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(A) The gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$.Here,$m_1 = m$

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$1$: $1$

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(A) The gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$.Here,$m_1 = m$ and $m_2 = (M - m)$.Substituting these values,we get $F = \frac{G m(M - m)}{r^2} = \frac{G}{r^2} (Mm - m^2)$.For the force to be maximum,the derivative of $F$ with respect to $m$ must be zero,i.e.,$\frac{dF}{dm} = 0$.$\frac{d}{dm} [\frac{G}{r^2} (Mm - m^2)] = 0$.$\frac{G}{r^2} (M - 2m) = 0$.Since $\frac{G}{r^2} 
eq 0$,we have $M - 2m = 0$,which implies $m = \frac{M}{2}$.Thus,the mass of the second piece is $(M - m) = M - \frac{M}{2} = \frac{M}{2}$.The ratio of the masses is $\frac{m}{M-m} = \frac{M/2}{M/2} = \frac{1}{1}$.

$A$ mass $M$ at rest is broken into two pieces having masses $m$ and $(M-m)$. The two masses are then separated by a distance $r$. The gravitational force between them will be maximum when the ratio of the masses $[m : (M-m)]$ of the two parts is:

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