Kinetic Theory Of Gases Question 41
Question: Consider a collection of a large number of dust particles each with speed v. The direction of velocity is randomly distributed in the collection. What is the magnitude of the relative velocity between a pairs in the collection?
Options:
A) $ \frac{3v}{\pi } $
B) $ \frac{4v}{\pi } $
C) $ \frac{2v}{\pi } $
D) $ \frac{v}{\pi } $
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Answer:
Correct Answer: B
Solution:
[b] Figure shows the particles each moving with same speed v but the different two particles having angle $ \theta $ between directions of their velocities. Then, $ \overrightarrow{v _{rel}}=\overrightarrow{v _{B}}-\overrightarrow{v _{A}} $ i.e.,
$ v _{rel}=\sqrt{v^{2}+v^{2}-2vv\cos \theta } $
$ \Rightarrow v _{rel}=\sqrt{2v^{2}(1-\cos \theta )}=2v\sin (\theta /2) $ So averaging $ v _{rel} $ over all pairs
$ {{\vec{v}} _{rel}}$
=$\frac{\int _{0}^{2\pi }{v _{rel}d\theta }}{\int _{0}^{2\pi }{d\theta }}$
=$\frac{\int _{0}^{2\pi }{2v\sin ( \theta /2 )}}{\int _{0}^{2\pi }{d\theta }} $
$ =\frac{2v\times 2[-\cos (\theta /2)] _{0}^{2\pi }}{2\pi } $
$ \Rightarrow {{\vec{v}} _{rel}}=( 4v/\pi )>v[ \text{ as 4/}\pi \text{1} ] $
BETA
