$\begin {array}{1,1}(a)\;3.16\; \text{times as fast as argon}\\(b)\;7.32\; \text{times as fast as argon}\\(c)\;1.58\; \text{times as fast as argon}\\(d)\;10\; \text{times as fast as argon}\end {array}$

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- Graham's law states that "at constant pressure and temperature the rate of diffusion or effusion of a gas is inversly proportional to the square root of its density" Rate of diffusion $\propto \large\frac{1}{\sqrt{\alpha}}$ If $r_1 \; and \; r_2$ represent the rates of diffusion of two gases and $d_1 \;and\;d_2$ are their respective densities then $\large\frac{r_1}{r_2} = \sqrt{\large\frac{d_2}{d_1}}$ (at constant P and T) , $\large\frac{r_1}{r_2} = \sqrt{\large\frac{M_2}{M_1}}\times \large\frac{P_1}{P_2}$ (at constant T) . $\large\frac{V_1\times t_2}{V_2\times t_1} = \sqrt{\large\frac{d_2}{d_1}} = \sqrt{\large\frac{M_2}{M_1}} \quad V\propto n$ (where n is no. of moles), $V_1\propto n_1 \;and\; V_2\propto n_2$.

$\large\frac{r_{He}}{r_{Ar}} = \sqrt{\large\frac{M_{Ar}}{M_{He}}}$

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

$\;\;\;\;\;\;\;\;= \sqrt{10}$

$\;\;\;\;\;\;\;\;\;=3.16$

Hence answer is (A)

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