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If R is the radius of earth, $ω$ is its angular velocity and g_p is the value of acceleration due to gravity at the poles, then effective value of acceleration due to gravity at the latitude $λ = 60^{\circ}$ will be equal to

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detailed solution

Correct option is A

g′=gp−Rω2cos2⁡λ,λ=60∘⋅g′=gp−Rω24

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If R is the radius of earth, ω is its angular velocity and gp is the value of acceleration due to gravity at the poles, then effective value of acceleration due to gravity at the latitude λ=60∘ will be equal to

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If R is the radius of earth, $ω$ is its angular velocity and g_p is the value of acceleration due to gravity at the poles, then effective value of acceleration due to gravity at the latitude $λ = 60^{\circ}$ will be equal to