JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- Numerical
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.......

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Answer & Solution
Correct Answer
(D) \(23\)
Step-by-step Solution
Detailed explanation
\(X _{ com }=\frac{ m _{1} x _{1}- m _{2} x _{2}}{ m _{1}- m _{2}}\)
where :
\(m _{1}=\) mass of complete disc
\(m _{2}=\) removed mass
Let \(\sigma=\) surface mass density of disc material
wrt \({ }^{\prime} O ^{\prime}: X_{ com }=\frac{\sigma \pi a ^{2}( O )-\sigma \cdot \frac{ a ^{2}}{4} \cdot d }{\sigma \pi a ^{2}-\sigma \frac{ a ^{2}}{4}}=\frac{-\frac{ a ^{2}}{4} d }{\pi a ^{2}-\frac{ a ^{2}}{4}}\)
\(=\frac{-d}{4 \pi-1}=-\frac{a}{2(4 \pi-1)}\)
So, \(X=2(4 \pi-1)=(8 \pi-2)=23.12\)
So, nearest integer value of \(X=23\)
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