Correct answer is (A)

We know,

V_rms = $\sqrt{\frac{3RT}{M}}$

Where M = molar mass of the gas.

Here temperature is 300 K for both the gas. So temperature is constant. R is also a constant.

$\therefore$   V_rms $\propto \sqrt{\frac{1}{M}}$

$\therefore$    $\frac{V_{rms}(helium|)}{V_{rms}\left(\arg an)\right)}=\sqrt{\frac{M_{Ar}}{M_{He}}}$

= $\sqrt{\frac{40}{4}}$

= $\sqrt{10}$

= 3.16

Correct answer is (B)

We know that $v_{rms}=\sqrt{\frac{3k_{B}T}{m}}$. Given, $k_B=1.4\times {10}^{-23}$ J/K; T = 300 K; m = 7 $\times$ 10^$-$27 kg. Therefore,

$v_{rms}=\sqrt{\frac{3\times 1.4\times {10}^{-23}\times 300}{7\times {10}^{-27}}}$

$=\sqrt{\frac{3\times 300\times 14\times {10}^{-23}}{7\times 10\times {10}^{-27}}}$

$=\sqrt{\frac{3^2\times {10}^2\times 2\times {10}^4}{10}}=3\times 10\times {10}^2\times \sqrt{\frac{2}{10}}$

$=3\times \sqrt{10}\times {10}^2\times \sqrt{2}=3\times \sqrt{2}\times \sqrt{5}\times {10}^2\times \sqrt{2}$

$=3\times 2\times \sqrt{5}\times {10}^2$

$v_{rms}=6\times {10}^2\times \sqrt{5}=13.41\times {10}^2$ or $v_{rms}=1.34\times {10}^3$

Correct answer is (A)

The speed of sound in a gas is given by $v=\sqrt{\frac{\gamma RT}{M}}$
$\therefore$ $\frac{v_{O_2}}{v_{He}}=\sqrt{\frac{{\gamma }_{O_2}}{M_{O_2}}\times \frac{M_{He}}{{\gamma }_{He}}}$
$=\sqrt{\frac{1.4}{32}\times \frac{4}{1.67}}=0.3237$
$\therefore$ $v_{He}=\frac{v_{O_2}}{0.3237}$
$=\frac{460}{0.3237}$
$=1421m/s$

Correct answer is (C)

Since pressure and volume are not changing, so temperature remains same.

Correct answer is (D)

$v_{rms}=$$\sqrt{\frac{RT}{M}}$

For $v_{rms}$ to be equal $\frac{T_{H_2}}{M_{H_2}}=\frac{T_{O_2}}{M_{O_2}}$

Here $M_{H_2}=2;M_{O_2}=32;$

$T_{O_2}=47+273=320K$

$\therefore$ $\frac{T_{H_2}}{2}=\frac{320}{32}$

$\Rightarrow T_{H_2}=20K$