To solve this problem, we will use the formula for the root mean square (rms) velocity of gas molecules:$\\ v_{rms} = \sqrt{\frac{3RT}{M}} \\$However, since we are comparing the same gas under different conditions, we can use the relation:$\\ \frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T_1}{T_2}} \\$where $v_{rms1}$ and $v_{rms2}$ are the rms velocities at temperatures $T_1$ and $T_2$ respectively.$\\$Given:$\\ v_{rms1} = 200$ ms$^{-1} \\ T_1 = 27^{\circ}C = 27 + 273 = 300$ K$\\ T_2 = 127^{\circ}C = 127 + 273 = 400$ K$\\$Substitute these values into the equation:$\\ \frac{200}{v_{rms2}} = \sqrt{\frac{300}{400}} \\$Simplify the square root:$\\ \frac{200}{v_{rms2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \\$Rearrange to solve for $v_{rms2}: \\ v_{rms2} = 200 \times \frac{2}{\sqrt{3}} \\ v_{rms2} = \frac{400}{\sqrt{3}} \\$Therefore, the rms velocity of the gas molecules at the new conditions is $\frac{400}{\sqrt{3}}$ ms$^{-1}. \\$This corresponds to Option 4.