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:

The weight of a body on the surface of the earth is $$100$$ N. We need to find the gravitational force at height $$h = \dfrac{R}{4}$$ from the surface.

$$g' = g \cdot \dfrac{R^2}{(R+h)^2}$$

$$g' = g \cdot \dfrac{R^2}{(R + R/4)^2} = g \cdot \dfrac{R^2}{(5R/4)^2} = g \cdot \dfrac{R^2}{25R^2/16} = g \cdot \dfrac{16}{25}$$

$$W' = mg' = mg \cdot \dfrac{16}{25} = 100 \times \dfrac{16}{25} = 64 \text{ N}$$

The gravitational force at height $$R/4$$ is $$64$$ N, which corresponds to Option A.