A square shaped hole of side$l=\frac{a}{2}$ is carved out at a distance $d=\frac{a}{2}$ from the centre '$O$' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from $O$ is $-\frac{a}{x}$, value of $X$ (to the nearest integer) is : ![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/648b5a67417cc3fb48d669bf/question_1__q_648b5a67417cc3fb48d669bf__cdn-question-pool.getmarks.app__2befcbc9-3b36-471f-a3ab-df15ed4ea7e1-image__bdd1aaa4d0_final_ppt_sync.png)

![JEE Main 2020 Physics, Center of Mass Momentum and Collision — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Center_of_Mass_Momentum_and_Collision/648b5a67417cc3fb48d669bf/solution_1__q_648b5a67417cc3fb48d669bf__cdn-question-pool.getmarks.app__f49b71ae-bd53-4dd1-8c06-b723c6f18b51-image__164708bf08_final_ppt_sync.png) ${X}_{a}=\frac{A\cdot X-A\cdot X1}{A-{A}_{1}}=\frac{{\pi a}^{2}\times 0-\frac{{a}^{2}}{4}\times \frac{a}{2}}{{\pi a}^{2}-\frac{{a}^{2}}{4}}$ $=\frac{-{a}^{3}/8}{(\pi -\frac{1}{4}){a}^{2}}=\frac{-a}{2(4\pi -1)}=\frac{-a}{8\pi -2}=-\frac{a}{23}$ $x=23$

A square shaped hole of sidel= a 2 is… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Mechanics

2020integerhard

A square shaped hole of side $l=\frac{a}{2}$ is carved out at a distance $d=\frac{a}{2}$ from the centre ' $O$ ' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from $O$ is $-\frac{a}{x}$ , value of $X$ (to the nearest integer) is :

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

Official previous-year question

Held on 2 Sept 2020 · Verified 6 Jul 2026.

Did you get this right?

Solution

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

${X}_{a}=\frac{A\cdot X-A\cdot X1}{A-{A}_{1}}=\frac{{\pi a}^{2}\times 0-\frac{{a}^{2}}{4}\times \frac{a}{2}}{{\pi a}^{2}-\frac{{a}^{2}}{4}}$

$=\frac{-{a}^{3}/8}{(\pi -\frac{1}{4}){a}^{2}}=\frac{-a}{2(4\pi -1)}=\frac{-a}{8\pi -2}=-\frac{a}{23}$

$x=23$

Your note

JEE Main Mechanics in other years

Mechanics 2026 Mechanics 2025 Mechanics 2024 Mechanics 2023 Mechanics 2022 Mechanics 2021 Mechanics 2019 Mechanics 2018 Mechanics 2017 Mechanics 2016 Mechanics 2015 Mechanics 2014

More JEE Main Mechanics questions

If $\epsilon, E$ and $t$ represent the free space permittivity, electric field and time respectively, then the unit of $\frac{\epsilon E}{t}$ will be :

2026easy

When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height $h$. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by $1 \%$ each, then the height of the liquid in the tube will change by $\%$.

2026medium

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s. The magnitude of acceleration at point $(1, 2, 4)$ is _______ m/s$^2$.

2026hard

Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^{2}$. The value of $x$ is $\_\_\_\_$. ![JEE Main 2026 Physics, Rotational Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Rotational_Motion/6983c19b8bfdbb24356396d6/question_1__q_6983c19b8bfdbb24356396d6__cdn-question-pool.getmarks.app__23S2_q_48_1__7376098484_final_ppt_sync.png)

2026hard

A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm. If $(15000-x) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is $\_\_\_\_$. ($\pi=22 / 7$)

2026medium

View all Mechanics questions