WAEC Physics May/June 2008-2024 Paper (1&2) Questions and Answers

Get free questions and answers on WAEC Physics Paper 1, May/June 2008, 11, 12...... 2024 WAEC Physics Paper 1, MAy/June.  WAEC Physics Nov/Dec. Questions and Answers.

WAEC Physics

WAEC has provided resources to aid students in grasping the standards required for the Physics final examination. 

These materials encompass general comments, performance analysis, noted strengths and weaknesses, and anticipated answers to questions. 

They serve as valuable guidance for students preparing for the exam, offering insights into effective question approaches and meeting expected standards. 

It's advisable for students to carefully review these resources to enhance their understanding and performance in the exam.

General WAEC physics instructions:

The standard of the two alternative papers (A & B) compared favorably with each other and with those of the previous years. The questions were well framed in simple language while marking schemes were quite adequate and clear.

The papers tested the candidates' ability to:

• Set up simple experiments
• Carry out the experimental procedure
• Collect and analyze data
• Make deductions from the analysis
• Show understanding of the theories behind the experiments

The performance of the candidates was quite similar to those in previous years. The candidates recorded a mean score of 25 and a standard deviation of 10.25 as against a mean score of 25 and standard deviation of 11.14 in 2007. 

WAEC Physics Paper 1, May/June Answers

Section A

Step	Description
1	Suspend metre rule horizontally on knife edge and record point of balance G
2	Suspend object W at 15 cm mark using thread
3	Suspend mass M (20 g) on other side of G and adjust until metre rule balances
4	Record position Y of mass M and distances L (between mass and G) and D (between W and G)
5	Repeat procedure for M = 30, 40, 50, and 60 g, keeping W constant at 15 cm mark and knife edge at G
6	Evaluate L⁻¹ in each case and tabulate readings
7	Plot graph of M against L⁻¹ and determine slope s
8	Evaluate s/D

Section B

Question	Answer
State principles of moments	Sum of clockwise moments equals sum of anticlockwise moments about a point
Define centre of gravity	Point through which resultant weight acts

Precautions

Precaution
Avoid draughts
Ensure parallax error is avoided when reading metre rule
Take repeated readings
Ensure suspended masses do not touch the table

Physics Paper 1 Question 2

Section A

Step	Description
1	Secure meter rule on bench with graduated side facing upwards
2	Position illuminated object at 0 cm end and screen at 100 cm end, ensuring d = 100 cm
3	Record d and calculate d^2
4	Place converging lens between object and screen, adjusting until sharp image formed
5	Read and record position l1 of lens
6	Move lens towards object until another sharp image formed, read and record l2
7	Calculate L = (l1 – l2), L^2, and D = d^2 - L^2
8	Repeat procedure for d = 85, 75, 65, and 55 cm, tabulate readings
9	Plot graph of D against d, determine slope s
10	Evaluate k = s/4
11	State two precautions taken to ensure accurate results

Section B

Question	Answer
Explain focal length of converging lens	Distance between optical center and principal focus
Differentiate between real and virtual image	Real: formed by actual rays, virtual: formed by imaginary rays

Precautions

Precaution
Avoid parallax error when reading meter rule
Ensure correct alignment of optical components
Ensure vertical placement of lens
Take repeated readings (must be shown on table)

Physics Paper 1 Question 3

Section A

Step	Description
1	Connect Rx, wire A, and apparatus as shown
2	Find and record balance point P, lx, and ly
3	Calculate R1 = (ly / lx) * Rx
4	Repeat for four other lengths of wire A
5	Repeat experiment with wire B, find balance points P, record lx and ly, and calculate R2
6	Tabulate results
7	Plot graph of R2 against R1, determine slope s, and calculate k = √s
8	List two precautions taken to ensure accurate results

Section B

Question	Answer
Advantages of potentiometer over voltmeter	High accuracy, null deflection, range can be increased, no friction in moving parts
Define internal resistance of a cell	Opposition to current flow through electrolyte of cell

