NEET 2023 — Physics solution

NEET2023Physics

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

A
$- \frac{12 G m}{R}$
B
$- \frac{16 G m}{R}$
C
$- \frac{20 G m}{R}$
D
$- \frac{8 G m}{R}$

Solution

Let the null-field point P lie at distance $r_{1}$ from mass $m$ and $r_{2} = R - r_{1}$ from mass $9 m$ . Setting fields equal in magnitude:

$\frac{G m}{r _{1}^{2}} = \frac{G ( 9 m )}{r _{2}^{2}} \Rightarrow \frac{r _{2}}{r _{1}} = 3 \Rightarrow r_{1} = \frac{R}{4}, r_{2} = \frac{3 R}{4}$ .

Potential at P (scalar sum):

$V_{P} = - \frac{G m}{R /4} - \frac{G ( 9 m )}{3 R /4} = - \frac{4 G m}{R} - \frac{12 G m}{R} = - \frac{16 G m}{R}$ .

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