Error Analysis in Simple Pendulum Experiment for Gravity

To determine the correct statements, we analyze the measurements and error propagation for a simple pendulum.

1. Error in measuring the time period $T$:
The time period $T$ is given by the total time $t$ divided by the number of oscillations $n$:
$T = \frac{t}{n}$

Given:

• Total time measured, $t = 40\text{ s}$
• Number of oscillations, $n = 20$
• Least count of the stopwatch, $\Delta t = 1\text{ s}$

The error $\Delta T$ in the measurement of the time period $T$ is:
$\Delta T = \frac{\Delta t}{n} = \frac{1\text{ s}}{20} = 0.05\text{ s}$

Thus, Option A is correct and Option B is incorrect.

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2. Percentage error in the determination of $g$:
The relationship between the time period $T$, length $L$, and acceleration due to gravity $g$ is:
$T = 2\pi\sqrt{\frac{L}{g}} \implies g = \frac{4\pi^2 L}{T^2}$

Since the length is specified to be "exactly $1\text{ m}$", there is no uncertainty in the length measurement ($\Delta L = 0$).

The fractional error in $g$ is obtained by differentiating:
$\frac{\Delta g}{g} = \left| -2\frac{\Delta T}{T} \right| = 2\frac{\Delta T}{T}$

Substituting $\frac{\Delta T}{T} = \frac{\Delta t / n}{t / n} = \frac{\Delta t}{t}$:
$\frac{\Delta g}{g} = 2\left(\frac{\Delta t}{t}\right) = 2 \times \left(\frac{1\text{ s}}{40\text{ s}}\right) = \frac{2}{40} = \frac{1}{20} = 0.05$

The percentage error in the determination of $g$ is:
$\text{Percentage error} = \frac{\Delta g}{g} \times 100\% = 0.05 \times 100\% = 5\%$

Thus, Option C is correct and Option D is incorrect.

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Conclusion:
The true statements are A and C.