A gas has an initial root-mean-square velocity of 900 m/s. If the pressure of the gas increases by a factor of nine while its density remains constant, what is the new root-mean-square velocity?

The given formula for pressure $P$ in terms of density $\rho$ and root-mean-square velocity $\bar{c}$ is:

The initial root-mean-square velocity is $\bar{c}_1 = 900 \text{ m s}^{-1}$.

The gas is heated such that its pressure increases by a factor of nine, so $P_2 = 9 P_1$.
The volume of the cylinder does not change, which means $V_1 = V_2$.
Since density $\rho = \frac{m}{V}$ and the mass $m$ of the gas remains constant, and the volume $V$ also remains constant, the density $\rho$ must be constant throughout the process ($\rho_1 = \rho_2 = \rho$).

For the initial state:

For the final state:

We know $P_2 = 9 P_1$. Substitute the expressions for $P_1$ and $P_2$:

Cancel out the common terms $\frac{1}{3}$ and $\rho$ from both sides:

Take the square root of both sides:

Now, substitute the value of $\bar{c}_1$:

The new numerical value for the root-mean-square velocity is $2700 \text{ m s}^{-1}$.

The final answer is $\text{A: 2700}$.