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If earth contracts to half of its present radius what would be the length of the day at equator ? |
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Answer» Solution :`I_(1) =2/5 MR^(2) rArr I_(2) = 2/5 M(R/2)^(2) rArr I_(2) = I_(1)/4` `L=I_(1)omega_(1) = I_(2)omega_(2)` or `I(2PI)/(T_(1)) =1/4(2pi)/(T_(2))` or `T_(2) = T_(1)/4 = 24/4 =6` hours. |
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