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

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$\frac{-16Gm}{r}$

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(B) Let the point where the gravitational field is zero be at a distance $x$ from the mass $m$. The gravitational field due to both masses must be equal in magnitude and opposite in direction.$E_1 = E_2 \Rightarrow \frac{Gm}{x^2} = \frac{G(9m)}{(r-x)^2}$Taking the square root on both sides: $\frac{1}{x} = \frac{3}{r-x}$$r-x = 3x \Rightarrow 4x = r \Rightarrow x = \frac{r}{4}$.The distance from mass $9m$ is $r-x = r - \frac{r}{4} = \frac{3r}{4}$.The gravitational potential $V$ at this point is the sum of the potentials due to both masses:$V = V_1 + V_2 = -\frac{Gm}{x} - \frac{G(9m)}{r-x}$$V = -\frac{Gm}{r/4} - \frac{9Gm}{3r/4} = -\frac{4Gm}{r} - \frac{12Gm}{r}$$V = -\frac{16Gm}{r}$.

Two bodies of masses $m$ and $9m$ are placed at a distance $r$. The gravitational potential at a point on the line joining them,where the gravitational field is zero,is ($G$ is the universal gravitational constant).

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