\( V_{\text{rms}} \text{(mono)} > V_{\text{rms}} \text{(dia)} > V_{\text{rms}} \text{(poly)} \)
\( V_{\text{rms}} \text{(dia)} < V_{\text{rms}} \text{(poly)} < V_{\text{rms}} \text{(mono)} \)
\( V_{\text{rms}} \text{(mono)} < V_{\text{rms}} \text{(dia)} < V_{\text{rms}} \text{(poly)} \)
\( V_{\text{rms}} \text{(mono)} = V_{\text{rms}} \text{(dia)} = V_{\text{rms}} \text{(poly)} \)
The root mean square speed \( V_{\text{rms}} \) is given by: \[ V_{\text{rms}} = \sqrt{\frac{3RT}{m}} \] Since the gases are at the same temperature and pressure, the root mean square speed depends on the molar mass \( m \). For neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic), the molar mass increases in the order: \[ V_{\text{rms}} \text{(mono)} > V_{\text{rms}} \text{(dia)} > V_{\text{rms}} \text{(poly)} \] Thus, the correct answer is \( V_{\text{rms}} \text{(mono)} > V_{\text{rms}} \text{(dia)} > V_{\text{rms}} \text{(poly)} \).
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 