InterviewSolution
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InterviewSolution
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If g is acceleration due to gravity at the equator when earth were at rest and `g_(1)` is acceleration due to gravity at the same place when earth spins with angular velocity `omega,` the relation between them isA. `g_(1)=g(1-(R omega^(2))/(g))`B. `g_(1)=g(1-R omega^(2))`C. `g_(1)=g-R^(2)omega`D. `g=g_(1)-R^(2)omega` |
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Answer» Correct Answer - a `g_(1)=g-Romega^(2)cos^(2)phi` `=g-Romega^(2)cos^(2)(0)=g-Romega^(2)xx1` `g_(1)=g-Romega^(2)` `=g[1-(Romega^(2))/(g)].` |
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