1. Given relation

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:

• $L = 20.0\,\text{cm}$
• accuracy in $L = 1\,\text{mm} = 0.1\,\text{cm}$

Therefore, $\frac{\Delta L}{L} = \frac{0.1}{20.0} = 0.005 = 0.5\%$

3. Error in time period measurement

Time for $100$ oscillations is measured as $t = 90\,\text{s}$ with a wrist watch of resolution $1\,\text{s}$.

So absolute error in measuring total time is $\Delta t = 1\,\text{s}$

Since $T = \frac{t}{100}$ we have $\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{90}$

Thus, $\frac{\Delta T}{T} = \frac{1}{90} \approx 0.0111 = 1.11\%$

4. Error in $g$

Now, $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$

Substituting, $\frac{\Delta g}{g} = 0.005 + 2\left(\frac{1}{90}\right)$ $= 0.005 + 0.0222$ $= 0.0272$

Therefore percentage error is $0.0272 \times 100 = 2.72\%$

This is approximately $3\%$

5. Option check

• A: $1\%$ ❌
• B: $5\%$ ❌
• C: $2\%$ ❌
• D: $3\%$ ✅

Therefore the correct answer is D.