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If a man at the equator would weigh `(3//5)`th of his weight, the angular speed of the earth isA. `sqrt(2/5 g/R)`B. `sqrt(g/R)`C. `sqrt(R/g)`D. `sqrt(2/5 R/g)` |
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Answer» Correct Answer - A `3/5mg=mg=mRomega^(2)` `=omega^(2)=g-3/5impliesomega=sqrt(2/5g/R)` |
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