Question

Consider two solid spheres of radii $$R_1 = 1m, \, R_2 = 2m$$ and masses $$M_1$$ and $$M_2$$, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $$\dfrac{M_1}{M_2}$$ is:

Answer 1

Correct option is A. $$\dfrac{1}{6}$$
Let $$E_1, E_2$$ be the gravitational field intensities on the surface of the two spheres.

$$E_1 = \dfrac{GM_1}{R_1^2}$$ and $$E_2 = \dfrac{GM_2}{R_2^2}$$

In a solid sphere, the gravitational field varies linearly from the center to the surface and is proportional to $$R^{-2}$$ outside the sphere.

From the figure, we have $$E_1 = 2$$ at the surface of the first spher and $$E_2 = 3$$ at the surface of the second sphere.

$$\therefore 2 = \dfrac{GM_1}{R_1^2} = \dfrac{GM_1}{1}, \, GM_1 = 2$$ ...(i)

$$\therefore 3 = \dfrac{GM_2}{R_2^2} = \dfrac{GM_2}{2^2}, \, GM_2 = 12$$ ...(ii)

Dividing (i) by (ii),

$$\dfrac{M_1}{M_2} = \dfrac{2}{12} = \dfrac{1}{16}$$