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If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is (g = acceleration due to gravity near the poles and R is the radius of earth) ( Ignore equatorial bulge)

$\begin {array} {1 1} (1)\;2 \pi \sqrt {\large\frac{2R}{g}} & \quad (2)\;2 \pi \sqrt{\large\frac{R}{2g}} \\ (3)\;2 \pi \sqrt{\large\frac{R}{3g}} & \quad (4)\;2 \pi \sqrt{ \large\frac{R}{g}} \end {array}$

1 Answer

(1) $ 2 \pi \sqrt{\large\frac{2R}{g}}$
answered Nov 7, 2013 by pady_1
 

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