Let $A$ and $B$ be two gases and given: $\frac{T_A}{M_A} = 4 \cdot \frac{T_B}{M_B}$,where $T$ is the temperature and $M$ is the molecular mass. If $C_A$ and $C_B$ are the $r.m.s.$ speeds,then the ratio $\frac{C_A}{C_B}$ will be equal to:

Two gases $A$ and $B$ are contained in two separate, but otherwise identical containers. Gas $A$ consists of monatomic molecules, each with atomic mass of $4 \ u$, whereas Gas $B$ consists of rigid diatomic molecules, each with atomic mass of $20 \ u$. If gas $A$ is kept at $27^{\circ} C$, at what temperature should gas $B$ be kept so that both have the same rms speed (in $^{\circ} C$)?

At $27\,^oC$,the $rms$ velocity of $H_2$ gas is $100\, m/s$. Find the $rms$ velocity of $O_2$ gas at $327\,^oC$.

The root mean square speed of smoke particles of mass $5 \times 10^{-17} \, kg$ in their Brownian motion in air at $NTP$ is approximately $....... \, mm \, s^{-1}$. [Given $k = 1.38 \times 10^{-23} \, J \, K^{-1}$ and $T = 293 \, K$]

The velocities of $4$ gas molecules are given by $1 \ km/s, 3 \ km/s, 5 \ km/s$ and $7 \ km/s$. Calculate the difference between the average velocity and the root mean square $(RMS)$ velocity. (in $km/s$)

Three closed vessels $A, B$ and $C$ are at the same temperature $T$ and contain gases. Vessel $A$ contains only $O_2$,vessel $B$ contains only $N_2$,and vessel $C$ contains a mixture of equal quantities of $O_2$ and $N_2$. If the rms speed of $O_2$ molecules in vessel $A$ is $v_1$ and that of $N_2$ molecules in vessel $B$ is $v_2$,then the rms speed of $O_2$ molecules in vessel $C$ is: