JEE Main 2022 · 26 July · morning · Single correct · +4 / −1

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be :
( Radius of earth =6400 km)

  1. (a) 1%
  2. (b) 3%
  3. (c) 4%
  4. (d) 0.5%
Answer and solution

Answer (a) 1%

Solution

Physics sir · 4 steps

32 km is tiny next to 6400 km, so use the short form: g drops by a fraction 2h/R.

  1. g at height. Weight is mg, and the mass stays the same. So the weight falls by the same percentage as g. At height h, g' = g(R/(R+h))².
  2. Small height. Here h/R = 32/6400 = 0.005, which is much less than 1. Expanding with the binomial approximation gives a simple result.

    g' ≈ g(1 - 2h/R) ⇒ (Δ W)/W× 100 = 2h/R× 100

  3. Put in numbers. (2 × 32)/6400 × 100 = 1%. The exact formula gives 1 - (6400/6432)² ≈ 0.99%, which agrees.
  4. So. The weight of the rocket decreases by about 1%, option (a).

The idea to keep. For heights much smaller than the Earth's radius, g falls by a fraction 2h/R. Going down a depth d, it falls by only d/R. Do not mix up the factor of 2.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 160 times in JEE Main since 2020. This question is from 2022.