Waec Physics Practical 2021 Answers – 24th September 2021

Waec timetable 2021
Waec timetable 2021

Waec Physics Practical Answers 2021: Waec Physics Practical Questions and Answers – See all the answers and questions on Physics Practical for this year’s AUGUST/SEPTEMBER Waec Physics Practical exams Expo here.

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NO 1

NO 1X GRAPH

 

(1bi)
The period of an oscillatory motion is the time taken for a body to complete one oscillation.

(1bii)
-TABULATE-
(i)Mass is a scalar quantity WHILE weight is a vector quantity.
(ii)The SI unit of mass is the kilogram (kg) WHILE the SI unit of weight is Newton (N).
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°′
(3av)
Tabulate

S/N ||1 || 2 || 3 || 4 || 5

d(cm) || 80.00|| 70.00 || 50.00 || 40.00 || 30.00

I(A) || 0.30||0.35 || 0.40|| 0.45||0.50 ||

(I)^-¹ ||3.333 ||2.857 ||2.500 || 2.222||2.000 ||

(3avii)
from the graph, Slope s, = ∆d/∆I^-¹
= 71 – 20 ÷ 3 – 1.75
= 51/1.25
= 40.8cmA
Or 0.408Am

(3aviii)
(i) I ensured tight connections
(ii) I avoided error due to parallax while reading the ammeter

(3bi)
(i) Straight at the magnetic field
(ii) The number of turns of it’s coils

(3bii)
Given:
Diameter d, = 0.6cm = 6×10^-³m
Resistivity, e = 1.0×10^-6 Ωm
Resistance R = 4Ω

Using R = eL/A
L=RA/e
= Rπd²/4e
= 4 × 22/7 × 36 × 10^-6 ÷ 4 × 1 × 10^-⁶
= 22/7 × 36
= 113.14m
:. L = 113.14metres


(3a)

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Waec Physics Practical Answers 2021 for 23rd September 2021

Friday, 24th September 2021

  • Physics 3 (Practical) (Alternative A) 9:30 am. – 12:15 pm. (1stSet) Physics 3 (Practical) (Alternative A) 12:45 pm.–3:30 pm.(2ndSet)

EXAM TYPE: Waec 2021

SUBJECT: Physics Practical

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Waec Physics Specimen 2021 (PHYSICS APPARATUS)

QUESTION 1
  • Two-metre rules
  • Two Retort stand and clamp
  • Thread (about 250cm long)
  • Stopwatch Or Clock
  • A weighing balance
  • Splits Corks
  • A set of Masses (Two 10g, two 20g and two 50g)
QUESTION 2
  • Ray box with illuminated Object (Crosswire)
  • Convex Lens(f=15cm)
  • Screen
  • Metre Rule
  • Lens Holder
QUESTION 3
  • Accumulator or two Dry Cells (2×1.5V)
  • Ammeter (0—1A)
  • Crocodile Clip
  • A 2Ω Resistor
  • Key
  • Constant Wire (About 1m long)
  • Connecting Wires

CONDUCTING THE PRACTICAL PHYSICS TEST

3. The following announcement should be made to the candidates at the beginning of the examination:

‘The Examiner does not want you to waste your time because you cannot proceed with an experiment. Any candidate, therefore, who cannot get on with the experiment after spending 15 minutes on it, may come to me and ask for help.’

4. The Physics Teacher should be allowed to give a ‘Lint to a candidate who is unable to proceed with the experiment.

5. The following regulations must be strictly adhered to.

(a) No hint should be announced to the candidates as a whole. (b) A candidate who is unable to proceed and requires assistance must come up to you and state his or her difficulty. The candidate should be told that the Examiner will be informed of any assistance given in this way. (c) A note must be made, on the Report Form, of any assistance given to any candidate, with the name and index number of the candidate.

REPORT FORMS

6. Report Forms are provided separately, on which you are required to state the following:

(a) detailed information about the apparatus where necessary; (b) any particular difficulties experienced by any candidate, especially if the Examiner would be unable to discover these from the scripts;

(c) any departure from the specified items which could not be avoided; (d) any assistance given to the candidates under the regulations explained in 5 above.

