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Q. 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 the maximum when he ratio of the masses $[m:(M-m)]$ of the two parts is

WBJEEWBJEE 2013Gravitation

Solution:

The gravitation force between two masses,
$F=\frac{G m_{1} m_{2}}{r}$
here, $m_{1}=m$
and $m_{2} =(M-m) $
$\therefore $ F $=\frac{G m(M-m)}{r^{2}}$
For maximum gravitational force, $\frac{d F}{d m}=0$
$\therefore \frac{d}{d m}[m(M-m)]=0$
By solving, we get $m=\frac{M}{2}$
So, $\frac{m}{(M-m)}=\frac{1}{1}$