Home EXAMS NECO Physics Questions and Answers 2022/2023 Expo

NECO Physics Questions and Answers 2022/2023 Expo

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NECO Physics Questions and Answers – Objective and Theory Expo 2022/2023.

NECO Physics Questions and Answers: Below is the National Examination Council (NECO) Physics & Answers to Questions – Complete Solution 2022/2023.

The NECO 2022 questions and answers for physics are set from SS1 to SS3 scheme of work. All the topics that this 2022 NECO physics theory and OBJ questions are set from are in the physics syllabus.

NECO Physics Questions and Answers

About NECO

The National Examination Council is an examination body in Nigeria that conducts the Senior Secondary Certificate Examination and the General Certificate in Education in June/July and November/December respectively.

Quick NECO Physics OBJ answers 2022.

The 2022 NECO Physics expo will be posted here during the NECO Physics examination. Keep checking and reloading this page for the answers.

PHYSICS-OBJ
1-10: BBBDAEDBEA
11-20: CBAEAEDDED
21-30: BCDCEECBCD
31-40: DCCCEBEACD
41-50: DEDDDAADED
51-60: AADCEBADAC
COMPLETED

PHYSICS-ESSAY-ANSWERS
PLEASE SCROLL DOWN FOR SECTION B ANSWERS
(1a)
(i) Tubes, fins or passages for conducting or directing the heat transfer fluid from the inlet to the outlet.
(ii) Abserver plate which may be flat, corrugated or grooved with tubes, fins or passages attached to it.

(1b)
NigeriaSat-1 is a micro satellite used in monitoring disasters managements and various mapping campaign WHILE NigercomSat-1 is a mega satellite responsible for creating an enabling environment for competition among operators in the industry as well as ensuring the provision of qualitative and efficient telecommunications services

(1c)
V = image
U = Object
V = 2u
1/f₀ = (1/V) + (1/u)
1/f₀ = (1/2u) + (1/u)
1/f₀ = (1+2)/2u
1/f₀ = 3/2u
f₀ = 2u/3
Magnification = (V/f₀ + 1)
M = [2u/(2u/3)] + 1
M = [(2u/1) x (3/2u)] + 1
M = 3 + 1
M = 4
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(2a)
scalar quantities;they have magnitude but no direction and they are added by ordinary algebraic methods. WHILE vector quantities; they have both magnitude and direction and they are added by geometrical methods.

(2b)
r²=3²+4²
=9+16
=25

r=✓25
r=5n
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

(3i)
The principle of floatation states that when an object floats in a liquid the buoyant force/upthrust that acts on the object is equal to the weight of the object.
Weight of the object (W) = upthrust (F)

(3ii)
(i) Volume of the object
(ii) Density of the liquid
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

(4i)
E= mgh = 40 X 10 X 0.3 X 6= 720J

(4ii)
P= mgh/t
m= PT/GH= 60 X 12 / 10 X 0.3 X 6 = 40g
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(5a).
Positive charge: This type of charge can be produced by rubbing a glass rod with silk.

(5b)
P₁ = 600mmHg
V₁ = 500cm³
T₁ = 23°C + 273.15 = 296.15K
T₂ = 37°C + 273.15 = 310.15K
P₂ = 720mmHg
V₂ =?
P₁V₁/T₁ = P₂V₂/T₂
V₂ = (P₁V₁T₂)/T₁P₂
V₂ = (600 x 500 x 310.15)/(296.15 x 720)
V₂ = (93045000)/213228
V₂ = 436.36cm³
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(6a)
In longitudinal waves, the particles transfer energy from one point to another by vibrating in the line of direction of propagation of the wave. The movement of particles is usually parallel to the movement of energy WHILE In transverse waves, the particles transfer energy from one point to another point by vibrating perpendicularly in the direction of the propagation of the wave. The particles are displaced perpendicular to the direction that the wave travels.

(6b)
(i) sound waves
(ii) seismic waves.
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

(7i)
The angle between the incident ray and the normal is equal to the angle between the reflected ray and the normal.

(7ii)
(i) Colorimeter
(ii) spectrophotometer
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

(8a)
Resistivity of a wire is the resistance per unit length and cross-sectional area. It is the property of the material that opposes the flow of charge or the flow of electric current. The unit of resistivity is ohm metre. We know that R = ρ L / A.

(8b).
R = ρ L / A.
ρ = R A /L
A = ρ L / R
A= 3.0 x 10^-6 x 4 / 0.5
A = 1.2 x 10^-5 / 0.5
A = 2.4 x 10^-5 m²

(12)

(12b)

(12c)

(13AandB)

(13C)

(15)

(15contd)

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