Two bodies of masses m and M are placed a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V, then

The correct option is: V = -G(√m + √M)²/d

Here is a step-by-step derivation to find the gravitational potential at the position where the net gravitational field is zero.

Step 1: Identify the position of zero gravitational field (null point)
Let two masses $m$ and $M$ be placed at a distance $d$ apart. Let the point where the net gravitational field is zero be at a distance $r_1$ from mass $m$ and a distance $r_2$ from mass $M$.
Since the point lies on the line joining the two masses, we have:
$r_1+r_2=d$

Step 2: Relate the distances using the zero-field condition
At the null point, the gravitational field strength due to mass $m$ must be equal in magnitude and opposite in direction to the field strength due to mass $M$:
$\frac{Gm}{r_1^2}=\frac{GM}{r_2^2}$
Taking the square root on both sides:
$\frac{\sqrt{m}}{r_1}=\frac{\sqrt{M}}{r_2}$
This gives:
$r_2=r_1\sqrt{\frac{M}{m}}$

Step 3: Solve for the individual distances
Substitute the expression for $r_2$ into the equation $r_1+r_2=d$:
$r_1+r_1\frac{\sqrt{M}}{\sqrt{m}}=d$
$r_1\left(\frac{\sqrt{m}+\sqrt{M}}{\sqrt{m}}\right)=d$
Therefore, the distances are:
$r_1=\frac{\sqrt{m}}{\sqrt{m}+\sqrt{M}}d$
And similarly:
$r_2=\frac{\sqrt{M}}{\sqrt{m}+\sqrt{M}}d$

Step 4: Calculate the gravitational potential V at this point
The total gravitational potential $V$ is the sum of the potentials due to each individual mass:
$V=-\frac{Gm}{r_1}-\frac{GM}{r_2}$
Substituting the expressions for $r_1$ and $r_2$:
$V=-\frac{Gm(\sqrt{m}+\sqrt{M})}{\sqrt{m}d}-\frac{GM(\sqrt{m}+\sqrt{M})}{\sqrt{M}d}$
Using the algebraic simplifications $\frac{m}{\sqrt{m}}=\sqrt{m}$ and $\frac{M}{\sqrt{M}}=\sqrt{M}$:
$V=-\frac{G\sqrt{m}(\sqrt{m}+\sqrt{M})}{d}-\frac{G\sqrt{M}(\sqrt{m}+\sqrt{M})}{d}$
Factoring out the common terms:
$V=-\frac{G}{d}(\sqrt{m}+\sqrt{M})(\sqrt{m}+\sqrt{M})$
$V=-\frac{G{(\sqrt{m}+\sqrt{M})}^2}{d}$