WAEC 2026 Physics Practical Question And Answers

Teachers Report Form

1a(xxi)
(PICK ANY TWO)
(i) Ensure the oscillations are small and strictly vertical.
(ii) Start and stop the stopwatch accurately as the mass passes the same reference point.
(iii) Avoid parallax error when taking readings.
(iv) Take readings in a draught-free environment.
(v) Repeat readings and obtain the average value.

(xvii)
s₁ = Change in m / Change in T²
s₁ = (m₂ − m₁) / (T₂² − T₁²)

(xix)
s₂ = Change in m / Change in T₀²
s₂ = (m₂ − m₁) / (T₀₂² − T₀₁²)

(xx)
y = s₂/s₁
Substitute your calculated values of s₁ and s₂ from the graph and evaluate.

1(bi)
As the number of springs increases, the frequency of oscillation increases.

(bii)
Given:
k = 125 Nm⁻¹
F = 8.0 N

Using Hooke’s Law:
F = ke
e = F/k
e = 8.0/125
e = 0.064 m

Answer:
e = 0.064 m

NUMBER 1

NUMBER 3

NUMBER 3 GRAPH UPDATED

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2a(xvii)
s = Change in P / Change in Q
s = (P₂ − P₁) / (Q₂ − Q₁)

(xviii)
(PICK ANY TWO)
(i) Stir the water continuously to ensure uniform temperature.
(ii) Take thermometer readings at eye level to avoid parallax error.
(iii) Transfer the heated pendulum quickly into beaker B to minimize heat loss.
(iv) Ensure the thermometer does not touch the sides or bottom of the beaker.
(v) Read the temperature immediately after it becomes steady.

2(bi)
Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass (1 kg) of a substance by 1 K (or 1°C).

(bii)
Given:
Mass of copper ball, m₁ = 50 g =
0.05 kg

Specific heat capacity of copper, c₁ = 400 J kg⁻¹ K⁻¹

Initial temperature of copper = 100°C

Initial temperature of water = 23°C

Final temperature of mixture = 62°C

Specific heat capacity of water, c₂ = 4200 J kg⁻¹ K⁻¹

Heat lost by copper = Heat gained by water

m₁c₁(100 − 62) = m₂c₂(62 − 23)

0.05 × 400 × 38 = m₂ × 4200 × 39

760 = 163800m₂

m₂ = 760/163800

m₂ = 0.00464 kg

Mass of water = 0.00464 kg or 4.64g