Solved: PHYSICS Question for NEET

NEET PHYSICSHard

Two bodies of mass $m$ and $9m$ are placed at a distance $R$ . The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( $G = \text{gravitational constant}$ )

$-\frac{8Gm}{R}$

$-\frac{12Gm}{R}$

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

$-\frac{20Gm}{R}$

Step-by-Step Solution

Let the point be at distance $x$ from mass $m$ . Field is zero when $\frac{Gm}{x^2} = \frac{G(9m)}{(R-x)^2} \implies \frac{1}{x} = \frac{3}{R-x} \implies R-x = 3x \implies x = R/4$ . The distance from $9m$ is $3R/4$ . Potential $V = -\frac{Gm}{x} - \frac{G(9m)}{R-x} = -\frac{Gm}{R/4} - \frac{9Gm}{3R/4} = -\frac{4Gm}{R} - \frac{12Gm}{R} = -\frac{16Gm}{R}$ .

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