7. In cases where several sets of apparatus are provided for a question and specific information about the appAratus is required by the Examiner, each piece of apparatus about which the information is required, and which may differ slightly for different candidates. must he clearly m, ked letters A, B, C, etc. and the corresponding information must be clearly reported, showing which apparatus was supplied to which candidate. The candidates should be instructed to record these letters in their scripts.

8. A completed Report Form must be enclosed in each envelope of scripts.

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WAEC Physics Practical Past Questions and Answers

Waec Past Questions and Answer

The West African Senior School Certificate Examination (WASSCE) is a type of standardized test taken in West Africa, mostly by students who wish to proceed to the higher institution. It is administered by the West African Examination Council (WAEC).

The resources below on Physics Practical have been provided by WAEC to assist you to understand the required standards expected in Physics Practical final Examination. Students’ performance in examination under review was done by the Chief examiner Questions.

The contents in each WASSCE Physics Practical question paper (for a specific year) are usually similar from one country to another. Questions on the WASSCE Physics Practical section may be specified to be answered by candidates.

The standard of the paper was good and did not deviate from those of the previous years.  The questions were straightforward, unambiguous, and spread to cover most aspects of the Syllabus.  The rubrics were clearly stated.

However, there has been so much significant improvement in the overall candidates’ performance compared with those of the previous years because of the Past question that student engages themselves.

WAEC Physics Practical PAST QUESTION AND ANSWERS FOR 2021/2021

Carefully read and take note of the following questions and try to answers by choosing the option that best suits the question after that use your textbook to confirm it. Let get started with the Waec Past Questions and Answer

2021 NECO PHYSICS-PRACTICAL

(1)

Tabulate

reading- 1,2,3,4,5

masses(m)- 50,53,55,58,60 depth(d)-5.50,5.80.6.00,6.30,6.60 ti(s)- 6.45,6.80,6.85,7.40,7.45 T=t/n(s)- 0.655,0.680,0.685,0.740,0.758 T^2(s^2)- 0.429,0.462,0.469,0.548,0.575 Li(g/cm)- 5.455,5.776,5.833,6.032,6.154 Mli(g)-30.0,33.5,35.0,38.0,40.0

1aix) i)I avoided parallax error in reading clock/weight balance ii)I ensue repeated readings are taken to avoid random error

1bi) i)weight ii)upthrust iii)viscous force/drag or liquid friction

1bii) The amplitude of oscillation of the loaded test tube decreases with time because of the viscous drag(force) which opposes the motion.

(1xii) (i)i avoided draught. (ii)i avoided error due to parallax when reading the meter rule. (iii) I ensured steady oscillation. (iv) I avoided zero error to the stop clock. =====================================

3)

Table W1 s/n-1,2,3,4,5,6 xn(cm)/d1=0.009m- 3.06,3.50,3.70,4.30,4.50,4.80 xm(cm)converted/- 15.30,16.50,18.50,21.50,22.50,24.40 xm(cm)converted d2=0.005m- 84.70,83.50,81.50,78.50,77.50,75.60 Ri(n)- 11.07,10.12,8.81,7.30,6.89,620

Tabulate W2 s/n-1,2,3,4,5,6 xmi(cm)- 1.40,1.65,1.80,2.20,2.55,2.94 converted/xmi(cm)7.00,8.25,9.00,11.00,12.75,14.70 xni(cm)- 93.00,91.75,91.00,89.00,87.25,85.3 R2i(n)- 26.57,22.24,20.22,16.18,13.69,11.61 slope dy/dx=26.57-13.69/11.07-6.89 =12.88/4.18 =3.08

3ix) k=d2/d1*squareroot5 =0.005/0.009*squareroot13.08 =0.56*1.75 =0.98

3x) i)i would ensure that the terminal are well tightened to avoid partial contact ii)i would obey the conect connection of positive to positive to negetive to negetive when connecting to circuit

3bi) when temperature increases the molecules tends to vibrate more intense there by opposing the flow of current

