Physics MCQs for NEET — Practice Questions with Answers

The formula for mean free path is $l = 1 / (\sqrt{2} \pi n d^2)$. If $n$ is doubled, the denominator becomes $2n$, making the mean free path half of its original value ($l' = 1 / (\sqrt{2} \pi (2n) d^2) = l/2$).

At constant pressure, according to the ideal gas law ($PV = N k_B T$), increasing temperature ($T$) leads to a decrease in number density ($n = N/V$). Since $l$ is inversely proportional to $n$, an increase in temperature (and thus a decrease in $n$) will lead to an increase in the mean free path.

The NCERT text states, 'In a highly evacuated tube n [number density] is rather small and the mean free path can be as large as the length of the tube'.

The NCERT 'POINTS TO PONDER' section notes, 'What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule.' A large mean free path implies that molecules travel significant distances before colliding, suggesting ample space and facilitating motion that can be described as approximately unhindered for these distances.

The mean free path $l$ is inversely proportional to $d^2$ ($l \propto 1/d^2$). If $d$ is halved, then $d^2$ becomes $(d/2)^2 = d^2/4$. Therefore, $l$ would be proportional to $1/(d^2/4) = 4/d^2$, meaning the mean free path would be quadrupled.

In Example 12.9, the calculation for air molecules gives $l = 2.9 \times 10^{-7} m$ and $d = 2 \times 10^{-10} m$. The ratio $l/d \approx (2.9 \times 10^{-7} m) / (2 \times 10^{-10} m) \approx 1450$, which is approximately 1500. The NCERT explicitly states '$l \approx 1500 d$' after the calculation.

The NCERT text explains, 'The top of a cloud of smoke holds together for hours. This happens because molecules in a gas have a finite though small size, so they are bound to undergo collisions. As a result, they cannot move straight unhindered; their paths keep getting incessantly deflected.' These incessant collisions, characterized by the mean free path, prevent rapid diffusion.

The rate of collisions is given by $n\pi d^2 $ according to the NCERT text. The time between two successive collisions (collision time) $\tau$ is the inverse of the collision rate, i.e., $\tau = 1 / (n\pi d^2 )$. The mean free path $l$ is then $ \tau$, showing a direct relationship between these quantities. The question asks what is DIRECTLY related to the rate of collisions; the time between collisions is inversely related to the rate, so shorter time means higher rate.

The NCERT text states, 'The average distance between two successive collisions, called the mean free path $l$, is : $l = \tau$'. Rearranging this gives $\tau = l / $.

The NCERT text states, 'In this derivation, we imagined the other molecules to be at rest. But actually all molecules are moving and the collision rate is determined by the average relative velocity of the molecules. Thus we need to replace $ $ by $ $ in Eq. (12.38). A more exact treatment gives $l = 1 / (\sqrt{2} n \pi d^2)$ (12.40).' The $\sqrt{2}$ factor arises from considering the relative motion of all molecules.