A) \[1.25\times {{10}^{-3}}\] rad/s
B) \[2.50\times {{10}^{-3}}\] rad/s
C) \[3.75\times {{10}^{-3}}\] rad/s
D) \[5.0\times {{10}^{-3}}\] rad/s
Correct Answer: B
Solution :
Acceleration due to gravity \[g=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \] \[0=g-{{\omega }^{2}}R{{\cos }^{2}}{{60}^{o}}\] \[0=g-\frac{{{\omega }^{2}}R}{4}\] \[\omega =2\sqrt{\frac{g}{R}}=2\sqrt{\frac{10}{6400\times 100}}\] Angular velocity \[\omega =\frac{1}{400}\] \[=2.5\times {{10}^{-3}}\] rad/s (Take \[g=10\,\,m/{{s}^{2}}\] for the acceleration due to gravity if the earth were at rest and radius of earth equal to 6400 km.)You need to login to perform this action.
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