3bii) P=I^R P=V^2/R power=1000W Voltage=200V R=? 1000=200^2/R R=40000/1000 R=40

(3xi) (i)I ensured tightened connections. (ii) I avoided error due to parallax on the ammeter. (iii) I removed key when reading was not taken (iv) Avoidance of repeated reading from table ====================================

CHOOSE ANY PERFECT TWO FROM EACH

More loading 👇👇👇

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2021 NECO PHYSICS-PRACTICAL

(1xii) (i)i avoided draught. (ii)i avoided error due to parallax when reading the meter rule. (iii) I ensured steady oscillation. (iv) I avoided zero error to the stop clock. =====================================

(3xi) (i)I ensured tightened connections. (ii) I avoided error due to parallax on the ammeter. (iii) I removed key when reading was not taken (iv) Avoidance of repeated reading from table ====================================

CHOOSE ANY PERFECT TWO FROM EACH

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(1xii) (i)i avoided drought by closing the window (ii)i avoided error due to parallax when reading the meter rule =====================================

(3xi) (i)I ensured tight connections (ii) I avoided error due to parallax on the ammeter ===================================== 1. Thermodynamics, science of the relationship between heat, work, temperature, and energy. In broad terms,thermodynamics deals with the transfer of energy from one place to another and from one form to another. 2. THERMO DYNAMIC SYSTEM: A system is defined as a region in space containing a quantity of matter whose behaviour is being investigated. This quantity of matter is separated from its surroundings by boundary, such as they walls of a vessel. (3a) (xii) (i)I ensured tight connections on all terminals (ii) I avoided error due to parallax on the ammeter

(3bi)Period refers to the amount of time it takes a wave to complete one full cycle of oscillation or vibration. Frequency, on the contrary, refers to the number of complete cycles or oscillations occur per second. Period is a quantity related to time, whereas frequency is related to rate. Period simply refers to the time for something to occur periodically, whereas frequency means how often that happens.

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More coming

COMPLETED…..

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USE THE DIAGRAM AS A GUIDE TO CARRY OUT THE FOLLOWING EXPERIMENT

📌 Trace the outline ABCD of the rectangular glass prism on the drawing paper provided. 📌 Remove the Prism. Select a point N on AB such that AN is about one quarter of AB. 📌 Draw the normal LMN. Also draw a line RN to make an angle ∅° 75° with AB at N 📌 Fix two pins at P1 and P2 on line RN. Replace the prism on its outline. 📌 Fix two other pins at P3 and P4 such that they appear to be in a straight line with the image of the pins at P1 and P2 when viewed through the prism from side DC. 📌 Remove the Prism and the pins at P3 and P4. Draw a line to join P3 and P4 Chat Exammovement for your runzs ,07062154881, 📌 Produce line P4 P3 to meet the DC at O. Draw a line to join NO. 📌 Evaluate ∅ = MO/NO and Cos∅ 📌 You will be given some values to find e.g. 30, 65, 100, and so on. Tabulate the values plot a graph vertical and horizontal determine the slope on the graph

📌 Note the precautions below👇🏿👇🏿

1. I ensured that pins are vertically erect 2. I ensured that pins are reasonably spaced 3. Ensured neat traces

Some Questions that are likely to come out 📌

I. Explain the term refractive index and give the mathematical expression for it in terms of wavelength.

II. State the conditions necessary for total internal reflection to occur for a given pair of media. III. State Snell’s law of refraction 📌📌📌

