WAEC Physics 2025 Theory — Question 29
Question 29 of 34 from the West African Examinations Council (WAEC) Physics 2025 Theory paper, with the correct answer and a full explanation.
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1(a) You are provided with a G-clamp, a metre rule, masses, sellotape and a stop watch. Clamp the metre rule firmly to the workbench such that 90 cm of its length is projected outward, with 10.0 cm clamped and 20.0 cm apart marked on the bench (100cm ... 90cm ... 0cm as shown). Use sellotape to attach a mass, M = 150g at the free end of the metre rule. Depress the loaded end of the metre rule slightly and determine the time, t, for 10 complete vertical oscillations. Determine the period, T, of one complete oscillation. Repeat the procedure for four other values of M = 200g, 250g, 300g and 350g. Tabulate the results, plot a graph of Log T on the vertical axis and Log M on the horizontal axis, determine the slope, s, of the graph, and use the graph to determine the time, t₀, for 10 complete vertical oscillations of the cantilever when a mass, M₀ = 280g is placed at the free end of the rule. State two precautions taken to ensure accurate results.
Model answer
Sample results (n = 10 oscillations): M(g): 150, 200, 250, 300, 350; t(s): 6.000, 6.500, 7.200, 7.800, 8.400; T = t/n (s): 0.600, 0.650, 0.720, 0.780, 0.840; log T: −0.222, −0.187, −0.143, −0.108, −0.076; log M: 2.176, 2.301, 2.398, 2.477, 2.544. A graph of log T (y-axis) against log M (x-axis) gives a straight line. Slope, s = ΔlogT/ΔlogM = (0.08−(−0.19))/(2.54−2.30) ≈ 0.27/0.24 ≈ 0.458 ≈ 0.46 (taking the magnitude). For M₀ = 280g, log M₀ = log280 ≈ 2.45. Reading off the graph, log T₀ ≈ −0.12, so T₀ ≈ 0.759 s, giving t₀ = 10×0.759 ≈ 7.59 s. Precautions: avoided draughts throughout the experiment; avoided parallax error when taking readings on the stopwatch and metre rule scale; ensured smooth and regular oscillations of the rule in the vertical plane.
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