A) \[\left( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right)t\]
B) \[\left( \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right)t\]
C) \[\frac{\left( \alpha +\beta \right)t}{\alpha \beta }\]
D) \[\frac{\alpha \beta t}{\alpha +\beta }\]
Correct Answer: D
Solution :
In fig, \[A{{A}_{1}}={{v}_{\max .}}=\alpha {{t}_{1}}=\beta {{t}_{2}}\] But \[t={{t}_{1}}+{{t}_{2}}=\frac{{{v}_{\max }}}{\alpha }+\frac{{{v}_{\max }}}{\beta }\] \[={{v}_{\max }}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)={{v}_{\max }}\left( \frac{\alpha +\beta }{\alpha \beta } \right)\] or \[{{v}_{\max }}=t\left( \frac{\alpha \beta }{\alpha +\beta } \right)\]You need to login to perform this action.
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