A) Zero
B) Greater than v
C) Less than v
D) v
Correct Answer: B
Solution :
[b] The magnitude of relative velocity between two particles, with their velocity vector at angle \[\theta \] is \[|\overset{\to }{\mathop{\Delta }}\,v|=\sqrt{{{v}^{2}}+{{v}^{2}}-2v.v\cos \theta }=2v\sin \left( \frac{\theta }{2} \right)\] Here \[\theta \] varies between \[0\] and \[2\pi \]. Therefore, average value of \[|\overset{\to }{\mathop{\Delta }}\,v|\] will be \[<|\overset{\to v}{\mathop{\Delta }}\,|>=\frac{\int\limits_{0}^{2\pi }{2v\,\sin \left( \frac{\theta }{2} \right)}}{2\pi }=\frac{4}{\pi }v>v\]You need to login to perform this action.
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