Precautions

Precaution
Key removed when not taking readings
Ensured tight connections/clean terminals
Avoided parallax errors in reading meter rule
Jockey not allowed to scratch resistance wire
Zero error of meter rule noted and corrected

Physics Paper 1 Question 4

Section A

Step	Description
1	Determine and record centre of gravity of meter rule
2	Fix 100 g mass at 80 cm mark, suspend another 100 g mass at 10 cm mark
3	Balance arrangement horizontally on knife edge, measure and record distance B
4	Repeat for four other values of A, measure and record corresponding values of B
5	Tabulate readings, plot graph of B against A, determine slope s and intercept c
6	Evaluate k₁ = 1 – 2s * 100, k₂ = 2c * 160
7	State two precautions taken to obtain accurate results

Section B

Question	Answer
Define moment of a force about a point	Product of force and perpendicular distance from point to line of action
Conditions for rigid body equilibrium	No rotation, forces must be concurrent, resultant force equals zero, algebraic sum of resolved components equals zero

Precautions

Precaution
Draught avoided
Ensure mass did not rest on table
Avoided error due to parallax when reading meter rule
Repeated readings shown on table
Zero error of meter rule noted and corrected

Physics Paper 2, Nov/Dec. 2011

Question 1

Part A

Step	Description
1	Stone projected vertically upward with speed 30 ms⁻¹ from top of 50 m tower
2	Determine time of flight on reaching ground, neglecting air resistance

Expected Answer

Step	Description
1	Time taken to reach maximum height
2	h = ut - 1/2gt², substitute values and solve for t
3	t = 7.4 s or -2.07 s (reject negative value)
4	OR, use V = u - gt to find time to reach maximum height
5	t = 3 s
6	Maximum height reached: V² = u² - 2gx, substitute values and solve for x
7	x = 45 m
8	Time taken to reach ground from maximum height: x = ut + 1/2gt²
9	t = √(90/5) = 4.24 s
10	Time of flight: 3 + 4.24 = 7.24 s

Observation

• This question was popular among candidates but some didn't understand the difference between projectile motion from a height and from level ground.
• Some candidates used the wrong equations or substituted values incorrectly.

Question 2

A body of mass 0.5 kg is thrown vertically upwards with a speed of 25 ms⁻¹. Calculate the potential energy of the body at the maximum height reached.

Comment:

This question tested candidates understanding on the concept of conservation of energy. Majority of the candidates were able to solve this problem. However, few candidates could not solve this question correctly because they lack the knowledge that at maximum height the potential energy P.E is equal to the kinetic energy K.E.

Expected Answer:

At max. Ht, P.E = initial K.E

= ½ mu²

= ½ × 0.5 × (25)²

= 156.25 J

OR

V² = u² - 2gh

At maximum height, v = 0

0 = 25² - 2 × 10h

h = ^25²/_{2 × 10}

h = 31.25m

P.E = mgh

= 0.5 × 10 × 31.25

= 156.25J

Question 12

(a) State the principle of conservation of energy.

(b) State three effects of heat on a substance.

(c) Explain why when the bulb of thermometer is dipped into ice-cold water the mercury level first rises before falling.

(d) Water of volume 200 cm³ in an aluminum container at 20°C is cooled to – 10°C when put in a freezer.

Calculate the quantity of heat energy extracted from the water.

If heat is extracted at a rate of 120 Js^-1 by the freezer, calculate the time taken to produce this ice.

State a reason why the calculated time will be less than the actual time.

Comments:

P) This was well answered by most candidates; however, few candidates omitted the phrase "in a closed system" or "isolated system" in their statement.

(b) The three effects of heat on a substance were generally well understood by candidates, with satisfactory performance.

Explanation of how the mercury level changes when dipped in ice-cold water proved difficult for most candidates, as they could not provide satisfactory explanation. The numerical section was poorly tackled by most candidates. The formula \( m = v \) eluded most of them. Additionally, applying the formula \( Q = mc \) appropriately was another challenge due to their poor understanding of the concepts of specific heat capacity and latent heat.

