91Ó°ÊÓ

Mass of the sphere at origin (m) = $20kg,$mass of the sphere at$20cmfromorigin=10kg.$

The formula to be used is : $\mathrm{F}=\mathrm{Gm}1\mathrm{m}2\mathrm{r}2.$

Where $\mathrm{m}1$and $\mathrm{m}2$are masses, $r$ is the distance between the centre of the masses and $\mathrm{G}$ is the universal gravitational constant.

Let the position of the small mass be $x=r,$and mass is $m.$Hence the force on the small mass due to $m1$is :$\mathrm{F}1=Gm1mr2=\mathrm{G}(20\mathrm{kg})mr2.$

And force due to$m2$is :

${\mathrm{F}}_2=\mathrm{Gm}2\mathrm{m}\left[\right(0.2\mathrm{m})-\mathrm{r}]2=\mathrm{G}(10\mathrm{kg})\mathrm{m}\left[\right(0.2\mathrm{m})-\mathrm{r}]2.$

The total force on the small mass will be zero if the above two forces are equal and opposite.

So, equate the above two forces and solve for $r.$

Comparing we get,

$$\begin{gathered} \mathrm{G}(20\mathrm{kg})m*r2=\mathrm{G}(10\mathrm{kg})\mathrm{m}\left[\right(0.2\mathrm{m})-\mathrm{r}]2 \\ ⇒(20\mathrm{kg})\mathrm{r}2=(10\mathrm{kg})\left[\right(0.2\mathrm{m})-\mathrm{r}]2. \\ ⇒\mathrm{r}=0.12\mathrm{m},0.68\mathrm{m}. \end{gathered}$$

The above two solutions show at which position forces will be equal.

Now, since the both forces are attractive forces, they will be opposite only when the small mass is in between the two masses.

Hence, the correct answer is$\mathrm{r}=0.12\mathrm{m}=12\mathrm{cm}.$

At the position of $x=12,$a small mass could be placed, such that the net gravitational force on it due to the spheres is zero.