Step-by-step Derivations

1. Analyze the given information:

• Length of the pendulum, L = 1 m. The word "exactly" implies that the error in length, ΔL, is zero.
• Number of oscillations, n = 20.
• Total time measured for n oscillations, t = 40 s.
• Least count of the stopwatch, which is the error in the time measurement, Δt = 1 s.

2. Calculate the time period (T): The time period T is the time taken for one complete oscillation. $T = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{t}{n}$ $T = \frac{40 \text{ s}}{20} = 2 \text{ s}$

3. Calculate the error in the time period (ΔT): The error in the time period ΔT can be calculated using the formula for error propagation. Since T = t/n, the relative error is: $\frac{\Delta T}{T} = \frac{\Delta t}{t} + \frac{\Delta n}{n}$ Since the number of oscillations n is an exact count, its error Δn is 0. $\frac{\Delta T}{T} = \frac{\Delta t}{t}$ Now, we can find the absolute error ΔT: $\Delta T = T \times \left( \frac{\Delta t}{t} \right)$ $\Delta T = 2 \text{ s} \times \left( \frac{1 \text{ s}}{40 \text{ s}} \right) = \frac{2}{40} \text{ s} = \frac{1}{20} \text{ s} = 0.05 \text{ s}$ Alternatively, ΔT = Δ(t/n) = (1/n)Δt = (1/20) * 1s = 0.05s.

• Evaluate Option A and B:
  • Option A states that the error ΔT in measuring T is 0.05 seconds. This matches our calculation. So, Option A is true.
  • Option B states that the error ΔT is 1 second. This is incorrect; 1 second is the error in the total time t, not the time period T. So, Option B is false.

4. Calculate the percentage error in the determination of g: The formula for the time period of a simple pendulum is: $T = 2\pi \sqrt{\frac{L}{g}}$ Squaring both sides and rearranging for g, we get: $g = \frac{4\pi^2 L}{T^2}$ The relative error in g is given by: $\frac{\Delta g}{g} = \frac{\Delta (4\pi^2)}{4\pi^2} + \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$ Since $4π^2$ is a constant, its error is zero. The problem states L is exactly 1m, so ΔL = 0. $\frac{\Delta g}{g} = 0 + 0 + 2 \frac{\Delta T}{T} = 2 \frac{\Delta T}{T}$ We can use the values we have: $\frac{\Delta T}{T} = \frac{0.05 \text{ s}}{2 \text{ s}} = 0.025$ So, the relative error in g is: $\frac{\Delta g}{g} = 2 \times 0.025 = 0.05$ The percentage error in the determination of g is: $\text{Percentage error} = \frac{\Delta g}{g} \times 100\% = 0.05 \times 100\% = 5\%$

• Evaluate Option C and D:
  • Option C states that the percentage error in the determination of g is 5%. This matches our calculation. So, Option C is true.
  • Option D states that the percentage error is 2.5%. This is incorrect; 2.5% is the percentage error in the time period T (0.025 * 100%), not in g. So, Option D is false.

Conclusion: Based on the calculations, statements A and C are true.