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The acceleration due to gravity at the poles is 10ms^-2 and equatorial radius is 6400km for the earth. Then the angular velocity of rotation of the earth about its axis so that the weight of a body at the equator reduces to 75% is

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$\frac{1}{1600} {rads}^{- 1}$

$\frac{1}{800} rads^{- 1}$

$\frac{1}{400} rads^{- 1}$

$\frac{1}{200} rads^{- 1}$

answer is A.

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$g_{ϕ} = g - {Rω}^{2} Given g_{ϕ} = \frac{75}{100} g$

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