Welcome to the NECO 2025 Physics essay and OBJ answers online page. This is where you’ll find the verified NECO Physics Questions and answers June/July solution to support your practical preparations. According to the latest update, the exam is scheduled for 15th July, 2025. The National Examinations Council (NECO), a key Nigerian examination body, oversees the conduct and administration of national secondary school exams. NECO is instrumental in evaluating and certifying students’ academic performance across multiple subjects.
NECO Physics examination are composed of multiple-choice and theory sections, covering topics such as mechanics, electricity, magnetism, waves, and modern physics. The questions assess students’ understanding of fundamental physics principles and their ability to apply these concepts to solve problems.
We understand you did well in your practical, but believe me, you can not pass except you also have good grade in this particular section of the subject.
The Physics expo answers in this page is solved according to the marking scheme. Make sure you use it and change some words while using it.
NECO Physics IS #1,000 And is Ready
CLICK HERE For NECO Physics SUBSCRIPTION
All students writing the Physics examination are advised to join us on WhatsApp for early solution. To join now, click on the link in this page.
In this article, NECO Physics June/July expo questions and answers will be revealed for free. Adequate guide on how to make use of the solutions are also provided.
Also Read: NECO Timetable
Is NECO Physics Questions 2025 Answers Out?
Yes, we wish to inform all science student that the National examination board Physics theory and Obj answers is out. The question papers have already been uploaded on this page.
Students sitting for this exam should review fundamental physics concepts, practice past questions, and manage their time effectively during the test to enhance performance. All students writing this subject can count on us for reliable support and a path to success.
NECO Physics Essay Answers
2025 NECO Physics answers loading Loading…Keep refreshing this page
Are you taking the Physics 2025 exam? If yes, you have nothing to worry about. Simply be at alert and keep refreshing this page as all answers will be made available before the next 30minutes
PHYSICS-OBJ
01 – 10: BCCBEDCDDB
11-20: DDEDBBCBCC
21-30: BBCCECDDDE
31-40: ECDEBBDDEB
41-50: AEAADCEEBE
51-60: CBAEBDDCAD
Solved by bingmat.com.ng
Completed!!!
More Answers loading…. Keep refreshing
PART I: ANSWER SIX(6) QUESTIONS ONLY
(1a)
(i) Drumhead (membrane)
(ii) Tension cords (strings)
(1b)
A camera captures and records images by focusing light from a scene onto film or a digital sensor. Its main purpose is to take photographs or videos for later use. A projector, on the other hand, displays images or videos by shining light through a lens to project the image onto a screen or surface. It is mainly used to show images to a larger audience.
(1c)
Capacity of petrol tanker = 263,000m³
Conversion factor: 1 barrel=159,000m³
Volume in barrels = (Capacity in m³)/(Volume of 1 barrel in m³)
Volume in barrels = 263,000/159,000
Volume in barrels = 1.654
===================================
(2a)
Both vectors and scalars are physical quantities that have magnitude an scan be measured.
(2b)
Given:
-Mass of block = m
-Angle of inclination = 30°
-Acceleration due to gravity = g = 10 m/s²
The component of weight along the incline is: F= mgsin θ
F= mgsin30°
F = m ×10 × 0.5
F = 5mN
===================================
(3a)
(i) A football kicked into the air
(ii) A basketball shot toward the hoop
(3b)
Given:
-Range (R) = 30 m
-Initial velocity (u) = 20 m/s
-Acceleration due to gravity (g) = 10 m/s²
Using the range formula: R = (u² sin(2θ))/g
30 = (20² sin(2θ))/10
30 = (400 sin(2θ))/10
300 = 400 sin(2θ)
sin(2θ) = 300/400
sin(2θ) = 0.75
2θ = sin⁻¹(0.75)
2θ = 48.59°
θ = 24.3°
===================================
(4a)
A thermocouple is a temperature-measuring device that utilizes the thermoelectric effect to generate a voltage proportional to a temperature difference. It consists of two dissimilar metal wires joined at one end, creating a junction.
(4b)
Given:
R₀ = 10 Ω (at 0°C)
R₁₀₀ = 15 Ω (at 100°C)
R = 11.50 Ω (at unknown temperature)
Change in resistance = R₁₀₀ – R₀ = 15 – 10 = 5 Ω
Change in resistance per degree Celsius = 5 Ω / 100°C = 0.05 Ω/°C
Temperature change = (R – R₀) / 0.05
= (11.50 – 10) / 0.05
= 1.50 / 0.05
= 30°C
===================================
(5a)
The statement “the velocity ratio of a block and tackle system is 4” means that the effort moves four times the distance the load moves. In other words, if the load is raised by 1 metre, the effort must move 4 metres.
(5b)
The 50 kg refers to the mass of the bag, which is the amount of matter it contains and remains constant regardless of location, while the 500 N refers to its weight, which is the force due to gravity and can change with gravitational strength.
