Question

The angular speed of earth in rad/s so that the objects on equator may appear weightless (taking $ g=10\text{ }m/{{s}^{2}} $ and radius of earth $ =6400 $ km) is

• A. $ 1.56 $
• B. $ 1.25\times {{10}^{-1}} $
• C. $ 1.56\times {{10}^{-3}} $
• D. $ 1.25\times {{10}^{-3}} $

Answer 1

Answer: (D)

: Weight of the body = Centripetal force $ \therefore $ $ mg=mR{{\omega }^{2}} $ or $ \omega =\sqrt{g/R}=\sqrt{\frac{10}{6400\times {{10}^{3}}}}=\frac{1}{8\times {{10}^{2}}} $ $ =1.25\times {{10}^{-3}}rad/\sec $