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 |
& |
\[\Rightarrow \] \[{{\nu }_{{{I}_{2}},\,\,{{I}_{1}}}}=\sqrt{{{(\nu -\nu \cos 2\beta )}^{2}}+{{(v\sin 2\beta )}^{2}}}\] |
\[=2\,\,\nu \sin \beta \] |
You need to login to perform this action.
You will be redirected in
3 sec