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(B) Let the point where the gravitational field is zero be at a distance $x$ from mass $m$. The gravitational field due to both masses must be equal i

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$-\frac{9Gm}{r}$

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(B) Let the point where the gravitational field is zero be at a distance $x$ from mass $m$. The gravitational field due to both masses must be equal in magnitude and opposite in direction.$\frac{Gm}{x^2} = \frac{G(4m)}{(r-x)^2}$Taking the square root on both sides:$\frac{1}{x} = \frac{2}{r-x} \Rightarrow r-x = 2x \Rightarrow 3x = r \Rightarrow x = \frac{r}{3}$The distance from mass $4m$ is $r-x = r - \frac{r}{3} = \frac{2r}{3}$.The gravitational potential $V$ at this point is the sum of potentials due to both masses:$V = -\frac{Gm}{x} - \frac{G(4m)}{r-x}$Substituting the values of $x$ and $r-x$:$V = -\frac{Gm}{r/3} - \frac{4Gm}{2r/3} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}$

Two bodies of masses $m$ and $4m$ are placed at a distance $r$. The gravitational potential at a point on the line joining them where the gravitational field is zero is

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