A) \[\frac{1}{2}\left( \frac{\alpha \beta }{\alpha +\beta } \right){{t}^{2}}\]
B) \[\frac{1}{2}\left( \frac{\alpha +\beta }{\alpha \beta } \right){{t}^{2}}\]
C) \[\frac{1}{2}\left( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right){{t}^{2}}\]
D) \[\frac{1}{2}\left( \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right){{t}^{2}}\]
Correct Answer: A
Solution :
Let us check this also dimensionally \[\frac{1}{2}\left( \frac{\alpha \beta }{\alpha +\beta } \right){{t}^{2}}=\frac{{{({{a}_{cc}})}^{2}}}{{{a}_{cc}}}{{t}^{2}}\] \[=({{a}_{cc}}){{t}^{2}}\] \[=[L{{T}^{-2}}][{{T}^{2}}]\] = [L] = distance
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