A) \[7.4\times {{10}^{-4}}\,rad/\sec \]
B) \[6.7\times {{10}^{-4}}\,rad/\sec \]
C) \[7.8\times {{10}^{-4}}\,rad/\sec \]
D) \[8.7\times {{10}^{-4}}\,rad/\sec \]
Correct Answer: C
Solution :
Weight of the body at equator = \[\frac{3}{5}\] of initial weight \ \[g'=\frac{3}{5}g\] (because mass remains constant) \[g'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \]Þ \[\frac{3}{5}g=g-{{\omega }^{2}}R{{\cos }^{2}}(0{}^\circ )\] Þ \[{{\omega }^{2}}=\frac{2g}{5R}\]Þ \[\omega =\sqrt{\frac{2g}{5R}}=\sqrt{\frac{2\times 10}{5\times 6400\times {{10}^{3}}}}\] = \[7.8\times {{10}^{-4}}\frac{rad}{\sec }\]
You need to login to perform this action.
You will be redirected in
3 sec
