A) \[Zero\]
B) \[\frac{1}{800}rad\,\,{{\sec }^{-1}}\]
C) \[\frac{1}{80}rad\,\,{{\sec }^{-1}}\]
D) \[\frac{1}{8}\,rad\,\,{{\sec }^{-1}}\]
Correct Answer: B
Solution :
[b] \[g'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \] To make effective acceleration due to gravity zero at equator \[\lambda =0\] and \[g'=0\] \[\therefore \] \[0=g-{{\omega }^{2}}\,R\]\[\Rightarrow \]\[\omega =\sqrt{\frac{g}{R}}=\frac{1}{800}\frac{rad}{s}\]
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