As shown in figure. When a spherical cavity (centred at $O$ ) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$ ), the centre of mass of remaining (shaded part of sphere is at $G,$ i.e., on the surface of the cavity. $R$ can be determined by the equation: ![JEE Main 2020 Physics, Center of Mass Momentum and Collision — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Center_of_Mass_Momentum_and_Collision/648b5a69417cc3fb48d66d32/question_1__q_648b5a69417cc3fb48d66d32__cdn-question-pool.getmarks.app__00f846d1-8478-4768-9814-42f8f3e3cd72__6cfd26e7b5_final_ppt_sync.png)

JEE Main 2020 — Physics Mechanics

2020mcqhard

As shown in figure. When a spherical cavity (centred at $O$ ) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$ ), the centre of mass of remaining (shaded part of sphere is at $G,$ i.e., on the surface of the cavity. $R$ can be determined by the equation:

JEE Main 2020 Physics, Center of Mass Momentum and Collision — question figure

Official previous-year question

Held on 8 Jan 2020 · Verified 6 Jul 2026.

Options

A
$({R}^{2}+R+1)(2-R)=1$
B
$({R}^{2}-R-1)(2-R)=1$
C
$({R}^{2}-R+1)(2-R)=1$
D
$({R}^{2}+R-1)(2-R)=1$

Did you get this right?

Solution

${M}_{1}=\frac{4}{3}\pi {R}^{3}\rho$

${M}_{2}=\frac{4}{3}\pi {(1)}^{3}(-\rho )$

${X}_{com}=\frac{{M}_{1}{X}_{1}+{M}_{2}{X}_{2}}{{M}_{1}+{M}_{2}}$

$\Rightarrow \frac{[\frac{4}{3}\pi {R}^{3}\rho ]0+[\frac{4}{3}\pi {(1)}^{3}(-\rho )][R-1]}{\frac{4}{3}\pi {R}^{3}\rho +\frac{4}{3}\pi {(1)}^{3}(-\rho )}-(2-R)$

$\Rightarrow \frac{(R-1)}{{(R}^{3}-1)}=(2-R)(R\neq 1)$

$\frac{(R-1)}{(R-1)({R}^{2}+R+1)}=2-1$

$({R}^{2}+R+1)(2-R)=1$

Alternative:

${M}_{remaining}(2-R)={M}_{cavity}(1-R)$

$\Rightarrow ({R}^{3}-{1}^{3})(2-R)={1}^{3}[R-1]$

$\Rightarrow ({R}^{2}+R+1)(2-R)=1$

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Solution

${M}_{1}=\frac{4}{3}\pi {R}^{3}\rho$

${M}_{2}=\frac{4}{3}\pi {(1)}^{3}(-\rho )$

${X}_{com}=\frac{{M}_{1}{X}_{1}+{M}_{2}{X}_{2}}{{M}_{1}+{M}_{2}}$

$\Rightarrow \frac{[\frac{4}{3}\pi {R}^{3}\rho ]0+[\frac{4}{3}\pi {(1)}^{3}(-\rho )][R-1]}{\frac{4}{3}\pi {R}^{3}\rho +\frac{4}{3}\pi {(1)}^{3}(-\rho )}-(2-R)$

$\Rightarrow \frac{(R-1)}{{(R}^{3}-1)}=(2-R)(R\neq 1)$

$\frac{(R-1)}{(R-1)({R}^{2}+R+1)}=2-1$

$({R}^{2}+R+1)(2-R)=1$

Alternative:

${M}_{remaining}(2-R)={M}_{cavity}(1-R)$

$\Rightarrow ({R}^{3}-{1}^{3})(2-R)={1}^{3}[R-1]$

$\Rightarrow ({R}^{2}+R+1)(2-R)=1$

JEE Main 2020 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

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JEE Main 2020 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

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