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Step-by-Step Solution
Step 1: Write down the formula for the period of a simple pendulum
The period of a simple pendulum of length $l$ under gravity $g$ is given by:
$$ T = 2\pi \sqrt{\frac{l}{g}}.$$
Step 2: Express $g$ in terms of $T$ and $l$
Rearranging the above formula to solve for $g$, we get:
$$ g = \frac{4 \pi^2 \, l}{T^2}. $$
Step 3: Identify the measured quantities and their uncertainties
In this problem:
• Length of the pendulum, $l = 25.0 \text{ cm}$ (or $0.25 \text{ m}$). Suppose its uncertainty in length is $\Delta l = 0.1 \text{ cm}$ (since it is given that each measurement might have at least a 0.1 cm resolution).
• Time for 40 oscillations, $t = 50\, \text{s}$ with a stopwatch of 1 s resolution. Hence the uncertainty in the total time measurement $\Delta t = 1\,\text{s}$.
• Therefore, the time period for one oscillation is
$$T = \frac{t}{40} = \frac{50\, \text{s}}{40} = 1.25\, \text{s}.$$
• The uncertainty in the time period $\Delta T$ can be taken as
$$\Delta T = \frac{\Delta t}{40} = \frac{1\, \text{s}}{40} = 0.025\, \text{s},$$
if we distribute the total uncertainty across 40 oscillations.
However, for a simplified calculation (as is often done in short problems), you can directly take $\Delta T = \frac{1\,\text{s}}{50}$ times 2, depending on how the question frames it. In the solution, uncertainty is being handled as a fraction of the total time measurement.
Step 4: Use the error-propagation formula
For a quantity $g$ depending on $l$ and $T$ as $g \propto \frac{l}{T^2}$, the fractional error in $g$ is given by:
$$
\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}.
$$
Step 5: Substitute the numerical values
According to the given (or typical) data:
• $\Delta l = 0.1 \text{ cm}, \quad l = 25.0 \text{ cm}$
• $\Delta T = \frac{1 \text{ s}}{50}, \quad T = \frac{50 \text{ s}}{40}$ in the simplified approach.
Thus,
$$
\frac{\Delta g}{g}
= \frac{0.1}{25.0} + 2 \times \frac{1}{50}
= 0.004 + 0.04
= 0.044.
$$
The percentage error is:
$$
0.044 \times 100\% = 4.40\%.
$$
Step 6: State the final answer
The accuracy (or percentage error) in measuring $g$ is 4.40%.
