The weight of a body on the surface of the earth is 100 ~N . The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

Gravitation PYQ — JEE Main 2023 · 6 Apr · Shift 2 · Q62 with Solution

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Correct answer: B

Given data
- Weight on the surface of Earth: $W = 100\,\text{N}$
- Height above the surface: $h = \dfrac{R}{4}$ , where $R$ is the radius of Earth
Formula for gravitational force at height $h$

The gravitational force varies inversely as the square of the distance from the center of Earth: $F_h = F_0\left(\frac{R}{R+h}\right)^2$ where:
- $F_0 = 100\,\text{N}$ is the force on the surface
- $R+h$ is the distance from the center at height $h$
Substitute $h = \dfrac{R}{4}$ $R+h = R + \frac{R}{4} = \frac{5R}{4}$

So, $F_h = 100\left(\frac{R}{5R/4}\right)^2$
Simplify $\frac{R}{5R/4} = \frac{4}{5}$

Therefore, $F_h = 100\left(\frac{4}{5}\right)^2$ $F_h = 100\cdot \frac{16}{25}$ $F_h = 64\,\text{N}$
Evaluate options
- A: $50\,\text{N}$ ❌
- B: $64\,\text{N}$ ✅
- C: $25\,\text{N}$ ❌
- D: $100\,\text{N}$ ❌

Final Answer: Option B, $64\,\text{N}$