2022 WAEC Physics Essay and Objective Questions and Answers Now Available

Wednesday, 15th June 2022
Physics 2 (Essay) 09:30am – 11:00am
Physics 1 (Objective) 11:00am – 12:15pm

2022-WAEC-PHYSICS-ESSAY-AND-OBJECTIVE-ANSWERS

PHYSICS-OBJ- Examcode.net
PHYSICS OBJ
01-10: BCABBDDCDA
11-20: CAACABBBBC
21-30: BDCCBDABDA
31-40: CABBCBACCD
41-50: DDBDDDCBDD

COMPLETED.
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Also Read: How To Receive NECO Questions and Answers Midnight Before the Exam Time.

PHYSICS-ESSAY- Examcode.net
INSTRUCTION: Answer Eight (8) Questions in All. Five Questions From Part I and Three Questions From Part II.
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PART I
Answer Any Five Questions From This Part.
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(1)
Range /PQ/ = u²sin2θ/g
u = 250ms-¹
θ =42°
= [250² x sin(2×42°)]/10
= (62,500 x sin84°)/10
= 6250 x 0.9945
= 6215.63m
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(2)
(i) They are pure semiconductors with impurities less than 1 and have teravalent crystals
(ii) Presence of mobile charge carriers is intrinsic property of material
(iii) No doping or addition of impurities required.
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(3)

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(4)
F = 50N
k = 100Nm-¹
F = ke
50 = 100e
e = 50/100 = 0.5m
Work done = (1/2)ke²
= 1/2 x 100 x 0.5²
= 25/2
= 12.5joules
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(5a)
A geostationary satellite is an earth orbiting satellite placed at an altitude of approximately 35,800km directly over the equator that revolves in the same direction which the earth rotates.

(5b)
(i) Gas lasers
(ii) Solid-state lasers
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(6a)
The S.I unit is watt per steradian (W/sr)

(6b)
(i) As the temperature of the black body increases, the peak wavelength decreases.
(ii) The total energy being radiated is the area under the graph.
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(7)
(i) Magnetic materials are materials having a magnetic domain and are attracted to an external magnetic field WHILE Non-magnetic materials are materials that are not attracted to an external magnetic field.
(ii) The magnetic domains of magnetic materials are aligned either parallel or antiparallel arrangements thus they can respond to a magnetic field when they are under the influence of an external magnetic field WHILE The magnetic domains of non-magnetic materials are arranged in a random manner in such a way that the magnetic moments of these domains are cancelled out. Thus, they do not respond to a magnetic field.
(iii) Magnetic materials are used to make permanent magnets are the parts of operating systems where magnetic properties are required WHILE Non-magnetic materials are used to make parts of some operating systems where no magnetic effect are expected and other things such as cases of compasses.
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PART II
Answer Three Questions From This Part.
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(8ai)
The law of inertia states that a body will continue in its present state of rest or if it is in motion, will continue to move with uniform speed in a straight line unless it is acted upon by a force.

(8aii)
When the brakes of the moving vehicle are suddenly applied, or the passengers jerks forward as thet tend to continue in a straight line. The seat belt helps to provide a backward force and reduces injuries.

(8bi)
The moment of a force about a fixed point is the product of the force and the perpendicular distance at the point to the line of action of the force.

(8bii)
(I) Draw the Diagram

(II) clock wise moment = Anticlock wise moment
T x 1.5 = 54 x 0.5
T = 54×0.5/1.5
T = 18N

(III) F + T = w
F = w – T
F = 54 – 8
F = 36N

(8c)
[TABULATE]

=STATIC FRICTION=
(i) Independent of surface area in contact
(ii) Depends on normal reaction ( of one solid surface on the other)
(iii) Occurs between solids surfaces in contact

=VISCOSITY=
(i) Depends in surface area in contact
(ii) Independent of normal reaction
(iii) occurs between fluid layers/solid and fluid in contact/in fluids.

(8d)
By increasing the number of pulleys.

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(9)

(9ai)
Coulomb’s law states that the electric force between two point charges, q₁ and q₂ separated by a distance r is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges.

(9aii)
Given r = 5.28×10-¹¹m
k = 9.0×10⁹mF-¹
q₂ = q₁ = 1.6×10⁻¹⁹C = q
Electrostatic force F = kq²/r²
F = [9.0×10⁹(1.6×10⁻¹⁹)²]/(5.2×10⁻¹¹)²
F = 0.852×10⁻¹⁶N
F = 8.52×10⁻¹⁷N

(9bi)
(i) Current I = Vin/R₁+R₂
I = Ir₁ =Ir₂
Vout = Vr₂ = Ir₁ x R₂
Vout = [Vin/(R₁+R₂)] x R₂
Vout = Vin[R₂/(R₁+R₂)]

(ii) Given;
Vin/Vout = (R₁+R₂)/R₂ = 2.5
Where R₂ = 30Ω
(R₁+30)/30 = 2.5
R₁+30 = 75
R₁ = 75 – 30
R₁ = 45Ω

(9bii)
The volt is the work done of one joule in taking one coulomb of positive charge from one point to another.

(9c)
A wooden core cannot provide path to the flux which is needed to link either the secondary or primary side

(9d)
Used in medicine for visual functions.
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(11a – 11c)