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A body on the surface of the earth at latitude $60 °$ will be under weight lessness state if the angular velocity of rotation of the earth becomes

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\sqrt{\frac{GM}{R^{3}}}

\sqrt{\frac{4GM}{R^{3}}}

\sqrt{\frac{2GM}{R^{3}}}

\sqrt{\frac{GM}{{2R}^{3}}}

answer is D.

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Detailed Solution

g^{|} = g - {Rw}^{2} \frac{1}{4}

${weight g}^{|} = 0$

$\frac{{Rw}^{2}}{4} = g - g^{|}$

$\frac{{Rw}^{2}}{4} = g$

$w^{2} = \frac{4 g}{R} = \frac{4GM}{R^{3}}$

$w = \sqrt{\frac{4GM}{R^{3}}}$

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