Physics Circle home Student Login

Home›Chapters›Gravitation PYQs›Acceleration Due to Gravity›Question 106

JEE Main 2020 (2 Sep, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2020 (2 Sep, Shift 2)

Q106JEE Main 2020 · 2 SEP · SHIFT 2Difficult

Acceleration Due to Gravity

The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

Show answer and solution

Answer: (b)

  1. Equal weight: g(1−hR)=gR2(R+h)2g\left(1 - \frac{h}{R}\right) = \frac{gR^2}{(R+h)^2} (exact formulas)
  2. Put x=hRx = \frac{h}{R}: (1−x)(1+x)2=1⇒x2+x−1=0(1-x)(1+x)^2 = 1 \Rightarrow x^2 + x - 1 = 0
  3. x=5−12⇒h=5R−R2x = \frac{\sqrt{5}-1}{2} \Rightarrow h = \frac{\sqrt{5}R - R}{2}
  4. So the answer is (b).
Report a mistake

More questions on Acceleration Due to Gravity