A) \[0\,\ rad\,{{\sec }^{-1}}\]
B) \[\frac{1}{800}rad\,se{{c}^{-1}}\]
C) \[\frac{1}{80}rad\,se{{c}^{-1}}\]
D) \[\frac{1}{8}rad\,se{{c}^{-1}}\]
Correct Answer: B
Solution :
\[{g}'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \,\] For weightlessness at equator \[\lambda =0\] and \[g'=0\] \ \[0=g-{{\omega }^{2}}R\] Þ \[\omega =\sqrt{\frac{g}{R}}=\frac{1}{800}\ \frac{rad}{s}\]
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