By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecules in a container of fixed volume is increased from $$5 \times 10^4$$ cm/s to $$10 \times 10^4$$ cm/s?

The root mean square velocity (RMS velocity) of gas molecules is given by the formula:

$$ v_{rms} = \sqrt{\frac{3RT}{M}} $$

where $$ R $$ is the universal gas constant, $$ T $$ is the absolute temperature, and $$ M $$ is the molar mass of the gas. Since the gas is the same, $$ M $$ is constant, and $$ R $$ is always constant. Therefore, the RMS velocity is proportional to the square root of the temperature:

$$ v_{rms} \propto \sqrt{T} $$

This means that the ratio of the RMS velocities is equal to the square root of the ratio of the temperatures. Let the initial RMS velocity be $$ v_1 = 5 \times 10^4 $$ cm/s and the final RMS velocity be $$ v_2 = 10 \times 10^4 $$ cm/s. Let the initial temperature be $$ T_1 $$ and the final temperature be $$ T_2 $$. Then:

$$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} $$

We need to find by how many folds the temperature increases, which is the ratio $$ \frac{T_2}{T_1} $$. To find this, we square both sides of the equation:

$$ \left( \frac{v_2}{v_1} \right)^2 = \frac{T_2}{T_1} $$

Now substitute the given values:

$$ v_2 = 10 \times 10^4 \text{ cm/s}, \quad v_1 = 5 \times 10^4 \text{ cm/s} $$

So:

$$ \frac{v_2}{v_1} = \frac{10 \times 10^4}{5 \times 10^4} = \frac{10}{5} = 2 $$

Then:

$$ \left( \frac{v_2}{v_1} \right)^2 = (2)^2 = 4 $$

Therefore:

$$ \frac{T_2}{T_1} = 4 $$

This means the temperature increases by a factor of 4, or four folds.

Hence, the correct answer is Option D.