1. Formula for a simple pendulum

For a simple pendulum, $T = 2\pi \sqrt{\frac{l}{g}}$ So, $g = \frac{4\pi^2 l}{T^2}$

Hence, the fractional error in $g$ is $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$

2. Error in length

Given length: $l = 25.0\text{ cm}$ Since it is written as $25.0\,$cm, the least count implied is $0.1\,$cm, so absolute error is $\Delta l = 0.1\text{ cm}$ Therefore, $\frac{\Delta l}{l} = \frac{0.1}{25.0} = 0.004 = 0.4\%$

3. Error in time period

Time for 40 oscillations is measured as $t = 50\text{ s}$ Stopwatch resolution is $1\,$s, so absolute error in total time is $\Delta t = 1\text{ s}$

Since $T = \frac{t}{40}$ we have $\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{50} = 0.02 = 2\%$

Thus, $2\frac{\Delta T}{T} = 4\%$

4. Total percentage error in $g$

$\frac{\Delta g}{g} = 0.4\% + 4\% = 4.4\%$

So, the accuracy/error in $g$ is $\boxed{4.40\%}$

5. Option check

• A: $4.40\%$ ✅
• B: $3.40\%$ ❌
• C: $2.40\%$ ❌
• D: $5.40\%$ ❌

Therefore, the correct option is A.