A solid sphere of radius R gravitationally attracts a particle
Question:

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 $\left(\frac{\mathrm{R}}{2}\right)$ is made in the sphere

(as shown in figure) and the force becomes $\mathrm{F}_{2}$. The value of $F_{1}: F_{2}$ is :

  1. $25: 36$

  2. $36: 25$

  3. $50: 41$

  4. $41: 50$


Correct Option: , 3

Solution:

Let initial mass of sphere is $\mathrm{m}^{\prime}$. Hence mass of removed portion will be $\mathrm{m}^{\prime} / 8$

$\mathrm{F}_{1}=\mathrm{m} \cdot \mathrm{E} .=\frac{\mathrm{m} \cdot \mathrm{Gm}^{\prime}}{9 \mathrm{R}^{2}}$

$\mathrm{F}_{2}=\mathrm{m}\left[\frac{\mathrm{G} \cdot \mathrm{m}^{\prime}}{(3 \mathrm{R})^{2}}-\frac{\mathrm{G} \cdot \mathrm{m}^{\prime} / 8}{(5 \mathrm{R} / 2)^{2}}\right]$

$=\frac{G m^{\prime}}{9 R^{2}}-\frac{G m^{\prime} \times 4}{8 \times 25}$

$=\left(\frac{1}{9}-\frac{1}{50}\right) \frac{\mathrm{Gm}^{\prime}}{\mathrm{R}^{2}}$

$\mathrm{F}_{2}=\frac{41}{50 \times 9} \cdot \frac{\mathrm{Gm}^{\prime}}{\mathrm{R}^{2}}$

$\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{1}{9} \times \frac{50 \times 9}{41}=\frac{50}{41}$

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