X mL of ${\mathrm{H}}_2$ gas effuses through a hole in a container is 5 second. The time taken for the effusion of the same volume of the gas specied below under identical conditions is

Let the time of diffusion for the same volume of two gases are '𝑡ଵ '  and '𝑡ଶ '  respectively then

$\frac{{\mathrm{t}}_2}{{\mathrm{t}}_1}=\sqrt{\frac{M_2}{M_1}}$

$\frac{{\mathrm{t}}_2}{5}=\sqrt{\frac{\cancel{4^2}}{\cancel{2_1}}}\Rightarrow t_2=5\times 1.414=7.07 sec$

$\frac{{\mathrm{t}}_2}{5}=\sqrt{\frac{\cancel{{35}^{16}}}{\cancel{2_1}}}\Rightarrow t_2=5\times 4=20 sec$

$\frac{{\mathrm{t}}_2}{5}=\sqrt{\frac{\cancel{{28}^{14}}}{\cancel{2_1}}}\Rightarrow t_2=5\times 3.74=18.7 sec$$\frac{{\mathrm{t}}_2}{5}=\sqrt{\frac{\cancel{{44}^{22}}}{\cancel{2_1}}}\Rightarrow t_2=5\times 4.69=23.45 sec$

20 second :$\mathit{ }{\mathrm{O}}_2$ is the correct answer