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JEE Main 2024 (29 Jan, Shift 1) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2024 (29 Jan, Shift 1)

Q42JEE Main 2024 · 29 JAN · SHIFT 1Medium

Acceleration Due to Gravity

At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as RR.)

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Answer: (c)

  1. Equal weight: g(1−hR)=gR2(R+h)2g\left(1 - \frac{h}{R}\right) = \frac{gR^2}{(R+h)^2} (exact formulas, hh not small)
  2. Put x=hRx = \frac{h}{R}: (1−x)(1+x)2=1⇒x−x2−x3=0(1-x)(1+x)^2 = 1 \Rightarrow x - x^2 - x^3 = 0
  3. x2+x−1=0⇒x=5−12x^2 + x - 1 = 0 \Rightarrow x = \frac{\sqrt{5}-1}{2}
  4. h=5R−R2h = \frac{\sqrt{5}R - R}{2}, so the answer is (c).
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