The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to [AMU 2012J (Take, g = 10 ms-2 for the acceleration due to gravity, if the earth were at rest, radius of earth equal to 6400 km and λ = 60)

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1.25 x 10-3 rad s-1

2.50 x 10-3 rads-1

3.75 X 10-3 rads-1

5 X 10-3 rad s-1

answer is B.

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Detailed Solution

Acceleration due to gravity, g=g-ω2Rcos2λ⇒ 0=g-ω2Rcos260°0=g-ω2R4⇒ ω=2gR=2106400×1000Angular velocity, ω=1400=2.50×10-3rads-1

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