Two bodies of masses $M_{1}$ and $M_{2}$ are placed at a distance ' $r$ '. The gravitational potential at the point where the gravitational field is zero

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$- \frac{G}{r} (M_{1} + 2 \sqrt{M_{1} M_{2}} + M_{2})$

$- 2 \frac{G}{r} (M_{1} + M_{2})$

$- \frac{G}{r} {(M_{1} + M_{2})}^{2}$

$- \frac{G}{r} (\sqrt{M_{1}} + \sqrt{M_{2}})$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{G M_{1}}{x^{2}} = \frac{G M_{2}}{(R - x)^{2}} \\ or (R - x) \sqrt{M_{1}} = x \sqrt{M_{2}} ∣ \\ or x = \frac{R \sqrt{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}} \\ and R - x = R - [\frac{R \sqrt{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}}] = \frac{R \sqrt{M_{2}}}{\sqrt{M_{1}} + \sqrt{M_{2}}} \\ So, \frac{1}{x} = \frac{\sqrt{M_{1}} + \sqrt{M_{2}}}{R \sqrt{M_{1}}} \\ and \frac{1}{R - x} = \frac{\sqrt{M_{1}} + \sqrt{M_{2}}}{R \sqrt{M_{2}}} \end{array}$

Now gravitational potential is given by,

$\begin{aligned} = - \frac{G M_{1}}{x} + [- \frac{G M_{2}}{R - x}] \\ = - G [\frac{M_{1}}{x} + \frac{M_{2}}{R - x}] \end{aligned}$

Putting the values

$\begin{array}{l} V = - G [M_{1} [\frac{\sqrt{M_{1}} + \sqrt{M_{2}}}{R \sqrt{M_{1}}} + M_{2} \frac{\sqrt{M_{1}} + \sqrt{M_{2}}}{R \sqrt{M_{2}}}]] \\ = - \frac{G}{R} [M_{1} + M_{2} + 2 \sqrt{M_{1} M_{2}}] \end{array}$