===================================
(6a)
Given:
Period (T) = 0.3 s
Time = 0.6 s
Number of cycles = Time / Period
= 0.6/0.3
= 2 cycles
(6b)
Given:
Speed (v) = 2 m/s
Period (T) = 0.3 s
Frequency (f) = 1/T
= 1/0.3
= 10/3 Hz
Wavelength (λ) = v/f
= 2/(10/3)
= 2 × 3/10
= 0.6m
===================================
(7a)
(i) In optical fibers used for communication.
(ii) In diamonds, causing their sparkle.
(7b)
Forced vibration occurs when an external force makes a system vibrate at the force’s frequency, while resonance happens when this frequency matches the system’s natural frequency, causing the vibration amplitude to increase greatly.
===================================
(8a)
Both capacitors and inductors are passive components that store energy in a circuit.
(8b)
Given:
F = 6.23 × 10⁻¹ N
q = 1.6 × 10⁻¹⁹ C
v = 3 × 10⁸ m/s
θ = 60°
Using the formula:
F = qvBsin(θ)
B = F/(qv sin(θ))
B = 6.23 × 10⁻¹/(1.6 × 10⁻¹⁹ × 3 × 10⁸ × sin(60°))
B = 6.23 × 10⁻¹/(1.6 × 10⁻¹⁹ × 3 × 10⁸ × 0.866)
B = 6.23 × 10⁻¹ / (4.1568 × 10⁻¹¹)
B = 1.5 × 10¹⁰ T
===================================
(9a)
A potentiometer is preferred to a voltmeter for measuring e.m.f. of a cell because it does not draw current from the cell, giving a more accurate reading.
(9b)
A conductor allows electric current to flow easily because it has free electrons, while an insulator resists the flow of electric current due to the absence of free charge carriers.
===================================
(10a)
Adhesion is the force of attraction between molecules of different substances, such as water molecules sticking to the walls of a glass tube while Cohesion is the force of attraction between molecules of the same substance, like water molecules attracting each other to form droplets.
(10b)
The level of mercury is depressed because the cohesive forces between mercury molecules are stronger than the adhesive forces between mercury and the glass tube.
===================================
(11a)
(i) The maximum kinetic energy of an ejected electron is determined by the frequency of the incident radiation.
Kₘₐₓ = hf – φ
(ii) The intensity of X-rays produced in an X-ray tube is determined by the current flowing through the cathode and the number of electrons hitting the target per second.
(11b)
Given:
λ = 6 × 10⁻⁷ m
h = 6.6 × 10⁻³⁴ Js
c = 3 × 10⁸ m/s
Using E = hc/λ:
E = (6.6 × 10⁻³⁴ × 3 × 10⁸) / (6 × 10⁻⁷)
E = 19.8 × 10⁻²⁶ / 6 × 10⁻⁷
E = 3.3 × 10⁻¹⁹ J
===================================
_PART II: ANSWER FOUR(4) QUESTIONS ONLY_
(12ai)
(i) A ceiling fan spinning about its axis.
(ii) A wheel of a moving bicycle.
(12aii)
(i) A cone resting on its base.
(ii) A book lying flat on a table.
(12b)
(12ci)
Total horizontal component:
Fₕ = -7.66 + 6.432 + 6.000
Fₕ = 4.679 N
(12cii)
Total vertical component:
Fᵥ = 6.432 + 10.392 – 13.595
Fᵥ = 3.229 N
(iii) Magnitude of resultant force:
F = √(Fₕ² + Fᵥ²)
F = √(4.679² + 3.229²)
F = √(21.893 + 10.426)
F = √32.319
F = 5.685 N
===================================
(13ai)
(i) Narrow capillary tube
(ii) Calibrated for human body temperature range from 35°C to 42°C.
(13aii)
The kink in a clinical thermometer prevents the mercury from flowing back into the bulb after taking a reading, allowing for accurate measurement.
(13b)
(i) Air pollution caused by burning fossil fuels.
(ii) Global warming due to increased greenhouse gas emissions.
(13ci)
Ice is a solid form of water, while steam is a gaseous form of water.
(13cii)
Both ice and steam are composed of water molecules (H₂O).
(13d)
Given:
Mass of steam = 100 g = 0.1 kg
Latent heat of vaporization = 2.26 × 10⁶ J/kg
Specific heat capacity of water = 4.2 × 10³ J/kg°C
Initial temperature = 100°C (boiling point of water)
Final temperature = 25°C
Heat lost = Heat lost during condensation + Heat lost during cooling
= mL + mcΔT
= (0.1 × 2.26 × 10⁶) + (0.1 × 4.2 × 10³ × (100 – 25))
= 226,000 + (0.1 × 4.2 × 10³ × 75)
= 226,000 + 31,500
= 257,500 J
===================================
(14a)
Diffraction is the bending or spreading of waves as they pass through an opening or around an obstacle, causing the waves to deviate from straight-line propagation without losing energy.
(14b)
(i) Concave mirror: Used in shaving mirrors to produce magnified images of the face.
(ii) Convex mirror: Used as rear-view mirrors in vehicles to provide a wider field of view.
