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A solid sphere of radius R gavitationally attracts a particle placed at 3R from its centre with a force F1. Now a spherical cavity radius \left (\frac{R}{2} \right ) is made in the sphere (as shown in the image) and the force becomes F2. The value F:F2 is:
Option: 1 41 :50
Option: 2 36:25
Option: 3 50:41
Option: 4 25 :36

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Let the initial mass of the sphere is M.

\begin{array}{l} {[\text { Using, Mass }=\text { Volume } \times \text { density }} \\ \text { we get } \left.\mathrm{M}=\frac{4 \pi \mathrm{R}^{3}}{3} \times \mathrm{ d}\right] \end{array}

If a spherical part of radius \frac{R}{2} is taken out from the big sphere, then the mass of the remaining sphere is

\begin{array}{l} =\left[\frac{4 \pi \mathrm{R}^{3}}{3}-\frac{4 \pi}{3}\left(\frac{\mathrm{R}}{2}\right)^{3}\right] \times \mathrm{d} =\frac{4 \pi \mathrm{R}^{3}}{3} \times\left(\frac{7}{8}\right) \times \mathrm{d} =\frac{7 \mathrm{M}}{8} \end{array}

Hence mass of removed portion will be  \frac{M}{8}

From the figure it is clear that

\mathrm{F}_{1}=\frac{\mathrm{GMm}}{(3\mathrm{R})^{2}}=\frac{\mathrm{GMm}}{9 \mathrm{R}^{2}}

From superposition principle,

\mathrm{F}_{1}=\mathrm{F}_{\mathrm{r}}+\mathrm{F}_{\mathrm{c}}
Here \mathrm{F}_{\mathrm{r}}= force due to remaining part =\mathrm{F}_{2}
\mathrm{F}_{\mathrm{c}}= force due to mass on the cavity

As \mathrm{F}_{\mathrm{c}}=\frac{\mathrm{G}\left(\frac{\mathrm{M}}{8}\right) \mathrm{m}}{\left(\frac{5}{2} \mathrm{R}\right)^{2}}=\frac{\mathrm{GMm}}{50 \mathrm{R}^{2}}

\begin{array}{l} \Rightarrow \mathrm{F}_{2}=\mathrm{F}_{1}-\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{GMm}}{9 \mathrm{R}^{2}}-\frac{\mathrm{GMm}}{50 \mathrm{R}^{2}}=\frac{41 \mathrm{GMm}}{450 \mathrm{R}^{2}} \\ \\ \Rightarrow \frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{50}{41} \end{array}

 

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avinash.dongre

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