Two point object of mass $2 x$ snd $3 x$ are separated by a distance r.Keeping the distance fixed,how much mass should be transferred from $3 x$ to $2 x$ ,so that gravitational force between them becomes maximum?

\frac{x}{4}

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\frac{x}{3}

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\frac{x}{2}

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\frac{2 x}{3}

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Solution

The correct option is C $\frac{x}{2}$
Let $y$ amount of mass should be transferred. Then:
$F^{'} = \frac{G (M - y) (m + y)}{r^{2}}$
$= \frac{G (3 x - y) (2 x + y)}{r^{2}}$
For this to be maximum, $\frac{d F^{'}}{d y} = 0$
$\frac{d F^{'}}{d y} = \frac{d [6 x^{2} + 3 x y - 2 x y - y^{2}]}{d y}$
$= 3 x - 2 x - 2 y = 0$
$x - 2 y = 0$
$y = \frac{x}{2}$

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Two point object of mass 2x snd 3x are separated by a distance r.Keeping the distance fixed,how much mass should be transferred from 3x to 2x,so that gravitational force between them becomes maximum?

The correct option is C x2Let y amount of mass should be transferred. Then:F′=G(M−y)(m+y)r2=G(3x−y)(2x+y)r2For this to be maximum, dF′dy=0dF′dy=d[6x2+3xy−2xy−y2]dy=3x−2x−2y=0x−2y=0y=x2

Two point object of mass $2 x$ snd $3 x$ are separated by a distance r.Keeping the distance fixed,how much mass should be transferred from $3 x$ to $2 x$ ,so that gravitational force between them becomes maximum?