Expected Answers:

(a) In an isolated/closed system, the total amount of energy is (always) constant, although energy may be transformed from one form to another.

(b) The effects of heat on a substance:

• Change in temperature
• Change in size (expansion/contraction)
• Change of phase/state

(c) The bulb comes in contact with the cold water before the mercury, causing the glass to cool first, contract, and squeeze the mercury up the stem until thermal equilibrium is reached. The mercury, being a better conductor of heat, cools faster than the glass, resulting in the fall in mercury level.

(d)(i) \( m = \rho V = \frac{1000 \times 200}{1,000,000} = 0.2 \, \text{kg} \)

\( Q = M_wC_w\Delta\theta_w + M_wL_f + M_iC_i\Delta\theta_i \)

\( = 0.2 \times 4200 \times (20-0) + 0.2 \times 3.36 \times 10^5 + 0.2 \times 2100 \times (0-(-10)) \)

\( = 16800 + 67200 + 4200 = 88200 \, \text{J} \)

(ii) \( Q = Pt \) or \( P = \frac{Q}{t} = \frac{88200}{120} = 735 \, \text{s} \) or 12.25 minutes

(iii) Why the calculated time is less than the actual time:

• No consideration is given to heat absorbed by the aluminum container
• Heat lost / absorbed from the surroundings
• Effect of impurities on the freezing point of the liquid.

Physics Paper 1, MAy/June. 2011

Question 4

(a) Diagram

You are provided with two retort stands, two metre rules, pieces of thread and other necessary apparatus.

1. Set up the apparatus as illustrated above ensuring that the strings are permanently 10 cm from either end of the rule.
2. Measure and record the length L = 80 cm of the two strings.
3. Hold both ends of the rule and displace the rule slightly, then release so that it oscillates about a vertical axis through its centre.
4. Determine and record the time t for 10 complete oscillations.
5. Determine the period T of oscillations.
6. Evaluate log T and log L.
7. Repeat the procedure for four other values of L = 70, 60, 50, and 40 cm.
8. Tabulate your readings.
9. Plot a graph with log T on the vertical axis and log L on the horizontal axis.
10. Determine the slope, s, and the intercept, c, on the vertical axis.
11. State two precautions taken to ensure accurate results.

(b)

(i) Define simple harmonic motion.

(ii) Determine the value of L corresponding to t = 12s from the graph in (a) above.

Observation

The observation and its tabulation were well handled by most responding candidates. However, the evaluation of Log L and Log T to the nearest decimal places was poorly tackled.

The plotting of the graph of log T against Log L was poorly handled. Evaluation of slope and determination of intercept on the vertical axis was fairly attempted. Candidates were able to state the precautions correctly in acceptable language.

The part b was fairly handled by most responding candidates as they were able to define simple harmonic motion but could not satisfactorily deduce the value of L from the graph when t = 12s probably because of the Logarithm involved.

In part (a) candidates are expected to:

• Measure and record five values of L to at least 1 d.p. and in trend. Trend: As L increases, t decreases
• Evaluate five values each of T, log L, and log I to at least 3 s.f.
• Record data in a composite table.
• Plot five points, correctly using reasonable scales and distinguished axes.
• Draw a line of best fit.
• Determine the intercept on the vertical axes and the slope of the graph using a large right-angled triangle.
• State any two of the following precautions in acceptable language:
  • Ensured supports of pendula were rigged
  • Avoided parallax error on meter rule/stopwatch
  • Zero error was noted and corrected on meter rule/stopwatch/clock
  • Ensured smooth and regular oscillations in a horizontal plane
  • Repeated readings shown on the table

Part b Expected Answers:

Simple harmonic motion is the motion of a body whose acceleration is always directed towards a fixed point and is proportional to the displacement from the fixed point.

T = 1.2 secs, Log 1.2 = 0.079

0.079 shown on the graph with corresponding log L read. L correctly determined.