A) \[2\,\nu \sin \,\,\beta \]
B) \[\nu \cos 2\,\,\beta \]
C) \[2\,\nu \sin \,\,\beta \]
D) \[\nu \,\sqrt{2}\]
Correct Answer: C
Solution :
| From image diagram |
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| \[\Rightarrow \] \[{{\nu }_{{{I}_{2}},\,\,{{I}_{1}}}}=\sqrt{{{(\nu -\nu \cos 2\beta )}^{2}}+{{(v\sin 2\beta )}^{2}}}\] |
| \[=2\,\,\nu \sin \beta \] |
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