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Ten small planes are flying at a speed of $150km{h^{ - 1}}$ in total darkness in an airspace that is $20 \times 20 \times 1.5$$k{m^3}$in volume. You are in one of the planes, flying at random within this space this no way of knowing where the other planes are. On the average about how long a time elapses between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius $10m$.a. $125h$b. $220h$c. $432h$d. $225h$

Last updated date: 20th Jun 2024
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Hint: This problem can be solved using the concept of law of equipartition of energy and the kinetic theory of gases. According to the Law of Equipartition of energy the total energy for the system is equally divided among the degrees of freedom, this is applicable for any dynamic system in thermal equilibrium. To understand the concept of kinetic theory of gases we need to understand what is the degree of freedom. It is the number of independent parameters that define its configuration or state. For a mono-atomic gas like Helium the degree of freedom is one where for a triatomic molecule it is three.

Complete step by step solution:
The formula which we are going to use is
$t = \dfrac{\lambda }{v}$
Since the value of Lambda is not given we have to find it. That is the mean free path of a plane
$\Rightarrow \lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}n}}$
Where d is the diameter and n is the number density.
The given values are
$N = 10m$ $v = 150km{h^{ - 1}}$ $V = 20 \times 20 \times 1.5k{m^3}$
$\Rightarrow n = \dfrac{N}{V}$
$\Rightarrow \dfrac{{10}}{{20 \times 20 \times 1.5}}$
$\Rightarrow n = 0.0167k{m^{ - 3}}$
Now we have to find time elapse before collision of two planes
$t = \dfrac{\lambda }{v}$
$\Rightarrow t = \dfrac{1}{{\sqrt 2 \pi {d^2}nv}}$
$\Rightarrow t = \dfrac{1}{{1.414 \times 3.14 \times {{\left( {20} \right)}^2} \times {{10}^{ - 6}} \times 0.0167 \times 150}}$
$\Rightarrow t = \dfrac{{{{10}^6}}}{{4449.5}}$
$\Rightarrow t = 225h$
So the correct answer is option d.