A) \[1.25\times {{10}^{-3}}rad/s\]
B) \[1.56\times {{10}^{-3}}rad/\sec \]
C) \[1.25\times {{10}^{-1}}rad/s\]
D) \[1.56\,\,rad/\sec \]
Correct Answer: A
Solution :
Using the relation \[W=m(g-R\,\,{{\omega }^{2}})\] \[0=m{{(g-R\,\,\omega )}^{2}}\] So, \[\omega =\sqrt{\frac{g}{R}}=\sqrt{\frac{10}{6.4\times {{10}^{6}}}}=\frac{1}{800}\] or \[\omega =1.25\times {{10}^{-3}}\,\,rad/\sec \]
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