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JEE Main 20202 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Evening Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2020 · 2 September · Evening Shift · Physics · Center of Mass, Momentum and Collision

A square shaped hole of side l = (a)/(2) is carved out at a...

A square shaped hole of sidel=a2l = \dfrac{a}{2} is carved out at a distance d=a2d = \dfrac{a}{2} from the centre 'OO' of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from OO is −ax- \dfrac{a}{x}, value of XX (to the nearest integer) is :

A square shaped hole of side l = (a)/(2) is carved out at a distance d = (a)/(2) from the centre ' O ' of a uniform

Correct answer

0

Step-by-step solution

A square shaped hole of side l = (a)/(2) is carved out at a distance d = (a)/(2) from the centre ' O ' of a uniform

Xa=A⋅X−A⋅X1A−A1=πa2×0−a24×a2πa2−a24{X}_{a} = \dfrac{A \cdot X - A \cdot X1}{A - {A}_{1}} = \dfrac{{\pi a}^{2} \times 0 - \dfrac{{a}^{2}}{4} \times \dfrac{a}{2}}{{\pi a}^{2} - \dfrac{{a}^{2}}{4}}
=−a3/8(π−14)a2=−a2(4π−1)=−a8π−2=−a23= \dfrac{- {a}^{3} / 8}{\left(\pi - \dfrac{1}{4}\right){a}^{2}} = \dfrac{- a}{2 ( 4\pi - 1 )} = \dfrac{- a}{8\pi - 2} = - \dfrac{a}{23}
x=23x = 23

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