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Q.

The number of air molecules per cm³ increased from $3 \times 10^{19} to 12 \times 10^{19}$ . The ratio of collision frequency of air molecules before and after the increase in number respectively is

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a

0.75

b

1.25

c

0.25

d

0.50

answer is C.

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Detailed Solution

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$Collision frequency (f) α (n number of molecules per unit volume)$

$So, \frac{f_{1}}{f_{2}} = \frac{n_{1}}{n_{2}}$

$= \frac{3 \times 10^{19}}{12 \times 10^{19}} = \frac{1}{4} = 0.25$

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