Question

Mass $m$ units is divided in two parts X and $(1-X).$ For a given separation the value of $X$ for which the gravitational force between them becomes maximum is :

• A. $1/2$
• B. $3/5$
• C. $1$
• D. $2$

Answer 1

The correct option is A $1/2$

Gravitational Force, $F=\frac{G{m}_1{m}_2}{r^2}=\frac{GXM(1-X)M}{r^2}=\frac{G{M}^2}{r^2}X(1-X)$

For maximum value of force, $\frac{dF}{dx}=0$

That is,

$\frac{d}{dx}\left[\frac{G{M}^{2}X}{r^2}\right(1-X\left)\right]=0$

$⟹\frac{d}{dx}(X-X^2)=0⟹1-2X=0⟹X=\frac{1}{2}$