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Consider two solid spheres of radii $R_{1} = 1 m$ , $R_{2} = 2 m$ and masses $M_{1}$ and $M_{3}$ respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $\frac{m_{1}}{m_{2}}$ is

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$\frac{1}{6}$

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{3}$

answer is B.

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Detailed Solution

Field on the surface of solid spheres is

$E = \frac{G M}{R^{2}}$

$\frac{E_{1}}{E_{2}} = \frac{M_{1}}{M_{2}} {(\frac{R_{2}}{R_{1}})}^{2} \Rightarrow \frac{2}{3} = \frac{M_{1}}{M_{2}} {(\frac{2}{1})}^{2}$

$\frac{M_{1}}{M_{2}} = \frac{1}{6}$

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