A) \[{{v}_{e}}=2v\]
B) \[{{v}_{e}}=v\]
C) \[{{v}_{e}}=v/2\]
D) \[{{v}_{e}}=\sqrt{3}v\]
Correct Answer: A
Solution :
[a] \[v=\omega R\] \[g={{g}_{0}}-{{\omega }^{2}}R\] [\[g=\]at equator, \[{{g}_{0}}=\]at poles] \[\frac{{{g}_{0}}}{2}={{g}_{0}}-{{\omega }^{2}}R;{{\omega }^{2}}R=\frac{{{g}_{0}}}{2}\] \[{{v}^{2}}=\frac{{{g}_{0}}R}{2}\] \[{{v}_{e}}=\sqrt{2{{g}_{0}}R}=\sqrt{4{{v}^{2}}}=2v\]
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