Correct answer is (A)

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

$\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\sqrt{\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}}$

$=$ let initial speed is $\mathrm{v}$

As speed is increased by $3$ times so final speed become $4v$

$\Rightarrow \frac{\mathrm{v}}{4\mathrm{v}}=\sqrt{\frac{223}{\mathrm{T}}}$

$\mathrm{T}=3568\mathrm{K}$

So temp. in ${}^{\circ }\mathrm{C}=3568-273={3295}^{\circ }\mathrm{C}$

Correct answer is (D)

All three vessels have equal volume and same temperature and pressure.

From ideal gas equation

$PV=nRT$

$nR=\frac{PV}{T}$

$n=\frac{PV}{RT}=$ constant

So, here all three vessels contains equal number of moles and number of gas molecules.

Now, $v_{rms}=\sqrt{\frac{3RT}{M}}\Rightarrow v_{rms}\propto \frac{1}{\sqrt{M}}$

Here, rms speed of helium is the largest.

Correct answer is (B)

Escape velocity from the Earth’s surface is vescape = 11200 m s^–1

Let at temperature T, rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere.

Also, V_rms = V_escape = $\sqrt{\frac{3k_{B}T}{m_{O_2}}}$

$\Rightarrow$ 11200 = $\sqrt{\frac{3\times 1.38\times {10}^{-23}\times T}{2.76\times {10}^{-26}}}$

$\Rightarrow$ T = 8.360 × 10⁴ K

Correct answer is (D)

It is observed that the rms velocity of molecule is directly proportional to temperature, so

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

$\Rightarrow v_{rms}^{{}^{\prime }}=v_{rms}\sqrt{\frac{T^{\prime }}{T}}$

Hence, $v_{rms}^{{}^{\prime }}=200\times \sqrt{\frac{400}{300}}=400/\sqrt{3}$