A body weight W , is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be :

Gravitation PYQ — JEE Main 2023 · 31 Jan · Shift 2 · Q52 with Solution

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Correct answer: C

Use the variation of gravitational force with distance from Earth's center

The weight of a body at a distance $r$ from the center of Earth is $W_r = \frac{GMm}{r^2}$

On the surface of Earth, where $r=R$ , $W = \frac{GMm}{R^2}$
Find the new distance from Earth's center

The body rises to a height equal to nine times Earth's radius: $h = 9R$

Therefore, its distance from the center becomes $r = R + h = R + 9R = 10R$
Find the new weight

Let the new weight be $W'$ . Then $W' = \frac{GMm}{(10R)^2} = \frac{GMm}{100R^2}$

Since $W = \frac{GMm}{R^2},$ we get $W' = \frac{W}{100}$
Check options
- A: $\frac{W}{91}$ ❌
- B: $\frac{W}{3}$ ❌
- C: $\frac{W}{100}$ ✅
- D: $\frac{W}{9}$ ❌

Therefore, the correct option is C.