Ncert Exercises & Solutions

Ten small planes are flying at a speed of 150 km/h in total darkness in an air space 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 with no way of knowing where the other planes are. On average about how long a time will elapse 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 10 m.

Hint: The average time of collision depends on the mean free path.

Step 1: Find the average time of the collision.

The situation can be considered as the time of relaxation, based on the kinetic theory of gases. It means the free path is the distance between two successive collisions, which will be the distance travelled by plane before it just avoids the collision. The safe radius is equivalent to the radius of the atom.
Hence, the required time,

t = \frac{l}{v}, l

= mean free path

= \frac{1}{\sqrt{2} {nπd}^{2}}

, n =number density

= \frac{N}{V}

n = \frac{N u m b e r o f a e r o p l a n e s (N)}{V o l u m e (V)}

= \frac{10}{20 \times 20 \times 1.5} = 0.0167 k m^{- 3}

t = \frac{1}{\sqrt{2} {πd}^{2} (N / V)} \times \frac{1}{v}

[v=velocity of the aeroplane]

By putting the given data,

t = \frac{1}{\sqrt{2} \times 3.14 \times {(20)}^{2} \times 0.0167 \times 10^{- 6} \times 150}

= \frac{10^{6}}{1776.25 \times 2.505}

= \frac{10^{6}}{4449.5} = 224.74 h

= 225 h

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