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Physics-ExpoBite.Com!
1BCEDBEDBBD 11DEDCAEDCEB 21DDBBCBDBDC 31BEDBBCAEDD 41EDACEECCED 51DCECECDBCB ==================== ====================
==================================== Physics-THEORY (1a) (i) A multimeter (ii) Reads if there is short circuit in the electric circuit (1b) (i)A simple microscope uses one lens while a compound microscope uses two lens (ii)A compound microscope produces a higher magnification than that of simple microscope (1c) Using M= ZIt I = M/Zt I = 450/3.30*10-⁴ *25*60 I = 909A ==================== ==================== (3a) Range (R) is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again (3b) Time of flight: T= 2usinθ/g sinθ = Tg/2u = 1.5*10/2*12 sinθ = 15/24 = 0.625 θ = sin-¹(0.625) θ = 38.7° ==================== ==================== (4a) (i)Flying bullet (ii)A moving car (4b) mgh = mc∆θ ∆θ = gh/c = 10*108/4200 = 0.257°C or 0.257K ==================== ==================== (6a) A stationary wave is produced by a superposition of two waves of the frequency and amplitude which are exactly out of phase with each other (6b) Wavelength = 4/2.5 = 1.6m ==================== ==================== (5a) Vaporization is the conversion of a substance from the liquid or solid phase into the gaseous (vapour) phase. (5b) (i) Deform (ii) No change ==================== ==================== (8a) Lost Volt refers to the amount of energy lost (in a battery, generally) to energy resistance of components. It is generally fairly negligible, but to calculate: V=IR where I is the current and R is resistance. (8b) (i) Secondary cell last long and can be used over again and again. (ii) A secondary cell is rechargeable meaning that the chemical reactions taking place in the cell are reversible and can be carried out in the opposite direction to charge the cell again. ==================== ==================== Expobite Section B. ==================== ==================== (12)

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(15)

==================================== We Remain Your Favourite Site. Ensure You Subscribe For Your Next Paper @ Www.ExpoBite.Com ===================================== On How To Get Your Live Question And Answer before Exam. Www.ExpoBite.Com We Remain The No1 ===================================== PHYSICS PRACTICAL NECO 2020 QUESTIONS AND ANSWERS ==================== ANSWERS LOADING…….. ==================== KEEP REFRESHING THIS PAGE ==================== (1) Tabulate. S/N; 1|2|3|4|5| M|g; 40.0|35.00|30.00|25.00|20.00| θ°; 41.80|35.70|30.00|24.60|19.47| ϕ=(90-θ°); 48.20|54.30|60.00|65.40|70.53| Cosϕ; 0.6665|0.5835|0.5000|0.4163|0.3333| (1xi) Slope , S=Δm|g/ΔCosϕ =(40-20)g/0.6665-0.3333 S= 20g/0.3332=60.024g (1xii) K=s/2 =60.024g/g= 30.012g (1xiii) -Precautions- (i) I avoided error due to parallax when reading from the protector (ii) I ensured that the force board is firmly and correctly clamped to the retort stand (1) Graph (1b) Coming check by 8;10am ==================================== (3) Tabulate S/N; 1|2|3|4|5| R/Ω; 2.00|4.00|6.00|8.00|10.00| L/cm; 50.00|66.70|75.00|80.00|83.30| B/Ω^-¹; 0.500|0.250|0.167|0.125|0.100| Δ/cm; 1.000|0.499|0.333|0.250|0.200| (3xiii) Slope ,S=ΔB/Ω^-¹/ΔΔ/cm =0.5-0.1/1.0-0.2=0.4/0.8=0.5 Slope =0.5Ω^-¹/cm (3xiv) K=S^-¹= 1/5=1/0.5Ω^-¹/cm =2Ωcm (3xv) -precautions- (i) I avoided error due to parallax when reading from the potentiometer (ii) I ensured that the wire are tightly connected. (3) Graph (3bi) (i) potentiometer give exact value of potential difference while voltmeter draws apart of the current (ii) potentiometer can be used for measuring the internal resistance of a cell which cannot be done with voltmeter

*NECO 2020 PHYSICS OBJ* 1BCBDBEEDCD 11DACBEBABDE 21CDDDABBBAC 31CDAEBBCCDC 41BEEBDBAEEC 51DDCDADDEBD

======================

*NECO 2020 SECTION A*

*ANSWERS ONLY 6 FROM THIS SECTION NUMBER 1 IS COMPULSORY* ===================================== PART I

(1a) (i)printing machine (ii)sterilizer

(1b) Diagram (step up transformer)

(1c) 1=1.26 t=48005 m=? z=3.3*10^-⁴g m=3.3*10^-⁴*1.26*4500 =1.996gram ≅2gram =====================================

(2a) A system is unstable when displaced, it experiences a net force in the same direction as the displacement from equilibrium. It tends to accelerate away from its equilibrium position if displaced even slightly while for neutral equilibrium, the system is independent of displacement from its original position

(2b) (ii)Low centre of gravity (ii)Large surface area in contact =====================================

(3a) Velocity is defined as the rate of change in the displacement of a body it is a vector quantity with the S.I unit of ms-¹

(3b) DIAGRAM =====================================

(4) At equilibrium position Point B, the body possess maxim kinetic energy. As it moves out to end point A and C. It stops momentarily before changing its direction.

