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QJEEM20Phf77c: JEE Main 2020 Physics - Class 11 JEEM20Phf77c Solution: A square shaped hole of side l = is carved out at a distance d…

A square shaped hole of side l = a2{a \over 2} is carved out at a distance d = a2{a \over 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 form O is −aX- {a \over X} , value of X (to the nearest integer) is : JEE Main 2020 (Online) 2nd September Evening Slot Physics - Center of Mass and Collision Question 72 English

Step-by-Step Verified Solution

Xcom = σπa2×0−σ.a24.dσπa2−σ.a24{{\sigma \pi {a^2} \times 0 - \sigma .{{{a^2}} \over 4}.d} \over {\sigma \pi {a^2} - \sigma .{{{a^2}} \over 4}}}

= −a24.dπa2−a24{{ - {{{a^2}} \over 4}.d} \over {\pi {a^2} - {{{a^2}} \over 4}}}

= −d4π−1{{ - d} \over {4\pi - 1}} = −a2(4π−1){{ - a} \over {2\left( {4\pi - 1} \right)}}

∴\therefore −aX- {a \over X} = −a2(4π−1){{ - a} \over {2\left( {4\pi - 1} \right)}}

⇒\Rightarrow X = 2(4π−1){2\left( {4\pi - 1} \right)} = (8π−2){\left( {8\pi - 2} \right)} = 23.12

So, nearest integer value of X = 23

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