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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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