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MHT CET 2023 (12 May, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2023 (12 May, Shift 2)

Q57MHT CET 2023 · 12 MAY · SHIFT 2Easy

Acceleration Due to Gravity

The depth at which acceleration due to gravity becomes g2n\frac{g}{2n} is (R=(R = radius of earth, g=g = acceleration due to gravity on earth's surface, nn is integer)

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Answer: (d)

  1. g(1−dR)=g2ng\left(1 - \frac{d}{R}\right) = \frac{g}{2n}
  2. dR=1−12n=2n−12n\frac{d}{R} = 1 - \frac{1}{2n} = \frac{2n-1}{2n}
  3. d=R(2n−1)2nd = \frac{R(2n-1)}{2n}
  4. So the answer is (d) R(2n−1)2n\frac{R(2n-1)}{2n}.
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