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Gravitation question

2022 · 24 Jun · Shift 1 · Q53
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  5. /2022 · 24 Jun · Shift 1 · Q53

Gravitation question

2022 · 24 Jun · Shift 1 · Q53

JEE MainPhysicsGravitationMCQ+4 / −1
The approximate height from the surface of earth at which the weight of the body becomes 13{1 \over 3}31​ of its weight on the surface of earth is : [Radius of earth R = 6400 km and 3\sqrt 33​ = 1.732]
  1. A
    3840 km
  2. B
    4685 km
  3. C
    2133 km
  4. D
    4267 km
View written solutionFree

Correct answer: B

  1. Weight at the surface of Earth

The weight of a body on the surface is W=GMmR2W = \frac{GMm}{R^2}W=R2GMm​ where:

  • GGG = gravitational constant
  • MMM = mass of Earth
  • mmm = mass of the body
  • RRR = radius of Earth
  1. Weight at height hhh above the surface

At a height hhh, the distance from Earth’s center becomes (R+h)(R+h)(R+h), so the weight is Wh=GMm(R+h)2W_h = \frac{GMm}{(R+h)^2}Wh​=(R+h)2GMm​

  1. Given condition

The weight becomes one-third of its surface value: Wh=13WW_h = \frac{1}{3}WWh​=31​W

Substitute the formulas: GMm(R+h)2=13⋅GMmR2\frac{GMm}{(R+h)^2} = \frac{1}{3}\cdot \frac{GMm}{R^2}(R+h)2GMm​=31​⋅R2GMm​

Cancel GMmGMmGMm: 1(R+h)2=13R2\frac{1}{(R+h)^2} = \frac{1}{3R^2}(R+h)21​=3R21​

Cross-multiply: (R+h)2=3R2(R+h)^2 = 3R^2(R+h)2=3R2

Take square root: R+h=3 RR+h = \sqrt{3}\,RR+h=3​R

Therefore, h=(3−1)Rh = (\sqrt{3}-1)Rh=(3​−1)R

  1. Substitute values

Given: R=6400 km,3=1.732R = 6400\text{ km}, \quad \sqrt{3} = 1.732R=6400 km,3​=1.732

So, h=(1.732−1)×6400h = (1.732 - 1)\times 6400h=(1.732−1)×6400 h=0.732×6400h = 0.732 \times 6400h=0.732×6400 h=4684.8 kmh = 4684.8\text{ km}h=4684.8 km

Thus, approximately, h≈4685 kmh \approx 4685\text{ km}h≈4685 km

  1. Check options
  • A: 384038403840 km
  • B: 468546854685 km ✅
  • C: 213321332133 km
  • D: 426742674267 km

So the correct option is B.

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