A simple pendulum is being used to determine the value of gravitational acceleration $$g$$ at a certain place. The length of the pendulum is 25.0 cm and a stopwatch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:

For a simple pendulum, the theoretical expression for the time-period is stated first:

$$T \;=\; 2\pi \sqrt{\dfrac{L}{g}}$$

where $$T$$ is the period of one oscillation, $$L$$ is the length of the pendulum and $$g$$ is the local acceleration due to gravity.

We rearrange this formula to write $$g$$ explicitly, because that is the quantity whose accuracy we have to evaluate:

$$T^2 \;=\; 4\pi^2 \dfrac{L}{g} \quad\Rightarrow\quad g \;=\; \dfrac{4\pi^2\,L}{T^2}$$

The measurement data are now introduced. The length is given as

$$L \;=\; 25.0 \text{ cm} \;=\; 0.250 \text{ m}.$$

The time for 40 complete oscillations is read on a stopwatch as

$$t_{\text{total}} \;=\; 50 \text{ s}.$$

Because this total time corresponds to 40 oscillations, the observed period of one oscillation is

$$T \;=\; \dfrac{t_{\text{total}}}{40} \;=\; \dfrac{50 \text{ s}}{40} \;=\; 1.25 \text{ s}.$$

Next we translate the resolutions of the measuring instruments into absolute uncertainties. The stopwatch has a resolution of $$1 \text{ s},$$ so the uncertainty in the total time is

$$\Delta t_{\text{total}} \;=\; \pm 1 \text{ s}.$$

Hence the uncertainty in the period is obtained by the same factor of division because the period is exactly the total time divided by 40:

$$\Delta T \;=\; \dfrac{\Delta t_{\text{total}}}{40} \;=\; \dfrac{1 \text{ s}}{40} \;=\; 0.025 \text{ s}.$$

So the relative (fractional) error in the period becomes

$$\dfrac{\Delta T}{T} \;=\; \dfrac{0.025}{1.25} \;=\; 0.020 \;=\; 2.0\%.$$

The pendulum length has been recorded as $$25.0 \text{ cm}.$$ Because a centimetre scale with one-millimetre (0.1 cm) least count is tacitly implied, the absolute uncertainty in the length is

$$\Delta L \;=\; \pm 0.1 \text{ cm} \;=\; 0.001 \text{ m}.$$

Therefore the relative error in the length measurement is

$$\dfrac{\Delta L}{L} \;=\; \dfrac{0.1}{25.0} \;=\; 0.004 \;=\; 0.4\%.$$

We now propagate these individual errors to the derived quantity $$g.$$ For a product-quotient relation such as $$g = 4\pi^2 L / T^2,$$ the standard rule of error propagation states:

$$\dfrac{\Delta g}{g} \;=\; \dfrac{\Delta L}{L} \;+\; 2\,\dfrac{\Delta T}{T}$$

because $$T$$ is present with the power 2 in the denominator.

Substituting the numerical values gives

$$\dfrac{\Delta g}{g} \;=\; 0.004 \;+\; 2(0.020) \;=\; 0.004 \;+\; 0.040 \;=\; 0.044.$$

To convert this fractional error into a percentage accuracy, we multiply by 100:

$$\text{Percentage error in } g \;=\; 0.044 \times 100\% \;=\; 4.4\%.$$

Hence, the correct answer is Option C.