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 measurement

Given:

• Length, $l = 55.0\,\text{cm}$
• Least count of meter scale $= 1\,\text{mm} = 0.1\,\text{cm}$

So, $\Delta l = 0.1\,\text{cm}$

Therefore, $\frac{\Delta l}{l} = \frac{0.1}{55.0} = 0.001818$

In percentage, $\frac{\Delta l}{l}\times 100 \approx 0.1818\%$

3. Error in time period measurement

Given:

• Time for 20 oscillations $= 30\,\text{s}$
• Least count of watch $= 1\,\text{s}$

So absolute error in measured time for 20 oscillations: $\Delta t = 1\,\text{s}$

Fractional error in time for 20 oscillations: $\frac{\Delta t}{t} = \frac{1}{30}$

Since $T = t/20$, the fractional error remains same: $\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{30}$

Thus, $2\frac{\Delta T}{T} = 2\times \frac{1}{30} = \frac{1}{15} = 0.0667$

In percentage, $0.0667\times 100 = 6.67\%$

4. Total percentage error in $g$

$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$

$= 0.001818 + 0.0667 = 0.0685$

In percentage, $0.0685\times 100 \approx 6.85\%$

So the percentage error is close to $\boxed{6.8\%}$

5. Option check

• A: $0.2\%$ ❌
• B: $3.5\%$ ❌
• C: $0.7\%$ ❌
• D: $6.8\%$ ✅

Therefore, the correct option is D.