The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on60° latitude becomes zero.

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The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on $60 °$ latitude becomes zero.

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2 \sqrt{\frac{g_{p}}{R}}

3 \sqrt{\frac{g_{p}}{R}}

4 \sqrt{\frac{g_{p}}{R}}

\sqrt{\frac{g_{p}}{R}}

answer is C.

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Detailed Solution

g_{ϕ} = g - R ω^{2} \cos^{2} ϕ

$0 = g - R ω^{2} \cos^{2} 60 °$

$∴ g = R ω^{2} {(\frac{1}{2})}^{2} = \frac{R ω^{2}}{4}$

$ω = \sqrt{\frac{4g}{R}} \Rightarrow ω = 2 \sqrt{\frac{g}{R}}$

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