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 $(\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes ${F}_{2}$. The value of ${F}_{1}:{F}_{2}$ is: ![JEE Main 2021 Physics, Gravitation — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Gravitation/648b5a6a417cc3fb48d6724a/question_1__q_648b5a6a417cc3fb48d6724a__cdn-question-pool.getmarks.app__752e8432-e36d-4c02-b770-3703f2eb95d5-image__dbbc54753a_final_ppt_sync.png)

Let the initial mass of the sphere is $m'.$ Hence, mass of a removed portion will be $m'/8$,${F}_{1}=m.E.=\frac{m.Gm'}{9{R}^{2}}$ ![JEE Main 2021 Physics, Gravitation — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Gravitation/648b5a6a417cc3fb48d6724a/solution_1__q_648b5a6a417cc3fb48d6724a__cdn-question-pool.getmarks.app__b1b97846-87e1-4ee0-9810-4a47bea4c847-image__5a76e9e50d_final_ppt_sync.png) ${F}_{2}=m[\frac{G\cdot m'}{{(3R)}^{2}}-\frac{G\cdot m'/8}{{(5R/2)}^{2}}]$$=\frac{Gm'}{9{R}^{2}}-\frac{Gm'\times 4}{8\times 25}$$=(\frac{1}{9}-\frac{1}{50})\frac{Gm'}{{R}^{2}}$ ${F}_{2}=\frac{41}{50\times 9}\cdot \frac{Gm'}{{R}^{2}}$$\Rightarrow \frac{{F}_{1}}{{F}_{2}}=\frac{1}{9}\times \frac{50\times 9}{41}=\frac{50}{41}$

A solid sphere of radius R… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Mechanics

2021mcqmedium

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

JEE Main 2021 Physics, Gravitation — question figure

Official previous-year question

Held on 25 Feb 2021 · Verified 6 Jul 2026.

Options

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

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Solution

Let the initial mass of the sphere is $m'.$ Hence, mass of

a removed portion will be $m'/8$ , ${F}_{1}=m.E.=\frac{m.Gm'}{9{R}^{2}}$

JEE Main 2021 Physics, Gravitation — solution figure

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

${F}_{2}=\frac{41}{50\times 9}\cdot \frac{Gm'}{{R}^{2}}$ $\Rightarrow \frac{{F}_{1}}{{F}_{2}}=\frac{1}{9}\times \frac{50\times 9}{41}=\frac{50}{41}$

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Solution

Let the initial mass of the sphere is $m'.$ Hence, mass of

a removed portion will be $m'/8$ , ${F}_{1}=m.E.=\frac{m.Gm'}{9{R}^{2}}$

JEE Main 2021 Physics, Gravitation — solution figure

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

${F}_{2}=\frac{41}{50\times 9}\cdot \frac{Gm'}{{R}^{2}}$ $\Rightarrow \frac{{F}_{1}}{{F}_{2}}=\frac{1}{9}\times \frac{50\times 9}{41}=\frac{50}{41}$

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Solution

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Options

Did you get this right?

Solution

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JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics