Consider two solid spheres of radii $R_{1}=1 m, R_{2}=2 m$ and masses $M_{1}$ and $M_{2}$, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $\frac{m_{1}}{m_{2}}$ is:
Correct Option: , 2
(2) Gravitation field at the surface
$E=\frac{G m}{r^{2}}$
$\therefore E_{1}=\frac{G m_{1}}{r_{1}^{2}}$ and $E_{2}=\frac{G m_{2}}{r_{2}^{2}}$
From the diagram given in question,
$\frac{E_{1}}{E_{2}}=\frac{2}{3}\left(r_{1}=1 \mathrm{~m}, R_{2}=2 m\right.$ given $)$
$\therefore \frac{E_{1}}{E_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{2}\left(\frac{m_{1}}{m_{2}}\right) \Rightarrow \frac{2}{3}=\left(\frac{2}{1}\right)^{2}\left(\frac{m_{1}}{m_{2}}\right)$
$\Rightarrow\left(\frac{m_{1}}{m_{2}}\right)=\frac{1}{6}$
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