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Q. A solid sphere of radius $R$ gravitationally attracts a particle placed at $3 R$ form its centre with a force $F _{1}$. Now a spherical cavity of radius $\left(\frac{ R }{2}\right)$ is made in the sphere (as shown in figure) and the force becomes $F _{2}$. The value of $F _{1}: F _{2}$ is
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JEE MainJEE Main 2021Gravitation

Solution:

Let initial mass of sphere is $m$ '. Hence mass of removed portion will be $m ^{\prime} / 8$ $F _{1}= m . E .=\frac{ m . Gm ^{\prime}}{9 R ^{2}}$
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$F _{2}= m \left[\frac{ G \cdot m ^{\prime}}{(3 R )^{2}}-\frac{ G . m ^{1} / 8}{(5 R / 2)^{2}}\right]$
$=\frac{ Gm ^{\prime}}{9 R ^{2}}-\frac{ Gm ^{\prime} \times 4}{8 \times 25}$
$=\left(\frac{1}{9}-\frac{1}{50}\right) \frac{ Gm ^{\prime}}{ R ^{2}}$
$F _{2}=\frac{41}{50 \times 9} \cdot \frac{ Gm ^{\prime}}{ R ^{2}}$
$\frac{ F _{1}}{ F _{2}}=\frac{1}{9} \times \frac{50 \times 9}{41}=\frac{50}{41}$