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What will be the answer of question number 45 and how it will be?

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  1. the answer is 7.8 * 10^-4 rad/sec

    True weight at the equator is w=mg and bserved weight is W’ = mg’ = 3/5 mg
    taking λ= 0.
    mg’ = mg – mRw^2 cosλ
    3/5mg = mg – mRw^2 cos0
    = mg – mRw^2
    or
    mRw^2 = 2/5 mg

    w = (2g/5R)^1/2
    = 7.8 * 10^-4 rad/sec

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