(Diagram) =====================================

(5a) They are both derived quantities

(Diagram)

(5b) 67-0/100-0 = x-25/100-25 67/100 = x-25/75 61*3 = 4(x-25) 201 = 4x-100 4x = 201+100 4x = 301 x = 301/4 = 75.25 ohms =====================================

(6a) (i)violin (ii)guitar

(6b) The depth of field of a lens camera is less than that of a pinhole camera =====================================

(7a) – An amplifier – A speaker

(7b) u = 16cm m = 4=v/u :. v =4u v = 4×16 = 64cm Recall, 1/f = 1/u + 1/v 1/f = 1/16 + 1/64 1/f = 4+1/64 1/f = 5/64 f = 64/5 = 12.8cm =====================================

(8a) The zinc atoms on the surface of the anode oxidizes I;e they give up both their valence electrons to become positively

(8b) Diagram =====================================

(9a) The power factor of a circuit is unity when the apparent power equals or some as the real power

(9b) V=Vr+Vc 15=12+Vc Vc=15-12 Vc=3v =====================================

(10a) (i)Viscosity (ii)weight (iii)upthrust

(10bi) (i)Viscosity; the force in between the layers of flurds this force would tend to reduce the velocity at which the stone moves down till it reach terminal velocity

(ii)weight; is the actual force acting of the stone in the liquid.

(iii)upthrust; is the upward force felt by reply objects in flurd, due to the area displace =====================================

(11i) A=9 (molecular mass) B=6 (atomic mass) C=0

(11ii) Z=Energy

(11iii) Nuclear fusion =====================================

 

8ai) Work is the measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.

(8aii) Energy is defined as the “ability to do work, which is the ability to exert a force causing displacement of an object.”

8d)At constant pressure, the number of collision per unit time is constant. An increase in temperature causes average Kinetic energy increases. The molecules then move faster. The volume therefore increases so as to maintain the number of collision per unit time.

(9ai) The momentum of a body is the product of its mass and its velocity.

(9aii) The total momentum of a system before collision equals to the total momentum after collision provided no external force acts. OR In a system of colliding bodies the total momentum is constant provided no external force acts.

9b) (b) Mp Up + Mq Uq = Mp Vp + Mq Vq

0.25 Up + 0 = 0.25 x 23 Up + 0.25 x 2; 0.25

(Up – 2/3 Up) = 0.50; 1/3

Up = 0.500.25 = 2m/s

Up = 2 x 3 = 6m/s

(9c) Newton’s second law of motion states that the time rate of change of linear momentum is directly proportional to the force causing the change and the change takes place in the direction of the force

(9d) Resultant force on the vehicle = 9600 – 200 = 7400N.

But F = ma

a = Fm=74003400 = 2.18ms−2 or 2.2ms−2

(10b)

Heat lost = 0.024 x 400 (230 – T).

Heat gained by water = 0.054 x 4200 (T – 31)

Heat gained by calorimeter = 0.060 x 400 (T – 31)

Heat lost = Heat gained = 0.024 x 400 (230 – T) = 0.054 x 4200

(T – 31) + 0.060 x 400 (T – 31) T = 38.34°C

 

*_AREA OF CONCENTRATION FOR PHYSICS WAEC_*

🔘Equilibrium of bodies

🔘Motion, Speed and Velocity

🔘Rectilinear Acceleration

🔘Simple Harmonic Motion

🔘Equilibrium of Forces

🔘Newtons Law of Motion

🔘Energy, Work and Power

🔘Heat Energy and Wave

🔘Optics and Fields

 


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