(14c)
(14d)
Given:
Distance from boy A to cliff = 16m
Time for echo = 0.1s
Distance between boy A and boy B = 5m
(i) Velocity of sound:
Total distance traveled by sound = 2 × 16 = 32m
Velocity = Distance/Time = 32/0.1 = 320m/s
(ii) Time for sound to reach cliff when B clapped:
Distance from boy B to cliff = 16 + 5 = 21m
Time = Distance / Velocity = 21/320 = 0.0656s
===================================
(15ai)
(i) Magnetic declination
(ii) Magnetic inclination
(iii) Horizontal component of Earth’s magnetic field
(15aii)
(i) Iron has high magnetic permeability.
(ii) Iron loses its magnetism quickly when the magnetizing force is removed.
(15b)
Given:
Q = C₁V₁ = C₂V₂
Total voltage V = V₁ + V₂
Q = C₁V₁ and Q = C₂V₂
V₁ = Q/C₁ and V₂ = Q/C₂
V = Q/C₁ + Q/C₂
V = Q(1/C₁ + 1/C₂)
Effective capacitance Cₑff = Q/V = Q / [Q(1/C₁ + 1/C₂)]
1/Cₑff = 1/C₁ + 1/C₂
(15c)
Given:
R = 500 Ω
L = 6 H
V = 220 V
f = 60 Hz
(i) Reactance of inductor:
Xₗ = 2πfL
= 2 × 3.14 × 60 × 6
= 2260.8Ω
(ii) Impedance in circuit:
Z = √(R² + Xₗ²)
= √(500² + 2260.8²)
= √(250,000 + 5,111,216.64)
= √5,361,216.64
= 2315.43Ω
===================================
(16a)
Diffusion is the process by which particles or molecules move from a region of higher concentration to a region of lower concentration, resulting in a gradual mixing without the need for bulk motion.
(16b)
(i) Pour equal volumes of liquids A and B into two separate graduated cylinders.
(ii) Drop the same ball bearing from the same height into each liquid.
(iii) Use a stopwatch to measure the time it takes the ball bearing to fall a fixed distance in each liquid.
(iv) The liquid in which the ball bearing takes more time to fall is more viscous.
Conclusion: The slower the fall, the higher the viscosity.
(16c)
Both x-rays and gamma rays are forms of electromagnetic radiation with high energy and penetrating power.
(16d)
Given:
f = 2.14 × 10¹⁵ Hz
Kₘₐₓ = 1.09 × 10⁻¹⁸ J
h = 6.6 × 10⁻³⁴ Js
(i) Energy of photon:
E = hf
= 6.6 × 10⁻³⁴ × 2.14 × 10¹⁵
= 1.4124 × 10⁻¹⁸ J
(ii) Work function:
Kₘₐₓ = hf – φ
φ = hf – Kₘₐₓ
= 1.4124 × 10⁻¹⁸ – 1.09 × 10⁻¹⁸
= 0.3224 × 10⁻¹⁸ J
Note: The given the frequency value of 2.14 × 10⁻¹⁵ Hz has a typo as Work function calculation would not be meaningful with such a low photon energy.
===================================
(17a)
(i) Lead dioxide (PbO₂) plate
(ii) Dilute sulphuric acid (H₂SO₄)
(17aii)
(i) Lead dioxide plate: Acts as the positive electrode (cathode) during discharge.
(ii) Dilute sulphuric acid: Serves as the electrolyte and provides ions for the chemical reaction.
(17b)
Solar collector is a device that absorbs solar radiation and converts it into heat energy, typically used for heating water or air while Solar panel is a device composed of photovoltaic cells that converts solar radiation directly into electrical energy.
(17c)
Repair refers to fixing a machine that is already damaged or not working properly while Maintenance involves regular servicing or checks to prevent breakdown and ensure smooth operation.
(17d)
Given:
f₀ = 2 cm
fₑ = 10 cm
u₀ = -3 cm
Distance from first image to eyepiece = 7.5 cm
(i) First image distance from the objective lens (v₀):
1/f₀ = 1/v₀ – 1/u₀
1/2 = 1/v₀ + 1/3
1/v₀ = 1/2 – 1/3 = 1/6
v₀ = 6 cm
(ii) Distance between the lenses:
Distance = v₀ + 7.5
= 6 + 7.5
= 13.5 cm
===================================
Keep refreshing this page
to load all answers
OBJ Answers Loading…
NECO Government -OBJ loading keep refreshing!
====================
====================
====================
Tips For passing NECO Physics exam
To pass the NECO physics examination, students should be adept at doing many calculations with a calculator and using the correct units in their answers. Students are also put to show an excellent grasp of vector and scalar quantities, knowing the differences between them and applying them appropriately to solve problems. With regular practice from past questions and a critical eye on these cardinal points, they will be bound to succeed.
Getting NECO Physics June/July questions And Answers 2025
Joining our WhatsApp group is the first step to take by anyone who wish to to get free answers here. Make sure you keep refreshing this page at interval as we are getting ready to release all answers here.
If you have any queries or require further assistance on WAEC Physics Questions 2025 Answers, do not hesitate to reach out to us.