Question 19 of 223

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

A$41:50$B$50:41$C$25:36$D$36:25$

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Answer: (B) $50:41$

Let the initial mass of the sphere is $m'.$ Hence, mass of
a removed portion will be $m'/8$,${F}_{1}=m.E.=\dfrac{m.Gm'}{9{R}^{2}}$

${F}_{2}=m\left[\dfrac{G\cdot m'}{{\left(3R\right)}^{2}}-\dfrac{G\cdot m'/8}{{\left(5R/2\right)}^{2}}\right]$$=\dfrac{Gm'}{9{R}^{2}}-\dfrac{Gm'\times 4}{8\times 25}$$=\left(\dfrac{1}{9}-\dfrac{1}{50}\right)\dfrac{Gm'}{{R}^{2}}$
${F}_{2}=\dfrac{41}{50\times 9}\cdot \dfrac{Gm'}{{R}^{2}}$$\Rightarrow \dfrac{{F}_{1}}{{F}_{2}}=\dfrac{1}{9}\times \dfrac{50\times 9}{41}=\dfrac{50}{41}$