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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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Two gases are sent through 2 ends of a 2m long pipe. If the gases reach at a distance of 1.5m from the end of the lighter gas, then what is the ratio of their masses (lighter: heavier), assuming pressure conditions to be the same?

(A) $\large\frac{1}{3}$ (B) $\large\frac{1}{9}$ (C) $\large\frac{1}{27}$ (D) $\large\frac{1}{\sqrt 3}$
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Since the gases react at a distance 1.5m from the lighter end, and the time taken should be the same,
$\Rightarrow$ Time $=\large\frac{0.5 \times 3}{r_{\text{lighter gas}}}$$=\large\frac{0.5 }{r_{\text{heavier gas}}}$
$\Rightarrow \large\frac{3}{\sqrt M_{\text{lighter gas}}} $$ = \large\frac{1}{\sqrt M_{heavier gas}} $$\rightarrow M_{\text_{Lighter gas}}: M_{\text_{Heavier Gas}} = $$ \large\frac{1}{3^2} $$= \large\frac{1}{9}$
answered Mar 11, 2014 by balaji.thirumalai

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