A mass M is split into two parts (M-m) and m, which are then separated by a distance r. For what ratio of m/M, will gravitational attraction between them be maximum:

= 0.20

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= 0.25 M

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= 0.40

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0.50

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Solution

The correct option is D m= $0.50$ M
Let r be the direction between the parts
$F = \frac{G m (M - m)}{r^{2}}$
= $\frac{G}{r^{2}} (m M - m 2)$

The force will be maximum

$\frac{d F}{d m}$ =0

ie $\frac{d}{d m} [\frac{G}{r^{2}} (m M - m^{2})]$ =0

$\frac{G}{r^{2}} (M - 2 m)$ =0

M−2m=0

M=2m

$\frac{M}{m}$ =2

or $\frac{m}{M} = \frac{1}{2}$

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A mass M is split into two parts (M-m) and m, which are then separated by a distance r. For what ratio of m/M, will gravitational attraction between them be maximum:

The correct option is D m=0.50MLet r be the direction between the partsF=Gm(M−m)r2=Gr2(mM−m2)The force will be maximumdFdm=0ieddm[Gr2(mM−m2)]=0Gr2(M−2m)=0M−2m=0M=2mMm=2or mM=12