectrum is a spectrum of light that shows the wavelengths of light absorbed by a substance. When light passes through a medium, such as a gas, liquid, or solid, certain wavelengths are absorbed by the substance, while others are transmitted or reflected.
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(5)
I ? (distance)² × mass
I ? (distance)² × mass
The dimensions of:
-- distance are L (length)
-- mass are M (mass)
So, the dimensions of I are:
I ? L² × M
I ? ML²
Therefore, the dimensions of moment of inertia are ML².
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(6)
a = g sin ?
= 10 m/s² × sin 30°
= 10 m/s² × 0.5
= 5 m/s²
10 = 0 + (1/2) × 5 × t²
10 = 2.5 × t²
t² = 10 / 2.5
t² = 4
t = ?4
t = 2s
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(8)
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(9ai)
(i) Sensible heat changes the temperature of a substance without changing its state, while latent heat changes the state of the substance without changing its temperature.
(ii) Sensible heat can be measured with a thermometer, but latent heat cannot be directly measured by temperature change.
(iii) Sensible heat is associated with heating or cooling, whereas latent heat is involved in phase changes such as melting, boiling, or condensation.
(9aii)
(i) The Sun
(ii) Geothermal energy from the Earth's interior
(9aiii)
(i) Mercury has a much higher boiling point, so it can measure higher temperatures than alcohol.
(ii) Mercury does not wet glass (gives a clear, continuous column and more precise readings) and is highly visible.
(9bi)
At 0?°C:?R? = 10??
At 100?°C:?R??? = 110??
Linear relationship:?R = R? + (?R/?T)·T, where ?R = 100?? over 100?°C --> 1??/°C.
For R = 75??:
T = (R – R?) / (?R/?T) = (75?? – 10??) / (1??/°C) = 65°C
(9bii)
Mass of water,?m = 150?g = 0.150?kg
Specific heat capacity,?c = 4200?J?kg?¹?K?¹
Temperature rise,??T = 107°C – 60°C = 47K
Energy required,?Q = mc?T = 0.150?kg × 4200?J?kg?¹?K?¹ × 47?K = 29610J
Heater power,?P = 60W (Js?¹)
Time,?t = Q/P = 29610/60 = 493.5s
Convert to minutes: 493.5?s ÷ 60?s/min = 8.2?minutes
(9bii)
(i) Both involve the transfer of thermal energy between a body and its surroundings.
(ii) Both are influenced by the material’s specific heat capacity (the amount of energy needed to change its temperature).
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(10)
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