The acceleration due to gravity at the poles is `10ms^(-2)` and equitorial radius is `6400 km` for the earth. Then the angular velocity of rotaiton of the earth about its axis so that the weight of a body at the equator reduces to `75%` isA. `(1)/(1600)"rad/s"`B. `(1)/(800)"rad/s"`C. `(1)/(400)"rad/s"`D. `(1)/(200)"rad/s"`

The acceleration due to gravity at the poles is `10ms^(-2)` and equito

1.	The acceleration due to gravity at the poles is `10ms^(-2)` and equitorial radius is `6400 km` for the earth. Then the angular velocity of rotaiton of the earth about its axis so that the weight of a body at the equator reduces to `75%` isA. `(1)/(1600)"rad/s"`B. `(1)/(800)"rad/s"`C. `(1)/(400)"rad/s"`D. `(1)/(200)"rad/s"`
Answer» Correct Answer - a `g_(phi)=g-Romega^(2)cos^(2)phi` `7.5=10-6.4xx10^(6)xx omega^(2)xx"1 (at equator)"` `6.4xx10^(6)omega^(2)=2.5` `omega^(2)=sqrt((25)/(64xx10^(6)))=(5)/(8)xx10^(-3)` `=8.25xx10^(-4)=(1)/(1600)"rad/s."`

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