\( \mathbf{X} \) and \( \mathbf{Y} \) are two volatile liquids with molar weights of \( 10 \mathrm{~g} \mathrm{~mol}^{-1} \) and \( 40 \mathrm{~g} \mathrm{~mol}^{-1} \) respectively. Two cotton plugs,
\( \mathrm{P} \) one soaked in \( \mathbf{X} \) and the other soaked in \( \mathbf{Y} \), are
\( \mathrm{VV} \) simultaneously placed at the ends of a tube of length \( \mathbf{L} \) \( =24 \mathrm{~cm} \), as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of \( 300 \mathrm{~K} \). Vapours of \( \mathbf{X} \) and \( \mathbf{Y} \) react to form a product which is first observed at a distance \( \mathbf{d ~ c m} \) from the plug soaked in \( \mathbf{X} \). Take \( \mathbf{X} \) and \( \mathbf{Y} \) to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.
The value of \( \mathbf{d} \) in \( \mathrm{cm} \) (shown in the figure), as estimated from Graham's law, is :
[JEE Advanced-2014]
(A) 8
(B) 12
(C) 16
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