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**Hint:**The first step is to find the point lying on the line joining the point masses where the net gravitational force is zero. Then the net gravitational potential at this point is found by adding the potentials by both masses.

**Complete step by step answer:**

Gravitational field intensity is defined as the gravitational force between a given mass and a unit mass. If the force of gravity between two objects is given by-

$F = \dfrac{{G{M_1}{M_2}}}{{{r^2}}}$

Where ${M_1}$ and ${M_2}$ represent the mass of two objects.

The distance between their centers is $r$,

And $G$ is the universal gravitational constant.

Then to define the Gravitational field intensity$({E_g})$ of an object, We can assume ${M_2} = 1$,

${\vec E_g} = \dfrac{{GM}}{{{r^2}}}$

It is given in the question that the mass of the first body (body A) is, $M$

And the mass of the second body (body B) is,$4M$

If a unit point mass is put along the line joining the centers of both the objects. Then, C can be defined as a neutral point, lying on this line where the point mass does not experience any gravitational force.

We know that both masses are separated by a distance$r$between them.

Then this neutral point C is situated at, $x$distance from A and $r - x$distance from B.

The gravitational field at C due to mass A is-

${\vec E_{gA}} = \dfrac{{GM}}{{{x^2}}}$

The gravitational field at C due to mass B is-

${\vec E_{gB}} = \dfrac{{4GM}}{{{{\left( {r - x} \right)}^2}}}$

Since the net gravitational field at this point is zero,

$ \Rightarrow {\vec E_{gA}} + {\vec E_{gB}} = 0$

The directions of the fields created at C by both masses are opposite, this is why when these are combined their resultant is zero.

$\dfrac{{4GM}}{{{{\left( {r - x} \right)}^2}}} = \dfrac{{GM}}{{{x^2}}}$

\[ \Rightarrow \dfrac{4}{{{{\left( {r - x} \right)}^2}}} = \dfrac{1}{{{x^2}}}\]

This expression can be rewritten as-

\[\dfrac{2}{{\left( {r - x} \right)}} = \dfrac{1}{x}\]

$ \Rightarrow r - x = 2x$

$ \Rightarrow r = 3x$

$ \Rightarrow x = \dfrac{r}{3}$

And $r - x = \dfrac{{2r}}{3}$

The gravitational potential at this point$\left( {{V_C}} \right)$, is given by-

${V_C} = {V_A} + {V_B}$

We know that,

$V = - \dfrac{{Gm}}{r}$

Gravitational potential of A at C,

${V_A} = - \dfrac{{GM}}{x}$

Substituting the value of$x$,

${V_A} = - \dfrac{{GM}}{{\dfrac{r}{3}}} = - \dfrac{{3GM}}{r}$

Gravitational potential of B at C,

${V_B} = - \dfrac{{4GM}}{{r - x}}$

Substituting the value of $x$,

$ \Rightarrow {V_B} = - \dfrac{{4GM}}{{r - \dfrac{r}{3}}} = - \dfrac{{3 \times 4GM}}{{2r}}$

$ \Rightarrow {V_B} = - \dfrac{{6GM}}{r}$

Therefore,

${V_C} = - \dfrac{{3GM}}{r} - \dfrac{{6GM}}{r}$

$ \Rightarrow {V_C} = - \dfrac{{9GM}}{r}$

**Option (A) is correct.**

**Note:**The gravitational potential has a negative sign because the gravitational force is always attractive. The gravitational potential energy is defined as the work done required to move a mass (say m) from infinity to a distance r from the main mass (say A). As the force of gravity is attractive, no work is done by the mass m, instead, it is done by the gravitational field, thus the negative value.

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