An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be:

The formula for the time period $$ T $$ of a simple pendulum is $$ T = 2\pi \sqrt{\frac{L}{g}} $$. Squaring both sides gives $$ T^2 = 4\pi^2 \frac{L}{g} $$, and rearranging for $$ g $$ yields $$ g = 4\pi^2 \frac{L}{T^2} $$.

To find the error in $$ g $$, we use the relative error formula. The relative error in $$ g $$ is given by $$ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} $$.

First, determine the error in length $$ L $$. The length $$ L = 20.0 \text{cm} $$ is measured with a meter scale of least count 1 mm = 0.1 cm. The absolute error $$ \Delta L $$ is the least count, so $$ \Delta L = 0.1 \text{cm} $$. The relative error in $$ L $$ is $$ \frac{\Delta L}{L} = \frac{0.1}{20.0} = 0.005 $$.

Next, determine the error in the time period $$ T $$. The time for 100 oscillations is $$ t = 90.0 \text{seconds} $$, measured with a watch of least count 1 second. The absolute error in the time for 100 oscillations $$ \Delta t $$ is the least count, so $$ \Delta t = 1 \text{second} $$. The time period $$ T $$ is the time for one oscillation, so $$ T = \frac{t}{100} = \frac{90.0}{100} = 0.9 \text{seconds} $$. The absolute error in $$ T $$ is $$ \Delta T = \frac{\Delta t}{100} = \frac{1}{100} = 0.01 \text{seconds} $$. The relative error in $$ T $$ is $$ \frac{\Delta T}{T} = \frac{0.01}{0.9} = \frac{1}{90} \approx 0.011111 $$.

Now substitute into the relative error formula for $$ g $$:

$$ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} = 0.005 + 2 \times \frac{1}{90} $$

Calculate the terms:

$$ 2 \times \frac{1}{90} = \frac{2}{90} = \frac{1}{45} \approx 0.022222 $$

$$ 0.005 + 0.022222 = 0.027222 $$

To express as a percentage, multiply by 100:

$$ 0.027222 \times 100 = 2.7222\% $$

Rounding to one decimal place gives 2.7%.

Hence, the error in the determination of $$ g $$ is 2.7%, which corresponds to Option D.

So, the answer is Option D.