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At present the acceleration due to gravity at latitude `45^(@)` on earth is `9.803ms^(-2)`. If earth stops rotating, the acceleration due to gravity at the same place would be`" "(Romega^(2)=0.034ms^(-2))`A. `9.837ms^(-2)`B. `9.82ms^(-2)`C. `9.81ms^(-2)`D. `9.786ms^(-2)` |
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Answer» Correct Answer - b `g_(phi)=g-Romega^(2)cos^(2)phi` `=g-Romega^(2)((1)/(sqrt2))^(2)=g-(Romega^(2))/(2)` `g=g_(phi)+(Romega^(2))/(2)=9.803+(0.034)/(2)` ltbRgt `=9.82ms^(-2)`. |
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