A) \[\frac{3v}{\pi }\]
B) \[\frac{4v}{\pi }\]
C) \[\frac{2v}{\pi }\]
D) \[\frac{v}{\pi }\]
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., \[{{v}_{rel}}=\sqrt{{{v}^{2}}+{{v}^{2}}-2vv\cos \theta }\] \[\Rightarrow {{v}_{rel}}=\sqrt{2{{v}^{2}}(1-\cos \theta )}=2v\sin (\theta /2)\] So averaging \[{{v}_{rel}}\]over all pairs \[{{\vec{v}}_{rel}}=\frac{\int_{0}^{2\pi }{{{v}_{rel}}d\theta }}{\int_{0}^{2\pi }{d\theta }}=\frac{\int_{0}^{2\pi }{2v\sin \left( \theta /2 \right)}}{\int_{0}^{2\pi }{d\theta }}\] \[=\frac{2v\times 2[-\cos (\theta /2)]_{0}^{2\pi }}{2\pi }\] \[\Rightarrow {{\vec{v}}_{rel}}=\left( 4v/\pi \right)>v\left[ \text{ as 4/}\pi \text{1} \right]\]
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